SOLUTION: A survey of 800 women shoppers found that 17% of them shop on impulse. What is the 98% confidence interval for the true proportion of women shoppers who shop on impulse? A. 0.136

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Question 1040399: A survey of 800 women shoppers found that 17% of them shop on impulse. What is the 98% confidence interval for the true proportion of women shoppers who shop on impulse?
A. 0.136 < p < 0.204
B. 0.139 < p < 0.201
C. 0.148 < p < 0.192
D. 0.144 < p < 0.196
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Use this table (or similar) to find that the critical z value is 2.326

Look at the confidence level of 98% (bottom of page). Then look directly above it to find the value 2.326

So z = 2.326

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Given information
phat = 0.17
n = 800

Standard Error:
SE = sqrt(phat*(1-phat)/n)
SE = sqrt(0.17*(1-0.17)/800)
SE = 0.0132806249853

Margin of error
ME = z*SE
ME = 2.326*0.0132806249853
ME = 0.0308907337158

Lower Bound (L) of the confidence interval
L = phat - ME
L = 0.17 - 0.0308907337158
L = 0.1391092662842
L = 0.139

Upper Bound (U) of the confidence interval
U = phat + ME
U = 0.17 + 0.0308907337158
U = 0.2008907337158
U = 0.201

The confidence interal is (L,U) = (0.139, 0.201) which is equilavent to saying L < p < U which turns into 0.139 < p < 0.201

So the answer is choice B) 0.139 < p < 0.201

The population proportion p is somewhere between 13.9% and 20.1% and we're 98% confident of this fact.