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27 July, 11:23

Nikhil asked 120 randomly chosenmoviegoers to watch a clip of an upcoming movie and choose the best title. The title "Everything" was chosen by 36 of the moviegoers as the preferred title. To the nearest percent, with a confidence level of 90% (z*-score 1.645), what is the confidence interval for the proportion of moviegoers who preferred the title "Everything"? E = z * and C = + E between 4% and 10% between 7% and 30% between 23% and 37% between 29% and 43%

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  1. 27 July, 11:30
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    C. between 23% and 37%

    Explanation:

    Given:

    The total viewers (n) = 120

    The total number that chooses the title "Everything" = 36

    Therefore,

    The mean proportion is p = 36/120 = 0.3.

    Let's find the standard deviation of the proportion

    =√ (p * (1-p) / n)

    = √ (0.3*0.7/120)

    = 0.0418.

    Next step is to multiply 0.0418 by the z-score of 1.645 to get the deviation

    The confidence interval

    = (0.3 - 0.0688, 0.3 + 0.0688)

    = (0.2312, 0.3688).

    = (23.12%, 36.88%)
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