A large supermarket chain commissioned a survey to find out whether people favoured extended opening hours at the weekend. Four hundred people were surveyed in Cork and 88% said that they were in favour of extended opening hours.
At the 95% confidence level,
1. Calculate the margin of error
2. Calculate the confidence interval.

Any one help/tips? :)

Question

A large supermarket chain commissioned a survey to find out whether people favoured extended opening hours at the weekend. Four hundred people were surveyed in Cork and 88% said that they were in favour of extended opening hours.
At the 95% confidence level,
1. Calculate the margin of error
2. Calculate the confidence interval.

Any one help/tips? :)

kropot72 · Answer

Using the binomial distribution, the sample mean for the numbers supporting extending the opening hours is
$$np=400\times 0.88$$
and the sample standard deviation is given by
$$\sigma=\sqrt{np(1-p)}=\sqrt{400\times 0.88\times 0.12}$$
The margin of error at the 95% confidence level is found by multiplying the sample standard deviation by 1.960

anonymous · Answer

Why 1.960?
Is it always like that?

kropot72 · Answer

1.960 applies to a 95% degree of confidence. Other values are possible if other degrees of confidence are required. These are found from inverse Normal techniques. For example 2.576 applies to 99% confidence.