QMB 3200 ch 8: Interval Estimation

What is zα/2 for a 95% confidence interval of the population mean?

1.96 Excel functions are = NORM.S.INV(1 − 0.05/2) = 1.96 or = NORM.INV(1 − 0.05/2,0,1) = 1.96

What is tα/2,df for a 95% confidence interval of the population mean based on a sample of 15 observations?

2.145 Excel function is = T.INV(1-0.05/2,15-1) = 2.145

What is tα/2,df for a 99% confidence interval of the population mean based on a sample of 25 observations?

2.797 Excel function is = T.INV(1-0.01/2,25-1) = 2.797

For a given confidence level (1 − α)100% and sample size n, which of the following is true in the interval estimation of the population mean when the population standard deviation σ is known?

The larger the population standard deviation, the wider the interval.

For a given confidence level and population standard deviation, which of the following is true in the interval estimation of the population mean?

The larger the sample size, the narrower the interval.

A machine that is programmed to package 1.20 pounds of cereal is being tested for its accuracy in a sample of 36 cereal boxes: the sample mean filling weight is calculated as 1.22 pounds. The population standard deviation is known to be 0.06 pounds. Which of the following conclusions is correct with a 95% confidence level?

The machine is operating properly because the interval contains the target.

An analyst takes a random sample of 25 firms in the telecommunications industry and constructs a confidence interval for the mean return for the prior year. Holding all else constant, if he increased the sample size to 30 firms, how are the standard error of the mean and the width of the confidence interval affected?

The standard error of the mean decreases and the confidence interval becomes narrower.

How do the tdf and z distributions differ?

The tdf distribution has broader tails (it is flatter around zero).

What conditions are required by the central limit theorem before a confidence interval of the population mean may be created?

The underlying population need not be normally distributed if the sample size is 30 or more.

A sample of 36 observations produces a mean of 27 and a standard deviation of 3.5. Develop a 95% confidence interval for the population mean.

[25.8158, 28.1842] Excel functions are Lower Limit =27-CONFIDENCE.T(0.05,3.5,36) = 25.8158 Upper Limit =27+CONFIDENCE.T(0.05,3.5,36) = 28.1842

We draw a random sample of size 36 from a population with standard deviation 3.5. If the sample mean is 27, what is a 95% confidence interval for the population mean?

[25.8567, 28.1433] Excel functions are Lower Limit = 27-CONFIDENCE.NORM(0.05,3.5,36) = 25.8567 Upper Limit = 27 + CONFIDENCE.NORM(0.05,3.5,36) = 28.1433

The mortgage foreclosure crisis that preceded the Great Recession impacted the U.S. economy in many ways, but it also impacted the foreclosure process itself as community activists better learned how to delay foreclosure and lenders became more wary of filing faulty documentation. Suppose the duration of the eight most recent foreclosures filed in the city of Boston (from the beginning of foreclosure proceedings to the filing of the foreclosure deed, transferring the property) has been 230 days, 420 days, 340 days, 367 days, 295 days, 314 days, 385 days, and 311 days. Assume the duration is normally distributed. Construct a 90% confidence interval for the mean duration of the foreclosure process in Boston.

[293.2229, 372.2771] Excel functions are Average =AVERAGE(230,420,340,367,295,314,385,311) = 332.75 Standard Deviation =STDEV.S(230,420,340,367,295,314,385,311) = 59.0103 Lower Limit =332.75-CONFIDENCE.T(0.1,59.0103,8) = 293.2229 Upper Limit =332.75+CONFIDENCE.T(0.1,59.0103,8) = 372.2771

A string quartet in Maui wants to better understand its income stream, which primarily comes from playing at weddings. The cellist records the number of weddings the quartet plays in six successive months. Construct a 90% confidence interval from her results: 3, 5, 7, 4, 4, and 5.

[3.5427, 5.7907] Excel functions are Average = AVERAGE(3,5,7,4,4,5) = 4.6667 Standard Deviation = STDEV.S(3,5,7,4,4,5) = 1.3663 Lower Limit = 4.6667-CONFIDENCE.T(0.1,1.3663,6) = 3.5427 Upper Limit = 4.6667 + CONFIDENCE.T(0.1,1.3663,6) = 5.7907

A medical engineering company creates x-ray machines. The machines the company sold in 1995 were expected to last six years before breaking. To test how long the machines actually lasted, the company took a simple random sample of six machines. The company got the following results (in years) for how long the x-ray machines lasted: 8, 6, 7, 9, 5, and 7. Assume the distribution of the longevity of x-ray machines is normally distributed. Construct a 98% confidence interval for the average longevity of x-ray machines.

[5.0573, 8.9427] Excel functions are Average = AVERAGE(8,6,7,9,5,7) = 7 Standard Deviation = STDEV.S(8,6,7,9,5,7) = 1.4142 Lower Limit = 7-CONFIDENCE.T(0.02,1.4142,6) = 5.0573 Upper Limit = 7-CONFIDENCE.T(0.02,1.4142,6) = 8.9427

The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan. A random sample of six employees reveals the following family dental expenses: $180, $260, $60, $40, $100, and $80. It is known that dental expenses follow a normal distribution. Construct a 90% confidence interval for the average family dental expenses for all employees of this corporation.

[50.9805, 189.0195] Excel functions are Average = AVERAGE(180,260,60,40,100,80) = 120 Standard Deviation = STDEV.S(180,260,60,40,100,80) = 83.9 Lower Limit = 120-CONFIDENCE.T(0.1,83.9,6) = 50.9805 Upper Limit = 120 + CONFIDENCE.T(0.1,83.9,6) = 189.0195

What is zα/2 for a 90% confidence interval of the population mean?

1.645 Excel functions are: = NORM.S.INV(1-0.1/2) = 1.645 or = NORM.INV(1-0.1/2,0,1) = 1.645

Like the z distribution, the tdf distribution is symmetric around 0, bell-shaped, and with tails that approach the horizontal axis and eventually cross it.

False

The ages of MBA students at a university are normally distributed with a known population variance of 10.24. Suppose you are asked to construct a 95% confidence interval for the population mean age if the mean of a sample of 36 students is 26.5 years. What is the margin of error for a 95% confidence interval for the population mean?

1.0453 Excel function is = NORM.S.INV(1-0.05/2)*(SQRT(10.24)/SQRT(36)) = 1.0453

The dean of the school of business at the university of Iowa administers a comprehensive examination to the candidate for B.S degrees in Finance. Ten examinations are selected at random and scored. The scores were 82,94, 96, 71, 83, 69, 88, 62, 79, and 91. Calculate the sample average score. Calculate the sample standard deviation. Develop a 98% confidence interval estimate for the population mean. Assume the population is normally distributed.

Excel functions are a. Sample average score: = AVERAGE(82,94,96,71,83,69,88,62,79,91) = 81.5 b. Sample standard Deviation: = STDEV.S(82,94,96,71,83,69,88,62,79,91) = 11.326 c. Lower Limit = 81.5-CONFIDENCE.T(0.02,11.326,10) = 71.395 Upper Limit = 81.5 + CONFIDENCE.T(0.02,11.326,10) = 91.605

Excel does not have a special function that allows to calculate t values.

False

A university researcher constructed a confidence interval for the mean salary of medical college graduates in Ann Arbor, Michigan. The researcher believes that the interval estimate is useless because it is too wide. What should the researcher do to correct this problem?

Increase the sample size

Students who graduated from college last year owed an average of $25,250 in student loans. An economist wants to determine if the average debt has changed. She takes a sample of 40 recent graduates and finds that their average debt is $27,500 with a standard deviation of $9,120. Use 90% confidence interval. Which of the following conclusions is correct?

The average debt has not changed.

When examining the possible outcome of an election, what type of confidence interval is most suitable for estimating the current support for a candidate?

The confidence interval for the population proportion

For a given sample size and population standard deviation, which of the following is true in the interval estimation of the population mean?

The greater the confidence level, the wider the interval.

What is the purpose of calculating a confidence interval?

To provide a range of values that, with a certain measure of confidence, contains the population parameter of interest.

For a given confidence level (1 − α)100% and population standard deviation σ, the smaller the sample size n, the wider the confidence interval for the population mean.

True

If a small segment of the population is sampled then an estimate will be less precise.

True

The main ingredient for developing a confidence interval is the sampling distribution of the underlying statistic.

True

The tdf distribution has broader tails than the z distribution.

True

Suppose taxi fare from Logan Airport to downtown Boston is known to be normally distributed with a standard deviation of $2.50. The last seven times John has taken a taxi from Logan to downtown Boston, the fares have been $22.10, $23.25, $21.35, $24.50, $21.90, $20.75, and $22.65. What is a 95% confidence interval for the population mean taxi fare?

[$20.46, $24.16] Excel functions are Average =AVERAGE(22.1,23.25,21.35,24.5,21.9,20.75) = 22.3083 Lower Limit =22.3083-CONFIDENCE.NORM(0.05,2.5,7) = 20.46 Upper Limit =22.3083+CONFIDENCE.NORM(0.05,2.5,7) = 24.16

The average natural gas bill for a random sample of 21 homes in a city in Michigan during the month of March was $311.90 with a standard deviation of $51.60. Construct a 90% confidence interval for the population average natural gas bill in the city.

[$292.48, $331.32] Excel functions are Lower Limit = 311.9-CONFIDENCE.T(0.1,51.6,21) = 292.48 Upper Limit = 311.9 + CONFIDENCE.T(0.1,51.6,21) = 331.32

In an examination of holiday spending (known to be normally distributed) of a sample of 16 holiday shoppers at a local mall, an average of $54 was spent per hour of shopping and the standard deviation was equal to $21. Find a 90% confidence interval for the population mean level of spending per hour.

[$44.796, $63.204] Excel functions are Lower Limit =54-CONFIDENCE.T(0.1,21,16) = 44.796 Upper Limit =54+CONFIDENCE.T(0.1,21,16) = 63.204

In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour of shopping. Based on previous years, the population standard deviation is thought to be $21 per hour of shopping. Assuming that the amount spent per hour of shopping is normally distributed, find a 90% confidence interval for the mean amount.

[$45.36, $62.64] Excel functions are Lower Limit = 54-CONFIDENCE.NORM(0.1,21,16) = 45.36 Upper Limit = 54 + CONFIDENCE.NORM(0.1,21,16) = 62.64

A machine that is programmed to package 1.20 pounds of cereal is being tested for its accuracy. In a sample of 36 cereal boxes, the mean filling weight is calculated as 1.22 pounds. The population standard deviation is known to be 0.06 pounds. Find the 95% confidence interval for the mean.

[1.20, 1.24] Excel functions are Lower Limit = 1.22-CONFIDENCE.NORM(0.05,0.06,36) = 1.20 Upper Limit = 1.22 + CONFIDENCE.NORM(0.05,0.06,36) = 1.24

A sample of the weights of 35 babies is taken from the local hospital maternity ward. The point estimate for the mean weight of the babies is 110.2 ounces with a sample standard deviation of 23.4 ounces. Construct a 90% confidence interval for the population mean baby weight at this hospital. Interpret this interval.

[103.5119, 116.8881] Lower Limit = 110.2-CONFIDENCE.T(0.1,23.4,35) = 103.5119 Upper Limit = 110.2 + CONFIDENCE.T(0.1,23.4,35) = 116.8881

A sample of size 25, drawn from a normal population, produces a mean of 12.5 and a variance of 2.4. Construct a 99% confidence interval for the population mean.

[11.6334, 13.3666] Excel functions are Lower Limit = 12.5 − CONFIDENCE.T(0.01,SQRT(2.4),25) = 11.6334 Upper Limit = 12.5 + CONFIDENCE.T(0.01,SQRT(2.4),25) = 13.3666

We draw a random sample of size 25 from the normal population with variance 2.4. If the sample mean is 12.5, what is a 99% confidence interval for the population mean?

[11.7019, 13.2981] Excel functions are Lower Limit = 12.5-CONFIDENCE.NORM(0.01,SQRT(2.4),25) = 11.7019 Upper Limit = 12.5 + CONFIDENCE.NORM(0.01,SQRT(2.4),25) = 13.2981

Professors of accountancy are in high demand at American universities. A random sample of 28 new accounting professors showed the average salary was $135 thousand with a standard deviation of $16 thousand. Assume the distribution of salaries is normally distributed. Construct a 95% confidence interval for the salary of new accounting professors. Answers are in thousands of dollars.

[128.7958, 141.2042] Excel functions are Lower Limit =135-CONFIDENCE.T(0.05,16,28) = 128.7958 Upper Limit =135+CONFIDENCE.T(0.05,16,28) = 141.2042

Professors of accountancy are in high demand at American universities. A random sample of 28 new accounting professors showed the average salary was $135 thousand with a standard deviation of $16 thousand. Assume the distribution of salaries is normally distributed. Construct a 90% confidence interval for the salary of new accounting professors. Answers are in thousands of dollars.

[129.8497, 140.1503] Excel functions are Lower Limit =135-CONFIDENCE.T(0.1,16,28) = 129.8497 Upper Limit =135+CONFIDENCE.T(0.1,16,28) = 140.1503

A company manager thinks he is overpaying for fertilizer at $174 per ton. To find out, he collects fertilizer prices per ton from five distributors. Construct a 90% confidence interval from his results: 128, 140, 170, 166, and 156.

[135.1059, 168.8941] Excel functions are Average = AVERAGE(128,140,170,166,156) = 152 Standard Deviation = STDEV.S(128,140,170,166,156) = 17.72 Lower Limit = 152-CONFIDENCE.T(0.1,17.72,5) = 135.1059 Upper Limit = 152 + CONFIDENCE.T(0.1,17.72,5) = 168.8941

At a particular academically challenging high school, the average GPA of a high school senior is known to be normally distributed. A sample of 20 seniors is taken, and shows an average GPA of 2.71 and a variance of 0.25. Find a 90% confidence interval for the population mean GPA.

[2.5167, 2.9033] Excel functions are Lower Limit =2.71-CONFIDENCE.T(0.1,SQRT(0.25),20) = 2.5167 Upper Limit =2.71+CONFIDENCE.T(0.1,SQRT(0.25),20) = 2.9033

At an academically challenging high school, the average GPA of a high school senior is known to be normally distributed with a variance of 0.25. A sample of 20 seniors is taken and their average GPA is found to be 2.71. Develop a 90% confidence interval for the population mean GPA.

[2.5261, 2.8939] Excel functions are Lower Limit = 2.71-CONFIDENCE.NORM(0.1,SQRT(0.25),20) = 2.5261 Upper Limit = 2.71 + CONFIDENCE.NORM(0.1,SQRT(0.25),20) = 2.8939

To estimate the mean earnings forecast for a large company, an investor looked up earnings forecasts from six financial analysts. The six forecasts he found were 200, 220, 300, 185, 210, and 210 (in millions). Suppose the investor knows the population standard deviation is 25 million. Calculate a 95% confidence interval of the population mean earnings forecast.

[200.8295, 240.8371] Excel functions are Average = AVERAGE(200,220,300,185,210,210) = 220.8333 (in millions) Lower Limit = 220.8333-CONFIDENCE.NORM(0.05,25,6) = 200.8295 (in millions) Upper Limit = 220.8333 + CONFIDENCE.NORM(0.05,25,6) = 240.8371 (in millions)

A basketball coach wants to know how many free throws an NBA player shoots during the course of an average practice. The coach takes a random sample of 43 players and finds the average number of free throws shot per practice was 225 with a standard deviation of 35. Construct a 99% confidence interval for the average number of free throws in practice.

[210.5992, 239.4008] Excel functions are Lower Limit =225-CONFIDENCE.T(0.01,35,43) = 210.5992 Upper Limit =225+CONFIDENCE.T(0.01,35,43) = 239.4008

An environmentalist is measuring the number of spotted leopard lizards in central California per acre. The environmentalist surveys six acres and finds signs of 24, 26, 30, 27, 28, and 27 lizards per acre. Construct a 95% confidence interval for the average number of leopard lizards per acre.

[24.9011, 29.0989] Excel functions are Average = AVERAGE(24,26,30,27,28,27) = 27 Standard Deviation = STDEV.S(24,26,30,27,28,27) = 2 Lower Limit = 27-CONFIDENCE.T(0.05,2,6) = 24.9011 Upper Limit = 27 + CONFIDENCE.T(0.05,2,6) = 29.0989

Each portion of the SAT exam is designed to be normally distributed such that it has a population standard deviation of 100 and a mean of 500. However, the mean has changed over the years as less selective schools began requiring the SAT, and because students later began to prepare more specifically for the exam. Construct a 90% confidence interval for the population mean from the following eight scores from the math portion, using the population standard deviation of 100: 450, 660, 760, 540, 420, 430, 640, and 580.

[501.8456, 618.1544] Excel functions are Average = AVERAGE(450,660,760,540,420,430,640,580) = 560 Lower Limit = 560-CONFIDENCE.NORM(0.1,100,8) = 501.8456 Upper Limit = 560 + CONFIDENCE.NORM(0.1,100,8) = 618.1544

A website advertises job openings on its website, but job seekers have to pay to access the list of job openings. The website recently completed a survey to estimate the number of days it takes to find a new job using its service. It took the last 30 customers an average of 60 days to find a job. Assume the population standard deviation is 10 days. Calculate a 95% confidence interval of the population mean number of days it takes to find a job.

[56.4216, 63.5784] Excel functions are Lower Limit = 60-CONFIDENCE.NORM(0.05,10,30) = 56.4216 Upper Limit = 60 + CONFIDENCE.NORM(0.05,10,30) = 63.5784

A company that produces computers recently tested the battery in its latest laptop in six separate trials. The battery lasted 8.23, 7.89, 8.14, 8.25, 8.30, and 7.95 hours before burning out in each of the tests. Assuming the battery duration is normally distributed, construct a 95% confidence interval for the mean battery life in the new model.

[7.9490, 8.3044] Excel functions are Average =AVERAGE(8.23,7.89,8.14,8.25,8.3,7.95) = 8.1267 Standard Deviation =STDEV.S(8.23,7.89,8.14,8.25,8.3,7.95) = 0.1693 Lower Limit =8.1267-CONFIDENCE.T(0.05,0.1693,6) = 7.9490 Upper Limit =8.1267+CONFIDENCE.T(0.05,0.1693,6) = 8.3044

The height of high school basketball players is known to be normally distributed with a standard deviation of 1.75 inches. In a random sample of eight high school basketball players, the heights (in inches) are recorded as 75, 82, 68, 74, 78, 70, 77, and 76. Construct a 95% confidence interval on the average height of all high school basketball players.

[73.7873, 76.2127] Excel functions are Average = AVERAGE(75,82,68,74,78,70,77,76) = 75 Lower Limit = 75-CONFIDENCE.NORM(0.05,1.75,8) = 73.7873 Upper Limit = 75 + CONFIDENCE.NORM(0.05,1.75,8) = 76.2127

Bob's average golf score at his local course is 93.6 based on a random sample of 34 rounds of golf. Assume that the population standard deviation for his golf score is 4.2. Develop a 90% confidence interval for the population mean.

[92.42, 94.78] The appropriate Excel functions are Lower Limit = 93.6-CONFIDENCE.NORM(0.1,4.2,34) = 92.42 Upper Limit = 93.6 + CONFIDENCE.NORM(0.1,4.2,34) = 94.78

A job candidate with an offer from a prominent investment bank wanted to estimate how many hours she would have to work per week during her first year at the bank. She took a sample of six first-year analysts, asking how many hours they worked in the last week. a. Construct a 95% confidence interval with her results: 64, 82, 74, 73, 78, and 87 hours. b. What will happen to the width of the confidence interval for the average hours of work if the job candidate uses a smaller confidence level than 95%? Explain.

a. Average = AVERAGE(64,82,74,73,78,87) = 76.3333 Standard Deviation = STDEV.S(64,82,74,73,78,87) = 7.9666 Lower Limit = 76.3333-CONFIDENCE.T(0.05,7.9666,6) = 67.9729 Upper Limit = 76.3333 + CONFIDENCE.T(0.05,7.9666,6) = 84.6937 b. Lower Limit = 76.3333 CONFIDENCE.T(0.1,7.9666,6) = 69.7797 Upper Limit = 76.3333 + CONFIDENCE.T(0.1,7.9666,6) = 82.8869

The confidence interval for the population proportion is based on __________.

the z distribution