Question

In: Statistics and Probability

Workers and​ senior-level bosses were asked if it was seriously unethical to monitor employee​ e-mail. The...

Workers and​ senior-level bosses were asked if it was seriously unethical to monitor employee​ e-mail. The results are summarized in the table to the right. Use a 0.05 significance level to test the claim that the response is independent of whether the subject is a worker or a boss. Yes No Workers 193 193 241 241 Bosses 36 36 86 86 a. State the null and the alternative hypotheses. Choose the correct answer below. A. The null​ hypothesis: Response is independent of whether the subject is a worker or a​ senior-level boss. The alternative​ hypothesis: There is some relationship between response and whether the subject is a worker or a​ senior-level boss.

Your answer is correct. B. The null​ hypothesis: There is some relationship between response and whether the subject is a worker or a​ senior-level boss. The alternative​ hypothesis: Response is independent of whether the subject is a worker or a​ senior-level boss. b. Assuming independence between the two​ variables, find the expected frequency for each cell of the table.

Table of expected frequencies Yes No Workers nothing nothing Bosses nothing nothing ​(Round to the nearest tenth as​ needed.)

Solutions

Expert Solution

Solution:

Here, we have to use chi square test for independence of two categorical variables.

Null hypothesis: H0: Response is independent of whether the subject is a worker or a​ senior-level boss.

Alternative hypothesis: Ha: There is some relationship between response and whether the subject is a worker or a​ senior-level boss.

We are given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

E = row total * column total / Grand total

We are given

Number of rows = r = 2

Number of columns = c = 2

Degrees of freedom = df = (r – 1)*(c – 1) = 1*1 = 1

α = 0.05

Critical value = 3.841459

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies

Response

Subject

Yes

No

Total

Workers

193

241

434

Bosses

36

86

122

Total

229

327

556

Expected Frequencies

Response

Subject

Yes

No

Total

Workers

178.7518

255.2482

434

Bosses

50.2482

71.7518

122

Total

229

327

556

Calculations

(O - E)

14.2482

-14.2482

-14.2482

14.2482

(O - E)^2/E

1.135716

0.795348

4.040169

2.829354

Chi square = ∑[(O – E)^2/E] = 8.800588

Chi square test statistic = 8.8

P-value = 0.003011

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that there is some relationship between response and whether the subject is a worker or senior-level boss.


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