Question

The following data was calculated during a study on loan borrowers. Use the following information, to find the test statistic and p-value at a 5% level of significance: - A bank collector like to test the claim that the percent of borrowers who are overdue on loan payments by at least 6 months is less than 26% - Sample size =35 borrowers - Sample proportion =0.20 Use the curve below to find the test statistic and p-value. Select the appropriate test by dragging the blue point to a rightleft-or two-tailed diagram, then set the sliders. Use the purple slider to set the significance level. Use the black sliders to set the information from the study described above.

Answer 1

The test statistic for the bank collector's claim is -1.33, and the p-value is approximately 0.0918. We fail to reject the null hypothesis at the 5% level of significance because the p-value is greater than 0.05.

Step-by-step explanation:

To find the test statistic and p-value, we will conduct a hypothesis test for the population proportion. The null hypothesis (H₀) will state that the proportion of borrowers who are overdue on loan payments by at least 6 months is equal to 0.26. The alternative hypothesis (H₁) will state that this proportion is less than 0.26. Given that the sample size is 35 and the sample proportion is 0.20, we can proceed with the z-test for proportions.

The test statistic for a one-sample z-test for the population proportion is calculated using the formula:
z = (p' - p) / sqrt((p*(1-p))/n)

where p' is the sample proportion, p is the hypothesized population proportion and n is the sample size. Substituting the values:
z = (0.20 - 0.26) / sqrt((0.26*(1-0.26))/35)
z = -1.33

The p-value associated with this test statistic can be found using a standard normal distribution table or calculator. Based on a significance level (alpha) of 0.05, if the p-value is less than 0.05, the null hypothesis is rejected.

Typically, using a calculator like the TI-83, TI-84, or an online statistics calculator, we can find that the p-value for our test statistic is approximately 0.0918. Since the p-value is greater than 0.05, we fail to reject the null hypothesis at the 5% level of significance.