# alternative hypothesis

Jun 17th, 2015
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As the alternative hypothesis is Ha: µ >45000, the given test is a one-tailed z-test.Since the sample size, n > 30 I will use z-test for mean to test the given hypothesis.

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Project Part B: Hypothesis Testing and Confidence Intervalsa. The average (mean) annual income was greater than \$45,000Null Hypothesis: The average annual income was greater than \$45,000H: > 45000Alternate Hypothesis: The average annual income was less than \$50,000.Ha: > 45000Analysis Plan: Significance Level, =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: > 45000, the given test is a one-tailed z-test.Critical Value and Decision Rule:The critical value for significance level, =0.05 for a lower-tailed z-test is given as-1.645.Decision Rule: Reject H, if z statistic, -1.645Test Statistic - minitabOne-Sample Z: Income (\$1000)Test of mu = 45 vs >45The assumed standard deviation = 14.5595% UpperVariable N Mean StDev SE Mean Bound Z PIncome (\$1000) 45 43.48 14.55 2.06 46.86 -3.17 0.001Interpretation 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 greater than \$45,000.Confidence Interval - minitabOne-Sample ZThe assumed standard deviation = 14.55N Mean SE Mean 95% CI45 43.48 2.06 (39.45, 47.51)The 95% upper confidence limit is 47.51. Since, 45.00 lies beyond the 95% upper confidence lim

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