/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q8BSC Test the given claim. Identify t... [FREE SOLUTION] | 91影视

91影视

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 鈥淏irths鈥 in Appendix B).

Short Answer

Expert verified

The hypotheses are as follows.

\(\begin{array}{l}{H_0}:\sigma = 660.2\\{H_1}:\sigma \ne 660.2\end{array}\)

The test statistic\({\chi ^2} = 45.78\), 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.

\(\begin{array}{l}{H_0}:\sigma = 660.2\\{H_1}:\sigma \ne 660.2\end{array}\)

Here, \(\sigma \) is the true standard deviation of the birth weights for girls.

03

State the test statistic

The test statistic\({\chi ^2}\)with\(\left( {n - 1} \right)\)degrees of freedom is given as follows.

\(\begin{array}{c}{\chi ^2} = \frac{{\left( {n - 1} \right){s^2}}}{{{\sigma ^2}}}\\ = \frac{{\left( {30 - 1} \right){{\left( {829.5} \right)}^2}}}{{{{\left( {660.2} \right)}^2}}}\\ = 45.7804\end{array}\).

The degree of freedom is computed as follows.

\(\begin{array}{c}df = n - 1\\ = 30 - 1\\ = 29\end{array}\)

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\({\chi ^2}_L,{\chi ^2}_R\),such that

\(\begin{array}{l}P\left( {{\chi ^2} < {\chi ^2}_L} \right) = \frac{{0.01}}{2}\left( {0.005} \right)\\P\left( {{\chi ^2} > {\chi ^2}_L} \right) = 0.995\\P\left( {{\chi ^2} > {\chi ^2}_R} \right) = \frac{{0.01}}{2}\left( {0.005} \right)\end{array}\)

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,

\(\begin{array}{l}\chi _L^2 = 13.121\\\chi _R^2 = 52.336\end{array}\)

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 91影视!

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

Final Conclusions. In Exercises 25鈥28, use a significance level of = 0.05 and use the given information for the following:

a. State a conclusion about the null hypothesis. (Reject H0or fail to reject H0.)

b. Without using technical terms or symbols, state a final conclusion that addresses the original claim.

Original claim: More than 58% of adults would erase all of their personal information online if they could. The hypothesis test results in a P-value of 0.3257.

In Exercises 1鈥4, use these results from a USA Today survey in which 510 people chose to respond to this question that was posted on the USA Today website: 鈥淪hould Americans replace passwords with biometric security (fingerprints, etc)?鈥 Among the respondents, 53% said 鈥測es.鈥 We want to test the claim that more than half of the population believes that passwords should be replaced with biometric security.

Number and Proportion

a. Identify the actual number of respondents who answered 鈥測es.鈥

b. Identify the sample proportion and the symbol used to represent it.

Estimates and Hypothesis Tests Data Set 3 鈥淏ody Temperatures鈥 in Appendix B includes sample body temperatures. We could use methods of Chapter 7 for making an estimate, or we could use those values to test the common belief that the mean body temperature is 98.6掳F. What is the difference between estimating and hypothesis testing?

In Exercises 9鈥12, refer to the exercise identified. Make subjective estimates to decide whether results are significantly low or significantly high, then state a conclusion about the original claim. For example, if the claim is that a coin favours heads and sample results consist of 11 heads in 20 flips, conclude that there is not sufficient evidence to support the claim that the coin favours heads (because it is easy to get 11 heads in 20 flips by chance with a fair coin).

Exercise 6 鈥淐ell Phone鈥

Explain how the P -value is obtained for a mean z-test in case of hypothesis test is

(a) left-tailed (b) right- tailed (c) two tailed

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.