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

Answer 1

`Z=-2.87`

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{(( \sigma_1 * \sigma_2 )/((1/n1 + 1/n2))}})`

Where

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

`\sigma_1=(( 65 + 133 ))/( ( 500 + 700 ))`

`\sigma_1=0.165`

And

`\sigma_2=1 - \sigma = 0.835`

Therefore

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

`Z=-2.87`