Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then the conclusion about the null hypothesis, as well as the final conclusion that address the original claim. Assume that a simple random sample is selected from a normally distributed population.

Birth Weights A simple random sample of birth weights of 30 girls has a standard deviation of 829.5 hg. Use a 0.01 significance level to test the claim that birth weights of girls have the same standard deviation as birth weights of boys, which is 660.2 hg (based on Data Set 4 “Births” in Appendix B).

Short Answer

Expert verified

The hypotheses are as follows.

H0:σ=660.2H1:σ660.2

The test statistic χ2=45.78, and the critical values are 52.336 and 13.121.

The null hypothesis is failed to be rejected.

There is sufficient evidence to support the claim that the standard deviation of the birth weights of girls is equal to that of the birth weights of boys.

Step by step solution

01

Given information 

The standard deviation of the birth weights for 30 girls is 829.5 hg.

The level of significance is 0.01 to test the claim that both girls and boys have the same standard deviation of birth weights, which is 660.2 hg.

02

State the hypotheses

To test the claim that the birth weights of girls have the same standard deviation as the birth weights of boys, the null and alternative hypotheses are formulated as follows.

H0:σ=660.2H1:σ660.2

Here, σis the true standard deviation of the birth weights for girls.

03

State the test statistic 

The test statistic χ2with degrees of freedom n-1is given as follows.

.χ2=n-1s2σ2=30-1829.52660.22=45.7804

The degree of freedom is computed as follows.

df=n-1=30-1=29

Thus, the test statistic is 45.78 with 29 degrees of freedom.

04

State the critical values 

The test is two-tailed.

The critical values are χ2L,χ2R, such that

Pχ2<χ2L=0.0120.005Pχ2>χ2L=0.995Pχ2>χ2R=0.0120.005

Using the chi-square table for 29 degrees of freedom and a 0.005 level of significance, the right-tailed critical value is 52.336, and the left-tailed one is 13.121.

Thus,

χL2=13.121χR2=52.336

05

State the decision 

The decision rule states the following:

If the test statistic lies between the critical values, the null hypothesis will fail to be rejected; otherwise, it will be rejected.

In this case, the test statistic lies between the critical values, and hence, the null hypothesis is failed to be rejected at a 0.01 level of significance.

Thus, it can be concluded that there is sufficient evidence to support the claim that the standard deviation of the birth weights of girls is equal to the standard deviation of the birth weights of boys.

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

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Drug Screening The company Drug Test Success provides a “1-Panel-THC” test for marijuana usage. Among 300 tested subjects, results from 27 subjects were wrong (either a false positive or a false negative). Use a 0.05 significance level to test the claim that less than 10% of the test results are wrong. Does the test appear to be good for most purposes?

Type I and Type II Errors. In Exercises 29–32, provide statements that identify the type I error and the type II error that correspond to the given claim. (Although conclusions are usually expressed in verbal form, the answers here can be expressed with statements that include symbolic expressions such as p = 0.1.).

The proportion of people who write with their left hand is equal to 0.1.

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Is the Diet Practical? When 40 people used the Weight Watchers diet for one year, their mean weight loss was 3.0 lb and the standard deviation was 4.9 lb (based on data from “Comparison of the Atkins, Ornish, Weight Watchers, and Zone Diets for Weight Loss and Heart Disease Reduction,” by Dansinger et al., Journal of the American Medical Association, Vol. 293, No. 1). Use a 0.01 significance level to test the claim that the mean weight loss is greater than 0. Based on these results, does the diet appear to have statistical significance? Does the diet appear to have practical significance?

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Earthquake Depths Data Set 21 “Earthquakes” in Appendix B lists earthquake depths, and the summary statistics are n = 600, x = 5.82 km, s = 4.93 km. Use a 0.01 significance level to test the claim of a seismologist that these earthquakes are from a population with a mean equal to 5.00 km.

Interpreting Power Chantix (varenicline) tablets are used as an aid to help people stop smoking. In a clinical trial, 129 subjects were treated with Chantix twice a day for 12 weeks, and 16 subjects experienced abdominal pain (based on data from Pfizer, Inc.). If someone claims that more than 8% of Chantix users experience abdominal pain, that claim is supported with a hypothesis test conducted with a 0.05 significance level. Using 0.18 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test.

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