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. Test whether there is a difference between these proportions at α = 0.05. What is the test statistic value? Group of answer choices

Answer 1

Answer:

Step-by-step explanation:

Step(i):-

Given first random sample size n₁ = 500

Given  Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer.

First sample proportion

                             $p^{-} _{1} = \frac{65}{500} = 0.13$

Given second sample size n₂ = 700

Given 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.

second sample proportion

                             $p^{-} _{2} = \frac{133}{700} = 0.19$

Level of significance = α = 0.05

critical value = 1.96

Step(ii):-

Null hypothesis : H₀: There  is no significance difference between these proportions

Alternative Hypothesis :H₁: There  is significance difference between these proportions

Test statistic

$Z = \frac{p_{1} ^{-}-p^{-} _{2} }{\sqrt{PQ(\frac{1}{n_{1} } +\frac{1}{n_{2} } )} }$

where

        $P = \frac{n_{1} p^{-} _{1}+n_{2} p^{-} _{2} }{n_{1}+ n_{2} } = \frac{500 X 0.13+700 X0.19 }{500 + 700 } = 0.165$

       Q = 1 - P = 1 - 0.165 = 0.835

$Z = \frac{0.13-0.19 }{\sqrt{0.165 X0.835(\frac{1}{500 } +\frac{1}{700 } )} }$

Z =  -2.76

|Z| = |-2.76| = 2.76 > 1.96 at 0.05 level of significance

Null hypothesis is rejected at 0.05 level of significance

Alternative hypothesis is accepted at 0.05 level of significance

Conclusion:-

There is there is a difference between these proportions at α = 0.05