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 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

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%.

\({H_0}:p = 0.24\)

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.

03

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.

04

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.

05

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.

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

In Exercises 9–12, 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 8 “Pulse Rates”

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.

Store Checkout-Scanner Accuracy In a study of store checkout-scanners, 1234 items were checked for pricing accuracy; 20 checked items were found to be overcharges, and 1214 checked items were not overcharges (based on data from “UPC Scanner Pricing Systems: Are They Accurate?” by Goodstein, Journal of Marketing, Vol. 58). Use a 0.05 significance level to test the claim that with scanners, 1% of sales are overcharges. (Before scanners were used, the overcharge rate was estimated to be about 1%.) Based on these results, do scanners appear to help consumers avoid overcharges?

P-Values. In Exercises 17–20, do the following:

a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.

b. Find the P-value. (See Figure 8-3 on page 364.)

c. Using a significance level of = 0.05, should we reject H0or should we fail to reject H0?

The test statistic of z = -1.94 is obtained when testing the claim that p=38 .

Testing Hypotheses. In Exercises 13–24, 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.

Is the Diet Practical? When 40 people used the Weight Watchers diet for one year, their mean weight loss was 3.0 lb and the standard deviation was 4.9 lb (based on data from “Comparison of the Atkins, Ornish, Weight Watchers, and Zone Diets for Weight Loss and Heart Disease Reduction,” by Dansinger et al., Journal of the American Medical Association, Vol. 293, No. 1). Use a 0.01 significance level to test the claim that the mean weight loss is greater than 0. Based on these results, does the diet appear to have statistical significance? Does the diet appear to have practical significance?

Cans of coke use the data and the claim given in exercise 1 to identify the null and alternative hypothesis and the test statistic. What is the sampling distribution of the test statistic?

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