There has been a trend toward less driving in the last few years, especially by young people. From 2001 to 2009, the annual vehicle miles tr

Question

There has been a trend toward less driving in the last few years, especially by young people. From 2001 to 2009, the annual vehicle miles traveled by people decreased. Assume the standard deviation was 2000 miles in 2009. Suppose you would like to conduct a survey to develop a 90% confidence interval of the annual vehicle-miles per person for people. A margin of error of 100 miles is desired. How large of a sample should be used for the survey?

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Harper 2 weeks 2021-11-25T12:21:40+00:00 1 Answer 0 views 0

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    2021-11-25T12:23:21+00:00

    Answer:

     The sample size is used for the survey (n) = 1082

    Step-by-step explanation:

    Explanation:-

    Step(i):-

    Given data There has been a trend toward less driving in the last few years, especially by young people. From 2001 to 2009

    Assume the standard deviation was 2000 miles in 2009

    Given the Population standard deviation ‘σ’ = 2000 miles

    Given the margin of error at 90% of confidence interval

         Margin of error = 100 miles

    The z- score value at 90% of confidence interval = 1.645

    Step(ii):-

    The Margin of error is determined by

    M.E = \frac{Z_{\alpha }S.D }{\sqrt{n} }

    M.E X \sqrt{n}  = Z_{\alpha } S.D

    Now the sample size \sqrt{n} = \frac{Z_{\alpha } S.D}{Marginerror} = \frac{1.645 X 2000}{100}

                                        √n  = 32.9

    squaring on both sides , we get  ‘ n ‘ = 1082.41

    Final answer:-

     The sample size is used for the survey (n) = 1082

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