MATH 225N MATHWeek 6 Confidence Intervals Questions and answers ALL ANSWERS CORRECT & VERI, Exams of Mathematics

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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.

(0.42 - 0.07), (0.42 + 0.07) = (0.35), (0.49)

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.

(31-3), 31+3) = (28), (34)

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.

(0.83 – 0.11), (0.83 + 0.11) = (0.72), (0.94)

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.

(.25 - .05). (.25 + .05) = (0.20), (.30)

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.

Since 95% falls in 2 SD’s the calculation would be (8.2 – 4.6) “4.6 is the margin of error”, (8.2 + 4.6) = (3.6) , (12.8)

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.

(0.58 – 3 x 0.08), (0.58 + 3 x 0.08) =(0.58 – .24), (0.58 + .24) = 0.34 + 0.

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

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use increasing order.

Use week 6 worksheet to get mean and SD.

Mean 23. Sample Standard Deviation 1.

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Confidence Level n Mean StDev pop stdev

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t Margin of Error Lower Limit Upper Limit

Lower margin of error = 22.69 and upper limit is 23.

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.

Confidence Level

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n Mean StDev pop stdev

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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 = 0.20 12

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.

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 =

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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

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Lower limit = 241.42 Upper Limit = 262.

Confidence 0.

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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.

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

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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

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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.

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. 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. Error 0.05 Write percentage as decimal

z-Value 1. 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

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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

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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.

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

zα 2 = 2.

(89.02, 94.98)

t or z Confidence Interval for μ

Confidence Level n Mean StDev pop stdev

SE

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z Margin of Error Lower Limit Upper Limit

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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.

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 52 days, 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 60 miles per hour, find a 98% confidence interval for the population mean.

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• Round the final answer to two decimal places.

Answer 56.36 – 63.

t or z Confidence Interval for μ

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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 6 minutes and an unknown population mean. A random sample

of 18 cyclists is taken and gives a sample mean of 146 minutes.

Find the confidence interval for the population mean with a 99% confidence level.

Answer: 142.36 – 149.

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 3 days, 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

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z-Value Minimum Sample Size 25

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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.

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.

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25.8 Mean 23. 22.2 Sample Standard Deviation 1.

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t or z Confidence Interval for μ

Confidence Level

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Mean

StDev pop stdev

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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.

59.015 Mean 59. 58.962 Sample Standard Deviation 0.

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t or z Confidence Interval for μ

Confidence Level

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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

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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 0.900

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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

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t Margin of Error Lower Limit

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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

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