Mathematics, 24.01.2020 23:31 marahkotelman

Shortly after september 11th 2001, a researcher wanted to determine if the proportion of females that favored war with iraq was significantly different from the proportion of males that favored war with iraq. in a sample of 73 females, 28 favored war with iraq. in a sample of 54 males, 29 favored war with iraq.

a) let pf represent the proportion of females that favor the war, pm represent the proportion of males that favor the war. what are the proper hypotheses?

a.)h0: pf = pm versus ha: pf > pm
b.)h0: pf < pm versus ha: pf = pm
c.)h0: pf = pm versus ha: pf < pm
d.)h0: pf = pm versus ha: pf ≠ pm

b) what is the test statistic? compute the statistic using male statistics subtracted from female statistics. give your answer to four decimal places.
c) what is the p-value for the test? give your answer to four decimal places.
d) using a 0.01 level of significance, what conclusion should be reached?

a.)the proportion of females that favor the war and the proportion of males that favor the war are significantly different because the p-value is less than 0.01.
b.)the proportion of females that favor the war and the proportion of males that favor the war are not significantly different because the p-value is greater than 0.01.
c.)the proportion of females that favor the war and the proportion of males that favor the war are significantly different because the p-value is greater than 0.01.
d.)the proportion of females that favor the war and the proportion of males that favor the war are not significantly different because the p-value is less than 0.01.

e) what is the lower endpoint of a 99% confidence interval for the difference between the proportion of females that favor the war and the proportion of males that favor the war? give your answer to four decimal places.
f) what is the upper endpoint of a 99% confidence interval for the difference between the proportion of females that favor the war and the proportion of males that favor the war? give your answer to four decimal places.

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Answer from: stefanylopez731

a) D.)H0: pF = pM versus Ha: pF ≠ pM

b) [ $z=\frac{0.537-0.384}{\sqrt{0.449(1-0.449)(\frac{1}{73}+\frac{1}{54})}}=1.7138$

c) p_v =2*P(Z1.7138)=0.0866

d) B.)The proportion of females that favor the war and the proportion of males that favor the war are not significantly different because the P-value is greater than 0.01.

e) $(0.537-0.384) - 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=-0.0755$

f) $(0.537-0.384) + 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=0.3815$

Step-by-step explanation:

1) Data given and notation

$X_{M}=28$ represent the number of men that favored war with Iraq

$X_{W}=29$ represent the number of women that favored war with Iraq

$n_{M}=73$ sample of male selected

$n_{W}=54$ sample of female selected

$p_{M}=\frac{28}{73}=0.384$ represent the proportion of men that favored war with Iraq

$p_{W}=\frac{29}{54}=0.537$ represent the proportion of women that favored war with Iraq

$\alpha=0.01$ represent the significance level

z would represent the statistic (variable of interest)

p_v represent the value for the test (variable of interest)

Part a

We need to conduct a hypothesis in order to checkif the proportion of females that favored war with Iraq was significantly different from the proportion of males that favored war with Iraq , the system of hypothesis would be:

Null hypothesis: $p_{M} = p_{W}$

Alternative hypothesis: $p_{M} \new p_{W}$

The best option is:

D.)H0: pF = pM versus Ha: pF ≠ pM

Part b

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{W}-p_{M}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}$ (1)

Where $\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{28+29}{73+54}=0.449$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.537-0.384}{\sqrt{0.449(1-0.449)(\frac{1}{73}+\frac{1}{54})}}=1.7138$

Part c

We have a significance level provided $\alpha=0.01$ , and now we can calculate the p value for this test.

Since is a one two sided test the p value would be:

p_v =2*P(Z1.7138)=0.0866

Part d

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the best conclusion would be:

B.)The proportion of females that favor the war and the proportion of males that favor the war are not significantly different because the P-value is greater than 0.01.

Part e

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_W -\hat p_M) \pm z_{\alpha/2} \sqrt{\frac{\hat W_A(1-\hat p_W)}{n_W} +\frac{\hat p_M (1-\hat p_M)}{n_M}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$(0.537-0.384) - 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=-0.0755$

Part f

$(0.537-0.384) + 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=0.3815$

Answer from: Quest

answer: for each 3 girls there is 2 boys . 3: 2

hope this ! (: mark me as brainliest if you can

Answer from: Quest

6e14 is the answer.

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