Math 533 Part B
PROJECT PART B: Hypothesis Testing and Confidence Intervals
Math-533 Applied Managerial Statistics
Prof. Jeffrey Frakes
December 8, 2014
Jared D Stock
A.) The average (mean) annual income was greater than $45,000
Null Hypothesis: The average (mean) annual income is greater than or equal to $45,000. Ho: u > $45,000
Alternative Hypothesis: The average (mean) annual income was less than $45,000 Ha: u < $45,000
I will use a = .05 as the significance level, and observing the sample size of n < 30 which tells me I need to use a Z-Test to find the mean of this test and the hypothesis. As the alternative hypothesis is Ha: u < $45,000, the given test is a one-tailed Z-Test.
The critical value for the significance level, α=0.05 for a one-tailed z-test is -1.645. Decision Rule: We must reject the hypothesis if z >1.645
Test Statistics from Minitab:
One-Sample Z: Income ($1000)
Test of mean = 45, < 45
The proposed standard deviation = 14.64
95% Upper
Variable N Mean StDev SE Mean Bound Z P
Income ($1000) 50 43.48 14.55 2.06 46.86 0.49 0.311
Confidence Intervals from MiniTab:
One-Sample Z
The assumed standard deviation = 14.64
N Mean SE Mean 95% CI
50 42.61 2.06 (39.45, 47.51)
Summary
Since the value of the test statistic is not in the critical I will fail to reject the null hypothesis. Therefore I can conclude that the data gives enough evidence to show the null hypothesis is true i.e. based on the sample data we can say with 95% confidence or 5% significance the claim that the average mean annual income was greater than $45,000, is false.
I am also able to test this hypothesis using a p-value approach.
B.) The true population proportion of customers who live in an urban area exceeds 45%.
Null Hypothesis: The
Math-533 Applied Managerial Statistics
Prof. Jeffrey Frakes
December 8, 2014
Jared D Stock
A.) The average (mean) annual income was greater than $45,000
Null Hypothesis: The average (mean) annual income is greater than or equal to $45,000. Ho: u > $45,000
Alternative Hypothesis: The average (mean) annual income was less than $45,000 Ha: u < $45,000
I will use a = .05 as the significance level, and observing the sample size of n < 30 which tells me I need to use a Z-Test to find the mean of this test and the hypothesis. As the alternative hypothesis is Ha: u < $45,000, the given test is a one-tailed Z-Test.
The critical value for the significance level, α=0.05 for a one-tailed z-test is -1.645. Decision Rule: We must reject the hypothesis if z >1.645
Test Statistics from Minitab:
One-Sample Z: Income ($1000)
Test of mean = 45, < 45
The proposed standard deviation = 14.64
95% Upper
Variable N Mean StDev SE Mean Bound Z P
Income ($1000) 50 43.48 14.55 2.06 46.86 0.49 0.311
Confidence Intervals from MiniTab:
One-Sample Z
The assumed standard deviation = 14.64
N Mean SE Mean 95% CI
50 42.61 2.06 (39.45, 47.51)
Summary
Since the value of the test statistic is not in the critical I will fail to reject the null hypothesis. Therefore I can conclude that the data gives enough evidence to show the null hypothesis is true i.e. based on the sample data we can say with 95% confidence or 5% significance the claim that the average mean annual income was greater than $45,000, is false.
I am also able to test this hypothesis using a p-value approach.
B.) The true population proportion of customers who live in an urban area exceeds 45%.
Null Hypothesis: The