MATH533 Project B
Chanelle Henderson
Math533 Applied Managerial Statistics
Prof. Lisabeth Goble
Week 6: Project B: Testing Hypothesis Testing and Confidence Intervals
A. The average annual income was less than $50,000
Null Hypothesis: The average annual income was greater than or equal to $50,000H₀: µ ≥ 50,000
Alternate Hypothesis: The average annual income was less than $50,000. (Claim)Ha: µ <50,000
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 <50,000, the given test is a one-sided z-test.
Critical Value and Decision Rule:The critical value for significance level, α=0.05 for a left -sided z-test = -1.645.
Decision Rule: Reject H₀, if z-test> -1.645
Test Statistic - Minitab
One-Sample Z: Income ($1000)Test of mu = 50 vs< 50
The assumed standard deviation = 14.64
95% Upper
N Mean SE Mean Bound Z P
50 43.74 2.07 47.15 -3.02 0.001
Interpretation of Results and Conclusion: Since the P-value (0.001) 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 and is the area to the left of the z test. 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 = 14.64
N Mean SE Mean 95% CI
50 43.74 2.07 (39.68, 47.80)
The 95% upper confidence limit is 43.74. 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.
B. The true population proportion of customers who live in an urban area exceeds40%
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
Math533 Applied Managerial Statistics
Prof. Lisabeth Goble
Week 6: Project B: Testing Hypothesis Testing and Confidence Intervals
A. The average annual income was less than $50,000
Null Hypothesis: The average annual income was greater than or equal to $50,000H₀: µ ≥ 50,000
Alternate Hypothesis: The average annual income was less than $50,000. (Claim)Ha: µ <50,000
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 <50,000, the given test is a one-sided z-test.
Critical Value and Decision Rule:The critical value for significance level, α=0.05 for a left -sided z-test = -1.645.
Decision Rule: Reject H₀, if z-test> -1.645
Test Statistic - Minitab
One-Sample Z: Income ($1000)Test of mu = 50 vs< 50
The assumed standard deviation = 14.64
95% Upper
N Mean SE Mean Bound Z P
50 43.74 2.07 47.15 -3.02 0.001
Interpretation of Results and Conclusion: Since the P-value (0.001) 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 and is the area to the left of the z test. 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 = 14.64
N Mean SE Mean 95% CI
50 43.74 2.07 (39.68, 47.80)
The 95% upper confidence limit is 43.74. 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.
B. The true population proportion of customers who live in an urban area exceeds40%
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