Respuesta :
Answer:
a) D.)H0: pF = pM versus Ha: pF ≠ pM
b) [[tex]z=\frac{0.537-0.384}{\sqrt{0.449(1-0.449)(\frac{1}{73}+\frac{1}{54})}}=1.7138[/tex]
c) [tex]p_v =2*P(Z>1.7138)=0.0866[/tex]
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) [tex](0.537-0.384) - 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=-0.0755[/tex]
f) [tex](0.537-0.384) + 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=0.3815[/tex]
Step-by-step explanation:
1) Data given and notation
[tex]X_{M}=28[/tex] represent the number of men that favored war with Iraq
[tex]X_{W}=29[/tex] represent the number of women that favored war with Iraq
[tex]n_{M}=73[/tex] sample of male selected
[tex]n_{W}=54[/tex] sample of female selected
[tex]p_{M}=\frac{28}{73}=0.384[/tex] represent the proportion of men that favored war with Iraq
[tex]p_{W}=\frac{29}{54}=0.537[/tex] represent the proportion of women that favored war with Iraq
[tex]\alpha=0.01[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] 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:[tex]p_{M} = p_{W}[/tex]
Alternative hypothesis:[tex]p_{M} \new p_{W}[/tex]
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:
[tex]z=\frac{p_{W}-p_{M}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{28+29}{73+54}=0.449[/tex]
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.537-0.384}{\sqrt{0.449(1-0.449)(\frac{1}{73}+\frac{1}{54})}}=1.7138[/tex]
Part c
We have a significance level provided [tex]\alpha=0.01[/tex], and now we can calculate the p value for this test.
Since is a one two sided test the p value would be:
[tex]p_v =2*P(Z>1.7138)=0.0866[/tex]
Part d
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] 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
[tex](\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}}[/tex]
For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
And replacing into the confidence interval formula we got:
[tex](0.537-0.384) - 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=-0.0755[/tex]
Part f
[tex](0.537-0.384) + 2.58 \sqrt{\frac{0.537(1-0.537)}{54} +\frac{0.384(1-0.384)}{73}}=0.3815[/tex]