In a recent survey of drinking laws, a random sample of 1000 women showed that 65% were in favor of increasing the legal drinking age. In a random sample of 1000 men, 60% favored increasing the legal drinking age. Test the hypothesis that the percentage of men and women favoring a higher legal drinking age is the same. Use α = 0.05.

Answer:The percentage of men and women favoring a higher legal drinking age is the sameStep-by-step explanation:A random sample of 1000 women showed that 65% were in favor of increasing the legal drinking agen = 1000 No. of females were in favor of increasing the legal drinking age = [tex]\frac{65}{100} \times 1000=650[/tex]y=650 In a random sample of 1000 men, 60% favored increasing the legal drinking agen = 1000 No. of males were in favor of increasing the legal drinking age = [tex]\frac{60}{100} \times 1000=600[/tex]y=600[tex]n_1=1000 , y_1=650\\n_2=1000 , y_2=600[/tex]We will use Comparing Two Proportions [tex]\widehat{p_1}=\frac{y_1}{n_1}[/tex][tex]\widehat{p_1}=\frac{650}{1000}[/tex][tex]\widehat{p_1}=0.65[/tex][tex]\widehat{p_2}=\frac{y_2}{n_2}[/tex][tex]\widehat{p_2}=\frac{600}{1000}[/tex][tex]\widehat{p_2}=0.6[/tex]Let p_1 and p_2 be the probabilities of men and women favoring a higher legal drinking age is the same.So, [tex]H_0:p_1=p_2\\H_a:p_1 \neq p_2[/tex][tex]\widehat{p}=\frac{y_1+y_2}{n_1+n_2} =\frac{600+650}{1000+1000}=0.625[/tex]Formula of test statistic :[tex]\frac{\widehat{p_1}-\widehat{p_2}}{\sqrt{\widehat{p}(1-\widehat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}[/tex] test statistic : [tex]\frac{0.65-0.6}{\sqrt{0.625(1-0.625)(\frac{1}{1000}+\frac{1}{1000})}}[/tex]test statistic : 2.3094Refer the z table for p value :p value : 0.9893α = 0.05.p value > α So, we failed to reject null hypothesis .So,  the percentage of men and women favoring a higher legal drinking age is the same