Marketers want to know about the differences between men\'s and women\'s use of
ID: 3155836 • Letter: M
Question
Marketers want to know about the differences between men's and women's use of the Internet. A research poll in April 2009 from a random sample of adults found that 2356 of 3008 men use the Internet, at least occasionally, while 2398 of 3156 women did. a) Find the proportions of men and women who said they use the Internet at least occasionally. b) What is the difference in proportions? c) What is the standard error of the difference? d) Find a 95% confidence interval for the difference between percentages of usage by men and women nationwide. a) Let p_1 be the sample proportion of men who use the Internet at least occasionally and let P_2 be the sample proportion of women. (b) p_1 = p_2 = [0.02] c) SE(p_1 - p_2) = d) The confidence interval is.Explanation / Answer
c.
Standard Error = Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
Probability Succuses( X1 )=2356
No.Of Observed (n1)=3008
P1= X1/n1=0.783
Proportion 2
Probability Succuses(X2)=2398
No.Of Observaed (n2)=3156
P2= X2/n2=0.76
Standard of Error = Za/2 * Sqrt ( (0.783)*(1-0.783)/3008 + 0.76(1-0.76)/3156 = Sqrt (0.000114) = 0.010677
d.
Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=2356
No.Of Observed (n1)=3008
P1= X1/n1=0.783245
Proportion 2
No. of chances(X2)=2398
No.Of Observed (n2)=3156
P2= X2/n2=0.759823
C.I = (0.783245-0.759823) ±Z a/2 * Sqrt( (0.783245*0.216755/3008) + (0.759823*0.240177/3156) )
=(0.783245-0.759823) ± 1.96* Sqrt(0.000114)
=0.023422-0.020951,0.023422+0.020951
=[ 0.0025,0.0444 ]