Question

A study is conducted to determine what percentage of students live on campus at a large university.

In a random sample found of 50 male students, it was found that that 27 of them lived on campus. In a sample of 55 female students, it was found that 40 lived on campus.  

At the .05 level of significance, does this sample data provide sufficient evidence to conclude that a difference exists between the proportion of male students who live on campus and the proportion of female students who live on campus?

Group of answer choices

Yes, there is sufficient evidence to conclude that the proportions are different because the test
value –2.09 is outside the interval (-1.96, 1.96).

Yes, there is sufficient evidence to conclude that the proportions are different because the test value –1.99 is outside the interval (-1.96, 1.96).

No, there is not sufficient evidence to reject the hypothesis that the proportions are the same because the test value –1.09 is outside the rejection region (-1.96, 1.96).

No, there is not sufficient information to reject the hypothesis that the proportions are the same because the test value –0.19 is not in the rejection region.

Answer 1

Given that,

n₁ = 50

x₁ = 27

$url=http://latex.codecogs.com/gif.latex?%5Chat%20p&t=1596246458&ymreqid=f4764063-0d40-e2b0-1cde-f002b5012f00&sig=NmAGBAyc5AQKuUSvJyqfxA--~D$₁ = 0.54

n₂ =55

x₂ = 40

$url=http://latex.codecogs.com/gif.latex?%5Chat%20p&t=1596246458&ymreqid=f4764063-0d40-e2b0-1cde-f002b5012f00&sig=NmAGBAyc5AQKuUSvJyqfxA--~D$₂ = 0.7273

Let $\bar p$ be the pooled proportion.

$\bar p$ = (x1 +x2)/(n1 + n2) = 0.6381

1 - $\bar p$ = 0.3619

The test statistic z is

z =  

1 - 92/VP*(1-P) * ((1/n1) + (1/n2))]

    = (0.54-0.7273)/$\sqrt{}$[0.6381*0.3619*((1/50)+(1/55))]

= -1.99

z = -1.99

$\alpha$ = 0.05

$\alpha$/2 = 0.25

Using z table ,

2a/2 = +1.96

Non rejection region is (-1.96 , 1.96)

-1.99 < -1.96

So , correct option is

Yes, there is sufficient evidence to conclude that the proportions are different because the test value –1.99 is outside the interval (-1.96, 1.96).