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.

M&Ms Data Set 27 “M&M Weights” in Appendix B lists data from 100 M&Ms, and 27% of them are blue. The Mars candy company claims that the percentage of blue M&Ms is equal to 24%. Use a 0.05 significance level to test that claim. Should the Mars company take corrective action?

Short Answer

Expert verified

Nullhypothesis: The proportion of blue M&Ms is equal to 0.24.

Alternativehypothesis: The proportion of blue M&Ms is not equal to 0.24.

Teststatistic: 0.702

Criticalvalue: 1.96

P-value: 0.4827

The null hypothesis is failed to reject.

There is not enough evidence to reject the claim that the proportion of blue M&Ms is equal to 0.24.

Since the proportion of blue M&Ms is equal to 24% per the claim of the company, the company does not need to take any corrective measure.

Step by step solution

01

Given information

The proportion of blue M&Ms in a sample of 100 M&Ms is equal to 27%.

02

Hypotheses

The null hypothesis is written as follows:

The proportion of blue M&Ms is equal to 24%.

H0:p=0.24

The alternative hypothesis is written as follows:

The proportion of blue M&Ms is not equal to 24%.

H1:p0.24

The test is two-tailed.

03

Sample size, sample proportion, and population proportion

The sample size is equal to n=100.

The sample proportion of blue M&Ms isas follows:

p^=27%=27100=0.27

The population proportion of blue M&Ms is equal to 0.27.

04

Test statistic

The value of the test statistic is computed below:

z=p^-ppqn=0.27-0.240.241-0.24100=0.702

Thus, z=0.702.

05

Critical value and p-value

Referring to the standard normal distribution table, the critical value of z at α=0.05 for a two-tailed test is equal to 1.96.

Referring to the standard normal distribution table, the p-value for the test statistic value of 2.694 is equal to 0.4827.

Since the p-value is greater than 0.05, the null hypothesis is failed to reject.

06

Conclusion of the test

There is not enough evidence to reject the claim that the proportion of blue M&Ms is equal to 0.24.

Since the proportion of blue M&Ms is equal to 24% per the claim of the company, the company does not need to take any corrective measure.

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Most popular questions from this chapter

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

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

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

Original claim: Fewer than 90% of adults have a cell phone. The hypothesis test results in a P-value of 0.0003.

Test Statistics. In Exercises 13–16, refer to the exercise identified and find the value of the test statistic. (Refer to Table 8-2 on page 362 to select the correct expression for evaluating the test statistic.)

Exercise 7 “Pulse Rates”

Finding P-values. In Exercises 5–8, either use technology to find the P-value or use Table A-3 to find a range of values for the P-value.

Airport Data Speeds: The claim that for Verizon data speeds at airports, the mean. The sample size is and the test statistic is

t =-1.625 .

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.

Medical Malpractice In a study of 1228 randomly selected medical malpractice lawsuits, it was found that 856 of them were dropped or dismissed (based on data from the Physicians Insurers Association of America). Use a 0.01 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed. Should this be comforting to physicians?

Interpreting Power For the sample data in Example 1 “Adult Sleep” from this section, Minitab and StatCrunch show that the hypothesis test has power of 0.4943 of supporting the claim that μ<7 hours of sleep when the actual population mean is 6.0 hours of sleep. Interpret this value of the power, then identify the value of βand interpret that value. (For the t test in this section, a “noncentrality parameter” makes calculations of power much more complicated than the process described in Section 8-1, so software is recommended for power calculations.)

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