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.

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?

Short answer

Null hypothesis: The proportion of items that are overcharged is equal to 1%.

Alternative hypothesis: The proportion of items that are overcharged is not equal to 1%.

Test statistic: 2.118

Critical value: 1.96

P-value: 0.0342

The null hypothesis is rejected.

There is enough evidence to reject the claim that the proportion of sales that are overcharged is equal to 1%.

As the sample proportion of overcharged items is greater than 1% (1.6%), the scanners detect the items that are overcharged accurately and thus, help the customers to avoid overcharges.

Step-by-step solution

Given information

In a sample of 1234 checked items, 20 checked items are overcharged. It is claimed that 1% of the items sold are overcharged.

Hypotheses

The null hypothesis is written as follows.

The proportion of overcharged items is equal to 1%.

$H_0:p=0.01$

The alternative hypothesis is written as follows.

The proportion of overcharged items is not equal to 1%.

$H_1:p\ne 0.01$

The test is two-tailed.

Sample size, sample proportion, and population proportion

The sample size is equal to n=1234.

The sample proportion of overcharged items is computed below.

$$\begin{gathered} \hat{p}=\frac{\text{Number}\text{of}\text{items}\text{that}\text{were}\text{overcharges}}{\text{Sample}\text{Size}} \\ =\frac{20}{1234} \\ =0.016 \end{gathered}$$

The population proportion of overcharged items is equal to 0.016.

Test statistic

The value of the test statistic is computed below.

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.016-0.01}{\sqrt{\frac{0.01\left(1-0.01\right)}{1234}}} \\ =2.118 \end{gathered}$$

Thus, z=2.118.

Critical value and p-value

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

Referring to the standard normal table, the p-value for the test statistic value of 2.118 is equal to 0.0342.

As the p-value is less than 0.05, the null hypothesis is rejected.

Conclusion of the test

There is enough evidence to reject the claim that the proportion of sales that are overcharged is equal to 1%.

It can be seen that the sample proportion of overcharged items isgreater than 1% (1.6%). Thus, it can be said that the scanners are working efficiently and helping in identifying items that are overcharged.

Thus, the scanners help the customers to avoid overcharges.