Answer:

From given information, we have following summarized data:

	CFO	CEO	Total
Service Firm	170	358	528
Manufacturing Firm	98	232	330
Total	268	590	858

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

Null hypothesis: H₀: Reporting department and type of firm are independent.

Alternative hypothesis: H_a: Reporting department and type of firm are not independent.

We assume level of significance = α = 0.05

a) Determine the χ2 statistic

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
	Column variable
Row variable	CFO	CEO	Total
Service	170	358	528
Manufacturing	98	232	330
Total	268	590	858

Expected Frequencies
	Column variable
Row variable	CFO	CEO	Total
Service	164.9231	363.0769	528
Manufacturing	103.0769	226.9231	330
Total	268	590	858

Calculations
(O - E)
5.076923	-5.07692
-5.07692	5.076923

(O - E)^2/E
0.156286	0.070991
0.250057	0.113585

Chi square = ∑[(O – E)^2/E] = 0.59092

b) Determine the p-value.

P-value = 0.442064

(By using Chi square table or excel)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that Reporting department and type of firm are independent.

c) Give 95% confidence interval for the difference in proportions of companies in which the CIO reports directly to CFO between service and manufacturing firms.

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

WE are given

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

N1 = 528, N2 = 330, X1 = 170, X2 = 98

P1 = X1/N1 = 170/528 = 0.321969697

P2 = X2/N2 = 98/330 = 0.296969697

P1 – P2 = 0.321969697 - 0.296969697 = 0.025

SE = sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

SE = sqrt[(0.321969697*(1 - 0.321969697)/528) + (0.296969697*(1 - 0.296969697)/330)]

SE = 0.0323

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Confidence interval = 0.025 ± 1.96*0.0323

Confidence interval = 0.025 ± 0.0634

Lower limit = 0.025 - 0.0634 = -0.0384

Upper limit = 0.025 + 0.0634 = 0.0884

Confidence interval = (-0.0384, 0.0884)