Question 1088021: 68% of students at a university live on campus. A random sample found that 24 of 40 male students and 43 of 55 of female students live on campus. At the 0.05 level of significance, is there sufficient evidence to support the claim that a difference exists between the proportions of male and female students who live on campus?
Yes, because the test value –20.26 is outside the noncritical region –1.96 < z < 1.96.
No, because the test value –0.96 is inside the noncritical region –1.96 < z < 1.96.
Yes, because the test value –4.21 is outside the noncritical region –1.96 < z < 1.96.
No, because the test value –1.92 is inside the noncritical region –1.96 < z < 1.96.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Define the following
p1 = population proportion of males that live on campus
x1 = number of successes for the males = number of males that live on campus
n1 = sample size of the males
phat1 = sample proportion of males that live on campus
.......
p2 = population proportion of females that live on campus
x2 = number of successes for the females = number of females that live on campus
n2 = sample size of the females
phat2 = sample proportion of females that live on campus
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Based on what we're given, we can determine that...
p1 = unknown
x1 = 24
n1 = 40
phat1 = x1/n1 = 24/40 = 0.6
.......
p2 = unknown
x2 = 43
n2 = 55
phat2 = x2/n2 = 0.7818 (approximate; accurate to 4 decimal places)
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Null Hypothesis:
H0: p1 = p2 (population proportions for male and female groups are the same)
Alternative Hypothesis
H1: p1 =/= p2 (population proportions for male and female groups are different)
This is a two tailed test. We will use a two proportion Z test.
We will reject H0 if the test statistic is in the critical region, which is outside the interval -1.96 < Z < 1.96
When the significance level is alpha = 0.05, the noncritical region is the interval -1.96 < Z < 1.96
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The pooled phat is calculated below
phat = (x1+x2)/(n1+n2)
phat = (24+43)/(40+55)
phat = 67/95
phat = 0.7053 (approximate; accurate to 4 decimal places)
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Standard Error (SE)
SE = sqrt(phat*(1-phat)*(1/n1+1/n2))
SE = sqrt(0.7053*(1-0.7053)*(1/40+1/55))
SE = 0.0947 (approximate; accurate to 4 decimal places)
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Z test statistic
Z = (phat1-phat2)/(SE)
Z = (0.6-0.7818)/(0.0947)
Z = -0.1818/0.0947
Z = -1.9197
Z = -1.92 (approximate; accurate to 2 decimal places)
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The Z test value is Z = -1.92 which is inside the noncritical region (aka the non-rejection region).
The value Z = -1.92 is in the interval -1.96 < Z < 1.96
So we fail to reject the null. We have to accept that the population proportions p1 and p2 are equal.
There is NOT enough sufficient evidence to support the claim that a difference exists between the proportions of male and female students who live on campus.
Answer is choice D
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