Week 6 Questions on Confidence Intervals
1. On a busy Sunday morning, a waitress randomly sampled customers about their preference for morning beverages, Specifically, she wanted to find out how many people preferre
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Week 6 Questions on Confidence Intervals
1. On a busy Sunday morning, a waitress randomly sampled customers about their preference for morning beverages, Specifically, she wanted to find out how many people preferred coffee over tea. The proportion of customers that preferred coffee was 0.42 with a margin of error 0.07.
Construct a confidence interval for the proportion of customers that preferred coffee.
2. A company sells juice in 1quart bottles. In a quality control test, the company found the mean volume of juice in a random sample of bottles was X = 31 ounces, with a marginal error of 3 ounces.
Construct a confidence interval for the mean number of ounces of juice bottled by this company.
3. Randomly selected employees at an office were asked to take part in a survey about overtime. The office manager wanted to find out how many employees worked overtime in the last week. The proportion of employees that worked overtime was 0.83, with a margin of error of 0.11.
4. A random sample or garter snakes were measured, and the proportion of snakes that were longer than 20 inches in length recorded. The measurements resulted in a sample proportion of p = 0.25 with a sampling standard deviation of Op = 0.05.
Write a 68% confidence interval for the true proportion of garter snakes that were over 20 inches in length.
5. The average number of onions needed to make French onion soup from the population of recipes is unknown. A random sample of recipes yields a sample mean of x = 8.2 onions. Assume the sampling distribution of the mean has a standard deviation of 2.3 onions.
Use the Empirical Rule to construct a 95% confidence interval for the true population mean number of onions.
6. In a survey, a random sample of adults were asked whether a tomato is a fruit or vegetable. The survey resulted in a sample proportion of 0.58 with a sampling standard deviation of 0.08 who stated a tomato is a fruit.
Write a 99.7 confidence interval for the true proportion of number of adults who stated the tomato is a fruit.
7. A college admissions director wishes to estimate the mean number of students currently enrolled. The age of random sample of 23 students is given below. Assume the ages are approximately normally distributed. Use Excel to construct a 90% confidence interval for the population mean age. Round your answer to 2 decimal places and use increasing order.
Use week 6 worksheet to get mean and SD.
Mean 23.1043
Sample Standard Deviation 1.3693
22.5
22.8
24.6
24
22.6
23.6
22.8
23.1
21.5
21.4
22.5
24.5
21.5
22.5
20.5
23
25.1
25.2
23.8
21.8
24.1
Confidence
Level
n Mean StDev pop stdev
SE
t Margin of
Error Lower Limit Upper Limit
Lower margin of error = 22.69 and upper limit is 23.51
8. Suppose that the scores of bowlers in a particular league follow a normal distribution such that a standard deviation of the population is 12. Find the 95% confidence interval of the mean score for all bowlers in this league using the accompanying data set of 40 random scores. Round your answers to 2 decimal places using ascending order.
Lower Limit = 90.78 Upper Limit = 98.22
Confidence
Level
n Mean StDev pop stdev
SE
z Margin of
Error Lower Limit Upper Limit
9. In the survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for the proportion of adults who regularly lie to people conducting surveys. Use excel to create the confidence interval rounding to 4 decimal places.
Lower Limit = 0.1238 Upper Limit = 0.2012
Confidence Level
n Number of Successes
Sample Proportion
SE
z Margin of Error Lower Limit Upper Limit
10. In a random sampling of 350 attendees at a minor league baseball game, 184 said that they bought food from the concession stand. Create a 95%confidence interval for the proportion of fans who bought food from the concession stand. Use excel to create the confidence interval rounding to 4 decimal places.
Lower limit = 0.4734 Upper Limit = 0.5780
Confidence Level
n Number of Successes
Sample Proportion
SE
z Margin of Error Lower Limit Upper Limit
11. Suppose that the weight of tight ends in a football league are normally distributed such that sigma squared =
1,369. A sample of 49 tight ends was randomly selected and the weights are given in the table below. Use Excel to create a 95% confidence interval for the mean weight of the tight ends in this league. Rounding your answers to 2 decimal places and using ascending order. (Have to get square root of 1369 which is 37). Population sample is yes .
Lower limit = 241.42 Upper Limit = 262.14
Confidence
Level
n Mean StDev pop stdev
SE
z Margin of
Error Lower Limit Upper Limit
12. Suppose heights, in inches of orangutans are normally distributed and have a known population standard deviation of 4 inches. A random sample of 16 orangutans is taken and gives a sample mean of 56 inches. Find the confidence interval of the population mean with a 95% confidence level.
Lower limit = 54.04 and Upper Limit = 57.96
13. The population standard deviation for the total snowfalls per year in a city is 13 inches. If we want to be 95% confident that the sample mean is within 3 inches of the true population mean, what is the minimum sample size that should be taken?
Answer: 73 snowfalls
Minimum Sample Size μ for population mean
Confidence Level
StDev Error
z-Value Minimum Sample Size
14. The population standard deviation for the body weights for employees of a company is 10 pounds. If we want
to be 95% confident that the sample mean is within 3 pounds of the true population mean, what is the minimum sample size that should be taken.
Answer: 43 employees
Minimum Sample Size μ for population mean
Confidence Level
StDev Error
z-Value Minimum Sample Size
15. The length, in words, of the essays written for a contest are normally distributed with a population standard
deviation of 442 words and an unknown population mean. If random sample of 24 essays is taken and results in a
sample mean of 1330 words, find a 99% confidence interval for the population mean. Round to two decimal places.
Answer: Lower limit = 1097.59 upper Limit = 1562.41
16. Brenda wants to estimate the percentage of people who eat fast food at least once per week. She wants to create a 95% Confidence interval which has an error bound of at most 2%. How many people should be polled to create the confidence interval?
Answer: 2401
Minimum Sample Size p for Proportion
Confidence Level Sample Proportion
Error
Enter decimal
If sample proportion unknown enter 0.5 Write percentage as decimal
z-Value Minimum Sample Size
17. Suppose a clothing store wants to determine the current percentage of customers who are over the age of forty.
How many customers should the company survey in order to be 92% confident that the estimated (sample) proportion is within 5% of the true population proportion of customers who are over the age of 40?
Answer: 307
Minimum Sample Size p for Proportion
Confidence Level 0.920 Enter decimal
Sample Proportion 0.5 If sample proportion unknown enter 0.5
Error 0.05 Write percentage as decimal
z-Value 1.751
Minimum Sample Size 307
18. Suppose the scores of a standardized test are normally distributed. If the population standard deviation is 2 points, what minimum sample size is needed to be 90% confident that the sample mean is within 1 point of the true population mean? Be sure to round up to the nearest integer.
Provide your answer below: 11
Minimum Sample Size μ for population mean
Confidence Level
StDev Error
z-Value Minimum Sample Size
19. The number of square feet per house are normally distributed with a population standard deviation
of 197 square feet and an unknown population mean. If a random sample of 25 houses is taken and results
in a sample mean of 1820 square feet, find a 99% confidence interval for the population mean. Round to 2 decimal places.
Answer: 1718.51 – 1921.49
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
z Margin of Error Lower Limit Upper Limit
20. Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 6 points and an unknown population mean. A random sample of 22 scores is taken and gives a sample mean of 92 points.
Identify the parameters needed to calculate a confidence interval at the 98% confidence level. Then find the confidence interval.
x = 92
σ = 6
n = 22
zα2= 2.326
(89.02, 94.98)
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
z Margin of Error Lower Limit Upper Limit
21. Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 6 points and an unknown population mean. A random sample of 22 scores is taken and gives a sample mean of 92 points.
What is the correct interpretation of the 95% confidence interval?
We can estimate that 98% of the time the test is taken, a student scores between 89.02 and 94.98 points.
We can estimate with 98% confidence that the true population mean score is between 89.02 and 94.98 points.
We can estimate with 98% confidence that the sample mean score is between 89.02 and 94.98 points
22. The weights of running shoes are normally distributed with a population standard deviation of 3 ounces and an unknown population mean. If a random sample of 23 running shoes is taken and results in a sample mean of 18 ounces, find a 90%confidence interval for the population mean. Round the final answer to two decimal places.
Answer: 16.97 – 19.03
23. The germination periods, in days, for grass seed are normally distributed with a population standard
deviation of 5 days and an unknown population mean. If a random sample of 17 types of grass seed is taken and results in a sample mean of 52days, find a 80% confidence interval for the population mean.
Select the correct answer below:
(50.45,53.55)
(50.01,53.99)
(49.85,54.15)
(49.62,54.38)
(49.18,54.82)
(48.88,55.12)
24. The speeds of vehicles traveling on a highway are normally distributed with a population standard
deviation of 7 miles per hour and an unknown population mean. If a random sample of 20 vehicles is taken and results in a sample mean of 60miles per hour, find a 98% confidence interval for the population mean.
• Round the final answer to two decimal places.
Answer 56.36 – 63.64
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
z Margin of Error Lower Limit Upper Limit
25. Suppose finishing time for cyclists in a race are normally distributed and have a known population standard deviation of 6minutes and an unknown population mean. A random sample of 18 cyclists is taken and gives a sample mean of 146minutes.
Find the confidence interval for the population mean with a 99% confidence level.
Answer: 142.36 – 149.64
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
z Margin of Error Lower Limit Upper Limit
26. Suppose the germination periods, in days, for grass seed are normally distributed. If the population
standard deviation is 3days, what minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean?
Answer: 25 seeds
Minimum Sample Size μ for population mean
Confidence Level
StDev Error
z-Value Minimum Sample Size
27. Suppose the number of square feet per house is normally distributed. If the population standard deviation is 155 square feet, what minimum sample size is needed to be 90% confident that the sample mean is within 47 square feet of the true population mean?
Answer: 30 houses
Minimum Sample Size μ for population mean
Confidence Level
StDev Error
z-Value Minimum Sample Size
28. In a survey of 1,000 adults in a country, 722 said that they had eaten fast food at least once in the past
month. Create a 95% confidence interval for the population proportion of adults who ate fast food at least once in the past month. Use Excel to create the confidence interval, rounding to four decimal places.
Answer: 0.6942 – 0.7498
Confidence Interval for p Proportions
Confidence Level
n Number of Successes
Sample Proportion
SE
z Margin of Error Lower Limit Upper Limit
29. A college admissions director wishes to estimate the mean age of all students currently enrolled. The age of a random sample of 23 students is given below. Assume the ages are approximately normally distributed. Use Excel to construct a 90% confidence interval for the population mean age. Round your answers to two decimal places and use increasing order.
Answer: 22.61 – 23.59
Data
25.8 Mean 23.1043
22.2 Sample Standard Deviation 1.3693
22.5
22.8
24.6
24
22.6
23.6
22.8
23.1
21.5
21.4
22.5
24.5
21.5
22.5
20.5
23
25.1
25.2
23.8
21.8
24.1
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
t Margin of Error Lower Limit
Upper Limit
30. The yearly incomes, in thousands, for 24 random married couples living in a city are given below. Assume the yearly incomes are approximately normally distributed. Use Excel to find the 95% confidence interval for the true mean, in thousands. Round your answers to three decimal places and use increasing order.
Answer: 58.984 – 59.026
59.015 Mean 59.0050
58.962 Sample Standard Deviation 0.0494
58.935
58.989
58.997
58.97
59
59.014
59.001
59.003
58.992
58.926
59.032
58.958
59.093
58.955
59.003
58.952
59.057
59.056
59.074
59.128
59.001
59.007
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
t Margin of Error Lower Limit
Upper Limit
31. A tax assessor wants to assess the mean property tax bill for all homeowners in a certain state. From a survey ten years ago, a sample of 28 property tax bills is given below. Assume the property tax bills are approximately normally distributed. Use Excel to construct a 95% confidence interval for the population mean property tax bill. Round your answers to two decimal places and use increasing order.
Answer: 1185.91 – 1595.59
32. The table below provides a random sample of 20 exam scores for a large geology class. Use Excel to construct a 90% confidence interval for the mean exam score of the class. Round your answers to one decimal place and use ascending order.
Answer: 79.7 – 88.5
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
t Margin of Error Lower Limit Upper Limit
33. Suppose scores on exams in statistics are normally distributed with an unknown population mean. A
sample of 26 scores is given below. Use Excel to find a 90% confidence interval for the true mean of statistics exam scores. Round your answers to one decimal place and use increasing order.
Answer: 67.2 – 69.4
Confidence
Level
n Mean StDev pop stdev
SE
t Margin of
Error Lower Limit Upper Limit
33. In a city, 22 coffee shops are randomly selected, and the temperature of the coffee sold at each shop is noted. Use Excel to find the 90% confidence interval for the population mean temperature.
Assume the temperatures are approximately normally distributed. Round your answers to two decimal places and use increasing order.
Answer: 153.21 – 161.43
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
t Margin of Error Lower Limit
Upper Limit
34. Weights, in pounds, of ten-year-old girls are collected from a neighborhood. A sample of 26 is given below. Assuming normality, use Excel to find the 98% confidence interval for the population mean weight μ. Round your answers to three decimal places and use increasing order.
Answer: 66.497 – 77.234
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
t Margin of Error Lower Limit Upper Limit
35. A sample of 22 test-tubes tested for number of times they can be heated on a Bunsen burner before they
crack is given below. Assume the counts are normally distributed. Use Excel to construct a 99% confidence interval for μ. Round your answers to two decimal places and use increasing order.
Answer: 1071.77 – 1477.33
36. The monthly incomes from a random sample of 20 workers in a factory is given below in dollars. Assume the population has a normal distribution and has standard deviation $518. Compute a 98% confidence interval for the mean of the population. Round your answers to the nearest dollar and use ascending order.
Answer: 11,833 – 12,372
37. Assume the distribution of commute times to a major city follows the normal probability distribution and the standard deviation is 4.5 minutes. A random sample of 104 commute times is given below in minutes. Use Excel to find the 98%confidence interval for the mean travel time in minutes. Round your answers to one decimal place and use ascending order.
Answer: 25.9 – 27.9
38. Installation of a certain hardware takes a random amount of time with a standard deviation of 7 minutes. A computer technician installs this hardware on 50 different computers. These times are given in the accompanying dataset. Compute a 95% confidence interval for the mean installation time. Round your answers to two decimal places and use ascending order.
Answer: 40.76 – 44.64
Confidence
Level
n Mean StDev pop stdev
SE
z Margin of
Error Lower Limit Upper Limit
39. Assume that farm sizes in a particular region are normally distributed with a population standard deviation of 150 acres. A random sample of 50 farm sizes in this region is given below in acres. Estimate the mean farm size for this region
with 90%confidence. Round your answers to two decimal places and use ascending order.
Answer: 474.87 – 544.65
40. The amounts of time that customers stay in a certain restaurant
for lunch is normally distributed with a standard deviation of 17 minutes. A random sample of 50 lunch customers was taken at this restaurant. Construct a 99% confidence interval for the true average amount of time customers spend in the restaurant for lunch. Round your answers to two decimal places and use ascending order.
Answer: 44.89 – 57.27
41. Recent studies have shown that out of 1,000 children, 885 children like ice cream. What is the 99% confidence interval for the true proportion of children who like ice cream, based on this sample? Round z⋆ to two decimal places and other answers to four decimal places.
Provide your answer below: .8590 - .9110
Confidence Interval for p Proportions
Confidence Level
n Number of Successes
Sample Proportion
SE
z Margin of Error Lower Limit Upper Limit
Enter decimal
42. A large company is concerned about the commute times of its employees. 333 employees were surveyed, and 131employees said that they had a daily commute longer than 30 minutes. Create a 95% confidence interval for the proportion of employees who have a daily commute longer
than 30 minutes. Use Excel to create the confidence interval, rounding to four decimal places.
Provide your answer below: .3409 - .4459
Confidence Interval for p Proportions
Confidence Level
n Number of Successes
Sample Proportion
SE
z Margin of Error Lower Limit Upper Limit
43. The following data represent a random sample for the ages of 41 players in a baseball league. Assume
that the population is normally distributed with a standard deviation of 2.1 years. Use Excel to find
the 98% confidence interval for the true mean age of players in this league. Round your answers to three decimal places and use ascending order.
Answer: 27.579 – 29.104
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
z Margin of Error Lower Limit Upper Limit
44. In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample
of 25 pieces of carry-on luggage was collected and weighed in pounds. Assume that the population is normally distributed with a standard deviation of 5 pounds. Find the 95% confidence interval of the mean weight in pounds. Round your answers to two decimal places and use ascending order.
Answer: 15.36 – 19.28
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
z Margin of Error Lower Limit Upper Limit
45. A company wants to determine a confidence interval for the average CPU time of its teleprocessing
transactions. A sample of 70 random transactions in milliseconds is given below. Assume that the transaction times follow a normal distribution with a standard deviation of 600 milliseconds. Use Excel to determine
a 98% confidence interval for the average CPU time in milliseconds. Round your answers to the nearest integer and use ascending order.
Answer: 5907 – 6240
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
z Margin of Error Lower Limit Upper Limit
46. The number of hours worked per year per adult in a state is normally distributed with a standard deviation
of 37. A sample of 115 adults is selected at random, and the number of hours worked per year per adult is given below. Use Excel to calculate the 98% confidence interval for the mean hours worked per year for adults in this state. Round your answers to two decimal places and use ascending order.
Answer: 2090.03 – 2106.09
47. An automobile shop manager timed 27 employees and recorded the time, in minutes, it took them to change a water pump. Assuming normality, use Excel to find the 99% confidence interval for the true mean. Round your answers to three decimal places and use increasing order.
Answer: 15.499 – 19.139
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
t Margin of Error Lower Limit Upper Limit
48. A type of golf ball is tested by dropping it onto a hard surface from a height of 1 meter. The height it
bounces is known to be normally: distributed. A sample of 25 balls is tested and the bounce heights are given below. Use Excel to find a 95%confidence interval for the mean bounce height of the golf ball. Round your answers to two decimal places and use increasing order.
Answer: 79.95 – 82.62
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
t Margin of Error Lower Limit Upper Limit
49. The heart rates for a group of 21 students taking a final exam are given below. Assume the heart rates are
normally distributed. Use Excel to find the 95% confidence interval for the true mean. Round your answers to two decimal places and use increasing order.
Answer: 91.31 – 95.17
t or z Confidence Interval for µ
Confidence Level
n Mean StDev pop stdev
SE
t Margin of Error
Lower Limit Upper Limit
50. Suppose a clothing store wants to determine the current percentage of customers who are over the age of forty.
How many customers should the company survey in order to be 90% confident that the estimated (sample) proportion is within 4percentage points of the true population proportion of customers who are over the age of forty?
Answer: 423
Minimum Sample Size p for Proportion
Confidence Level Sample Proportion
Error
z-Value Minimum Sample Size
51. Virginia wants to estimate the percentage of students who live more than three miles from the school. She
wants to create a 95% confidence interval which has an error bound of at most 5%. How many students should be polled to create the confidence interval?
Answer: 385
Minimum Sample Size p for Proportion
Confidence Level 0.950
Sample Proportion 0.5
Error 0.05
z-Value 1.960
Minimum Sample Size 385
52. Suppose an automotive repair company wants to determine the current percentage of customers who keep up with regular vehicle maintenance. How many customers should the company survey in order to
be 95% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion of customers who keep up with regular vehicle maintenance?
Answer: 601
Minimum Sample Size p for Proportion
Confidence Level Sample Proportion
Error
z-Value Minimum Sample Size
53. Suppose a clothing store wants to determine the current percentage of customers who are over the age of
forty. How many customers should the company survey in order to be 92% confident that the estimated
(sample) proportion is within 5percentage points of the true population proportion of customers who are over the age of forty?
Answer: 307
Minimum Sample Size p for Proportion
Confidence Level Sample Proportion
Error
z-Value Minimum Sample Size
54. The average height of a population is unknown. A random sample from the population yields a sample mean of x¯=66.3inches. Assume the sampling distribution of the mean has a standard deviation of σx¯=0.8 inches.
Use the Empirical Rule to construct a 95% confidence interval for the true population mean height. Provide your answer below: 64.7 – 67.9
55. In a random sample of 30 young bears, the average weight at the age of breeding is 312 pounds. Assuming the population ages are normally distributed with a population standard deviation is 30 pounds, use the Empirical Rule to construct a 68%confidence interval for the population average of young bears at
the age of breeding. Do not round intermediate calculations. Round only the final answer to the nearest pound. Remember to enter the smaller value first, then the larger number.
Answer: 307 – 317
56. In a food questionnaire, a random sample of teenagers were asked whether they like pineapple pizza. The questionnaire resulted in a sample proportion of p′=0.43, with a sampling standard deviation of σp′=0.06, who like this type of pizza.
Write a 99.7% confidence interval using the Empirical Rule for the true proportion of teenagers who like pineapple pizza.
Answer: 0.25 - 0.61
57. A marine biologist is interested in whether the Chinook salmon, a particular species of salmon in
the Pacific Northwest, are getting smaller within the last decade. In a random sample of this species of salmon, she found the mean length was x¯=36inches with a margin of error of 9 inches.
Construct a confidence interval for the mean length of Chinook salmon.
Answer: 27 - 45
58. A researcher is trying to estimate the population mean for a certain set of data. The sample mean is 45, and the error bound for the mean is 9, at a 99.7% confidence level. (So, x¯=45 and EBM = 9.) Find and
interpret the confidence interval estimate.
Answer: We can estimate, with 99.7% confidence that the true value of the population mean is between 36 and 54.
59. A random sample of registered voters were asked about an issue on the ballot of an upcoming election. The proportion of those surveyed who plan to vote "Yes" on the issue is 0.54, with a margin of error of 0.06.
Construct a confidence interval for the proportion of registered voters that plan to vote "Yes" on the issue.
Answer: .48 - .60
(0.54-0.06 ; 0.54 + 0.06)
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