Mathematics, 09.07.2020 05:01 chocolate12377

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

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Answer from: savannaflorian5141

The test statistics value will be "Z = - 2.76".

According to the question,

Women,

$\hat p_1 = \frac{65}{500}$

= 0.13

Men,

$\hat p_2=\frac{300}{700}$

= 0.19

Now,

The pooled estimate of the proportion will be:

→ $\hat p=\frac{x_1+x_2}{n_1+n_2}$

By substituting the values,

$= \frac{65+133}{500+700}$

= 0.165

then,

→ $\hat q = 1- \hat p$

= 1- 0.165

= 0.835

hence,

The test statistics:

→ $Z = \frac{\hat p_1- \hat p_2}{\sqrt{\hat p\times \hat q\times (\frac{1}{n_1} +\frac{1}{n_2} )} }$

By putting the values, we get

$= \frac{0.13-0.19}{\sqrt{0.165\times 0.835\times (\frac{1}{500} +\frac{1}{700} )} }$

= -2.76

Thus the above response is right.

Learn more about statistic value here:

link

Answer from: timiaparker

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

Answer from: Quest

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )

Answer from: Quest

8doors on the second floor

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