Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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—the point-spread error—was 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 α = 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 : σ = 15

The true standard deviation of points spread errors is 15.

Alternative hypothesis:

Ha : σ > 15

The true standard deviation of points spread errors exceeds 15.

The level of significance is α = 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 α = 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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Beta value of a stock. The “beta coefficient” of a stock is a measure of the stock’s 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.

b. Establish the appropriate test statistic and rejection region for the test. Use a=0.10.

c. What assumptions are necessary to ensure the validity of the test?

d. Calculate the test statistic and state your conclusion.

e. What is the approximate p-value associated with this test? Interpret it.

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

For each of the following rejection regions, sketch the sampling distribution for z and indicate the location of the rejection region.

a. \({H_0}:\mu \le {\mu _0}\) and \({H_a}:\mu > {\mu _0};\alpha = 0.1\)

b. \({H_0}:\mu \le {\mu _0}\) and \({H_a}:\mu > {\mu _0};\alpha = 0.05\)

c. \({H_0}:\mu \ge {\mu _0}\) and \({H_a}:\mu < {\mu _0};\alpha = 0.01\)

d. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.05\)

e. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.1\)

f. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.01\)

g. For each rejection region specified in parts a–f, state the probability notation in z and its respective Type I error value.

Question: Water distillation with solar energy. In countries with a water shortage, converting salt water to potable water is big business. The standard method of water distillation is with a single-slope solar still. Several enhanced solar energy water distillation systems were investigated in Applied Solar Energy (Vol. 46, 2010). One new system employs a sun-tracking meter and a step-wise basin. The new system was tested over 3 randomly selected days at a location in Amman, Jordan. The daily amounts of distilled water collected by the new system over the 3 days were 5.07, 5.45, and 5.21 litres per square meter ( l / m2 ). Suppose it is known that the mean daily amount of distilled water collected by the standard method at the same location in Jordan is µ = 1.4 / I m2.

a. Set up the null and alternative hypotheses for determining whether the mean daily amount of distilled water collected by the new system is greater than 1.4.

b. For this test, give a practical interpretation of the value α = 0.10.

c. Find the mean and standard deviation of the distilled water amounts for the sample of 3 days.

d. Use the information from part c to calculate the test statistic.

e. Find the observed significance level (p-value) of the test.

f. State, practically, the appropriate conclusion.

Performance of stock screeners. Recall, from Exercise 6.36 (p. 350), that stock screeners are automated tools used by investment companies to help clients select a portfolio of stocks to invest in. The data on the annualized percentage return on investment (as compared to the Standard & Poor’s 500 Index) for 13 randomly selected stock screeners provided by the American Association of Individual Investors (AAII) are repeated in the accompanying table. You want to determine whether \(\mu \) , the average annualized return for all AAII stock screeners, is positive (which implies that the stock screeners perform better, on average, than the S&P 500). An XLSTAT printout of the analysis is shown on the top of page 418.

9.0 -.1 -1.6 14.6 16.0 7.7 19.9 9.8 3.2 24.8 17.6 10.7 9.1

  1. State \({H_0}\,and\,{H_a}\) for this test

Hotels’ use of ecolabels. Refer to the Journal of Vacation Marketing (January 2016) study of travelers’ familiarity with ecolabels used by hotels, Exercise 2.64 (p. 104). Recall that adult travelers were shown a list of 6 different ecolabels, and asked, “Suppose the response is measured on a continuous scale from 10 (not familiar at all) to 50 (very familiar).” The mean and standard deviation for the Energy Star ecolabel are 44 and 1.5, respectively. Assume the distribution of the responses is approximately normally distributed.

a. Find the probability that a response to Energy Star exceeds 43.

b. Find the probability that a response to Energy Star falls between 42 and 45. c. If you observe a response of 35 to an ecolabel, do you think it is likely that the ecolabel was Energy Star? Explain.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free