Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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 visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.