The owner of a retail store wanted to determine the proporti

The owner of a retail store wanted to determine the proportion of customers who use a credit card as payment. The owner surveyed 133 customers and found that 82 payed with a credit card.
For full marks your answer should be accurate to at least three decimal places.

a) Estimate the value of the population proportion.

Population proportion: 0

b) Develop a 95% confidence interval for the population proportion.

Confidence interval: (0,0)

The owner of a retail store wanted to determine the proportion of customers who use a credit card as payment. The owner surveyed 133 customers and found that 82 payed with a credit card. For full marks your answer should be accurate to at least three decimal places. a) Estimate the value of the population proportion. Population proportion: 0 b) Develop a 95% confidence interval for the population proportion. Confidence interval: (0,0)

Solution

(a) p=82/133 = 0.6165414

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(b) Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

p - Z*sqrt(p*(1-p)/n) =0.6165414 -1.96*sqrt(0.6165414*(1-0.6165414)/133)

=0.5339051

So the upper bound is

p + Z*sqrt(p*(1-p)/n) =0.6165414 +1.96*sqrt(0.6165414*(1-0.6165414)/133)

=0.6991777