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
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 >...