SOLUTION: A sample of 200 customers at a supermarket showed that 28 used a debit card to pay for their purchases. (a) Find the 95 percent confidence interval for the population proportion.

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Question 1175556: A sample of 200 customers at a supermarket showed that 28 used a debit card to pay for their purchases.
(a) Find the 95 percent confidence interval for the population proportion. [2]
(b) Why is it OK to assume normality in this case? [2]
(c) What sample size would be needed to estimate the population proportion with 90 percent confidence and an error of +/- 0.03? [2
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
 n = 200, 28 used debit card
 sample p = 28/200 = .14
ME = z%2Asqrt%28%28p%281-p%29%29%2Fn%29
95% confidence interval
ME = 1.96%2Asqrt%28%28.14%28.86%29%29%2F200%29 = ± .0481
CI = .14 ± .0481

b) Yes, np = 28 > 5

c) n = %28z%2FME%29%5E2+%28p%281-p%29%29%29
 90% confidence interval
  n =%281.645%2F.03%29%5E2+%28.14%28.86%29%29%29 = 362.006
Sample Size 362 needed.
Wish You the Best in your Studies.
 = CI	z = value
90%	z =1.645
92%	z = 1.751
95%	z = 1.96
98%	z = 2.326
99%	z = 2.576