A Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer; a sample of 700 men

Question

A Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer; a sample of 700 men (unrelated to the women) ages 18-29 found that 133 men said that they had the most say when purchasing a computer. What is the 99% confidence interval for the difference of the two proportions

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Cherry 4 years 2021-07-28T10:56:49+00:00 1 Answers 13 views 0

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  1. thanhha
    0
    2021-07-28T10:57:59+00:00
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    Answer:

    Z=-2.87

    Step-by-step explanation:

    From the question we are told that:

    Probability on women

    P(W)=65 / 500

    P(W) = 0.13

    Probability on women

    P(M)=133 / 700

    P(M) = 0.19

    Confidence Interval CI=99\%

    Generally the equation for momentum is mathematically given by

    Z = \frac{( P(W) - P(M) )}{\sqrt{(\frac{ \sigma_1 * \sigma_2 }{(1/n1 + 1/n2)}}})

    Where

    \sigma_1=(x_1+x_2)(n_1+n_2)

    \sigma_1=\frac{( 65 + 133 )}{ ( 500 + 700 )}

    \sigma_1=0.165

    And

    \sigma_2=1 - \sigma = 0.835

    Therefore

    Z = \frac{( 0.13 - 0.19)}{\sqrt{\frac{( 0.165 * 0.835}{ (500 + 700) )}}}

    Z=-2.87

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