I have tried to do it on Minitab but it just keeps saying error. A past survey o

Question

I have tried to do it on Minitab but it just keeps saying error.

A past survey of 1, 068, 000 students taking a standardized test revealed that 9.8% of the students were planning on studying engineering in college. In a recent survey of 1, 476, 000 students taking the SAT, 9.2% of the students were planning to study engineering. Construct a 95% confidence interval for the difference between proportions p_1 - p_2 by using the following inequality. Assume the samples are random and independent. (p_1 - p_2) - z_c squareroot p_1 q_1/n_1 + p_2 q-2/n_2

Explanation / Answer

z alpha/2 for 95%=1.96

p1^=9.8%=9.8/100=0.098

q1^=1-p1^=100-9.8=90.2%=0.902

p2^=9.2%=9.2/100=0.092

q2^=1-p2^=100-9.2=90.8%=90.8/100=0.908

n1=1068000

n2=1476000

95% confidence interval for difference in proprtion is

(0.098-0.092)-1.96sqrt(0.098*0.902/1068000+(0.092*0.908/1476000)

=0.005,0.007

lower limi0.005

upper limit=0.007

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