Answer:
Step-by-step explanation:
Given that:
Let sample size of women be [tex]n_1[/tex] = 1000
Let the proportion of the women be [tex]p_1[/tex] = 0.65
Let the sample size of the men be [tex]n_2[/tex] = 1000
Let the proportion of the mem be [tex]p_2[/tex] = 0.60
The null and the alternative hypothesis can be computed as follows:
[tex]H_0: p_1 = p_2[/tex]
[tex]H_0a: p_1 \neq p_2[/tex]
Thus from the alternative hypothesis we can realize that this is a two tailed test.
However, the pooled sample proportion p = [tex]\dfrac{p_1n_1+p_2n_2 } {n_1 +n_2}[/tex]
p =[tex]\dfrac{0.65 * 1000+0.60*1000 } {1000 +1000}[/tex]
p = [tex]\dfrac{650+600 } {2000}[/tex]
p = 0.625
The standard error of the test can be computed as follows:
[tex]SE = \sqrt{p(1-p) ( \dfrac{1} {n_1}+ \dfrac{1}{n_2} )}[/tex]
[tex]SE = \sqrt{0.625(1-0.625) ( \dfrac{1} {1000}+ \dfrac{1}{1000} )}[/tex]
[tex]SE = \sqrt{0.625(0.375) ( 0.001+0.001 )}[/tex]
[tex]SE = \sqrt{0.234375 (0.002)}[/tex]
[tex]SE = \sqrt{4.6875 * 10^{-4}}[/tex]
[tex]SE = 0.02165[/tex]
The test statistics is :
[tex]z =\dfrac{p_1-p_2}{S.E}[/tex]
[tex]z =\dfrac{0.65-0.60}{0.02165}[/tex]
[tex]z =\dfrac{0.05}{0.02165}[/tex]
[tex]z =2.31[/tex]
At level of significance of 0.05 the critical value for the z test will be in the region between - 1.96 and 1.96
Rejection region: To reject the null hypothesis if z < -1.96 or z > 1.96
Conclusion: Since the value of z is greater than 1.96, it lies in the region region. Therefore we reject the null hypothesis and we conclude that the percentage of men and women favoring a higher legal drinking age is different.
