A recent report suggests that Chief Information Officers (CIO\'s) who report dir

Question

A recent report suggests that Chief Information Officers (CIO's) who report directly to Chief Financial Officers (CFO's) rather than Chief Executive Officers (CEO's) are more likely to have IT agendas that deal with cost cutting and compliance. In a random sample of 842 companies, it was found that CIO's reported directly to CFO's in 165 out of 530 service firms and in 96 out of 312 manufacturing companies

1) Determine the x^2 statistic.

2) Find the degrees of freedom

3) What is the p value?

4) Give a 95%confidence interval for the difference p1p2,where p1and p2 are the proportions of CIO's who report directly to the CFO in service firms and manufacturing companies, respectively.

Explanation / Answer

1 ) Here first we hav to create contengncy table from given information .

842             

From that we can find the expected values

The formula for the X2 test statistics is

                           X2 =   sum [( Oi - Ei )2 / Ei ]

From above we can find this sum as 0.012088363

So, X2 test statistics = 0.012088363

2) Degrees of freedom = (number of rows - 1 ) *(number of columns -1) = (2-1 ) * (2-1) = 1

3) We can find p value using Excel function as =chidist( test statistic value , degrees of freedom)

P value =chidist( 0.012088363 , 1)

                                                                           = 0.91245

4) Confidence interval :

   Formula for confidence interval : (p1^ - p2^) - /+ Z(1-alpha/2)*sqrt[ ((p1^ * ( 1 - p1^))/n1) + ((p2^ * ( 1 - p2^))/n2)]

where p1^ = 165/530 = 0.31132 , n1 =530

           p2^ = 96/312 = 0.3077 , n2 = 312

(p1^ - p2^) = 0.003628

(p1^ * ( 1 - p1^))/n1 = 0.0004045

(p2^ * ( 1 - p2^))/n2 =0.0006827

For confidence level 0.95 , Z(1-alpha/2) = 1.96

So, confidence interval

Lower limit = 0.003628 - 1.96 * sqrt( 0.0004045 + 0.0006827)

                                     = 0.003628 - 0.06463

                                    = - 0.06100

                  Upper limit = 0.003628 + 1.96 * sqrt( 0.0004045 + 0.0006827)

                                     = 0.003628 + 0.06463

                                    = 0.06826

Confidence interval = ( -0.06100 , 0.06826)

service firms Manufacturing companies Total reported 165 96 261 not reported 365 216 581 Total 530 312

842             

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