By the end of this chapter, the student should be able to:
Calculate and interpret confidence intervals for one population average and one population proportion.
Interpret the studentt probability distribution as the sample size changes.
Discriminate between problems applying the normal and the studentt distributions.
Suppose you are trying to determine the average rent of a twobedroom apartment in your town. You might look in the classified section of the newspaper, write down several rents listed, and average them together. You would have obtained a point estimate of the true mean. If you are trying to determine the percent of times you make a basket when shooting a basketball, you might count the number of shots you make and divide that by the number of shots you attempted. In this case, you would have obtained a point estimate for the true proportion.
We use sample data to make generalizations about an unknown population. This part of statistics is called inferential statistics. The sample data help us to make an estimate of a population parameter . We realize that the point estimate is most likely not the exact value of the population parameter, but close to it. After calculating point estimates, we construct confidence intervals in which we believe the parameter lies.
In this chapter, you will learn to construct and interpret confidence intervals. You will also learn a new distribution, the Studentt, and how it is used with these intervals.
If you worked in the marketing department of an entertainment company, you might be interested in the average number of compact discs (CD’s) a consumer buys per month. If so, you could conduct a survey and calculate the sample average, , and the sample standard deviation, s . You would use to estimate the population mean and s to estimate the population standard deviation. The sample mean, , is the point estimate for the population mean, μ . The sample standard deviation, s , is the point estimate for the population standard deviation, σ .
Each of and s is also called a statistic.
A confidence interval is another type of estimate but, instead of being just one number, it is an interval of numbers. The interval of numbers is an estimated range of values calculated from a given set of sample data. The confidence interval is likely to include an unknown population parameter.
Suppose for the CD example we do not know the population mean μ but we do know that the population standard deviation is σ = 1 and our sample size is 100. Then by the Central Limit Theorem, the standard deviation for the sample mean is
.
The Empirical Rule, which applies to bellshaped distributions, says that in approximately 95% of the samples, the sample mean, , will be within two standard deviations of the population mean μ . For our CD example, two standard deviations is (2)(0.1) = 0.2. The sample mean is within 0.2 units of μ .
Because is within 0.2 units of μ , which is unknown, then μ is within 0.2 units of in 95% of the samples. The population mean μ is contained in an interval whose lower number is calculated by taking the sample mean and subtracting two standard deviations ((2)(0.1)) and whose upper number is calculated by taking the sample mean and adding two standard deviations. In other words, μ is between and in 95% of all the samples.
For the CD example, suppose that a sample produced a sample mean . Then the unknown population mean μ is between
and
We say that we are 95% confident that the unknown population mean number of CDs is between 1.8 and 2.2. The 95% confidence interval is (1.8, 2.2).
The 95% confidence interval implies two possibilities. Either the interval (1.8, 2.2) contains the true mean μ or our sample produced an that is not within 0.2 units of the true mean μ . The second possibility happens for only 5% of all the samples (100%  95%).
Remember that a confidence interval is created for an unknown population parameter like the population mean, μ . A confidence interval has the form
(point estimate  margin of error, point estimate + margin of error)
The margin of error depends on the confidence level or percentage of confidence.
Have your instructor record the number of meals each student in your class eats out in a week. Assume that the standard deviation is known to be 3 meals. Construct an approximate 95% confidence interval for the true average number of meals students eat out each week.
Calculate the sample mean.
σ = 3 and n = the number of students surveyed.
Construct the interval
We say we are approximately 95% confident that the true average number of meals that students eat out in a week is between __________ and ___________.
An interval estimate for an unknown population parameter. This depends on:
The desired confidence level.
Information that is known about the distribution (for example, known standard deviation).
The sample and its size.
Also called statistical inference or inductive statistics. This facet of statistics deals with estimating a population parameter based on a sample statistic. For example, if 4 out of the 100 calculators sampled are defective we might infer that 4 percent of the production is defective.
A numerical characteristic of the population.
A single number computed from a sample and used to estimate a population parameter.
To construct a confidence interval for a single unknown population mean μ , where the population standard deviation is known, we need as an estimate for μ and we need the margin of error. Here, the margin of error is called the error bound for a population mean (abbreviated EBM). The sample mean is the point estimate of the unknown population mean μ
(point estimate  error bound, point estimate + error bound) or, in symbols, 
The margin of error depends on the confidence level (abbreviated CL). The confidence level is the probability that the confidence interval estimate that we will calculate will contain the true population parameter. Most often, it is the choice of the person constructing the confidence interval to choose a confidence level of 90% or higher because he wants to be reasonably certain of his conclusions.
There is another probability called alpha ( α ). α is related to the confidence level CL. α is the probability that the sample produced a point estimate that is not within the appropriate margin of error of the unknown population parameter.
Example 8.1.
Suppose we have collected data from a sample. We know the sample average but we do not know the average for the entire population. 
The sample mean is 7 and the error bound for the mean is 2.5. 
7 and EBM = 2.5.
The confidence interval is ( 7 – 2.5 , 7 + 2.5 ) ; calculating the values gives ( 4.5 , 9.5 ) .
If the confidence level (CL) is 95%, then we say that “We estimate with 95% confidence that the true value of the population mean is between 4.5 and 9.5.”
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of and we have constructed the 90% confidence interval (5, 15) where EBM = 5 .
To get a 90% confidence interval, we must include the central 90% of the probability of the normal distribution. If we include the central 90%, we leave out a total of 10% in both tails, or 5% in each tail, of the normal distribution.
To capture the central 90%, we must go out 1.645 “standard deviations” on either side of the calculated sample mean. 1.645 is the zscore from a Standard Normal probability distribution that puts an area of 0.90 in the center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right tail.
It is important that the “standard deviation” used must be appropriate for the parameter we are estimating. So in this section, we need to use the standard deviation that applies to sample means, which is . is commonly called the “standard error of the mean” in order to clearly distinguish the standard deviation for a mean from the population standard deviation σ .
In summary, as a result of the Central Limit Theorem:
is normally distributed, that is, ~
When the population standard deviation σ is known, we use a Normal distribution to calculate the error bound.
To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps to construct and interpret the confidence interval are:
Calculate the sample mean from the sample data. Remember, in this section, we already know the population standard deviation σ .
Find the Zscore that corresponds to the confidence level.
Calculate the error bound EBM
Construct the confidence interval
Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain what the confidence interval means, in the words of the problem.)
We will first examine each step in more detail, and then illustrate the process with some examples.
When we know the population standard deviation σ, we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z~N(0,1).
The confidence level, CL, is he area in the middle of the standard normal distribution. CL = 1 – α . So α is the area that is split equally between the two tails. Each of the tails contains an area equal to .
The zscore that has an area to the right of is denoted by
For example, when CL = 0.95 then α = 0.05 and ; we write
The area to the right of z _{.025} is 0.025 and the area to the left of z _{.025} is 10.025 = 0.975
, using a calculator, computer or a Standard Normal probability table.
Using the TI83, TI83+ or TI84+ calculator:
invNorm
(.975,0,1) = 1.96
CALCULATOR NOTE: Remember to use area to the LEFT of ; in this chapter the last two inputs in the invnorm command are 0,1 because you are using a Standard Normal Distribution Z~N(0,1)
The error bound formula for an unknown population mean μ when the population standard deviation σ is known is
Constructing the Confidence Interval
The confidence interval estimate has the format .
The graph gives a picture of the entire situation.
.
The interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated (here, a population mean or average ), and should state the confidence interval (both endpoints). “We estimate with ___% confidence that the true population average (include context of the problem) is between ___ and ___ (include appopriate units).”
Example 8.2.
Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of 3 points. A random sample of 36 scores is taken and gives a sample mean (sample average score) of 68. Find a confidence interval estimate for the population mean exam score (the average score on all exams).
Problem
Find a 90% confidence interval for the true (population) mean of statistics exam scores.
Solution
You can use technology to directly calculate the confidence interval
The first solution is shown stepbystep (Solution A).
The second solution uses the TI83, 83+ and 84+ calculators (Solution B).
To find the confidence interval, you need the sample mean, , and the EBM.
σ = 3 ; n = 36 ; The confidence level is 90% (CL=0.90) 
CL = 0.90 so α = 1 – CL = 1 – 0.90 = 0.10
The area to the right of z _{.05} is 0.05 and the area to the left of z _{.05} is 1−0.05=0.95
using invnorm(.95,0,1) on the TI83,83+,84+ calculators. This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the Standard Normal distribution.
The 90% confidence interval is (67.1775, 68.8225).
Using a function of the TI83, TI83+ or TI84 calculators:
Press STAT
and arrow over to TESTS
.
Arrow down to 7:ZInterval
.
Press ENTER
.
Arrow to Stats
and press ENTER
.
Arrow down and enter 3 for
σ
, 68 for
, 36 for
n
, and .90 for Clevel
.
Arrow down to Calculate
and
press ENTER
.
The confidence interval is (to 3 decimal places) (67.178, 68.822).
We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82.
90% of all confidence intervals constructed in this way contain the true average statistics exam score. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score.
Example 8.3. Changing the Confidence Level
Problem
Suppose we change the original problem by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score.
Solution
To find the confidence interval, you need the sample mean, , and the EBM.
σ = 3 ; n = 36 ; The confidence level is 95% (CL=0.95) 
CL = 0.95 so α = 1 – CL = 1 – 0.95 = 0.05
The area to the right of z _{.025} is 0.025 and the area to the left of z _{.025} is 1−0.025=0.975
using invnorm(.975,0,1) on the TI83,83+,84+ calculators. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the Standard Normal distribution.)
We estimate with 95 % confidence that the true population average for all statistics exam scores is between 67.02 and 68.98.
95% of all confidence intervals constructed in this way contain the true value of the population average statistics exam score.
The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider.
Figure 8.1.
Summary: Effect of Changing the Confidence Level
Increasing the confidence level increases the error bound, making the confidence interval wider.
Decreasing the confidence level decreases the error bound, making the confidence interval narrower.
Example 8.4. Changing the Sample Size:
Suppose we change the original problem to see what happens to the error bound if the sample size is changed.
Problem
Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the error bound and the confidence interval if we increase the sample size and use n=100 instead of n=36? What happens if we decrease the sample size to n=25 instead of n=36?
σ = 3 ; The confidence level is 90% (CL=0.90) ;
Solution
If we increase the sample size n to 100, we decrease the error bound.
When n = 100 :
Solution
If we decrease the sample size n to 25, we increase the error bound.
When n = 25 :
Summary: Effect of Changing the Sample Size
Increasing the sample size causes the error bound to decrease, making the confidence interval narrower.
Decreasing the sample size causes the error bound to increase, making the confidence interval wider.
When we calculate a confidence interval, we find the sample mean and calculate the error bound and use them to calculate the confidence interval. But sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean.
Finding the Error Bound
From the upper value for the interval, subtract the sample mean
OR, From the upper value for the interval, subtract the lower value. Then divide the difference by 2.
Finding the Sample Mean
Subtract the error bound from the upper value of the confidence interval
OR, Average the upper and lower endpoints of the confidence interval
Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know.
Example 8.5.
Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68. Or perhaps our source only gave the confidence interval and did not tell us the value of the the sample mean.
Calculate the Error Bound:
If we know that the sample mean is 68:
If we don’t know the sample mean:
Calculate the Sample Mean:
If we know the error bound:
If we don’t know the error bound:
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.
The error bound formula for a population mean when the population standard deviation is known is
The formula for sample size is , found by solving the error bound formula for n
In this formula, z is , corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study.
Example 8.6.
The population standard deviation for the age of Foothill College students is 15 years. If we want to be 95% confident that the sample mean age is within 2 years of the true population mean age of Foothill College students , how many randomly selected Foothill College students must be surveyed?
From the problem, we know that σ = 15 and EBM=2 
z = z _{.025} = 1.96 , because the confidence level is 95%. 
= =216.09 using the sample size equation. 
Use n = 217: Always round the answer UP to the next higher integer to ensure that the sample size is large enough. 
Therefore, 217 Foothill College students should be surveyed in order to be 95% confident that we are within 2 years of the true population age of Foothill College students.
**With contributions from Roberta Bloom
An interval estimate for an unknown population parameter. This depends on:
The desired confidence level.
Information that is known about the distribution (for example, known standard deviation).
The sample and its size.
The percent expression for the probability that the confidence interval contains the true population parameter. For example, if the CL = 90%, then in 90 out of 100 samples the interval estimate will enclose the true population parameter.
The margin of error. Depends on the confidence level, sample size, and known or estimated population standard deviation.
In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.
William S. Gossett (18761937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to “discover” what is called the Studentt distribution. The name comes from the fact that Gosset wrote under the pen name “Student.”
Up until the mid 1990s, statisticians used the normal distribution approximation for large sample sizes and only used the Studentt distribution for sample sizes of at most 30. With the common use of graphing calculators and computers, the practice is to use the Studentt distribution whenever s is used as an estimate for σ .
If you draw a simple random sample of size n from a population that has approximately a normal distribution with mean μ and unknown population standard deviation σ and calculate the tscore , then the tscores follow a Studentt distribution with n – 1 degrees of freedom. The tscore has the same interpretation as the zscore. It measures how far is from its mean μ . For each sample size n , there is a different Studentt distribution.
The degrees of freedom, n – 1 , come from the calculation of the sample standard deviation s . In Chapter 2, we used n deviations to calculate s . Because the sum of the deviations is 0, we can find the last deviation once we know the other n – 1 deviations. The other n – 1 deviations can change or vary freely. We call the number n – 1 the degrees of freedom (df).
Properties of the Studentt Distribution
The graph for the Studentt distribution is similar to the Standard Normal curve.
The mean for the Studentt distribution is 0 and the distribution is symmetric about 0.
The Studentt distribution has more probability in its tails than the Standard Normal distribution because the spread of the t distribution is greater than the spread of the Standard Normal. So the graph of the Studentt distribution will be thicker in the tails and shorter in the center than the graph of the Standard Normal distribution.
The exact shape of the Studentt distribution depends on the “degrees of freedom”. As the degrees of freedom increases, the graph Studentt distribution becomes more like the graph of the Standard Normal distribution.
The underlying population of individual observations is assumed to be normally distributed with unknown population mean μ and unknown population standard deviation σ . In the real world, however, as long as the underlying population is large and bellshaped, and the data are a simple random sample, practitioners often consider the assumptions met.
Calculators and computers can easily calculate any Studentt probabilities. The TI83,83+,84+ have a tcdf function to find the probability for given values of t. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However for confidence intervals, we need to use inverse probability to find the value of t when we know the probability.
For the TI84+ you can use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm. The invT command requires two inputs: invT(area to the left, degrees of freedom) The output is the tscore that corresponds to the area we specified. The TI83 and 83+ do not have the invT command. (The TI89 has an inverse T command.)
A probability table for the Studentt distribution can also be used. The table gives tscores that correspond to the confidence level (column) and degrees of freedom (row). (The TI86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Studentt distribution.) When using ttable, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails. A Studentt table (See the Table of Contents 15. Tables) gives tscores given the degrees of freedom and the righttailed probability. The table is very limited. Calculators and computers can easily calculate any Studentt probabilities.
The notation for the Studentt distribution is (using T as the random variable) is
T ~ t _{df} where df = n – 1.
For example, if we have a sample of size n=20 items, then we calculate the degrees of freedom as df=n−1=20−1=19 and we write the distribution as T ~ t _{19}
If the population standard deviation is not known, the error bound for a population mean is:
is the tscore with area to the right equal to
use df = n – 1 degrees of freedom
s = sample standard deviation
The format for the confidence interval is:
.
The TI83, 83+ and 84 calculators have a function that calculates the confidence interval directly. To get to it,
Press STAT
Arrow over to TESTS
.
Arrow down to 8:Tinterval
and press ENTER
(or just press 8
).
Example 8.7.
Problem
Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given below. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data. The solution is shown stepbystep and by using the TI83, 83+ and 84+ calculators.
8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1; 5.5; 6.9
Solution
You can use technology to directly calculate the confidence interval.
The first solution is stepbystep (Solution A).
The second solution uses the Ti83+ and Ti84 calculators (Solution B).
To find the confidence interval, you need the sample mean, , and the EBM.
df = 15 – 1 = 14
so
The area to the right of t _{.025} is 0.025 and the area to the left of t _{.025} is 1−0.025=0.975
using invT(.975,14) on the TI84+ calculator.
The 95% confidence interval is (7.30, 9.15).
We estimate with 95% confidence that the true population average sensory rate is between 7.30 and 9.15.
Using a function of the TI83, TI83+ or TI84 calculators:
Press STAT
and arrow over to TESTS
.
Arrow down to 8:TInterval
and press ENTER
(or you can just press 8
).
Arrow to Data
and press ENTER
.
Arrow down to List
and enter the list name where you put the data.
Arrow down to Freq
and enter 1.
Arrow down to Clevel
and enter .95
Arrow down to Calculate
and press ENTER
.
The 95% confidence interval is (7.3006, 9.1527)
When calculating the error bound, a probability table for the Studentt distribution can also be used to find the value of t. The table gives tscores that correspond to the confidence level (column) and degrees of freedom (row); the tscore is found where the row and column intersect in the table.
**With contributions from Roberta Bloom
An interval estimate for an unknown population parameter. This depends on:
The desired confidence level.
Information that is known about the distribution (for example, known standard deviation).
The sample and its size.
The percent expression for the probability that the confidence interval contains the true population parameter. For example, if the CL = 90%, then in 90 out of 100 samples the interval estimate will enclose the true population parameter.
The number of objects in a sample that are free to vary.
The margin of error. Depends on the confidence level, sample size, and known or estimated population standard deviation.
A continuous random variable (RV) with pdf , where μ is the mean of the distribution and σ is the standard deviation. Notation: X ~ N (μ, σ). If μ = 0 and σ = 1, the RV is called the standard normal distribution.
A number that is equal to the square root of the variance and measures how far data values are from their mean. Notation: s for sample standard deviation and σ for population standard deviation.
Investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student. The major characteristics of the random variable (RV) are:
It is continuous and assumes any real values.
The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution.
It approaches the standard normal distribution as n gets larger.
There is a “family” of t distributions: every representative of the family is completely defined by the number of degrees of freedom which is one less than the number of data.
During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within 3 percentage points. Often, election polls are calculated with 95% confidence. So, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43 : ( 0.40 – 0.03 , 0.40 + 0.03 ) .
Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.
The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean. The formulas are different.
How do you know you are dealing with a proportion problem? First, the underlying distribution is binomial. (There is no mention of a mean or average.) If X is a binomial random variable, then X ~ B(n,p) where n = the number of trials and p = the probability of a success. To form a proportion, take X , the random variable for the number of successes and divide it by n , the number of trials (or the sample size). The random variable P ‘ (read “P prime”) is that proportion,
(Sometimes the random variable is denoted as , read “P hat”.)
When n is large, we can use the normal distribution to approximate the binomial.
X ~
If we divide the random variable by n , the mean by n , and the standard deviation by n , we get a normal distribution of proportions with P ‘ , called the estimated proportion, as the random variable. (Recall that a proportion = the number of successes divided by n .)
~
Using algebra to simplify :
P ‘ follows a normal distribution for proportions: P ‘ ~
The confidence interval has the form (p ‘ – EBP,p ‘ + EBP) .
p ‘ = the estimated proportion of successes ( p ‘ is a point estimate for p , the true proportion)
x = the number of successes.
n = the size of the sample
The error bound for a proportion is
This formula is similar to the error bound formula for a mean, except that the “appropriate standard deviation” is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is . For a proportion, the appropriate standard deviation is .
However, in the error bound formula, we use as the standard deviation, instead of
However, in the error bound formula, the standard deviation is .
In the error bound formula, the sample proportions p ‘ and q ‘ are estimates of the unknown population proportions p and q . The estimated proportions p ‘ and q ‘ are used because p and q are not known. p ‘ and q ‘ are calculated from the data. p ‘ is the estimated proportion of successes. q ‘ is the estimated proportion of failures.
For the normal distribution of proportions, the zscore formula is as follows.
If P ‘ ~ then the zscore formula is
Example 8.8.
Problem
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500 randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes  they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adults residents of this city who have cell phones.
Solution
You can use technology to directly calculate the confidence interval.
The first solution is stepbystep (Solution A).
The second solution uses a function of the TI83, 83+ or 84 calculators (Solution B).
Solution
Let X = the number of people in the sample who have cell phones. X is binomial. X ~ .
To calculate the confidence interval, you must find p ‘ , q ‘ , and EBP.
= the number of successes = 421
p ‘ = 0.842 is the sample proportion; this is the point estimate of the population proportion.
q ‘ = 1 – p ‘ = 1 – 0.842 = 0.158
Since CL = 0.95 , then .
Then
Use the TI83, 83+ or 84+ calculator command invnorm(.975,0,1) to find z _{.025} . Remember that the area to the right of z _{.025} is 0.025 and the area to the left of z _{.025} is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.
p ‘ – EBP = 0.842 – 0.032 = 0.81
p ‘ + EBP = 0.842 + 0.032 = 0.874
The confidence interval for the true binomial population proportion is (p ‘ – EBP,p ‘ + EBP) = (0.810,0.874) .
We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.
95% of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.
Solution
Press STAT
and arrow over to TESTS
.
Arrow down to A:PropZint
. Press ENTER
.
Arrow down to
x
and enter 421.
Arrow down to
n
and enter 500.
Arrow down to CLevel
and enter .95.
Arrow down to Calculate
and press ENTER
.
The confidence interval is (0.81003, 0.87397).
Example 8.9.
Problem
For a class project, a political science student at a large university wants to determine the percent of students that are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students that are registered voters and interpret the confidence interval.
Solution
You can use technology to directly calculate the confidence interval.
The first solution is stepbystep (Solution A).
The second solution uses a function of the TI83, 83+ or 84 calculators (Solution B).
x = 300 and n = 500.
q ‘ = 1 – p ‘ = 1 – 0.600 = 0.400
Since CL = 0.90 , then .
Use the TI83, 83+ or 84+ calculator command invnorm(.95,0,1) to find z _{.05} . Remember that the area to the right of z _{.05} is 0.05 and the area to the left of z _{.05} is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.
p ‘ – EBP = 0.60 – 0.036 = 0.564 p ‘ + EBP = 0.60 + 0.036 = 0.636
The confidence interval for the true binomial population proportion is (p ‘ – EBP,p ‘ + EBP) = (0.564,0.636) .
Interpretation:
We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%.
Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters.
90% of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters.
Using a function of the TI83, 83+ or 84 calculators:
Press STAT
and arrow over to TESTS
.
Arrow down to A:PropZint
. Press ENTER
.
Arrow down to
x
and enter 300.
Arrow down to
n
and enter 500.
Arrow down to CLevel
and enter .90.
Arrow down to Calculate
and press ENTER
.
The confidence interval is (0.564, 0.636).
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.
The error bound formula for a population proportion is
Solving for n gives you an equation for the sample size.
Example 8.10.
Suppose a mobile phone company wants to determine the current percentage of customers aged 50+ that use text messaging on their cell phone. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of customers aged 50+ that use text messaging on their cell phone.
From the problem, we know that EBP=0.03 (3%=0.03) and because the confidence level is 90%
However, in order to find n , we need to know the estimated (sample) proportion p’. Remember that q’=1p’. But, we do not know p’ yet. Since we multiply p’ and q’ together, we make them both equal to 0.5 because p’q’= (.5)(.5)=.25 results in the largest possible product. (Try other products: (.6)(.4)=.24; (.3)(.7)=.21; (.2)(.8)=.16 and so on). The largest possible product gives us the largest n. This gives us a large enough sample so that we can be 90% confident that we are within 3 percentage points of the true population proportion. To calculate the sample size n, use the formula and make the substitutions.
gives =751.7
Round the answer to the next higher value. The sample size should be 758 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of all customers aged 50+ that use text messaging on their cell phone.
**With contributions from Roberta Bloom.
A discrete random variable (RV) which arises from Bernoulli trials. There are a fixed number, n , of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X is defined as the number of successes in n trials. The notation is: X ~ B ( n , p ) . The mean is μ = np and the standard deviation is . The probability of exactly x successes in n trials is .
An interval estimate for an unknown population parameter. This depends on:
The desired confidence level.
Information that is known about the distribution (for example, known standard deviation).
The sample and its size.
The percent expression for the probability that the confidence interval contains the true population parameter. For example, if the CL = 90%, then in 90 out of 100 samples the interval estimate will enclose the true population parameter.
The margin of error. Depends on the confidence level, sample size, and the estimated (from the sample) proportion of successes.
A continuous random variable (RV) with pdf , where μ is the mean of the distribution and σ is the standard deviation. Notation: X ~ N (μ, σ). If μ = 0 and σ = 1, the RV is called the standard normal distribution.
Formula 8.1. General form of a confidence interval
( lower value , upper value ) = ( point estimate – error bound , point estimate + error bound )
Formula 8.2. To find the error bound when you know the confidence interval
OR
Formula 8.3. Single Population Mean, Known Standard Deviation, Normal Distribution
Use the Normal Distribution for Means
The confidence interval has the format .
Formula 8.4. Single Population Mean, Unknown Standard Deviation, Studentt Distribution
Use the Studentt Distribution with degrees of freedom df = n – 1 .
Formula 8.5. Single Population Proportion, Normal Distribution
Use the Normal Distribution for a single population proportion
The confidence interval has the format ( p ‘ – EBP , p ‘ + EBP ) .
Formula 8.6. Point Estimates
is a point estimate for μ
p ‘ is a point estimate for ρ
s is a point estimate for σ
The student will explore the properties of Confidence Intervals for Averages when the population standard deviation is known.
The average age for all Foothill College students for Fall 2005 was 32.7. The population standard deviation has been pretty consistent at 15. Twentyfive Winter 2006 students were randomly selected. The average age for the sample was 30.4. We are interested in the true average age for Winter 2006 Foothill College students. (http://research.fhda.edu/factbook/FHdemofs/Fact_sheet_fh_2005f.pdf)
Let X = the age of a Winter 2006 Foothill College student
Exercise 8.6.7. (Go to Solution)
As a result of your answer to (4), state the exact distribution to use when calculating the Confidence Interval.
Construct a 95% Confidence Interval for the true average age of Winter 2006 Foothill College students.
Exercise 8.6.10. (Go to Solution)
Identify the following specifications:
a. lower limit = 
b. upper limit = 
c. error bound = 
Exercise 8.6.12.
Figure 8.2. Fill in the blanks on the graph with the areas, upper and lower limits of the Confidence Interval, and the sample mean.
Exercise 8.6.13.
In one complete sentence, explain what the interval means.
Exercise 8.6.14.
Using the same mean, standard deviation and level of confidence, suppose that n were 69 instead of 25. Would the error bound become larger or smaller? How do you know?
Exercise 8.6.15.
Using the same mean, standard deviation and sample size, how would the error bound change if the confidence level were reduced to 90%? Why?
Solution to Exercise 8.6.4. (Return to Exercise)
the average age of 25 randomly selected Winter 2006 Foothill students
The student will explore the properties of confidence intervals for averages when the population standard deviation is unknown.
The following real data are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true average number of colors on a national flag. Let X = the number of colors on a national flag.
X  Freq. 

1  1 
2  7 
3  18 
4  7 
5  6 
Exercise 8.7.5. (Go to Solution)
As a result of your answer to (4), state the exact distribution to use when calculating the Confidence Interval.
Construct a 95% Confidence Interval for the true average number of colors on national flags.
Exercise 8.7.8. (Go to Solution)
Calculate the following:
a. lower limit = 
b. upper limit = 
c. error bound = 
Exercise 8.7.10.
Fill in the blanks on the graph with the areas, upper and lower limits of the Confidence Interval and the sample mean.
Figure 8.3.
Exercise 8.7.11.
In one complete sentence, explain what the interval means.
Exercise 8.7.12.
Using the same , s _{ x } , and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know?
Exercise 8.7.13.
Using the same , s _{ x } , and n = 39, how would the error bound change if the confidence level were reduced to 90%? Why?
The student will explore the properties of the confidence intervals for proportions.
The Ice Chalet offers dozens of different beginning iceskating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8  12, beginning iceskating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8  12, in all beginning iceskating classes at the Ice Chalet.
Exercise 8.8.1.
What is being counted?
Exercise 8.8.7.
State the estimated distribution of P’ . P ‘ ~
Construct a 92% Confidence Interval for the true proportion of girls in the age 8  12 beginning iceskating classes at the Ice Chalet.
Exercise 8.8.10. (Go to Solution)
Calculate the following:
a. lower limit = 
b. upper limit = 
c. error bound = 
Exercise 8.8.12.
Figure 8.4. Fill in the blanks on the graph with the areas, upper and lower limits of the Confidence Interval, and the sample proportion.
Exercise 8.8.13.
In one complete sentence, explain what the interval means.
Exercise 8.8.14.
Using the same p’ and level of confidence, suppose that n were increased to 100. Would the error bound become larger or smaller? How do you know?
Exercise 8.8.15.
Using the same p’ and n = 80, how would the error bound change if the confidence level were increased to 98%? Why?
Exercise 8.8.16.
If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)?
Solution to Exercise 8.8.2. (Return to Exercise)
The number of girls, age 812, in the beginning ice skating class
Solution to Exercise 8.8.6. (Return to Exercise)
The proportion of girls, age 812, in the beginning ice skating class.
If you are using a studentt distribution for a homework problem below, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Exercise 8.9.1. (Go to Solution)
Among various ethnic groups, the standard deviation of heights is known to be approximately 3 inches. We wish to construct a 95% confidence interval for the average height of male Swedes. 48 male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches.
a.
 
b. Define the Random Variables X and , in words.  
c. Which distribution should you use for this problem? Explain your choice.  
d. Construct a 95% confidence interval for the population average height of male Swedes.
 
e. What will happen to the level of confidence obtained if 1000 male Swedes are surveyed instead of 48? Why? 
Exercise 8.9.2.
In six packages of “The Flintstones® Real Fruit Snacks” there were 5 BamBam snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of BamBam snack pieces.
a. Define the Random Variables X and P’ , in words.  
b. Which distribution should you use for this problem? Explain your choice  
c. Calculate p’ .  
d. Construct a 96% confidence interval for the population proportion of BamBam snack pieces per bag.
 
e. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not? 
Exercise 8.9.3. (Go to Solution)
A random survey of enrollment at 35 community colleges across the United States yielded the following figures (source: Microsoft Bookshelf): 6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285; 28165; 5080; 11622. Assume the underlying population is normal.
a.
 
b. Define the Random Variables X and , in words.  
c. Which distribution should you use for this problem? Explain your choice.  
d. Construct a 95% confidence interval for the population average enrollment at community colleges in the United States.
 
e. What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why? 
Exercise 8.9.4.
From a stack of IEEE Spectrum magazines, announcements for 84 upcoming engineering conferences were randomly picked. The average length of the conferences was 3.94 days, with a standard deviation of 1.28 days. Assume the underlying population is normal.
a. Define the Random Variables X and , in words.  
b. Which distribution should you use for this problem? Explain your choice.  
c. Construct a 95% confidence interval for the population average length of engineering conferences.

Exercise 8.9.5. (Go to Solution)
Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the average amount of time individuals waste at the courthouse waiting to be called for service. The committee randomly surveyed 81 people. The sample average was 8 hours with a sample standard deviation of 4 hours.
a.
 
b. Define the Random Variables X and , in words.  
c. Which distribution should you use for this problem? Explain your choice.  
d. Construct a 95% confidence interval for the population average time wasted.
 
e. Explain in a complete sentence what the confidence interval means. 
Exercise 8.9.6.
Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample average is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal.
a.
 
b. Define the Random Variables X and , in words.  
c. Which distribution should you use for this problem? Explain your choice.  
d. Construct a 90% confidence interval for the population average time to complete the tax forms.
 
e. If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make?  
f. If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why?  
g. Suppose that the firm decided that it needed to be at least 96% confident of the population average length of time to within 1 hour. How would the number of people the firm surveys change? Why? 
Exercise 8.9.7. (Go to Solution)
A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was 2 ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce.
a.
 
b. Define the Random Variable X , in words.  
c. Define the Random Variable , in words.  
d. Which distribution should you use for this problem? Explain your choice.  
e. Construct a 90% confidence interval for the population average weight of the candies.
 
f. Construct a 98% confidence interval for the population average weight of the candies.
 
g. In complete sentences, explain why the confidence interval in (f) is larger than the confidence interval in (e).  
h. In complete sentences, give an interpretation of what the interval in (f) means. 
Exercise 8.9.8.
A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of 9 patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4 .
a.
 
b. Define the Random Variable X , in words.  
c. Define the Random Variable , in words.  
d. Which distribution should you use for this problem? Explain your choice.  
e. Construct a 95% confidence interval for the population average length of time.
 
f. What does it mean to be “95% confident” in this problem? 
Exercise 8.9.9. (Go to Solution)
Suppose that 14 children were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of 6 months with a sample standard deviation of 3 months. Assume that the underlying population distribution is normal.
a.
 
b. Define the Random Variable X , in words.  
c. Define the Random Variable , in words.  
d. Which distribution should you use for this problem? Explain your choice.  
e. Construct a 99% confidence interval for the population average length of time using training wheels.
 
f. Why would the error bound change if the confidence level was lowered to 90%? 
Exercise 8.9.10.
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car.
a. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95% confident that the population proportion is estimated to within 0.03? 
b. If it was later determined that it was important to be more than 95% confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why? 
Exercise 8.9.11. (Go to Solution)
Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed to always buckle up. We are interested in the population proportion of drivers who claim to always buckle up.
a.
 
b. Define the Random Variables X and P’ , in words.  
c. Which distribution should you use for this problem? Explain your choice.  
d. Construct a 95% confidence interval for the population proportion that claim to always buckle up.
 
e. If this survey were done by telephone, list 3 difficulties the companies might have in obtaining random results. 
Exercise 8.9.12.
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats.
a.
 
b. Define the Random Variables X and , in words.  
c. Which distribution should you use for this problem? Explain your choice.  
d. Construct a 92% confidence interval for the population average number of unoccupied seats per flight.

Exercise 8.9.13. (Go to Solution)
According to a recent survey of 1200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job.
a. Define the Random Variables X and P’ , in words.  
b. Which distribution should you use for this problem? Explain your choice.  
c. Construct a 90% confidence interval for the population proportion of people who feel the president is doing an acceptable job.

Exercise 8.9.14.
A survey of the average amount of cents off that coupons give was done by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal.
a.
 
b. Define the Random Variables X and , in words.  
c. Which distribution should you use for this problem? Explain your choice.  
d. Construct a 95% confidence interval for the population average worth of coupons.
 
e. If many random samples were taken of size 14, what percent of the confident intervals constructed should contain the population average worth of coupons? Explain why. 
Exercise 8.9.15. (Go to Solution)
An article regarding interracial dating and marriage recently appeared in the Washington Post. Of the 1709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. In this survey, 86% of blacks said that their families would welcome a white person into their families. Among Asians, 77% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.
a. We are interested in finding the 95% confidence interval for the percent of black families that would welcome a white person into their families. Define the Random Variables X and P’ , in words.  
b. Which distribution should you use for this problem? Explain your choice.  
c. Construct a 95% confidence interval

Exercise 8.9.16.
Refer to the problem above.
a. Construct the 95% confidence intervals for the three Asian responses. 
b. Even though the three point estimates are different, do any of the confidence intervals overlap? Which? 
c. For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions? 
d. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions? 
Exercise 8.9.17. (Go to Solution)
A camp director is interested in the average number of letters each child sends during his/her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The average from the sample is 7.9 with a sample standard deviation of 2.8.
a.
 
b. Define the Random Variables X and , in words.  
c. Which distribution should you use for this problem? Explain your choice.  
d. Construct a 90% confidence interval for the population average number of letters campers send home.
 
e. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why? 
Exercise 8.9.18.
Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight–year period.
a. Define the Random Variables X and P’ , in words.  
b. Which distribution should you use for this problem? Explain your choice.  
c. Construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight–year period.
 
d. Explain what a “97% confidence interval” means for this study. 
Exercise 8.9.19. (Go to Solution)
In a recent sample of 84 used cars sales costs, the sample mean was $6425 with a standard deviation of $3156. Assume the underlying distribution is approximately normal.
a. Which distribution should you use for this problem? Explain your choice.  
b. Define the Random Variable , in words.  
c. Construct a 95% confidence interval for the population average cost of a used car.
 
d. Explain what a “95% confidence interval” means for this study. 
Exercise 8.9.20.
A telephone poll of 1000 adult Americans was reported in an issue of Time Magazine. One of the questions asked was “What is the main problem facing the country?” 20% answered “crime”. We are interested in the population proportion of adult Americans who feel that crime is the main problem.
a. Define the Random Variables X and P’ , in words.  
b. Which distribution should you use for this problem? Explain your choice.  
c. Construct a 95% confidence interval for the population proportion of adult Americans who feel that crime is the main problem.
 
d. Suppose we want to lower the sampling error. What is one way to accomplish that?  
e. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ± 3%. In 13 complete sentences, explain what the ± 3% represents. 
Exercise 8.9.21. (Go to Solution)
Refer to the above problem. Another question in the poll was “[How much are] you worried about the quality of education in our schools?” 63% responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools.
Define the Random Variables X and P’ , in words.
Which distribution should you use for this problem? Explain your choice.
Construct a 95% confidence interval for the population proportion of adult Americans worried a lot about the quality of education in our schools.
i. State the confidence interval. 
ii. Sketch the graph. 
iii. Calculate the error bound. 
The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ± 3%. In 13 complete sentences, explain what the ± 3% represents.
Exercise 8.9.22.
Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is approximately normal.
a. Calculate a 90% confidence interval for the population average grams of fat per serving of chocolate chip cookies sold in supermarkets.
 
b. If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done?  
c. Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies.  
d. Calculate the average.  
e. Is the average within the interval you calculated in part (a)? Did you expect it to be? Why or why not? 
Exercise 8.9.23.
A confidence interval for a proportion is given to be (– 0.22, 0.34). Why doesn’t the lower limit of the confidence interval make practical sense? How should it be changed? Why?
The next three problems refer to the following: According a Field Poll conducted February 8 – 17, 2005, 79% of California adults (actual results are 400 out of 506 surveyed) feel that “education and our schools” is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education and the schools is one of the top issues facing California.
Exercise 8.9.24. (Go to Solution)
A point estimate for the true population proportion is:
A. 0.90 
B. 1.27 
C. 0.79 
D. 400 
Exercise 8.9.25. (Go to Solution)
A 90% confidence interval for the population proportion is:
A. (0.761, 0.820) 
B. (0.125, 0.188) 
C. (0.755, 0.826) 
D. (0.130, 0.183) 
Exercise 8.9.26. (Go to Solution)
The error bound is approximately
A. 1.581 
B. 0.791 
C. 0.059 
D. 0.030 
The next two problems refer to the following:
A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample average is 13.30 with a sample standard deviation is 1.55. Assume the underlying population is normally distributed.
Exercise 8.9.27. (Go to Solution)
Find the 95% Confidence Interval for the true population mean for the amount of soda served.
A. (12.42, 14.18) 
B. (12.32, 14.29) 
C. (12.50, 14.10) 
D. Impossible to determine 
Exercise 8.9.29. (Go to Solution)
What is meant by the term “90% confident” when constructing a confidence interval for a mean?
A. If we took repeated samples, approximately 90% of the samples would produce the same confidence interval. 
B. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the sample mean. 
C. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean. 
D. If we took repeated samples, the sample mean would equal the population mean in approximately 90% of the samples. 
The next two problems refer to the following:
Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventythree (173) of the homes surveyed met the minimum recommendations for earthquake preparedness and 338 did not.
Exercise 8.9.30. (Go to Solution)
Find the Confidence Interval at the 90% Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness.
A. (0.2975, 0.3796) 
B. (0.6270, 6959) 
C. (0.3041, 0.3730) 
D. (0.6204, 0.7025) 
Exercise 8.9.31. (Go to Solution)
The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is:
A. 0.6614 
B. 0.3386 
C. 173 
D. 338 
Solution to Exercise 8.9.1. (Return to Exercise)
a.
 
c. N  
d.

Solution to Exercise 8.9.3. (Return to Exercise)
a.
 
c. t _{ 34 }  
d.
 
e. It will become smaller 
Solution to Exercise 8.9.5. (Return to Exercise)
a.
 
c. t _{ 80 }  
d.

Solution to Exercise 8.9.7. (Return to Exercise)
a.
 
b. the weight of 1 small bag of candies  
c. the average weight of 16 small bags of candies  
d. N  
e.
 
f.

Solution to Exercise 8.9.9. (Return to Exercise)
a.
 
b. the time for a child to remove his training wheels  
c. the average time for 14 children to remove their training wheels.  
d. t _{ 13 }  
e.

Solution to Exercise 8.9.11. (Return to Exercise)
a.
 
c.  
d.

Solution to Exercise 8.9.17. (Return to Exercise)
a.
 
c.  
d.

Solution to Exercise 8.9.19. (Return to Exercise)
a. t _{ 83 }  
b. average cost of 84 used cars  
c.

The next three problems refer to the following situation: Suppose that a sample of 15 randomly chosen people were put on a special weight loss diet. The amount of weight lost, in pounds, follows an unknown distribution with mean equal to 12 pounds and standard deviation equal to 3 pounds.
Exercise 8.10.1. (Go to Solution)
To find the probability that the average of the 15 people lose no more than 14 pounds, the random variable should be:
A. The number of people who lost weight on the special weight loss diet 
B. The number of people who were on the diet 
C. The average amount of weight lost by 15 people on the special weight loss diet 
D. The total amount of weight lost by 15 people on the special weight loss diet 
Exercise 8.10.3. (Go to Solution)
Find the 90th percentile for the average amount of weight lost by 15 people.
The next three questions refer to the following situation: The time of occurrence of the first accident during rushhour traffic at a major intersection is uniformly distributed between the three hour interval 4 p.m. to 7 p.m. Let X = the amount of time (hours) it takes for the first accident to occur.
So, if an accident occurs at 4 p.m., the amount of time, in hours, it took for the accident to occur is _______.
μ = _______
σ ^{ 2 } = _______
Exercise 8.10.4. (Go to Solution)
What is the probability that the time of occurrence is within the first halfhour or the last hour of the period from 4 to 7 p.m.?
A. Cannot be determined from the information given 
B. 
C. 
D. 
Exercise 8.10.5. (Go to Solution)
The 20th percentile occurs after how many hours?
A. 0.20 
B. 0.60 
C. 0.50 
D. 1 
Exercise 8.10.6. (Go to Solution)
Assume Ramon has kept track of the times for the first accidents to occur for 40 different days. Let C = the total cumulative time. Then C follows which distribution?
A. U ( 0,3 ) 
B. 
C. N ( 60 , 30 ) 
D. N ( 1 . 5,0 . 01875 ) 
Exercise 8.10.7. (Go to Solution)
Using the information in question #6, find the probability that the total time for all first accidents to occur is more than 43 hours.
The next two questions refer to the following situation: The length of time a parent must wait for his children to clean their rooms is uniformly distributed in the time interval from 1 to 15 days.
Exercise 8.10.8. (Go to Solution)
How long must a parent expect to wait for his children to clean their rooms?
A. 8 days 
B. 3 days 
C. 14 days 
D. 6 days 
Exercise 8.10.9. (Go to Solution)
What is the probability that a parent will wait more than 6 days given that the parent has already waited more than 3 days?
A. 0.5174 
B. 0.0174 
C. 0.7500 
D. 0.2143 
The next five problems refer to the following study: Twenty percent of the students at a local community college live in within five miles of the campus. Thirty percent of the students at the same community college receive some kind of financial aid. Of those who live within five miles of the campus, 75% receive some kind of financial aid.
Exercise 8.10.10. (Go to Solution)
Find the probability that a randomly chosen student at the local community college does not live within five miles of the campus.
A. 80% 
B. 20% 
C. 30% 
D. Cannot be determined 
Exercise 8.10.11. (Go to Solution)
Find the probability that a randomly chosen student at the local community college lives within five miles of the campus or receives some kind of financial aid.
A. 50% 
B. 35% 
C. 27.5% 
D. 75% 
Exercise 8.10.12. (Go to Solution)
Based upon the above information, are living in student housing within five miles of the campus and receiving some kind of financial aid mutually exclusive?
A. Yes 
B. No 
C. Cannot be determined 
Exercise 8.10.13. (Go to Solution)
The interest rate charged on the financial aid is _______ data.
A. quantitative discrete 
B. quantitative continuous 
C. qualitative discrete 
D. qualitative 
Exercise 8.10.14. (Go to Solution)
What follows is information about the students who receive financial aid at the local community college.
1st quartile = $250
2nd quartile = $700
3rd quartile = $1200
(These amounts are for the school year.) If a sample of 200 students is taken, how many are expected to receive $250 or more?
A. 50 
B. 250 
C. 150 
D. Cannot be determined 
The next two problems refer to the following information: P ( A ) = 0 . 2 , P ( B ) = 0 . 3 , A and B are independent events.
Exercise 8.10.17. (Go to Solution)
If H and D are mutually exclusive events, P ( H ) = 0 . 25 , P ( D ) = 0 . 15 , then P(H D )
A. 1 
B. 0 
C. 0.40 
D. 0.0375 
Class Time:
Names:
The student will calculate the 90% confidence interval for the average cost of a home in the area in which this school is located.
The student will interpret confidence intervals.
The student will examine the effects that changing conditions has on the confidence interval.
Check the Real Estate section in your local newspaper. (Note: many papers only list them one day per week. Also, we will assume that homes come up for sale randomly.) Record the sales prices for 35 randomly selected homes recently listed in the county.
Complete the table:
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
Compute the following:
a. = 
b. s _{ x } = 
c. n = 
Define the Random Variable , in words. =
State the estimated distribution to use. Use both words and symbols.
Calculate the confidence interval and the error bound.
a. Confidence Interval: 
b. Error Bound: 
How much area is in both tails (combined)? α =
How much area is in each tail? =
Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the upper and lower limits of the confidence interval and the sample mean.
Figure 8.5.
Some students think that a 90% confidence interval contains 90% of the data. Use the list of data on the first page and count how many of the data values lie within the confidence interval. What percent is this? Is this percent close to 90%? Explain why this percent should or should not be close to 90%.
In two to three complete sentences, explain what a Confidence Interval means (in general), as if you were talking to someone who has not taken statistics.
In one to two complete sentences, explain what this Confidence Interval means for this particular study.
Using the above information, construct a confidence interval for each confidence level given.
Confidence level  EBM / Error Bound  Confidence Interval 

50%  
80%  
95%  
99% 
What happens to the EBM as the confidence level increases? Does the width of the confidence interval increase or decrease? Explain why this happens.
Class Time:
Names:
The student will calculate the 90% confidence interval for proportion of students in this school that were born in this state.
The student will interpret confidence intervals.
The student will examine the effects that changing conditions have on the confidence interval.
Survey the students in your class, asking them if they were born in this state. Let X = the number that were born in this state.
a. n =____________ 
b. x =____________ 
Define the Random Variable P’ in words.
State the estimated distribution to use.
Calculate the confidence interval and the error bound.
a. Confidence Interval: 
b. Error Bound: 
How much area is in both tails (combined)? α =
How much area is in each tail? =
Fill in the blanks on the graph with the area in each section. Then, fill in the number line with the upper and lower limits of the confidence interval and the sample proportion.
Figure 8.6.
In two to three complete sentences, explain what a Confidence Interval means (in general), as if you were talking to someone who has not taken statistics.
In one to two complete sentences, explain what this Confidence Interval means for this particular study.
Using the above information, construct a confidence interval for each given confidence level given.
Confidence level  EBP / Error Bound  Confidence Interval 

50%  
80%  
95%  
99% 
What happens to the EBP as the confidence level increases? Does the width of the confidence interval increase or decrease? Explain why this happens.
Class Time:
Names:
The student will calculate a 90% confidence interval using the given data.
The student will examine the relationship between the confidence level and the percent of constructed intervals that contain the population average.
59.4  71.6  69.3  65.0  62.9 
66.5  61.7  55.2  67.5  67.2 
63.8  62.9  63.0  63.9  68.7 
65.5  61.9  69.6  58.7  63.4 
61.8  60.6  69.8  60.0  64.9 
66.1  66.8  60.6  65.6  63.8 
61.3  59.2  64.1  59.3  64.9 
62.4  63.5  60.9  63.3  66.3 
61.5  64.3  62.9  60.6  63.8 
58.8  64.9  65.7  62.5  70.9 
62.9  63.1  62.2  58.7  64.7 
66.0  60.5  64.7  65.4  60.2 
65.0  64.1  61.1  65.3  64.6 
59.2  61.4  62.0  63.5  61.4 
65.5  62.3  65.5  64.7  58.8 
66.1  64.9  66.9  57.9  69.8 
58.5  63.4  69.2  65.9  62.2 
60.0  58.1  62.5  62.4  59.1 
66.4  61.2  60.4  58.7  66.7 
67.5  63.2  56.6  67.7  62.5 
Listed above are the heights of 100 women. Use a random number generator to randomly select 10 data values.
Calculate the sample mean and sample standard deviation. Assume that the population standard deviation is known to be 3.3 inches. With these values, construct a 90% confidence interval for your sample of 10 values. Write the confidence interval you obtained in the first space of the table below.
Now write your confidence interval on the board. As others in the class write their confidence intervals on the board, copy them into the table below:
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
__________  __________  __________  __________  __________ 
The actual population mean for the 100 heights given above is μ = 63.4. Using the class listing of confidence intervals, count how many of them contain the population mean μ ; i.e., for how many intervals does the value of μ lie between the endpoints of the confidence interval?
Divide this number by the total number of confidence intervals generated by the class to determine the percent of confidence intervals that contains the mean μ . Write this percent below.
Is the percent of confidence intervals that contain the population mean μ close to 90%?
Suppose we had generated 100 confidence intervals. What do you think would happen to the percent of confidence intervals that contained the population mean?
When we construct a 90% confidence interval, we say that we are 90% confident that the true population mean lies within the confidence interval. Using complete sentences, explain what we mean by this phrase.
Some students think that a 90% confidence interval contains 90% of the data. Use the list of data given (the heights of women) and count how many of the data values lie within the confidence interval that you generated on that page. How many of the 100 data values lie within your confidence interval? What percent is this? Is this percent close to 90%?
Explain why it does not make sense to count data values that lie in a confidence interval. Think about the random variable that is being used in the problem.
Suppose you obtained the heights of 10 women and calculated a confidence interval from this information. Without knowing the population mean μ , would you have any way of knowing for certain if your interval actually contained the value of μ ? Explain.