Question

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

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

Given information

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

Hypotheses

The null hypothesis is written as follows:

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

$H_0:p=0.24$

The alternative hypothesis is written as follows:

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

$H_1:p\ne 0.24$

The test is two-tailed.

Sample size, sample proportion, and population proportion

The sample size is equal to n=100.

The sample proportion of blue M&Ms isas follows:

$$\begin{gathered} \hat{p}=27\% \\ =\frac{27}{100} \\ =0.27 \end{gathered}$$

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

Test statistic

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.27-0.24}{\sqrt{\frac{0.24\left(1-0.24\right)}{100}}} \\ =0.702 \end{gathered}$$

Thus, z=0.702.

Critical value and p-value

Referring to the standard normal distribution table, the critical value of z at $\alpha =0.05$ 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.

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.