MATH SOLVE

2 months ago

Q:
# Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina. A random sample of 40 adults in Virginia had an average credit score of 699 and a random sample of 35 adults in North Carolina had an average credit score of 682. It is believed that the population standard deviation for credit scores is 44 and 41 for Virginia and North Carolina residents, respectively. Experian would like to set α = 0.05. Define Population 1 as Virginia and Population 2 as North Carolina anda. use the critical value approach to test this hypothesis. b. Construct a 95% confidence interval for the difference in population mean and interpret the results.

Accepted Solution

A:

Answer:a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means [tex]\mu_{1}-\mu_{2}[/tex] is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.Step-by-step explanation:Let [tex]\mu_{1}-\mu_{2}[/tex] be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes [tex]n_{1} = 40[/tex] and [tex]n_{2} = 35[/tex], the unbiased point estimate for [tex]\mu_{1}-\mu_{2}[/tex] is [tex]\bar{x}_{1} - \bar{x}_{2}[/tex], i.e., 699-682 = 17.The standard error is given by [tex]\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}[/tex], i.e.,[tex]\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}[/tex] = 9.8198.a. We want to test [tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} \neq 0[/tex] (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is [tex]Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}[/tex] and the observed value is [tex]z_{0} = \frac{17}{9.8198} = 1.7312[/tex]. Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.b. The endpoints for a 95% confidence interval for [tex]\mu_{1}-\mu_{2}[/tex] is given by [tex]17\pm (z_{0.05/2})9.8198[/tex], i.e., [tex]17\pm (z_{0.025})9.8198[/tex] where [tex]z_{0.025}[/tex] is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means [tex]\mu_{1}-\mu_{2}[/tex] is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.