Question

I took a sample of 400 Commerce students and found the proportion using Apple computers was 60%. In a sample of 450 Nursing students, the proportion using Apple computers was 50%. At the twelve percent level of significance, the null hypothesis that the difference between the proportion of Commerce and Nursing students using Apple computers (calculated as Commerce - Nursing) is equal to 0.05 against the two tailed alternative is O not rejected, with a calculated test value of 1.47 and a pvalue of 0.071 O rejected, with a calculated test value of -3.03 and a pvalue of 0.001 O rejected, with a calculated test value of 3.09 and a pvalue of 2.341 O not rejected, with a calculated test value of -1.13 and a pvalue of 0.004

Answer 1

Ans:

pooled proportion=(0.6*400+0.5*450)/(400+450)=0.5471

Test statistic:

z=((0.6-0.5)-0.05)/SQRT(0.5471*(1-0.5471)*((1/400)+(1/450)))

z=1.47

p-value(2 tailed)=2*P(z>1.47)=2*0.071=0.142

alpha=0.12

As,p-value>0.12,we fail to reject the null hypothesis.

not rejected,with a calculated test statistic value of 1.47 and p-value=0.071