# MATH 533 Hypothesis Testing

Apr 28th, 2015
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In this report, I will be performing hypothesis testing to each requirement using seven elements of Test Hypothesis with a = 0.02 in order to provide more insights for my manager. I will also include the interpretation of results so that non-statistical people can understand the meaning of number.

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Class: MATH 533Project Part BIn this report, I will be performing hypothesis testing to each requirement using seven elements of Test Hypothesis with a = 0.02 in order to provide more insights for my manager. I will also include the interpretation of results so that non-statistical people can understand the meaning of number. a) The average (mean) annual income of all customers was less than \$48,000I found the average annual income is \$43.74 and the standard deviation is \$14.64.Step1: Null and Alternative Hypothesis Set upNull Hypothesis: The average (mean) annual income was greater than or equal to \$48000Ho: >= 48Alternate Hypothesis: The average (mean) annual income was less than \$48000.Ha: < 48Step2: For =0.02. Since the sample size,Error: Reference source not found n > 30 we will use z-test for mean to test the given hypothesis. As the alternative hypothesis is Ha: < 48Error: Reference source not found, the given test is a one-tailed (lower-tailed) z-test.Step3: The critical value for significance level, =0.02, I found that z = 1-.960. Therefore, the rejection region will be z < -1.960Step4: Test Statistic (MINITAB OUTPUT):Test of mu = 48 vs < 48The assumed standard deviation = 14.64 98% UpperVariable N Mean StDev SE Mean Bound Z PIncome (\$1000) 50 43.74 14.64 2.07 47.99 -2.06 0.020Step5: Interpretation Since the P-value 0.02 is less the significa

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