Answered

At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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]

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.