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Chapter 9: Q. 9.28 (page 393)

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 2109college basketball games. The mean point-spread error was found to be -0.2point with a standard deviation of 10.9points. For a particular game, a point-spread error of 0indicates 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).

Short Answer

Expert verified

Part (a): The mean point-spread error is 0.

Part (b): The test results are not statistically significant at 5% level of significance.

Part (c): There is no sufficient evidence to conclude that the mean point-spread error differ from 0.

Step by step solution

01

Part (a) Step 1. Given information.

Consider the given question,

The mean point-spread error is -0.2.

The standard deviation is 10.9.

The significance level is α=0.05.

02

Part (a) Step 2. Determine the mean point-spread error.

Consider the given question,

The mean point-spread error is 0 if the oddsmakers are estimating correctly.

03

Part (b) Step 1. Check whether or not the mean point-spread error differs from 0.

The null hypothesis is given below,

H0:μ=0

The mean point-spread error is0.

The alternative hypothesis is given below,

Ha:μ≠0

The mean point-spread error differs from 0.

04

Part (b) Step 2. Compute the value of the test statistics and P-value.

On computing the value of the test statistics and P-value,

  1. Choose Stat>Basic Statistics>1-Sample t.
  2. In Summarized data, enter the sample size 2109 and mean -0.2.
  3. In Standard deviation, enter 10.9.
  4. In Perform hypothesis test, enter the test mean as 0.
  5. Check Options, enter Confidence level as 95.
  6. Choose not equal in alternative.
  7. Click OK in all dialogue boxes.


Hence, from the MINITAB output, the value of test statistics is -0.84 and the P-value is 0.4.

05

Part (b) Step 3. Write the conclusion.

If P≤α, then reject the null hypothesis.

The P-value is 0.4which is greater than the level of significance that is P=0.4>α=0.05.

Therefore, the null hypothesis is not rejected at 5% level.

It can be concluded that the test results are not statistically significant at 5% level of significance.

06

Part (c) Step 1. Write the interpretation.

Consider the part (b),

On interpreting, we can say that there is no sufficient evidence to conclude that the mean point-spread error differ from0.

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Most popular questions from this chapter

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.

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 \(9.87\)

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 \(48.16in 2009 . Last year's local monthly bills, in dollars, for a random sample of 75 cell phone users are given on the WeissStats site. Use the technology of your choice to do the following.
a. Obtain a normal probability plot, boxplot, histogram, and stemand-leaf diagram of the data.
b. At the 5%significance level, do the data provide sufficient evidence to conclude that last year's mean local monthly bill for cell phone users decreased from the 2009 mean of \)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.

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.

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 1.91gallons 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?
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