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nickersia

  • one year 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? :)

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  1. kropot72
    • one year ago
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    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

  2. nickersia
    • one year ago
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    Why 1.960? Is it always like that?

  3. kropot72
    • one year ago
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    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.

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