/*! 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} Q19 Testing Hypotheses. In Exercises... [FREE SOLUTION] | 91影视

91影视

Testing Hypotheses. In Exercises 13鈥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.

Cans of Coke Data Set 26 鈥淐ola Weights and Volumes鈥 in Appendix B includes volumes (ounces) of a sample of cans of regular Coke. The summary statistics are n = 36, x = 12.19 oz, s = 0.11 oz. Use a 0.05 significance level to test the claim that cans of Coke have a mean volume of 12.00 ounces. Does it appear that consumers are being cheated?

Short Answer

Expert verified

The hypotheses are stated as follows.

H0:=12ozH1:12oz

The test statistic t=10.364. The critical values are 2.030.

The null hypothesis is rejected. It can be concluded that there is not sufficient evidence to support the claim. The consumers do not appear to be cheated.

Step by step solution

01

Given information

The summary statistics for the volumes of sample cans of regular Coke are as follows.

Sample size n=36.

Sample mean x=12.19oz.

Sample standard deviation s=0.11oz.

Level of significance =0.05.

The claim states that the Coke volume has a mean volume of 12.00 ounces.

02

State the hypotheses

Null hypothesis H0: The cans of Coke have a mean volume of 12 ounces.

Alternative hypothesis H1: The cans of Coke do not have a mean volume of 12 ounces.

The hypotheses are formulated as follows.

H0:=12ozH1:12oz

03

Compute the test statistic

The simple random sample is collected from a normally distributed population. For an unknown population standard deviation, use student t-distribution.

The test statistic is given as follows.

t=x-sn=12.19-120.1136=10.3636

The test statistic is 10.3636.

04

Compute the critical value

The level of significance is =0.05.

The degree of freedom is computed as follows.

df=n-1=36-1=35

Refer to the t-table for the critical value corresponding to 35 degrees of freedom and a level of significance of 0.05 for the two-tailed test, which is t0.05=2.030.

Thus, the critical values are , and the non-rejection area is t:-2.030<t<2.030.

05

State the decision rule

The decision rule states the following:

If the test statistic is greater than the critical value, the null hypothesis will be rejected.

If the test statistic is not greater than the critical value, the null hypothesis will fail to be rejected.

Here, the test statistic is 10.364, which is greater than 2.030. Thus, the null hypothesis is rejected at a 0.05 level of significance.

06

Conclusion

As the null hypothesis is rejected, it can be concluded that there is insufficient evidence to support the claim that the mean volume of the Coke cans is 12.00 ounces.

As the sample mean is 12.19 ounces, which is greater than the claimed volume of 12.00 ounces, the consumers will get slightly more than the claimed amount. Thus, the consumers will not be cheated.

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

Testing Hypotheses. In Exercises 13鈥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.

Car Booster Seats The National Highway Traffic Safety Administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified requirement?

774 649 1210 546 431 612

Testing Hypotheses. In Exercises 13鈥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.

Insomnia Treatment A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. Before treatment with zopiclone, 16 subjects had a mean wake time of 102.8 min. After treatment with zopiclone, the 16 subjects had a mean wake time of 98.9 min and a standard deviation of 42.3 min (based on data from 鈥淐ognitive Behavioral Therapy vs Zopiclone for Treatment of Chronic Primary Insomnia in Older Adults,鈥 by Sivertsenet al.,Journal of the American Medical Association, Vol. 295, No. 24). Assume that the 16 sample values appear to be from a normally distributed population, and test the claim that after treatment with zopiclone, subjects have a mean wake time of less than 102.8 min. Does zopiclone appear to be effective?

PowerFor a hypothesis test with a specified significance level , the probability of a type I error is, whereas the probability of a type II error depends on the particular value ofpthat is used as an alternative to the null hypothesis.

a.Using an alternative hypothesis ofp< 0.4, using a sample size ofn= 50, and assumingthat the true value ofpis 0.25, find the power of the test. See Exercise 34 鈥淐alculating Power鈥漣n Section 8-1. [Hint:Use the valuesp= 0.25 andpq/n= (0.25)(0.75)/50.]

b.Find the value of , the probability of making a type II error.

c.Given the conditions cited in part (a), find the power of the test. What does the power tell us about the effectiveness of the test?

Hypothesis Test with Known How do the results from Exercise 13 鈥淐ourse Evaluations鈥 change if is known to be 0.53? Does the knowledge of have much of an effect?

Type I and Type II Errors. In Exercises 29鈥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 adults who use the internet is greater than 0.87.

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.