Question

A survey of customers at a local supermarket asks whether they found shopping at this market more pleasing than at a nearby store. Of the 2,500 forms distributed to customers, 421 were filled in and 296 of these said that the experience was more pleasing. With 90% confidence, use the plus-four confidence interval to estimate the true proportion of customers that consider this market more pleasing than the other store, and write it in a complete sentence related to the scenario.

Answer 1

Solution :

Given that,

n = 421

x = 296

Point estimate = sample proportion = = x / n = 296/421=0.703

1 - = 1-0.703=0.297

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.703*0.297) /421 )

= 0.037

A 90% confidence interval for proportion p is ,

- E < p < + E

0.703-0.037 < p <0.703+0.037

0.666< p < 0.740

The 90% confidence interval for the proportion p is :0.666 , 0.740

estimate the true proportion of customers that consider this market more pleasing than the other store ( 0.666 to 0.740)