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

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 σ=6.9lb. 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.

Short Answer

Expert verified

Part (a): The data provide sufficient evidence to conclude that last year's mean cheese consumption for all Americans has increased over the 2010 mean.

Part (b): There are two types of error.

Type I error is error committed in rejecting a null hypothesis H0, when it is true.

Type I error is error committed in rejecting a null hypothesis H0, when it is false.

Step by step solution

01

Part (a) Step 1. Given information.

Consider the given question,

02

Part (a) Step 2. State the null and alternative hypothesis.

The null hypothesis is given below,

H0:μ=33lb

There is no evidence that last year's mean cheese consumption for Americans has increased over the 2010mean.

The alternative hypothesis is given below,

Ha:μ>33lb

Last year's mean cheese consumption for Americans has increased over the 2010mean.

03

Part (a) Step 3. Compute the confidence interval.

On using the MINITAB procedure,

  1. Choose Stat>Basic Statistics>1-Sample Z.
  2. In Samples in Column, enter the column of Consumption.
  3. In Standard deviation, enter 6.9.
  4. In Perform hypothesis test, enter the test mean as 33.
  5. Check Options, enter Confidence level as 90.
  6. Choose greater than in alternative.
  7. Click OK in all dialogue boxes.

Hence, from the MINITAB output, the value of test statistics is 1.54 and the P-value is 0.061.

Critical value is given below,

From "Table II Areas under the standard normal curve" the required value of za2 with 95% confidence level is ±1.96.

P-value is given below,

From the MINITAB output, the P-value is 0.061.

04

Part (a) Step 4. Use the critical value and P-value approach.

Using the rejection rule on the critical value approach,

If the value of test statistic falls in the rejection region, reject the null hypothesis H0.

Here, the value of test statistics fall sin the rejection region. Therefore, the null hypothesis is rejected at 10% level.

It can be concluded that the test results are statistically significant at 10% level of significance.

Using the P-value approach,

If Pα, then reject the null hypothesis.

The P-value is 0.061, which is greater than the level of significance, that is P=0.061<α=0.1.

Therefore, the null hypothesis is rejected at 5% level.

It can be concluded that the test results are statistically significant at 10% level of significance.

On interpreting, we can say that the data provide sufficient evidence to conclude that last year's mean cheese consumption for all Americans has increased over the 2010mean.

05

Part (b) Step 1. Identify the type of error.

Consider the type I error,

This is the error committed in rejecting a null hypothesis H0, when it is true.

Consider the type II error,

This is the error committed in rejecting a null hypothesis H0, when it is false.

In this situation, the Type I error would be appropriate because the null hypothesis is rejected in part (a).

The conclusion is that last year's mean cheese consumption for all Americans has increased over the 2010 mean but in reality the last year's mean cheese consumption for all Americans does not increased over the2010mean.

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.

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

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 α, the null hypothesis H0:μ=μ0will be rejected in favor of the alternative hypothesis Ha:μ<μ0if and only if μ0is greater than or equal to the (1-α)-level upper confidence bound for μ. 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

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.

12791457202016821273
22232233219216111734
26882029216618602444
18441765226715222012
19901751211322021712

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.)

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).

The normal probability curve and histogram of the data are shown in figure; σis known.

Perform Hypothesis test for mean of the population from which data is obtained and decide whether to use z-test, t-test or neither. Explain your answer.

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 \(9.87\)

See all solutions

Recommended explanations on Math Textbooks

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