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You may need to use the appropriate technology to this on A Pew Research study conducted in 2017 found that approximately 75of Americans believe that rooms and computers might one day do any of the jobs currently done by people. Suppose we have the following data collected from u a utor, and fast food workers in which a higher score means the person for his or her job is more likely to be automated, Fast-Food (a) U -.05 to test to detesthe t ta pero's job is to be tomated for the three professions State the null and alternative these OHN two of the population means are equal Nt least two of the promenaderent O w ww Not all the population means are al H P Prat O rtowe

Answer 1

Answer Date: 11/11/2019 To test the hypothesis is that there is a difference in mean scores for the three professionals at 5% level of significance. The null and alternative hypothesis is, H:/4 = \l, = 42 H: at least one uis unequal. The answer is 2 The F-test statistics is, 1) By using Excel software, find F-test statistics with the help of following steps: Import the data. Choose ANOVA: Single Factor from Data Analysis in Data menu. Choose Input Range and Column in Groups by. Choose Output Range. Click Ok.

Anova: Single Factor Count Sum SUMMARY Groups Nurse Tax auditor Fast-food worker 8 8 Average Variance 30 3.75 1.357143 344 .25 1.357143 48 6 2.285714 df F ANOVA Source of Variation Between Groups Within Groups Total P -value 6.70 0.005616 F crit 3.466800112 SS 22.33333333 35.00 57.33333333 MS 2 11.16667 21 1.666667 23 From the Excel output, the F-test statistics is 6.70 The p-value for this test is, From the Excel output, the p value for this test is 0.006). Decision

The conclusion is that p-value in this context is less than 0.05 which is 0.006, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there is a difference in mean scores for the three professionals. The result is statistically significant. The Fisher's LSD value is, First, compute t critical value then find Fisher's LSD value. The t critical value is, The sample size is small and two-tailed test. Look in the column headed a = 0.05 and the row headed in the t distribution table by using degree of freedom within is 21 and the t critical value is 2.080.

The critical value of Fisher's LSD is, LSD=t,DFW i 1 MSW x -+- nA ng = 2.080 x (11) 1.6667 x -+- 18 8 = 1.34 The critical value of Fisher's LSD is 1.341. The mean difference table is, Treatments Nurse-Tax auditors Mean 1 3.75 Mean 2 | Difference | Absolute 4.25 -0.50 0.50 Decision The statistical decision is that the mean difference is less than critical value of Fisher's LSD which is 1.34, so the null hypothesis is not rejected at 5% level of significance. There is not a significance difference between the means for nurses and tax auditors. The result is not statistically significant.

The answer option is 2