A.) The average (mean) annual income was greater than $45,000

Null Hypothesis: The average (mean) annual income is greater than or equal to $45,000.

Ho: u > $45,000

Alternative Hypothesis: The average (mean) annual income was less than $45,000

Ha: u < $45,000

I will use a = .05 as the significance level, and observing the sample size of n < 30 which tells me I need to use a Z-Test to find the mean of this test and the hypothesis. As the alternative hypothesis is Ha: u < $45,000, the given test is a one-tailed Z-Test.

The critical value for the significance level, α=0.05 for a one-tailed z-test is -1.645.

Decision Rule: We must reject the hypothesis if z >1.645

Test Statistics from Minitab:

One-Sample Z: Income ($1000)

Test of mean = 45, < 45

The proposed standard deviation = 14.64

95% Upper Variable N Mean StDev SE Mean Bound Z P Income ($1000) 50 43.48 14.55

2.06 46.86 0.49 0.311 Confidence Intervals from MiniTab:

One-Sample Z

The assumed standard deviation = 14.64

N Mean SE Mean 95% CI

50 42.61 2.06 (39.45, 47.51)

Summary

Since the value of the test statistic is not in the critical I will fail to reject the null hypothesis. Therefore I can conclude that the data gives enough evidence to show the null hypothesis is true i.e. based on the sample data we can say with 95% confidence or 5% significance the claim that the average mean annual income was greater than $45,000, is false.

I am also able to test this hypothesis using a p-value approach.

B.) The true population proportion of customers who live in an urban area exceeds 45%.

Null Hypothesis: The true population proportion of customers who live in an urban area is greater than or equal to 45%.

Ho: p > .45

Alternate Hypothesis: The true population proportion of customers who live in an urban area is less than 45%. Ha: p < .45

I will use a Significance Level of, α=0.05. I will use a Z-Test of proportions to test the given hypothesis. And with the alternative hypothesis being Ho: p > .45, the given test is a one-tailed Z-Test.

The critical value for significance level, α=0.05 for a one-tailed Z-Test is given as 1.645.

Decision Rule: To Reject the null hypothesis if z 0.45

95% Lower Sample X N Sample p Bound Z-Value P-Value 1 21 50 0.420000 0.406599 -2.13 0.017

Confidence Interval from MiniTab:

Test and CI for One Proportion

Sample X N Sample p 95% CI 1 21 50 0.406599 (0.283195, 0.556805)

Summary

Since the value of the test statistic is in the critical region I am going to reject the null hypothesis. Therefore I can determine the data gives me enough evidence to show the null hypothesis is false. I can say with 95% confidence or 5% significance, the true population proportion of customers who live in a suburban area is less than 45%, is true.

So, here the p-value for this test is 0.017. Which is smaller than the significance level 0.05, thus I am rejecting the null hypothesis based on the p-value approach as well.

The 95% Upper Confidence Bound in this case is 0.406599; which tells me, with a probability of 0.95, the true population proportion of customers who live in a suburban area is less than 40.6599%.

C.) The average (mean) number of years lived in the current home is less than 13 years.

Null Hypothesis: The average (mean) number of years lived in the current home is less than or equal to 8 years. Ho: u < 8

Alternate Hypothesis: The average (mean) number of years lived in the current home is greater than 8 years.

Ha: u > 8

I will utilize, α=0.05. Because the sample size is less than 30, a Z-Test will be utilized to find the mean of the hypothesis. As the alternative hypothesis is Ha: u > 8, the given test is a one-tailed Z-Test.

The critical value for significance level, α=0.05 for a lower-tailed z-test is given as 1.645. Decision Rule: Reject null hypothesis z >1.645

Test Statistic from Minitab:

One-Sample Z: Years

Test of mu = < 8, > 8

The proposed standard deviation = 4.4855

95% Upper Variable N Mean StDev SE Mean Bound Z P Years 50 12.380 4.4855 0.722 8.557 2.52 0.006

Summary

Since the value of the test statistic is in the critical region, I am rejecting the null hypothesis. Therefore, I can conclude the data gives enough evidence to show the null hypothesis is false. I can say with 95% confidence or 5% significance the claim, the average (mean) number of years lived in the current home is greater than 8 years, is true.

So the p-value for this test is 0.006. Which is smaller than the significance level 0.05, as a result I am rejecting the null hypothesis based on the p-value approach as well.

The 95% Lower Confidence Bound in this case is 8.557; which tells me, with a probability of 0.95, the average (mean) number of years lived in the current home greater than 8.557.

D.) The average (mean) credit balance for suburban customers is more than $4300.

Null Hypothesis: The average (mean) credit balance for suburban customers is greater than or equal to $3200

Ho: u > 3200

Alternative Hypothesis: The average (mean) credit balance for suburban customers is less than $3200.

Ha: u < 3200

I will use the Significance Level from the case, α=0.05.

Because the sample size is 13 a t-test will be used for mean to test the hypothesis. As the alternative hypothesis is Ha: u < 3200, the test is a one-tailed t-test.

The critical value for given alpha, α=0.05 for a one-tailed z-test is 1.645.

Decision Rule: Reject the null hypothesis if z

Now the test statistic can be computed by the following formula which gives, Z =

Z = -1.35 with corresponding p-value 0.1.

And the 95% Upper Bound = 3251.

Summary

Since the value of the test statistic is not in the critical region I am going to fail to reject the null hypothesis. Therefore I can conclude the data gives enough evidence to show the null hypothesis is true. I can say, with 95% confidence or 5% significance, the claim, the average (mean) credit balance for rural customers is less than $3200, is false.

So the p-value for this test is 0.1. Which is larger than the significance level 0.05, so I am failing to reject the null hypothesis based on the p-value approach as well.

The 95% Upper Confidence Bound in this case is 3251; which tells me, with probability 0.95, I can claim the average (mean) credit balance for rural customers is less than 3251.