Hypothesis Tests for One Population MEan

Data on salaries in the public school system are published annually in Ranking of the States and Estimates of School Statistics by the National Education Association. The mean annual salary of (public) classroom teachers is $\$55.4$ thousand. A hypothesis test is to be performed to decide whether the mean annual salary of classroom teachers in Ohio is greater than the national mean.

Refer to Exercise 9.15. Explain what each of the following would mean.

(a) Type I error

(b) Type II error

(c) Correct decision

Now suppose that the results of carrying out the hypothesis test lead to nonrejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean cadmium level in Boletus Pinicola mushrooms.

(d) equals the safety limit of $0.5$ppm.

(e) exceeds the safety limit of $0.5$ppm.

Refer to Exercise 9.16. Explain what each of the following would mean.

(a) Type I error.

(b) Type II error.

(c) Correct decision.

Now suppose that the results of carrying out the hypothesis test lead to the rejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean lactation period of grey seals.

(d) equals 23 days.

(e) differs from 23 days.

Refer to Exercise 9.17. Explain what each of the following would mean.

(a) Type I error.

(b) Type II error.

(c) Correct decision.

Now suppose that the results of carrying out the hypothesis test lead to the rejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean iron intake of all adult females under the age of 51 years.

(d) equals the RDA of 18 mg per day.

(e) is less than the RDA of 18 mg per day.

Cadmium, a heavy metal, is toxic to animals. Mushrooms, however are able to absorb and accumulate cadmium at high concentrations. The Czech and Slovak governments have set a safety limit for cadmium in dry vegetables at \(0.5\) part per million M. Melgar et al. measured the cadmium levels in a random sample of the edible mushroom Boletus pinicola and published the results in the paper "Influence of Some Factors in Toxicity and Accumulation of Cd from Edible Wild Macrofungi in NW Spain. A hypothesis test is to be performed to decide whether the mean cadmium level in Boletus pinicola mushrooms is greater than the government's recommended limit. 

a. determine the null hypothesis

b. determine the alternative hypothesis

c. classify the hypothesis test as two tailed, left tailed or right tailed.

Refer to Exercise 9.18. Explain what each of the following would mean.

(a) Type I error.

(b) Type II error.

(c) Correct decision.

Now suppose that the results of carrying out the hypothesis test lead to the rejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean age at diagnosis of all people with early-onset dementia.

(d) is 55 years old.

(e) is less than 55 years old.

Refer to Exercise 9.19. Explain what each of the following would mean.

(a) Type I error.

(b) Type II error.

(c) Correct decision.

Now suppose that the results of carrying out the hypothesis test lead to the rejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean length of imprisonment for motor-vehicle-theft offenders in Sydney. 

(d) equals the national mean of 16.7 months.

(e) differs from the national mean of 16.7 months.

In each of Exercises 9.41-9.46 ,determine the critical values for a one-mean z-test. For each exercise, draw a graph that illustrates your answer

A left - tailed test with$\alpha = 0.01$

In each of Exercises 9.41-9.46 ,determine the critical values for a one-mean z-test. For each exercise, draw a graph that illustrates your answer

A left-tailed test with$\alpha = 0.05$

In each of Exercises 9.41-9.46 ,determine the critical values for a one-mean z-test. For each exercise, draw a graph that illustrates your answer

A right- tailed test with$\alpha = 0.01$

In each of Exercises 9.41-9.46 ,determine the critical values for a one-mean z-test. For each exercise, draw a graph that illustrates your answer

A two-tailed test with $\alpha = 0.05$

State two reasons why including the P-value is prudent when you are reporting the results of a hypothesis test.

We have been provided a sample mean, sample size, and population standard deviation. In the given case, use the one-mean z-test to perform the required hypothesis test at the $5\%$  significance level.

$\overline{x}=24,n=15,\sigma =4,H_0:\mu =22,H_a:\mu >22$

We have been provided a sample mean, sample size, and population standard deviation. In the given case, use the one-mean z-test to perform the required hypothesis test at the $5\%$  significance level.

$\overline{x}=23,n=15,\sigma =4,H_0:\mu =22,H_a:\mu >22$

We have been provided a sample mean, sample size, and population standard deviation. In the given case, use the one-mean z-test to perform the required hypothesis test at the $5\%$  significance level.

$\overline{x}=23,n=24,\sigma =4,H_0:\mu =22,H_a:\mu \ne 22$

In the given exercise, we have provided a sample mean, sample size, and population standard. In each case, use the one-mean z-test to perform the required hypothesis test at the 5% significance level.

Dementia is the loss of the in actual and social abilities severe enough to interfere with judging behavior and daily functioning. Alzheimer's disease is the most common type of dementia. In the article "Living with Early Onsite dementia: Exploring the Experience and Developing Evidence Guidelines for Practice", P Harris and J Keady explored the experiment struggles of people diagnosed with dementia and their familiar simple random sample \(21\) people with early-onset dementia the following data on age at diagnosis in years.

At the \(1%\) significance level, do the data provide sufficient evidence to conclude that the mean age at diagnosis of all people with early onset dementia is less than \(55\) years old? Assume that the population is standard deviation is \(6.8\) years.

Serving Time. According to the Bureau of Crime Statistics and Research of Australia, as reported on Lawlink, the mean length of imprisonment for motor-vehicle theft offenders in Australia is 16.7 months. One hundred randomly selected motor-vehicle-theft offenders in Sydney, Australia, had a mean length of imprisonment of 17.8 months. At the $5\%$ significance level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle theft offenders in Sydney differs from the national mean in Australia? Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in Sydney is 6.0 months.

Job Gains and Losses. In the article "Business Employment Dynamics: New Data on Gross Job Gains and Losses" (Monthly Labor Review, Vol. 127, Issue 4, pp. 29-42), J. Spletzer et al. examined gross job gains and losses as a percentage of the average of previous and current employment figures. A simple random sample of 20 quarters provided the net percentage gains (losses are negative gains) for jobs as presented on the WeissStats site. Use the technology of your choice to do the following.

a. Decide whether, on average, the net percentage gain for jobs exceeds 0.2. Assume a population standard deviation of 0.42. Apply

which we provide on the WeissStats site. Use the technology of your choice to do the following.

a. Obtain a normal probability plot, boxplot, histogram, and stemand-leaf diagram of the data.

b. Based on your results from part (a), can you reasonably apply the one-mean $z$-test to the data? Explain your reasoning.

c. At the $1\%$ significance level, do the data provide sufficient evidence to conclude that the mean body temperature of healthy humans differs from $98.6°\mathrm{F}$ ? Assume that $\sigma =0.63°\mathrm{F}$.

According to the Bureau of Crime Statistics and Research of Australia, as reported on Lawlink, the mean length of imprisonment for motor-vehicle-theft offenders in Australia is 16.7 months. One hundred randomly selected motor-vehicle-theft of-fenders in Sydney, Australia, had a mean length of imprisonment of 17.8 months. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in Sydney differs from the national mean in Australia? Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in Sydney is 6months.

In the article "Business Employment Dynamics: New data on Gross Job Gains and Losses", J. Spletzer et al. examined gross job gains and losses as a percentage of the average of previous and current employment figures. A simple random sample of 20 quarters provided the net percentage gains for jobs as presented on the WeissStats site. Use the technology of your choice to do the following.

Part (a): Decide whether, on average, the net percentage gain for jobs exceeds 0.2. Assume a population standard deviation of 0.42. Apply the one-mean z-test with a 5% significance level.

Part (b): Obtain a normal probability plot, boxplot, histogram and stem-and-leaf diagram of the data.

Part (c): Remove the outliers from the data and then repeat part (a).

Part (d): Comment on the advisability of using the z-test here.

The daily charges, in dollars, for a sample of 15 hotels and motels operating in South Carolina are provided on the WeissStats site. The data were found in the report South Carolina Statistical Abstract, sponsored by the South Carolina Budget and Control Board.

Part (a): Use the one-mean z-test to decide, at the 5% significance level, whether the data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than \(75. Assume a population standard deviation of \)22.40.

Part (b): Obtain a normal probability plot, boxplot, histogram and stem-and-leaf diagram of the data.

Part (c): Remove the outliers (if any) from the data and then repeat part (a).

Part (d): Comment on the advisability of using the z-test here.

As we mentioned on page 378, the following relationship holds between hypothesis test and confidence intervals for one-mean z-procedures: For a two-tailed hypothesis test at the significance level $\alpha$, the null hypothesis $H_0:\mu ={\mu }_0$ will be rejected in favor of the alternative hypothesis $H_a:\mu \ne {\mu }_0$ if and only if ${\mu }_0$ lies outside the $\left(1-\alpha \right)$-level confidence interval for $\mu$. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean z-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.84

Part (b): Exercise 9.87

Serving Time. According to the Bureau of Crime Statistics and Research of Australia, as reported on Lawlink, the mean length of imprisonment for motor-vehicle-theft offenders in Australia is $16.7$ months. One hundred randomly selected motor-vehicle-theft offenders in Sydney. Australia, had a mean length of imprisonment of $17.8$ months. At the $5\%$ significance level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in Sydney differs from the national mean in Australia? Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in Sydney is $6.0$ months.

9.89 Job Gains and Losses. In the article "Business Employment Dynamics: New Data on Gross Job Gains and Losses" (Monthly Labor Review, Vol. 127. Issue 4. pp. 29-42). J. Spletzer et al. examined gross job gains and losses as a percentage of the average of previous and current employment figures. A simple random sample of 20 quarters provided the net percentage gains (losses are negative gains) for jobs as presented on the WeissStats site. Use the technology of your choice to do the following.
a. Decide whether, on average, the net percentage gain for jobs exceeds 0.2. Assume a population standard deviation of 0.42. Apply the one-mean z-test with a $5\%$ significance level.
b. Obtain a normal probability plot, boxplot, histogram, and stem-and-leaf diagram of the data.
c. Remove the outliers (if any) from the data and then repeat part (a).
d. Comment on the advisability of using the z-test here.

9.90 Hotels and Motels. The daily charges, in dollars, for a sample of 15 hotels and motels operating in South Carolina are provided on the WeissStats site. The data were found in the report South Caroline Statistical Abstract, sponsored by the South Carolina Budget and Control Board.
a. Use the one-mean $z$-test to decide, at the $5\%$ significance level, whether the data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than $\backslash (75$. Assume a population standard deviation of $\backslash )22.40$.

b. Obtain a normal probability plot, boxplot, histogram, and stem-and-leaf diagram of the data.
c. Remove the outliers (if any) from the data and then repeat part (a).
d. Comment on the advisability of using the $z$-test here.

9.93 Cell Phones. The number of cell phone users has increased dramatically since $1987$. According to the Semi-annual Wireless Survey, published by the Cellular Telecommunications & Internet Association, the mean local monthly bill for cell phone users in the United States was $\backslash (48.16$ in $2009$ . Last year's local monthly bills, in dollars, for a random sample of $75$ cell phone users are given on the WeissStats site. Use the technology of your choice to do the following.
a. Obtain a normal probability plot, boxplot, histogram, and stemand-leaf diagram of the data.
b. At the $5\%$ significance level, do the data provide sufficient evidence to conclude that last year's mean local monthly bill for cell phone users decreased from the $2009$ mean of $\backslash )48.16?$ Assume that the population standard deviation of last year's local monthly bills for cell phone users is $\$25$.
c. Remove the two outliers from the data and repeat parts (a) and (b).
d. State your conclusions regarding the hypothesis test.

9.95 Two-Tailed Hypothesis Tests and CIs. As we mentioned on page 378 , the following relationship holds between hypothesis tests and confidence intervals for one-mean $z$-procedures: For a two-tailed hypothesis test at the significance level $\alpha$, the null hypothesis $H_0:\mu ={\mu }_0$ will be rejected in favor of the alternative hypothesis $H_a:\mu \ne {\mu }_0$ if and only if ${\mu }_0$ lies outside the $(1-\alpha )$-level confidence interval for $\mu$. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean $z$-interval (Procedure 8.1 on page 322 ) and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise 9.84
b. Exercise 9.87

9.96 Left-Tailed Hypothesis Tests and Cls. In Exercise 8.105 on page 335, we introduced one-sided one-mean z-intervals. The following relationship holds between hypothesis tests and confidence intervals for one-mean $z$-procedures: For a left-tailed hypothesis test at the significance level $\alpha$, the null hypothesis $H_0:\mu ={\mu }_0$ will be rejected in favor of the alternative hypothesis $H_a:\mu <{\mu }_0$ if and only if ${\mu }_0$ is greater than or equal to the $(1-\alpha )$-level upper confidence bound for $\mu$. In each case, illustrate the preceding relationship by obtaining the appropriate upper confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise $9.85$
b. Exercise $9.86$

As we mentioned on page \(378\), the following relationship holds between hypothesis tests and confidence intervals for one mean \(z-\)procedures: For a two-tailed hypothesis test at the significance level \(\alpha\), the null hypothesis \(H_{0}:\mu =\mu_{0}\) will be rejected in favor of the alternative hypothesis \(H_{a}:\mu \neq \mu_{0}\) if and only if \(\mu_{0}\) lies outside the \((1-\infty)\) level confidence interval for \(\mu\). In each case, illustrate the preceding relationship by obtaining the appropriate one-mean \(z-\)interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

a. Exercise \(9.84\)

b. Exercise \(9.87\)

Apparel and Services. According to the document Consumer Expenditures, a publication of the Bureau of Labor Statistics, the average consumer unit spent \( 1736 on apparel and services in 2012  That same year, 25 consumer units in the Northeast had the following annual expenditures, in dollars, on apparel and services.

At the 5 % significance level, do the data provide sufficient evidence to conclude that the 2012 mean annual expenditure on apparel and services for consumer units in the Northeast differed from the national mean of \)1736? (Note: The sample mean and sample standard deviation of the data are \(1922.76 and \)350.90, respectively.)

Ankle Brachial Index. The ankle brachial index (ABI) compares the blood pressure of a patient's arm to the blood pressure of the patient's leg. The ABI can be an indicator of different diseases, including arterial diseases. A healthy (or normal) ABI is 0.9 or greater. In a study by M. McDermott et al. titled "Sex Differences in Peripheral Arterial Disease: Leg Symptoms and Physical Functioning" (Journal of the American Geriatrics Society, Vol. 51, No. 2, Pp. 222-228), the researchers obtained the ABI of 187 women with peripheral arterial disease. The results were a mean ABI of 0.64 with a standard deviation of 0.15 At the 1 % significance level, do the data provide sufficient evidence to conclude that, on average, women with peripheral arterial disease have an unhealthy ABI?

Dirt Bikes. Dirt bikes are simpler and lighter motorcysles that are designed for off-road events. Specifications for dirt bikes can be found through Motorcycle USA on their website www.motorcycle-usa.com. A random sample of 30 dirt bikes have a mean fuel capacity of $1.91$ gallons with a standard deviation of 0.74 gallons. At the $10\%$ significance level, do the data provide sufficient evidence to conclude that the mean fuel tank capacity of all dirt bikes is less than 2 gallons?

How Far People Drive. In 2011, the average car in the United States was driven 13.5 thousand miles, as reported by the Federal Highway Administration in Highway Statistics. On the WeissStats site, we provide last year's distance driven, in thousands of miles, by each of 500 randomly selected cars. Use the technology of your choice to do the following.

a. Obtain a normal probability plot and histogram of the data.

b. Based on your results from part (a), can you reasonably apply the one-mean t-test to the data? Explain your reasoning.

c. At the 5 % significance level, do the data provide sufficient evidence to conclude that the mean distance driven last year differs from that in 2011?

Fair Market Rent. According to the document Out of Reach published by the National Low Income Housing Coalition, the fai market rent (FMR) for a two-bedroom unit in the United States is  949 A sample of 100 randomly selected two-bedroom units yielded the data on monthly rents, in dollars, given on the WeissStats site. Use the technology of your choice to do the following.

a. At the 5 % significance level, do the data provide sufficient evidence to conclude that the mean monthly rent for two-bedroom units is greater than the FMR of $949? Apply the one-mean t-test.

b. Remove the outlier from the data and repeat the hypothesis test in part (a).

c. Comment on the effect that removing the outlier has on the hypothesis test.

d. State your conclusion regarding the hypothesis test and explain your answer.

The following relationship holds between hypothesis tests and confidence intervals for one-mean t-procedures: For a two-tailed hypothesis test at the significance level $\alpha$, the null hypothesis $H_0:\mu ={\mu }_0$ will be rejected in favor of the alternative hypothesis $H_a:\mu >{\mu }_0$ if and only if ${\mu }_0$ lies outside the $\left(1-\alpha \right)$-level confidence interval for $\mu$. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean t-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.113

Part (b): Exercise 9.116

In Exercise 8.146 on page 345, we introduced one-sided one-mean t-intervals. The following relationship holds between hypothesis tests and confidence intervals for one-mean t-procedures: For a left-tailed hypothesis test at the significance level $\alpha$, the null hypothesis $H_0:\mu ={\mu }_0$ will be rejected in favor of the alternative hypothesis $H_a:\mu <{\mu }_0$ if and only if ${\mu }_0$ is greater than or equal to the $\left(1-\alpha \right)$- level upper confidence bound for $\mu$. In each case, illustrate the preceding relationship by obtaining the appropriate upper confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.117

Part (b): Exercise 9.118

In Exercise 8.146 on page 345, we introduced one-sided one mean-t-intervals. The following relationship holds between hypothesis test and confidence intervals for one-mean t-procedures: For a right-tailed hypothesis test at the significance level $\alpha$, the null hypothesis $H_0:\mu ={\mu }_0$ will be rejected in favor of the alternative hypothesis data-custom-editor="chemistry" $H_a:\mu >{\mu }_0$ if and only if ${\mu }_0$ is less than or equal to the $\left(1-\alpha \right)$-level lower confidence bound for $\mu$. In each case, illustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.114 (both parts)

Part (b) Exercise 9.115

9.128 Two-Tailed Hypothesis Tests and CIs. The following relationship holds between hypothesis tests and confidence intervals for one-mean $t$-procedures: For a two-tailed hypothesis test at the significance level $\alpha$, the null hypothesis $H_0:\mu ={\mu }_0$ will be rejected in favor of the alternative hypothesis $H_2:\mu >{\mu }_0$ if and only if ${\mu }_0$ lies outside the $(1-\alpha )$-level confidence interval for $\mu$. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean $t$-interval (Procedure 8.2 on page 338 ) and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise 9.113
b. Exercise 9.116

Determine the critical value(s) for a one-mean z-test at the 1 % significance level if the test is

a. right tailed.

b. left tailed.

c. two tailed.

The following graph portrays the decision criterion for a onemean z-test, using the critical-value approach to hypothesis testing. The curve in the graph is the normal curve for the test statistic under the assumption that the null hypothesis is true.

Determine the

a. rejection region.

b. nonrejection region.

c. critical value(s).

d. significance level.

e. Draw a graph that depicts the answers that you obtained in parts (a)-(d).

f. Classify the hypothesis test as two tailed, left tailed, or right tailed.

Define the $P$- value of the hypothesis test.

In each part, we have given the value obtained for the test statistic, $z$, in a one-mean $z-$test. We have also specified whether the test is two tailed, left tailed, or right tailed. Determine the $P-$value in each case and decide whether, at the $5\%$ significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.

a. $z=-1.25$; left-tailed test

b. $z=2.36$; right-tailed test

c. $z=1.83$; two-tailed test

College basketball, and particularly the NCAA basketball tournament is a popular venue for gambling, from novices in office betting pools to high rollers. To encourage uniform betting across teams, Las Vegas oddsmakers assign a point spread to each game. The point spread is the oddsmakers' prediction for the number of points by which the favored team will win. If you bet on the favorite, you win the bet provided the favorite wins by more than the point spread; otherwise, you lose the bet. Is the point spread a good measure of the relative ability of the two teams? H. Stern and B. Mock addressed this question in the paper "College Basketball Upsets: Will a 16-Seed Ever Beat a 1-Speed? They obtained the difference between the actual margin of victory and the point spread, called the point-spread error, for 2109 college basketball games. The mean point-spread error was found to be -0.2 point with a standard deviation of 10.9 points. For a particular game, a point-spread error of 0 indicates that the point spread was a perfect estimate of the two teams' relative abilities.

Part (a): If, on average, the oddsmakers are estimating correctly, what is the (population) mean point-spread error?

Part (b): Use the data to decide, at the 5% significance level, whether the (population) mean point-spread error?

Part (c): Interpret your answer in part (b).

Cheese Consumption. Refer to Problem $24$. The following table provides last year's cheese consumption: in pounds, 35 randomly selected Americans.

$$\begin{gathered} \begin{array}{lllllll}46& 29& 33& 38& 42& 40& 34\\ 33& 32& 36& 28& 47& 26& 42\\ 36& 32& 45& 24& 39& 28& 33\\ 44& 33& 26& 37& 27& 31& 36\end{array} \\ \begin{array}{lllllll}37& 37& 36& 22& 44& 36& 29\end{array} \end{gathered}$$

1. At the $10\%$ significance level, do the data provide sufficient evidence to conclude that last year's mean cheese consumption for all Americans has increased over the 2010 mean? Assume that $\sigma =6.9\mathrm{lb}$. Use a z-test. (Note: The sum of the data is $1218\mathrm{lb}$.)
2. Given the conclusion in part (a), if an error has been made, what type must it be? Explain your answer.

Refer to Problem 24. The following table provides last year's cheese consumption, in pounds, for 35 randomly selected Americans.

Part (a): At the 10% significance level, do the data provide sufficient evidence to conclude that last year's mean cheese consumption for all Americans has increased over the 2010 mean? Assume that $\sigma =6.9 lb$. Use a z-test.

Part (b): Given the conclusion in part (a), if an error has been made, what type must it be? Explain your answer.

Purse Snatching. The Federal Bureau of Investigation (FBI) compiles information on robbery and property crimes by type and selected characteristic and publishes its findings in Uniform Crime Reports. According to that document, the mean value lost to purse snatching was $468$ in $2012$ . For last year, $12$ randomly selected purse-snatching offenses yielded the following values lost, to the nearest dollar.

Use a t-test to decide, at the$5\%$ significance level, whether last year's mean value lost to purse snatching has decreased from the $2012$ mean. The mean and standard deviation of the data are $455.0$ and $86.8$, respectively. 

The Federal Bureau of Investigation (FBI) compiles information on robbery and property crimes by type and selected characteristic and publishes its finding in Uniform Crime Reports. According to that document, the mean value lost to purse snatching was \(468 in 2012. For last year, 12 randomly selected purse-snatching offenses yielded the following values lost, to the nearest dollar.

Use a t-test tp decide, at the 5% significance level, whether last year's mean value lost to purse snatching has decreased from 2012 mean. The mean and standard deviation of the data are \)455 and $86.80 respectively.

 Betting the Spreads. College basketball, and particularly the NCAA basketball tournament, is a popular venue for gambling, from novices in office betting pools to high rollers. To encourage uniforn betting across teams, Las Vegas oddsmakers assign a point spread to each game. The point spread is the oddsmakers" prediction for th number of points by which the favored team will win. If you bet of the favorite, you win the bet provided the favorite wins by more than the point spread; otherwise, you lose the bet. Is the point spread a good measure of the relative ability of the two teams? H. Stern and B. Mock addressed this question in the paper "College Basketball Upsets: Will a $16$-Seed Ever Beat a $1-$Seed?" (Chance, Vol. 11(1), pp. $27-31$). They obtained the difference between the actual margin of victory and the point spread, called the point-spread error, for 2109 college basketball games. The mean point-spread error was found to be −$0.2$ point with a standard deviation of $10.9$ points. For a particular game, a point-spread error of 0 indicates that the point spread was a perfect estimate of the two teams' relative abilities.
(a) If, on average, the oddsmakers are estimating correctly, what is the (population) mean point-spread error?
(b) Use the data to decide, at the $5\%$ significance level, whether the (population) mean point-spread error differs from $0$ .
c) Interpret your answer in part (b).