# Math 533 Part B

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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…show more content…
Ho: u < 8 Alternate Hypothesis: The average (mean) number of years lived in the current home is greater than 8 years. Ha: u > 8 I will utilize, α=0.05. Because the sample size is less than 30, a Z-Test will be utilized to find the mean of the hypothesis. As the alternative hypothesis is Ha: u > 8, the given test is a one-tailed Z-Test. The critical value for significance level, α=0.05 for a lower-tailed z-test is given as 1.645. Decision Rule: Reject null hypothesis z >1.645 Test Statistic from Minitab: One-Sample Z: Years Test of mu = < 8, > 8 The proposed standard deviation = 4.4855 95% Upper Variable N Mean StDev SE Mean Bound Z P Years 50 12.380 4.4855 0.722 8.557 2.52 0.006 Summary Since the value of the test statistic is in the critical region, I am rejecting the null hypothesis. Therefore, I can conclude the data gives enough evidence to show the null hypothesis is false. I can say with 95% confidence or 5% significance the claim, the average (mean) number of years lived in the current home is greater than 8 years, is true. So the p-value for this test is 0.006. Which is smaller than the significance level 0.05, as a result I am rejecting the null hypothesis based on the p-value approach as well. The 95% Lower Confidence Bound in this case is 8.557;