Respuesta :
Answer:
The percentage of men and women favoring a higher legal drinking age is the same
Step-by-step explanation:
A random sample of 1000 women showed that 65% were in favor of increasing the legal drinking age
n = 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 age
n = 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 : 2.3094
Refer the z table for p value :
p value : 0.9893
α = 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
