A 2005 study found that 296 women held board seats out of a total of 1,104 seats

Question

A 2005 study found that 296 women held board seats out of a total of 1,104 seats in the Fortune 100 companies. A 2003 study found that 791 women held board seats out of a total of 5,714 seats in the Fortune 500 companies. (a-1) Treating these as random samples (board seat assignments change often), if we were to test whether Fortune 100 companies have a greater proportion of women board members than the Fortune 500, choose the appropriate hypotheses. Assume is the proportion of women board members in Fortune 100 companies and 2 is the proportion of women board members in Fortune 500 companies. (a2) Specify the decision rule at -05 (Round your answer to 3 decimal places.) Reject the null hypothesis if zcalc > (b) Calculate the sample proportions. (Round your answers to 4 decimal places.) Sample proportion Women board members in Fortune 100 companies Women board members in Fortune 500 companies (c-1) Find the test statistic Zcalc and p-value. (Do not round the intermediate calculations. Round proportions to 4 decimal places. Round your answers to 4 decimal places.) Zcalc

Explanation / Answer

Answer to the question is as follows:

a-1)

The claim goes into the alternate hypothesis. So,

H1: pi1-pi2 >0
Ho: pi1-pi2<=0

B. is correct option

a-2) For a alpha = .05 , Zcalc > 1.645
To calcualte this , open the Z tables and check where 1-alpha = 1-.05 = .95 ies.

b. The sample proportions for :
F100 companies is x1/n1 = 296/1104 = .268
F500 companies is x2/n2 = 791/5714 = .138

c-1)
Zcalc = (x1/n1 - x2/n2)/sqrt( p*(1-p)*(1/n1 + 1/n2))

p = x1+x2/n1+n2 = (296+791)/(1104+5714) = 0.159

Zcalc = (.268-.138)/sqrt((.159*.841)*((1/1104) + (1/5714)) = 10.8137
The p-vaue associated with this Zcalc is 0.00

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