Statistics – 8.1 Confidence Interval Practice.
1. The science behind lie detectors is based on the assumption that lying will be reflected in physiological changes in
the nervous system and are not under the voluntary control of the subject. When a person is telling the truth, the
galvanic skin response scores have a normal distribution with a mean of 49.4 and a standard deviation of 3.
a. What is the probability that a person will have a score less than 45?
7.12%
b. Suppose a simple random sample of 36 people is taken, describe the distribution of Ë?
Normally distributed because n = 36 and 36 30. Ë = 49.4, Ë = 8.23
c. What is the probability that the average score will be more than 50?
42.07%
2. You work for a consumer advocate agency and want to find the mean repair cost for a washing machine. As part of
your study, you randomly select 40 repair costs and find the mean to be $100.00 with a sample standard deviation
of $17.50. Construct a 90% confidence interval for the unknown mean repair costs. Interpret the result.
t-distribution Lower Bound: 95.45; Upper Bound: 104.55.
I am 90% confident that the mean repair costs, , is between $95.45 and $104.55.
3. Fill in the blanks with on of the following: increases, decreases, or stays the same. Then explain your choice.
a. As the sample size increases, the margin of error _decreases_.
Explain. We divide by
. As n gets larger, the denominator increases and the margin of error shrinks
b. As the confidence level increases, the margin of error _increases_.
Explain. As the width of the confidence interval increases, so does the margin of error.
c. As the standard deviation increases, the margin of error _increases_.
Explain.  is in the numerator. As the numerator gets larger, the margin of error also gets larger.
4. A company that produces white bread is concerned about the amount of sodium in its bread. The company takes a
simple random sample of 100 slices of bread and computes the sample mean to be 103 milligrams of sodium per
slice with a standard deviation of 10 milligrams. Construct a 99% confidence interval for the unknown mean sodium
level. Interpret the result.
t-distributionLower Bound: 100.42; Upper Bound: 105.58.
I am 99% sure that the mean sodium level, , is between 100.42 mg/slice and 105.58 mg/slice.
5. Suppose you want to rent an unfurnished one-bedroom apartment for next semester. The mean monthly rent for a
random sample of 30 apartments advertised locally is $540. With a standard deviation of $31.28.
a. Find a 90% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments in Forest
Grove. Interpret your interval.
Lower Bound: 530.61; Upper Bound: 549.39.
I am 90% sure that the mean cost for a 1 bedroom apartment, , is between $530.61 and $549.39.
i. Find the margin of error. Interpret the margin of error.
$9.39 The mean cost for a 1 bedroom apartment, , is within $9.39 of $540.
b. Find a 95% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments in Forest
Grove. Interpret your interval.
Lower Bound: 528.81; Upper Bound: 551.19.
t-distribution
I am 95% sure that the mean cost for a 1 bedroom apartment, , is between $528.81 and $551.19.
i. Find the margin of error. Interpret the margin of error.
$11.19 The mean cost for a 1 bedroom apartment, , is within $11.19 of $540.
c. Find a 99% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments in Forest
Grove. Interpret your interval.
Lower Bound: 529.29; Upper Bound: 554.71.
I am 99% confident that the mean cost for a 1 bedroom apartment, , is between $529.29 and $554.71.
i. Find the margin of error. Interpret the margin of error.
$14.71 The mean cost for a 1 bedroom apartment, , is within $14.71 of $540.
d.
Suppose you want a 90% confidence interval with a width of no more than $10. What sample size would you
need?
You would need 106 apartments in your sample.
6. Your company sells exercise clothing and equipment on the Internet. To design the clothing, you collect data on the
physical characteristics of you customers. Here are the weights, in kilograms, for a sample of 24 male runners.
Suppose the standard deviation of the normally distributed population is known to be 4.5 kg.
67.8 61.9 63.0 53.1 62.3 59.7 55.4 58.9 60.9 69.2 63.7 68.3 64.7 65.6 56.0 57.8 66.0 62.9 53.6 65.0 55.8 60.4 69.3 61.7
a.
Give a 95% confidence interval for the mean of the population from which the sample is drawn.
Lower Bound: 59.99; Upper Bound: 63.59.
t-distribution
I am 95% certain that the average weight of the male runners is between 59.99 kg and 63.59 kg.
b. Will the interval contain the weights of approximately 95% of all similar runners? Explain.
Your thoughts? No. Can you explain why not? Hint: Think about 95% of the population & the sample mean.
c. Find the margin of error for the 95% confidence interval found in part a. Interpret the margin of error.
1.80 kg. The mean weight of male runners, , is within 1.80 kg of 61.79 kg.
d. Does the confidence interval give us information about the statistic or the parameter? What is the difference?
Parameter. We are using the sample to predict the population mean. A statistic, ex: Ë, describes a sample
and a parameter, ex: , describes the population.
e. Suppose an addition male runner had a reported weight of 92.3 kg. Find the 95% interval for the new data and
compare it to part a.
Lower Bound: 61.25; Upper Bound: 64.78.
The mean is increased, the confidence intrerval’s width is smaller, the margin of error is smaller.
f. Find a 99% confidence interval for the mean weight of the population before adding in the weight of the runner
in part c. – when n = 24. Is the 99% interval wider or narrower than the 95% interval found in part a? Explain in
simple language why this is true.
Lower Bound: 59.43; Upper Bound: 64.16.
The 99% interval is wider than the 95% interval. Your thoughts as to why?
g. Suppose you want the 99% interval to be the same width as the 95% interval from part a. How many additional
runners would you need to include in your sample?
You would need 42 people in your sample.