Respuesta :
Answer:
There is no enough evidence to claim that there is a difference between the two population proportions.
Step-by-step explanation:
We have to perform an hypothesis testing for a difference between two population proportions.
The null hypothesis will state that both proportions are the same, and the alternative hypothesis will state that they differ. This would be than a two-side hypothesis test.
We can write this as:
[tex]H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq0[/tex]
The significance level for this test is 0.05.
The sample of city residents with school-age children has a sample size n1=230 and a sample proportion p1=0.41
[tex]p_1=X_1/N_1=94/230=0.41[/tex]
The sample of city residents without school-age children has a sample size n2=341 and a sample proportion p2=0.51
[tex]p_2=X_2/N_2=147/341=0.43[/tex]
The weighted p, needed to calculate the standard error, is the weighted average of both sample proportions:
[tex]p=\dfrac{n_1p_1+n_2p_2}{n_1+n_2}=\dfrac{230*0.41+341*0.51}{230+341}=\dfrac{94+147}{571} =\dfrac{241}{571}=0.42[/tex]
The standard error of the difference of proportions can now be calculated as:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.42*0.58}{230}+\dfrac{0.42*0.58}{341}}\\\\\\\sigma_p=\sqrt{\dfrac{0.2436}{230}+\dfrac{0.2436}{341}}=\sqrt{0.00106+0.00071}=\sqrt{0.00177}\\\\\\\sigma_p=0.042[/tex]
The test statistic z is:
[tex]z=\dfrac{p_1-p_2}{\sigma_p}=\dfrac{0.41-0.43}{0.042}= \dfrac{0.020}{0.042}= 0.476[/tex]
The P-value for this two side test and this value of the z-statistic is:
[tex]P-value=2*P(z>0.476)=0.634[/tex]
The P-value is bigger than the significance level, so the effect is not significant. The null hypothesis failed to be rejected.
There is no enough evidence to claim that there is a difference between the two population proportions.
The null hypothesis is failed to be rejected.Hence there is no convincing statistical evidence of a difference between the two population proportions at the significance level of 0.05.
Given-
Significance level of 0.05.
Sample proportion [tex]p_1[/tex] for the sample size [tex]N_1[/tex] which is 230.
[tex]p_1=\dfrac{X_1}{N_1}[/tex]
[tex]p_1=\dfrac{97}{230} =0.41[/tex]
Sample proportion [tex]p_2[/tex] for the sample size [tex]N_2[/tex] which is 341.
[tex]p_1=\dfrac{X_2}{N_2}[/tex]
[tex]p_1=\dfrac{147}{341}=0.43[/tex]
Calculate average value of p,
[tex]p=\dfrac{n_1p_1+n_2p_2}{n_1+n_2}[/tex]
[tex]p=\dfrac{230\times0.41+341\times0.51}{230+341}[/tex]
[tex]p=0.42[/tex]
Calculate the standard difference,
[tex]\sigma _p=\sqrt{\dfrac{p(1-p)}{n_2} +\dfrac{p(1-p)}{n_2} }[/tex]
[tex]\sigma _p=\sqrt{\dfrac{0.42\times 0.58}{230} +\dfrac{0.42\times 0.58}{341} }[/tex]
[tex]\sigma _p=0.042[/tex]
Now calculate the test statics for the z test,
[tex]z=\dfrac{p_1-p_2}{\sigma _p}[/tex]
[tex]z=\dfrac{0.43-0.41}{0.042}[/tex]
[tex]z=0.0476[/tex]
For this z score the value of p value for two test is,
p-value[tex]=0.643[/tex]
Significance level given in question is 0.05. This value of significance level is less than the p value 0.643.Thus the null hypothesis is failed to be rejected.Hence there is no convincing statistical evidence of a difference between the two population proportions at the significance level of 0.05.
For more about the p value follow the link below,
https://brainly.com/question/14723549
