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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

Sagot :

okpalawalter8

Answer:

[tex]Z=-2.87[/tex]

Step-by-step explanation:

From the question we are told that:

Probability on women

[tex]P(W)=65 / 500[/tex]

[tex]P(W) = 0.13[/tex]

Probability on women

[tex]P(M)=133 / 700[/tex]

[tex]P(M) = 0.19[/tex]

Confidence Interval [tex]CI=99\%[/tex]

Generally the equation for momentum is mathematically given by

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

Where

[tex]\sigma_1=(x_1+x_2)(n_1+n_2)[/tex]

[tex]\sigma_1=\frac{( 65 + 133 )}{ ( 500 + 700 )}[/tex]

[tex]\sigma_1=0.165[/tex]

And

[tex]\sigma_2=1 - \sigma = 0.835[/tex]

Therefore

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

[tex]Z=-2.87[/tex]

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