Math 533 Part B Aj Davis Dept. Store

a. The mediocre (mean) one-year income was little than $50,000 Solution: Step1: State the count and Alternate Hypothesis: Null Hypothesis: The add up (mean) annual income was greater than or equal to $50,000 [pic] Alternate Hypothesis: The intermediate (mean) annual income was less than $50,000. [pic] Step2: Analysis Plan: We shall use tax write-off Level, ?=0.05. Since the sample size,[pic] we shall use z-test for mean to test the given hypothesis. As the alternative hypothesis is[pic], the given test is a one-tailed (lower-tailed) z-test. Step3: critical Value and Decision Rule: The critical value for upshot level, ?=0.05 for a lower-tailed z-test is given as-1.645. [pic] Step4: Test Statistic (MINITAB OUTPUT): One-Sample Z: Income ($1000) Test of mu = 50 vs < 50 The anticipate standard deviation = 14.55 95% swiftness Variable N mean StDev SE squiffy march Z P Income ($1000) 50 43.48 14.55 2.06 46.86 -3.17 0.

001 Step5: Interpretation of Results and Conclusion: Since the P-value (0.0001) is small than the significance level (0.05), we jib the null hypothesis. The p-value implies the probability of rejecting a true null hypothesis. Thus, at a significance level of 0.05, there is sufficient reason to support the claim that the average (mean) annual income was less than $50,000. self-ass ertion Interval (MINITAB OUTPUT): One-Sam! ple Z The assumed standard deviation = 14.55 N Mean SE Mean 95% CI 50 43.48 2.06 (39.45, 47.51) The 95% swiftness combine limit is 47.51. Since, 50.00 lies beyond the 95% upper confidence limit, hence we can support the claim that the average (mean) annual income was less than $50,000. b. The true population proportion of...If you want to repulse a full essay, order it on our website: OrderEssay.net

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