Question

A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The p-value associated with the test statistic in this problem is approximately equal to:

a 0.0100
b 0.0051
c 0.0026
d 0.0013

Answer 1

The p-value associated with the test statistic in this problem is approximately 0.0026.

Step-by-step explanation:

To determine the p-value associated with the test statistic in this problem, we can use a two-proportion z-test. The null hypothesis is that the proportion of households equipped with VCRs is less than or equal to 5000/20000 = 0.25. The alternate hypothesis is that the proportion is greater than 0.25.

The test statistic can be calculated as z = (p1 - p2) / sqrt(p_pool * (1 - p_pool) * (1/n1 + 1/n2)), where p1 is the proportion in the sample, p2 is the proportion under the null hypothesis, n1 is the sample size, n2 is the total population size, and p_pool is the pooled proportion.

In this case, the sample proportion is 96/300 = 0.32, and the population proportion is 5000/20000 = 0.25. Plugging these values into the formula, we find that the test statistic is approximately 2.944. Looking up this value in the z-table or using a calculator function, we find that the p-value is approximately 0.0026.