Question 1039187: A researcher finds that of 1,000 people who said they attend a religious service at least once a week, 31 stopped to help a person with car trouble. Of 1,200 people interviewed who had not attended a religious service at least once a month, 22 stopped to help a person with car trouble. At the 0.05 significance level, test the claim that the two proportions are equal.
HO:
H1:
Test statistic:
Critical Value:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Ho:religious service frequency makes no difference, p1=p2; p1=1x/week, p2=not once a month
Ha:there is a difference p1 NE p2
alpha=0.05
test statistic is a 2 sample proportion;
This is normally distributed with a critical value at the 5% level of |z|>1.96
Pooled sample proportion. Since the null hypothesis states that P1=P2, we use a pooled sample proportion (p) to compute the standard error of the sampling distribution.
p = [(p1 * n1)+(p2 * n2)] / (n1 + n2), here is
p1 is the sample proportion from population 1, p2 is the sample proportion from population 2, n1 is the size of sample 1, and n2 is the size of sample 2.
The standard error (SE) of the sampling distribution difference between two proportions.
SE= sqrt{[ p*( 1- p )* (1/n1) + (1/n2) }
Test statistic. The test statistic is a z-score (z) defined by the following equation.
z = (p1-p2) / SE
p1=0.031;p2=0.01833
pooled is (31+22)/2200=0.0241
SE is sqrt (0.0241)(0.9759)=0.006566. Be sure to compute the (1/1000)+(1/1200) before multiplying that result by the rest.
(pi-p2)=0.0031-0.01833=0.01267
divide by SE=1.9294, which is z.
Fail to reject the null hypothesis, p=0.0537
This is checked in a calculator and is essentially the same result.
Stat/Test/2 proportion z-test
Not pooling the variance is one cause of a difference, although any test statistic greater than 1.96 (or less than -1.96) will have a p-value less than 0.05
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