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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? :)
 11 months ago
 11 months ago
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? :)
 11 months ago
 11 months ago

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kropot72Best ResponseYou've already chosen the best response.1
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 \[\sigma=\sqrt{np(1p)}=\sqrt{400\times 0.88\times 0.12}\] The margin of error at the 95% confidence level is found by multiplying the sample standard deviation by 1.960
 11 months ago

nickersiaBest ResponseYou've already chosen the best response.0
Why 1.960? Is it always like that?
 11 months ago

kropot72Best ResponseYou've already chosen the best response.1
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.
 11 months ago
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