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.

Ho: µ ≥ 50,000

Alternate Hypothesis is the average annual income is < than $50,000.

Ha: µ < 50,000

Therefore:

Ho: µ 50000

Ha: µ < 50000 Claim

The significance level is: alpha = 0.05

Since n > 30, the z test was used to test the hypothesis

Alternative Hypothesis Ha: µ < 50000, 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 versus < 50

- Standard Deviation = 14.64 using

- Confidence Level = 95%

- Alternative = not equal

Results from Minitab:

- N = 50

- Mean = 43.74

- Standard Deviation = 14.64

- SE Mean = 2.07

- 95% CI (39.68, 47.80)

what this means is that:

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 47.80. And this is because 50 lies beyond the 95% upper confidence limit. We can therefore, 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%

Ho: p ≤ 40% or 0.40

Alternative Hypothesis the true population of customers who live in an urban area is >...