A survey of 912 tablet owners was conducted. In response to a survey question ab

Question

A survey of 912 tablet owners was conducted. In response to a survey question about? shopping, 34?% of tablet owners said they use mobile devices for? payment, 19?% said they use such devices to make social media comments about their? purchases, and 7?% said they use such devices to retrieve mobile coupons. Complete parts? (a) through? (d) below. a. Construct a 95?% confidence interval estimate of the population proportion ? of tablet owners who said they use mobile devices for payment. __???__ ?(Round to three decimal places as? needed.)

b. Construct a 95?% confidence interval estimate of the population proportion ? of tablet owners who said they use mobile devices to make social media comments about their purchases. ___???___ ?(Round to three decimal places as? needed.)

c. Construct a95?% confidence interval estimate of the population proportion ? of tablet owners who said they use mobile devices to retrieve mobile coupons. ___???___?(Round to three decimal places as? needed.)

d. You have been asked to update the results of this study. Determine the sample size necessary to? estimate, with 95?% ?confidence, the population proportions in? (a) through? (c) to within ±0.03.

For part? (a), the sample size is =

For part? (b), the sample size is

For part? (c), the sample size is . ?(Round up to the nearest? integer.)

Explanation / Answer

a)

95?% confidence interval estimate of the population proportion ? of tablet owners who said they use mobile devices for payment

n= 912
p-hat = X/n= 0.340

ALPHA= 5.00%
z(a/2)
z(0.05/2)
1.960

ME= z(a/2)*sqrt(p*(1-p)/n)
1.96*SQRT(0.34*(1-0.34)/912)
0.0307

CI = p +- z(a/2)*sqrt(p*(1-p)/n)
lower =0.34-0.0307= 0.309
upper =0.34+0.0307= 0.371

b)

95?% confidence interval estimate of the population proportion ? of tablet owners who said they use mobile devices to make social media comments about their purchases

n= 912
p-hat = X/n= 0.190

ALPHA= 5.00%
z(a/2)
z(0.05/2)
1.960

ME= z(a/2)*sqrt(p*(1-p)/n)
1.96*SQRT(0.19*(1-0.19)/912)
0.0255

CI = p +- z(a/2)*sqrt(p*(1-p)/n)
lower =0.19-0.0255= 0.165
upper =0.19+0.0255= 0.216

c)

n= 912
p-hat = X/n= 0.070

ALPHA= 5.00%
z(a/2)
z(0.05/2)
1.960

ME= z(a/2)*sqrt(p*(1-p)/n)
1.96*SQRT(0.07*(1-0.07)/912)
0.0166

CI = p +- z(a/2)*sqrt(p*(1-p)/n)
lower =0.07-0.0166= 0.053
upper =0.07+0.0166= 0.087

d)

for part a)
p= 34%
ME= 3.00%
alpha= 5%
z(a/2)= z(0.05/2) =NORMSINV(1-0.05/2) 1.96

n = z(a/2)^2*p*(1-p)/ME^2
n= 1.96^2*0.34*(1-0.34)/0.03^2
957.8389333
958

for part b)
p= 19%
ME= 3.00%
alpha= 5%
z(a/2)= z(0.05/2) =NORMSINV(1-0.05/2) 1.96

n = z(a/2)^2*p*(1-p)/ME^2
n= 1.96^2*0.19*(1-0.19)/0.03^2
656.9136
657

for part c)
p= 7%
ME= 3.00%
alpha= 5%
z(a/2)= z(0.05/2) =NORMSINV(1-0.05/2) 1.96

n = z(a/2)^2*p*(1-p)/ME^2
n= 1.96^2*0.07*(1-0.07)/0.03^2
277.8757333
278

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