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Chapter 7: Q 131SE (page 443)

Question: Point spreads of NFL games. Refer to the Chance (Fall 1998) study of point-spread errors in NFL games, Exercise 7.41 (p. 411). Recall that the difference between the actual game outcome and the point spread established by odds makers鈥攖he point-spread error鈥攚as calculated for 240 NFL games. The results are summarized as follows: . Suppose the researcher wants to know whether the true standard deviation of the point spread errors exceeds 15. Conduct the analysis using 伪 = 0.10.

Short Answer

Expert verified

The true standard deviation of the points spread errors is 15.

Step by step solution

01

Given information

The given problem explains the point-spread error in NFL games. The point-spread error was calculated for 240 NFL games.

From the sample, the sample mean is -1.6, and the sample standard deviation is 13.3.

Here, the test is based on the true standard deviation of the point-spread error.

02

 Concept of testing of hypothesis

The null hypothesis, which can be further tested with the use of specific statistical tests, is an assumption that must be made before making any choice regarding the true value of the parameters. The choice is then made based on the calculated value of the parameter and critical, which means that if the computed value falls within the critical zone, the null hypothesis is rejected; otherwise, it is accepted.

03

Testing the #FA3273hypothesis

The null and alternative hypotheses are

Null hypothesis:

H0 : 蟽 = 15

The true standard deviation of points spread errors is 15.

Alternative hypothesis:

Ha : 蟽 > 15

The true standard deviation of points spread errors exceeds 15.

The level of significance is 伪 = 0.10

Since the test statistic is given by

From the given data sample size n is 240.

By substituting the values,

From the table of for df = 239 and 伪 = 0.10 the critical value for the test is,

The calculated statistics value is less than the critical value. That is .

So, the null hypothesis is rejected at the given level of significance, and we conclude that the true standard deviation of the points spread errors is 15.

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Most popular questions from this chapter

A t-test is conducted for the null hypothesis H0:渭=10versus the alternative Ha:渭>10for a random sample of n = 17 observations. The test results are t = 1.174, p-value = 0.1288.

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Suppose the sample in Exercise 7.64 has produced \(\hat p = .83\) and we wish to test \({H_0}:P = 0.9\) against the alternative \({H_a}:p < .9\)

a. Calculate the value of the z-statistic for this test.

FDA certification of new drugs. Pharmaceutical companies spend billions of dollars per year on research and development of new drugs. The pharmaceutical company must subject each new drug to lengthy and involved testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. The FDA鈥檚 policy is that the pharmaceutical company must provide substantial evidence that a new drug is safe prior to receiving FDA approval, so that the FDA can confidently certify the safety of the drug to potential consumers.

a. If the new drug testing were to be placed in a test of hypothesis framework, would the null hypothesis be that the drug is safe or unsafe? The alternative hypothesis?

A simple random sample of 25 observations was selected from a normal population. The mean and standard deviation of this sample are 20 and 5, respectively.

a. Test H0:渭=22against Ha:渭鈮�22at the 10% significance level.

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Beta value of a stock. The 鈥渂eta coefficient鈥� of a stock is a measure of the stock鈥檚 volatility (or risk) relative to the market as a whole. Stocks with beta coefficients greater than 1 generally bear the greater risk (more volatility) than the market, whereas stocks with beta coefficients less than 1 are less risky (less volatile) than the overall market (Alexander, Sharpe, and Bailey, Fundamentals of Investments, 2000). A random sample of 15 high-technology stocks was selected at the end of 2009, and the mean and standard deviation of the beta coefficients were calculated: x=1.23s = .37.

a. Set up the appropriate null and alternative hypotheses to test whether the average high-technology stock is riskier than the market as a whole.

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f. Conduct a test to determine if the variance of the stock beta values differs from .15. Use a=0.05.
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