QMB 3200 Exam 3

Which of the following meets the requirements of a cluster sample?

A population can be divided into 50 city blocks. The sample will include residents from two randomly chosen city blocks.

Which of the following meets the requirements of a stratified random sample of size 6?

A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include two people chosen at random under the age of 25 and four people chosen at random over 25.

In cluster sampling, the population is first divided up into mutually exclusive and collectively exhaustive groups, called clusters. A cluster sample includes randomly selected observations from each cluster, which are proportional to the cluster's size in the population.

False

We use a population parameter to make inferences about a sample statistic.

False

According to the IRS, the average refund in the 2011 tax year was $3,109. Assuming that the standard deviation for these refunds was $874, what is the standard error of the average refund for a random sample of 50 tax returns?

$123.60 =874/SQRT(50) = $123.60

According to the Bureau of Labor Statistics it takes an average of 16 weeks for young workers to find a new job. Assume that the probability distribution is normal and that the standard deviation is two weeks. What is the probability that 20 young workers average less than 15 weeks to find a job?

0.0127 Excel function is =NORM.DIST(15,16,2/SQRT(20),TRUE) = 0.0127

Suppose the local Best Buy store averages 522 customers every day entering the facility with a standard deviation of 124 customers. A random sample of 40 business days was selected. What is the probability that the average number of customers in the sample is between 530 and 540?

0.1623 Excel function is =NORM.DIST(540,522,124/SQRT(40),TRUE)-NORM.DIST(530,522,124/SQRT(40),TRUE) = 0.1623

A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If this is to happen, it will have to move one of its most viewed shows to another slot. The network conducts a survey asking its viewers which show they would rather watch. The network receives 857 responses, of which 435 indicate that they would like to see the new show in the lineup. The test statistic for this hypothesis would be __________.

0.44

The heights of fences in the city of Provo, Utah, are normally distributed with a mean of 52 inches and a standard deviation of 4 inches. What is the probability that the average height of 10 randomly selected fences is between 50 and 54 inches?

0.8862 Excel function is =NORM.DIST(54,52,4/SQRT(10),TRUE)- NORM.DIST(50,52,4/SQRT(10),TRUE) = 0.8862

The heights of fences in the city of Provo, Utah, are normally distributed with a mean of 52 inches and a standard deviation of 4 inches. What is the probability that the average height of 16 randomly selected fences is at least 50 inches?

0.9773 Excel function is =1-NORM.DIST(50,52,4/SQRT(16),TRUE) = 0.9773

According to the Bureau of Labor Statistics it takes an average of 22 weeks for someone over 55 to find a new job. Assume that the probability distribution is normal and that the standard deviation is two weeks. What is the probability that eight workers over the age of 55 take an average of more than 20 weeks to find a job?

0.9977 Excel function is =1-NORM.DIST(20,22,2/SQRT(8),TRUE) = 0.9977

Suppose the average math SAT score for students enrolled at a local community college is 490.4 with a standard deviation of 63.7. A random sample of 49 students has been selected. The standard error of the average score for this sample is __________.

9.1 =63.7/SQRT(49) = 9.1.

A parameter is a random variable, whereas a sample statistic is a constant.

False

Cluster sampling is preferred when the objective is to increase precision.

False

In order to estimate the mean 30-year fixed mortgage rate for a home loan in the United States, a random sample of 14 recent loans is taken. The average calculated from this sample is 7.35%. It can be assumed that 30-year fixed mortgage rates are normally distributed with a population standard deviation of 0.60%. Compute 90% and 99% confidence intervals for the population mean 30-year fixed mortgage rate.

Confidence Level: Confidence Interval 90%: 7.09% to 7.61% In Excel: Lower Limit: =7.35-CONFIDENCE.NORM(0.1,0.6,14) Upper Limit: =7.35+CONFIDENCE.NORM(0.1,0.6,14) 99%: 6.94% to 7.76% In Excel: Lower Limit: =7.35-CONFIDENCE.NORM(0.01,0.6,14) Upper Limit: =7.35+CONFIDENCE.NORM(0.01,0.6,14)

The Chartered Financial Analyst (CFA) designation is fast becoming a requirement for serious investment professionals. Although it requires a successful completion of three levels of grueling exams, it also entails promising careers with lucrative salaries. A student of finance is curious about the average salary of a CFA charterholder. He takes a random sample of 25 recent charterholders and computes a mean salary of $195,000 with a standard deviation of $40,000. Use this sample information to determine the 90% confidence interval for the average salary of a CFA charterholder.

Confidence interval: $181,313 to $208,687 In Excel: Lower Limit: =195000-CONFIDENCE.T(0.1,40000,25) Upper Limit: =195000+CONFIDENCE.T(0.1,40000,25)

The accompanying data file reports the number of customers who were served at a restaurant on six randomly selected weekdays. Assume that the number of customers served on weekdays follows a normal distribution. Construct the 95% confidence interval for the average number of customers served on weekdays.

Confidence interval: 93.93 to 249.40 In Excel: Lower Limit: =AVERAGE(A2:A7)-T.INV(0.975, 5)*STDEV.S(A2:A7)/SQRT(6) Upper Limit: =AVERAGE(A2:A7)+T.INV(0.975, 5)*STDEV.S(A2:A7)/SQRT(6) In Excel: Lower Limit: =171.666666666667-CONFIDENCE.T(0.05,74.072037009027,6) Upper Limit: =171.666666666667+CONFIDENCE.T(0.05,74.072037009027,6)

Which of the following is not a form of bias?

Information from the sample is typical of information in the population.

John would like to conduct a survey in his neighborhood to get homeowners' opinion on the Delmarva proposal to switch to natural gas. Which of the following is an example of a stratified sample?

John divides the population into two-story, split-level, and ranch houses. Then, he selects a proportional number of houses from each group.

Peggy would like to conduct a survey in her neighborhood to get homeowners' opinions on the Delmarva proposal to switch to natural gas. Which of the following is an example of a convenience sample?

Peggy selects the first 25 homes that she passes as she walks into the entrance of the development.

A survey is designed to collect data on students' evaluations of their instructor's teaching performance. Which of the following situations most likely results in social-desirability biases?

The survey is administered in class with the instructor in the room the entire time.

A simple random sample is a sample of n observations that has the same probability of being selected from the population as any other sample of n observations.

True

Social-desirability bias refers to systematic difference between a group's "socially acceptable" responses to a survey or poll and this group's ultimate choice.

True

When a statistic is used to estimate a parameter, the statistic is referred to as an estimator. A particular value of the estimator is called an estimate.

True

Consider the following hypotheses: H0: μ ≥ 205 HA: μ < 205 A sample of 83 observations results in a sample mean of 202. The population standard deviation is known to be 33. a-1. Calculate the value of the test statistic. a-2. Find the p-value. b. Does the sample evidence enable us to reject the null hypothesis at α = 0.10? c. Does the sample evidence enable us to reject the null hypothesis at α = 0.05? d. Interpret the results at α = 0.05.

a-1. Test statistic: -0.83 a-2: p-value ≥ 0.10 b. No since the p-value is greater than the significance level. c. No since the p-value is greater than the significance level. d. We cannot conclude that the population mean is less than 205.

Consider the following hypotheses: H0: μ = 240 HA: μ ≠ 240 The population is normally distributed with a population standard deviation of 69. a-1. Calculate the value of the test statistic with X̄ = 250 and n = 45. a-2. What is the conclusion to the hypothesis test at the 1% significance level? a-3. Interpret the results at α = 0.01. b-1. Calculate the value of the test statistic with X̄ = 227 and n = 45. b-2. What is the conclusion to the hypothesis test at the 10% significance level? b-3. Interpret the results at α = 0.10.

a-1. Test statistic: 0.97 a-2. Do not reject H0 since the p-value is greater than the significance level. a-3. We cannot conclude that the population mean differs from 240. b-1. Test statistic: -1.26 b-2. Do not reject H0 since the p-value is greater than the significance level. b-3. We cannot conclude that the population mean differs from 240.

A sample of 47 observations yields a sample mean of 59.7. Assume that the sample is drawn from a normal population with a population standard deviation of 5.7. a-1. Find the p-value. a-2. What is the conclusion if α = 0.10? a-3. Interpret the results at α = 0.10. b-1. Calculate the p-value if the above sample mean was based on a sample of 126 observations. b-2. What is the conclusion if α = 0.10? b-3. Interpret the results at α = 0.10.

a-1. p-value ≥ 0.10 a-2. Do not reject H0 since the p-value is greater than α. a-3. We cannot conclude that the population mean is greater than 58.7. b-1. 0.01 ≤ p-value < 0.025 b-2. Reject H0 since the p-value is smaller than α. b-3. We conclude that the population mean is greater than 58.7.

The accompanying data file shows the monthly closing stock price (in $) for a firm over the past six months. a. Calculate the sample mean and the sample standard deviation. b. Calculate the 99% confidence interval for the mean stock price of the firm assuming that the stock price is normally distributed. c. What happens to the margin of error if a higher confidence level is used for the interval estimate?

a. Sample mean: 165.17 Sample standard deviation: 3.54 In Excel: =AVERAGE(A2:A7) =STDEV.S(A2:A7) b. Confidence interval: $159.33 to $171.00 In Excel: Lower Limit: =AVERAGE(A2:A7)-T.INV(0.995, 5)*STDEV.S(A2:A7)/SQRT(6) Upper Limit: =AVERAGE(A2:A7)+T.INV(0.995, 5)*STDEV.S(A2:A7)/SQRT(6) In Excel: Lower Limit: =165.17-CONFIDENCE.T(0.01,3.54494945897211,6) Upper Limit: =165.17+CONFIDENCE.T(0.01,3.54494945897211,6) c. The margin of error increases as the confidence level increases.

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 12, 25, 26, 13, 20, 23, 16, 21. a. Calculate the sample mean and the sample standard deviation. b. Construct the 90% confidence interval for the population mean. c. Construct the 99% confidence interval for the population mean. d. What happens to the margin of error as the confidence level increases from 90% to 99%?

a. Sample mean: 19.50 Sample standard deviation: 5.32 b. Confidence interval: 15.93 to 23.06 In Excel: Lower Limit: =19.5-CONFIDENCE.T(0.1,5.31843156256751,8) Upper Limit: =19.5+CONFIDENCE.T(0.1,5.31843156256751,8) c. Confidence interval: 12.92 to 26.08 In Excel: Lower Limit: =19.5-CONFIDENCE.T(0.01,5.31843156256751,8) Upper Limit: =19.5+CONFIDENCE.T(0.01,5.31843156256751,8) d. As the confidence level increases, the margin of error becomes larger. Correct

The accompanying data file lists average monthly debt payments (Debt in $) for 26 metropolitan areas. a. Calculate the mean and standard deviation for debt payments. b. Construct the 90% and the 95% confidence intervals for the population mean for debt payments.

a. Sample mean: 981.62 Sample standard deviation: 127.83 b. Confidence Level: Confidence Interval 90%: $938.79 to $1,024.44 In Excel: Lower Limit: =981.615384615385-CONFIDENCE.T(0.1,127.831475599111,26) Upper Limit: =981.615384615385+CONFIDENCE.T(0.1,127.831475599111,26) 95%: $929.98 to $1,033.25 Lower Limit: =981.615384615385-CONFIDENCE.T(0.05,127.831475599111,26) Upper Limit: =981.615384615385+CONFIDENCE.T(0.05,127.831475599111,26)

Consider the following hypotheses: H0: μ ≤ 520 HA: μ > 520 Find the p-value for this hypothesis test based on the following sample information. a. X̄ = 533; s = 40; n = 19 b. X̄ = 533; s = 40; n = 38 c. X̄ = 533; s = 32; n = 30 d. X̄ = 530; s = 32; n = 30

a. 0.05 ≤ p-value < 0.10 b. 0.025 ≤ p-value < 0.05 c. 0.01 ≤ p-value < 0.025 d. 0.025 ≤ p-value < 0.05

According to a survey, high school students average 100 text messages daily. Assume that the population standard deviation is 20 text messages. Suppose a random sample of 36 high school students is taken. a. What is the probability that the sample mean is more than 105? b. What is the probability that the sample mean is less than 95? c. What is the probability that the sample mean is between 95 and 105?

a. 0.0668 In Excel: =1 - NORM.DIST(105, 100, 20/SQRT(36), TRUE) b. 0.0668 In Excel: =NORM.DIST(95, 100, 20/SQRT(36), TRUE) c. 0.8664 In Excel: =NORM.DIST(105, 100, 20/SQRT(36), TRUE) - NORM.DIST(95, 100, 20/SQRT(36), TRUE)

Beer bottles are filled so that they contain an average of 345 ml of beer in each bottle. Suppose that the amount of beer in a bottle is normally distributed with a standard deviation of 7 ml. a. What is the probability that a randomly selected bottle will have less than 339 ml of beer? b. What is the probability that a randomly selected 6-pack of beer will have a mean amount less than 339 ml? c. What is the probability that a randomly selected 12-pack of beer will have a mean amount less than 339 ml?

a. 0.1957 In Excel: =NORM.DIST(339, 345, 7, TRUE) b. 0.0179 In Excel: =NORM.DIST(339, 345, 7/SQRT(6), TRUE) c. 0.0015 In Excel: =NORM.DIST(339, 345, 7/SQRT(12), TRUE)

A random sample of 15 observations is used to estimate the population mean. The sample mean and the sample standard deviation are calculated as 152.5 and 28.90, respectively. Assume that the population is normally distributed. a. Construct the 95% confidence interval for the population mean. b. Construct the 99% confidence interval for the population mean. c. Use your answers to discuss the impact of the confidence level on the width of the interval.

a. 136.50 to 168.50 In Excel: Lower Limit: =152.5-CONFIDENCE.T(0.05,28.9,15) Upper Limit: =152.5+CONFIDENCE.T(0.05,28.9,15) b. 130.29 to 174.71 In Excel: Lower Limit: =152.5-CONFIDENCE.T(0.01,28.9,15) Upper Limit: =152.5+CONFIDENCE.T(0.01,28.9,15) c. As the confidence level increases, the interval becomes wider.

Consider a normal population with an unknown population standard deviation. A random sample results in X̄ = 52.23 and s2 = 22.09. a. Compute the 90% confidence interval for μ if X̄ and s2 were obtained from a sample of 15 observations. b. Compute the 90% confidence interval for μ if X̄ and s2 were obtained from a sample of 24 observations. c. Use your answers to discuss the impact of the sample size on the width of the interval.

a. 50.09 to 54.37 In Excel: Lower Limit: =52.23-CONFIDENCE.T(0.1,SQRT(22.09),15) Upper Limit: =52.23+CONFIDENCE.T(0.1,SQRT(22.09),15) b. 50.59 to 53.87 In Excel: Lower Limit: =52.23-CONFIDENCE.T(0.1,SQRT(22.09),24) Upper Limit: =52.23+CONFIDENCE.T(0.1,SQRT(22.09),24) c. The bigger sample size will lead to a smaller interval width and therefore a more precise interval. Correct

The accompanying data file reports the sale price (in $1,000s) for six single-family homes in San Luis Obispo County in California. a. Construct the 90% confidence interval for the mean sale price in San Luis Obispo County. b. What assumption have you made when constructing this confidence interval?

a. Confidence interval: 677.28 to 776.72 (in $1,000s) b. Single family homes in San Luis Obispo follow a normal distribution. In Excel: Lower Limit: =AVERAGE(A2:A7)-T.INV(0.95, 5)*STDEV.S(A2:A7)/SQRT(6) Upper Limit: =AVERAGE(A2:A7)+T.INV(0.95, 5)*STDEV.S(A2:A7)/SQRT(6)

The accompanying data file shows the annual return for a stock over the past 25 years (Return in %). An investor wants to test whether the average return on the stock is greater than 12%. Assume returns are normally distributed with a population standard deviation of 30%. a. Specify the null and the alternative hypotheses for the test b-1. Calculate the value of the test statistic. b-2. Find the p-value. c-1. At α = 0.05, what is the conclusion to the hypothesis test? c-2. Is the average stock return greater than 12%?

a. H0 :µ ≤ 12; versus HA: µ > 12 b-1. Test statistic: 0.53 b-2. p-value ≥ 0.10 c-1. Do not reject H0 since the p-value is greater than significance level. c-2. No

It is advertised that the average braking distance for a small car traveling at 75 miles per hour equals 120 feet. A transportation researcher wants to determine if the statement made in the advertisement is false. She randomly test drives 36 small cars at 75 miles per hour and records the braking distance. The sample average braking distance is computed as 115 feet. Assume that the population standard deviation is 23 feet. a. Specify the null and the alternative hypotheses for the test. b. Calculate the value of the test statistic and the p-value. c. Use α = 0.05 to determine if the average breaking distance differs from 120 feet.

a. H0: µ = 120; versus HA: µ ≠ 120 b. Test statistic: -1.30 Find the p-value: p-value ≥ 0.10 c. The average breaking distance does not differ from 120 feet.

A local bottler in Hawaii wishes to ensure that an average of 20 ounces of passion fruit juice is used to fill each bottle. In order to analyze the accuracy of the bottling process, she takes a random sample of 37 bottles. The mean weight of the passion fruit juice in the sample is 19.40 ounces. Assume that the population standard deviation is 1.34 ounce. a. Specify the null and the alternative hypotheses to test whether the bottling process is inaccurate. b-1. Calculate the value of the test statistic. b-2. Find the p-value. c-1. What is the conclusion at α = 0.10? c-2. Make a recommendation to the bottler.

a. H0: µ = 20; versus HA: µ ≠ 20 b-1. Test statistic: -2.72 b-2. p-value < 0.01 In Excel: =2*NORM.DIST(-2.7236, 0, 1, TRUE) = 0.0065. c-1. Reject H0 since the value of the p-value is less than the significance level. c-2. The accuracy of the bottling process is compromised

The accompanying data file shows hourly wages (Wage in $) for 50 employees. An economist wants to test if the average hourly wage is less than $29. Assume that the population standard deviation is $8. a. Specify the null and the alternative hypotheses for the test. b-1. Find the value of the test statistic. b-2. Find the p-value. c. At α = 0.10, what is the conclusion to the hypothesis test?

a. H0: µ ≥ 29 versus HA: µ < 29 b-1. Test statistic: -0.71 b-2. p-value ≥ 0.10 c. Do not reject H0,the hourly wage is not less than 29

A promising start-up wants to compete in the cell phone market. It understands that the lead product has a battery life of approximately 12 hours. The start-up claims that while its new cell phone is more expensive, its battery life is more than twice as long as that of the leading product. In order to test the claim, a researcher samples 45 units of the new cell phone and finds that the sample battery life averages 25.3 hours with a sample standard deviation of 3.1 hours. a. Specify the relevant null and the alternative hypotheses. b-1. Calculate the value of the test statistic. b-2. Find the p-value. c. At the 5% significance level, what is the conclusion?

a. H0:µ ≤ 24; versus HA:µ > 24 b-1. Test statistic: 2.813 b-2. p-value < 0.01 c. Reject H0; the claim is supported by the data.

The accompanying data file shows the selling price (in $1,000s) for 36 recent house sales in Mission Viejo, California. A Realtor believes that the average price of a house is more than $500 (in $1,000s). Assume the population standard deviation is $100 (in $1,000s). a. Specify the null and the alternative hypotheses for the test. b-1. What is the value of the test statistic? b-2. Find the p-value. c. At α = 0.05, what is the conclusion to the hypothesis test?

a. H0:µ ≤ t500; versus HA:µ > 500 b-1. Test statistic: 1.16 b-2. p-value > 0.10 c. Do not reject H0; realtor's claim is not supported by the data.

The accompanying data file shows hourly wages (Wage in $) for 50 employees. An economist wants to test if the average hourly wage is less than $29. Assume that the population standard deviation is $7. a. Specify the null and the alternative hypotheses for the test. b-1. Find the value of the test statistic. b-2. Find the p-value. c. At α = 0.05, what is the conclusion to the hypothesis test?

a. H0:µ ≥ 29; versus HA:µ < 29 b-1. Test statistic: -1.32 b-2. 0.05 ≤ p-value < 0.10 c. Do not reject H0,the hourly wage is not less than 29

A researcher wants to determine if Americans are sleeping less than the recommended 7 hours of sleep on weekdays. He takes a random sample of 100 Americans and computes the average sleep time of 6.7 hours on weekdays. Assume that the population is normally distributed with a known standard deviation of 2.1 hours. Test the researcher's claim at α = 0.01. a. Specify the null and the alternative hypotheses for the test. b. Calculate the value of the test statistic. c. Find the p-value. d. What is the conclusion at α = 0.01? e. Make an inference.

a. H0:µ ≥ 7; versus HA:µ < 7 b. Test statistic: -1.43 c. 0.05 ≤ p-value < 0.10 d. Do not reject H0 since the p-value is greater than α. e. There is insufficient evidence to suggest that Americans sleep less than the recommended 7 hours of sleep.

Consider the following hypothesis test: H0: μ ≤ 75 HA: μ > 75 A random sample of 100 observations yields a sample mean of 88. The population standard deviation is 70. a. What is the value of the test statistic? b. Find the p-value. c. What is the conclusion to the hypothesis test at the 10% significance level?

a. Test statistic: 1.857 b. 0.025 ≤ p-value < 0.05 c. Reject H0, the population mean is greater than 75.

One of the best-selling products for a specialty grocery store is a 16-ounce package of almonds. Because it is impossible to pack exactly 16 ounces in each packet, a researcher has determined that the weight of almonds in each packet is normally distributed with a mean and a standard deviation equal to 16.01 ounces and 0.08 ounce, respectively. a. Is the sampling distribution of the sample mean approximately normal based on any given sample size? b. Find the probability that a random sample of 56 bags of almonds will average less than 16 ounces. c. Suppose your cereal recipe calls for no less than 48 ounces of almonds. What is the probability that three packets of almonds will meet your requirement?

a. Yes b. 0.1748 In Excel: =NORM.DIST(16, 16.01, 0.08/SQRT(56), TRUE) c. 0.5857 In Excel: =1 - NORM.DIST(16, 16.01, 0.08/SQRT(3), TRUE)

Consider the following hypotheses: H0: μ = 27 HA: μ ≠ 27 Find the p-value for this hypothesis test based on the following sample information. a. X̄ = 25; s = 6.6; n = 31 b. X̄ = 29; s = 6.6; n = 31 c. X̄ = 25; s = 6.3; n = 40 d. X̄ = 26; s = 6.3; n = 52

a. p-value ≥ 0.10 b. p-value ≥ 0.10 c. 0.05 bad media p-value < 0.10 d. p-value ≥ 0.10

Consider the following hypotheses: H0: μ = 6,300 HA: μ ≠ 6,300 The population is normally distributed with a population standard deviation of 670. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level.

a. x̄ = 6,360; n = 135 Test statistic: 1.04 p-value: 0.2990 Do not reject H0 b. x̄ = 6,360; n = 245 Test statistic: 1.40 p-value: 0.1616 Do not reject H0 c. x̄ = 6,090; n = 43 Test statistic: -2.06 p-value: 0.0430 reject H0 d. x̄ = 6,170; n = 43 Test statistic: -1.27 p-value: 0.2069 Do not reject H0

A simple random sample of 11 observations is derived from a normally distributed population with a population standard deviation of 2.2. a. Is the condition that X̄ is normally distributed satisfied? b. Compute the margin of error with 95% confidence. c. Compute the margin of error with 90% confidence. d. Which of the two margins of error will lead to a wider interval?

a. yes b. 1.30 In Excel: =NORM.INV(0.975, 0, 1)*2.2/SQRT(11) In Excel: =CONFIDENCE.NORM(0.05,2.2,11) c. 1.09 In Excel: =NORM.INV(0.950, 0, 1)*2.2/SQRT(11) In Excel: =CONFIDENCE.NORM(0.1,2.2,11) d. The margin of error with 95% confidence.

Consider the following competing hypotheses: Ho: μ = 0, HA: μ ≠ 0. The value of the test statistic is z = −1.60. If we choose α = 0.01, then we __________.

do not reject the null hypothesis; therefore we cannot conclude that the population mean is significantly different from zero

A school teacher is worried that the concentration of dangerous, cancer-causing radon gas in her classroom is greater than the safe level of 6 pCi/L. The school samples the air for 25 days and finds an average concentration of 6.4 pCi/L with a standard deviation of 1 pCi/L. The value of the test statistic is __________

t24 = 2.00

A newly hired basketball coach promised a high-paced attack that will put more points on the board than the team's previously tepid offense historically managed. After a few months, the team owner looks at the data to test the coach's claim. He takes a sample of 49 of the team's games under the new coach and finds that they scored an average of 105 points with a standard deviation of 8 points. Over the past 10 years, the team had averaged 103 points. What is the value of the appropriate test statistic to test the new coach's claim at the 1% significance level?

t48 = 1.75

Bias can occur in sampling. Bias refers to __________.

the tendency of a sample statistic to systematically overestimate or underestimate a population parameter

Nonresponse bias occurs when __________.

those responding to a survey or poll differ systematically from the nonrespondents

The Boston public school district has had difficulty maintaining on-time bus service for its students ("A Year Later, School Buses Still Late," Boston Globe, October 5). Suppose the district develops a new bus schedule to help combat chronic lateness on a particularly woeful route. Historically, the bus service on the route has been, on average, 10 minutes late. After the schedule adjustment, the first 49 bus runs showed an average of eight minutes late. As a result, the Boston public school district claimed that the schedule adjustment was an improvement—students were not as late. Assume the population standard deviation for bus arrival time was 10 minutes. The value of the test statistic is __________.

z = −1.40