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 isnot 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 isasfollows:

\[\begin{array}{c}\hat p = 27\% \\ = \frac{{27}}{{100}}\\ = 0.27\end{array}\]

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

Test statistic

The value of the test statistic is computed below:

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

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.