Download Worksheet 8 (Chapter 5): Confidence Intervals 1. File MPG.xlsx

Worksheet 8 (Chapter 5): Confidence Intervals
Name: ______________________________________________
Section: _________________________
1. File MPG.xlsx contains the overall miles per gallon of a random sample of 2011 family sedans and a
random sample of small SUVs.
a. Construct a 95% confidence interval estimate by hand (using the formula) for the population
mean MPG of 2011 family sedans, assuming a normal distribution. (Be sure to check the
appropriate conditions for conducting a confidence interval of this type). [You may use EXCEL
to calculate mean and standard deviation of data]
Conditions: 1. Random Sample – stated in problem; 2. Sample mean normally distributed – met because
population is normally distributed.
𝑥𝑥̅ ± 𝑡𝑡𝛼𝛼⁄2
𝑠𝑠
√𝑛𝑛
25.78571 ± (2.160)
(21.542, 30.03)
7.350338
√14
b. Use DDXL to construct the interval you found in part (a). (Insert the resulting output onto this
document or attach in appendix – must give output from DDXL).
c. Interpret the interval constructed in part (a) and (b).
We are 95% confident that the population mean miles per gallon of 2011 family sedans lies between
21.542 and 30.030 miles per gallon.
d. Construct a 95% confidence interval estimate for the population mean MPG of 2011 small
SUVs, assuming a normal distribution.
Conditions: 1. Random Sample – stated in problem; 2. Sample mean normally distributed – met because
population is normally distributed.
𝑥𝑥̅ ± 𝑡𝑡𝛼𝛼⁄2
𝑠𝑠
√𝑛𝑛
20.4 ± (2.064)
3.8297
√25
(18.819, 21.981)
Worksheet 8 (Chapter 5): Confidence Intervals
Name: ______________________________________________
Section: _________________________
e. Use DDXL to construct the interval you found in part (d). (Insert the resulting output onto this
document or attach in appendix - must give output from DDXL).
f. Interpret constructed in (d).
We are 95% confident that the population mean miles per gallon of 2011 small SUVs lies
between 18.819 and 21.989 miles per gallon.
g. Based on your intervals found in parts (a/b) and (d/e) would you say the population mean miles
per gallon for family sedans is less than the population mean miles per gallons for small SUVs?
Explain.
Not necessarily – the two intervals overlap and therefore it is possible the population mean mpg
for family sedans and small SUVs could be equivalent or mpg for small SUVS could be larger
than mpg for family sedans.
2. Have you ever negotiated a pay raise? According to an Accenture survey, 52% of US workers have.
a. Suppose that the survey (of randomly selected workers) had a sample size of n = 50. Construct
by hand (using the formula) a 99% confidence interval for the proportion of all US workers who
have negotiated a pay raise. (Be sure to check the appropriate conditions for conducting a
confidence interval of this type).
Conditions: 1. Random Sample – stated in the problem; 2. 𝑛𝑛𝑝𝑝̂ ≥ 15 𝑎𝑎𝑎𝑎𝑎𝑎 𝑛𝑛𝑞𝑞� ≥ 15  (50)(0.52) = 26
≥15 and (50)(0.48) = 24 ≥15
𝑝𝑝̂ (1 − 𝑝𝑝̂ )
𝑝𝑝̂ ± 𝑧𝑧𝛼𝛼⁄2 �
𝑛𝑛
0.52(1 − 0.52)
0.52 ± (2.575)�
50
(0.338,0.702)
Worksheet 8 (Chapter 5): Confidence Intervals
Name: ______________________________________________
Section: _________________________
b. Create an Excel document of the raw data from the sample survey conducted. Use DDXL to
construct the 99% confidence interval found in part (a). (Insert the resulting output onto this
document of attach in appendix – must give output from DDXL).
c. Interpret the interval found in parts (a) and (b).
We are 99% confident that the population proportion of US workers who have negotiated a pay raise
lies between 0.338 and 0.702.
d. Based on the interval found in parts (a) and (b), can you claim that more than half of all US
workers have negotiated a pay raise? Explain.
Not necessarily – although the interval includes 0.5, in order to ensure “more than half” of all US
workers have negotiated a pay raise the entire interval would need be greater than 0.5.
e. Repeat part (a) assuming that the survey had a sample size of n = 5000. (Use the same resulting
sample proportion).
Conditions: 1. Random Sample – stated in the problem; 2. 𝑛𝑛𝑝𝑝̂ ≥ 15 𝑎𝑎𝑎𝑎𝑎𝑎 𝑛𝑛𝑞𝑞� ≥ 15  (5000)(0.52) =
2600 ≥15 and (5000)(0.48) = 2400 ≥15
𝑝𝑝̂ (1 − 𝑝𝑝̂ )
𝑝𝑝̂ ± 𝑧𝑧𝛼𝛼⁄2 �
𝑛𝑛
0.52(1 − 0.52)
0.52 ± (2.575)�
5000
(0.5018,0.5382)
f. Based on the interval found in part (e), can you claim that more than half of all US workers have
negotiated a pay raise? Explain.
Yes – the entire interval is greater than 0.50 therefore we can claim that more than half of all US
workers have negotiated a pay raise.
g. Discuss the effect of sample size on confidence interval estimation.
The increase in sample size decreased the width of the interval.
Worksheet 8 (Chapter 5): Confidence Intervals
Name: ______________________________________________
Section: _________________________
3. A simple random sample of 100 customers were surveyed to determine the satisfaction with the updated
version of the automatic teller machine at the bank. Of those surveyed 12 were unsatisfied with the
updated machine. Use the appropriate methods to construct a 98% confidence interval for the
population proportion of customers that are unsatisfied with the updated machine. Interpret the
interval found. (Be sure to check the appropriate conditions for conducting a confidence interval
of this type).
Conditions: 1. Random Sample – stated in the problem; 2. 𝑛𝑛𝑝𝑝̂ ≥ 15 𝑎𝑎𝑎𝑎𝑎𝑎 𝑛𝑛𝑞𝑞� ≥ 15  12 < 15 and 10012 = 88 ≥ 15 Will need to do Wilson’s adjustment.
𝑝𝑝� =
14
𝑥𝑥 + 2
=
= 0.1346
𝑛𝑛 + 4 104
𝑝𝑝�(1 − 𝑝𝑝�)
𝑝𝑝� ± 𝑧𝑧𝛼𝛼⁄2 �
𝑛𝑛 + 4
0.1346(1 − 0.1346)
0.1346 ± (2.33)�
104
(0.0566,0.2126)
We are 98% confident that the population proportion of customers that are unsatisfied with the updated
machine lies between 0.0566 and 0.2126.
4. If the manager of a paint supply store wants to estimate, with 95% confidence, the mean amount of paint
in a 1-gallon can to within ± 0.004 gallon and also assumes that the standard deviation is 0.02 gallon,
what sample size is needed?
𝑛𝑛 =
2
�𝑧𝑧𝛼𝛼⁄2 � 𝜎𝜎 2
(𝑆𝑆𝑆𝑆)2
=
(1.96)2 (0.02)2
(0.004)2
= 96.04  97 1 gallon cans
5. In response to the question “How do you judge a company?” 84% said the most important way was how a
company responded to a crisis.
a. If you conduct a follow-up study to estimate the population proportion of individuals who said that the
most import way to judge a company was how the company responded to a crisis, would you use a p of
0.84 or 0.50 in the sample size formula? Discuss.
p=0.84 because the previous study gives us some indication of a value for p.
b. Using your answer to (a), find the sample size necessary to estimate, with 95% certainty, the population
proportion to within ± 0.03.
𝑛𝑛 =
2
�𝑧𝑧𝛼𝛼⁄2 � (𝑝𝑝)(𝑞𝑞)
(𝑆𝑆𝑆𝑆)2
=
(1.96)2 (0.84)(0.16)
(0.03)2
= 573.68  574 people