Respuesta :
Answer:
Null Hypothesis, [tex]H_0[/tex] : [tex]p_1=p_2[/tex] or [tex]p_1-p_2=0[/tex]
Alternate Hypothesis, [tex]H_a[/tex] : [tex]p_1\neq p_2[/tex] or [tex]p_1-p_2\neq 0[/tex]
Yes, normal sampling distribution can be used.
Step-by-step explanation:
We are given that the sample statistics are from independent random samples. Claim: [tex]p_1[/tex] = [tex]p_2[/tex] , alpha(α) = 0.10 Sample statistics: [tex]x_1[/tex] =59, [tex]n_1[/tex] =171 and [tex]x_2[/tex] =36, [tex]n_2[/tex] = 203
We have to test the claim about the difference between two population proportions [tex]p_1[/tex] and [tex]p_2[/tex] at the given level of significance alpha(α) using the given sample statistics.
Let [tex]p_1[/tex] = population proportion of first group
[tex]p_2[/tex] = population proportion of second group
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]p_1=p_2[/tex] or [tex]p_1-p_2=0[/tex]
Alternate Hypothesis, [tex]H_a[/tex] : [tex]p_1\neq p_2[/tex] or [tex]p_1-p_2\neq 0[/tex]
The test statistics that will be used here is Two-sample z proportion statistics;
T.S. = [tex]\frac{(\hat p_1 - \hat p_2)-(p_1-p_2)}{ \sqrt{\frac{\hat p_1(1-\hat p_1) }{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2}} }[/tex] ~ N(0,1)
where, [tex]\hat p_1 = \frac{x_1}{n_1}[/tex] = [tex]\frac{59}{171}[/tex] = sample proportion of first group
[tex]\hat p_2 = \frac{x_2}{n_2}[/tex] = [tex]\frac{36}{203}[/tex] = sample proportion of second group
[tex]n_1[/tex] = sample size of first group = 171
[tex]n_2[/tex] = sample size of second group = 203
So, Yes here normal sampling distribution can be used because the proportion test is approximately followed by a normal z distribution.
