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?

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

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.

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.

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.

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.

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.