Project Part B: Hypothesis Testing and Confidence Intervals

A. The average (mean) annual income was less than $50,000

• Null Hypothesis is the average annual income is ≥ to $50,000. o Ho: µ ≥ 50,000

• Alternate Hypothesis is the average annual income is < than $50,000. o Ha: µ < 50,000

• Analysis Plan significance level is: a = 0.05

• n > 30 the z test was used to test the hypothesis

• Alternative Hypothesis Ha: µ ≥ 50,000, this means the test is a one tailed z test.

Critical Value and Decision Rule

• C.V: a = 0.05 the lower tailed z test is 1.645

• D.R: reject Ho if z – statistic is -1.645

Sample Z using Minitab:

• Income ($1000)

• Variable I

• Test of µ = 50 verse < 50

• Standard Deviation = 14.55 using

• Confidence Level = 95%

• Alternative = not equal

Results from Minitab:

• N = 50

• Mean = 43.48

• Standard Deviation = 14.55

• SE Mean = 2.06

• 95% CI (39.45, 47.51)

Interpretation of Results:

We reject the Null Hypothesis since the P-value .0001 is smaller than the significance level 0.05. The p-value indicates the probability of rejecting a true Null Hypothesis. There is significant support to claim that the average annual income was less than $50,000 since there is a significance level of 0.05. The 95% upper confidence limit is 47.51. Because 50 lies beyond the 95% upper confidence limit we can state that the average annual income was less than $50,000.

B. The true population proportion of customers who live in an urban area exceeds 40%.

• Null Hypothesis The true population of customers who live in an urban area is ≤ 40% o Ho: p ≤ 40%

• Alternative Hypothesis The true population of customers who live in an urban area is > than 40% o Ha: p > 40%

• Analysis Plan significance level is: a = 0.05 and the z –test for proportion to test the given hypothesis

Critical Value and Decision Rule

• C.V: a = 0.05 the upper-tailed z test is 1.645

• D.R: reject Ho if z – statistic is > 1.645

Test Statistic using minitab

• Number of events (Urban Customers) = 21

• Number of Trials (Customers) = 50

• Test and CI for one proportion\

• Test of p = .4 vs p > .4

• 95% lower

Results from Minitab

• N= 50

• Number of Events = 21

• Sample P = .42

• 95% Lower Bound = .301384

• Z Value = .29

• P Value = .386

• 95% CI (.281882, .56794)

Interpretation of Results:

We fail to reject the null hypothesis since the p-value (.568) is greater than the significance level of (.05). The p-value implies the probability of rejecting a true null hypothesis. There is not enough evidence to support the claim that the true population of customers who live in an urban area is greater than 40% if there is a significance level of .05.

The 95% lower confidence limit is.28. Since the sample –p is .42 is greater than the 95% lower confidence limit we can support the claim that the true population proportion of customers who live in an urban area is greater than 40%.

C. The average number (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 < 13 years o Ho: µ > 13

• Alternate Hypothesis – The average (mean) number of years lived in the current home is < 13 years. o Ha: µ < 13

o Using the z test for mean to test the hypothesis

o Alternative Hypothesis Ha: µ 1.645

Results from Minitab

• Test of µ; 13 vs 4300

• Analysis Plan: Significance Level, a = 0.05

• n > 30 the z test for mean to test the given hypothesis o Alternate hypothesis is Ha: µ > 4300 and a one tailed z test

Critical Value and Decision Rule:

• C.V: a = .05 upper tail z test is 1.645

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