part b
Project Part B: Hypothesis Testing and Confidence Intervals
a. The average (mean) annual income was less than $50,000
Null Hypothesis: The average annual income was greater than or equal to $50,000
H₀: µ > 50000
Alternate Hypothesis: The average annual income was less than $50,000.
Ha: µ > 50000
Analysis Plan: Significance Level, α=0.05.
Since the sample size, n > 30 I will use z-test for mean to test the given hypothesis.
As the alternative hypothesis is Ha: µ > 50000, the given test is a one-tailed z-test.
Critical Value and Decision Rule:
The critical value for significance level, α=0.05 for a lower-tailed z-test is given as-1.645.
Decision Rule: Reject H₀, if z – statistic, -1.645
Test Statistic - minitab
One-Sample Z: Income ($1000)
Test of mu = 50 vs < 50
The assumed standard deviation = 14.55
95% Upper
Variable N Mean StDev SE Mean Bound Z P
Income ($1000) 50 43.48 14.55 2.06 46.86 -3.17 0.001
Interpretation of Results and Conclusion:
Since the P-value (0.0001) is smaller than the significance level (0.05), we reject the null hypothesis. The p-value implies the probability of rejecting a true null hypothesis.
The significance level of 0.05, there is enough evidence to support the claim that the average annual income was less than $50,000.
Confidence Interval - minitab
One-Sample Z
The assumed standard deviation = 14.55
N Mean SE Mean 95% CI
50 43.48 2.06 (39.45, 47.51)
The 95% upper confidence limit is 47.51. Since, 50.00 lies beyond the 95% upper confidence limit, we can support the claim 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 proportion of customers who live in an urban area is less than or equal to 40%.
H₀: p < 0.40
Alternate Hypothesis: The true population proportion of customers who live in an urban area is greater than 40%.
Ha: p > 0.40
Analysis Plan: Significance Level, α=0.05
a. The average (mean) annual income was less than $50,000
Null Hypothesis: The average annual income was greater than or equal to $50,000
H₀: µ > 50000
Alternate Hypothesis: The average annual income was less than $50,000.
Ha: µ > 50000
Analysis Plan: Significance Level, α=0.05.
Since the sample size, n > 30 I will use z-test for mean to test the given hypothesis.
As the alternative hypothesis is Ha: µ > 50000, the given test is a one-tailed z-test.
Critical Value and Decision Rule:
The critical value for significance level, α=0.05 for a lower-tailed z-test is given as-1.645.
Decision Rule: Reject H₀, if z – statistic, -1.645
Test Statistic - minitab
One-Sample Z: Income ($1000)
Test of mu = 50 vs < 50
The assumed standard deviation = 14.55
95% Upper
Variable N Mean StDev SE Mean Bound Z P
Income ($1000) 50 43.48 14.55 2.06 46.86 -3.17 0.001
Interpretation of Results and Conclusion:
Since the P-value (0.0001) is smaller than the significance level (0.05), we reject the null hypothesis. The p-value implies the probability of rejecting a true null hypothesis.
The significance level of 0.05, there is enough evidence to support the claim that the average annual income was less than $50,000.
Confidence Interval - minitab
One-Sample Z
The assumed standard deviation = 14.55
N Mean SE Mean 95% CI
50 43.48 2.06 (39.45, 47.51)
The 95% upper confidence limit is 47.51. Since, 50.00 lies beyond the 95% upper confidence limit, we can support the claim 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 proportion of customers who live in an urban area is less than or equal to 40%.
H₀: p < 0.40
Alternate Hypothesis: The true population proportion of customers who live in an urban area is greater than 40%.
Ha: p > 0.40
Analysis Plan: Significance Level, α=0.05