Question

In: Statistics and Probability

Researchers from the Pew Forum on Religion and Public Life interviewed two random samples of people....


Researchers from the Pew Forum on Religion and Public Life interviewed two random samples of people. Both samples had 1500 people. In 2002, 645 people expressed support for stem-cell research. In 2009, 725 expressed support. If group 1 is the 2002 sample and group 2 is the 2009 sample, run a hypothesis test at the 0.01 level of significance to determine if the proportion of people that support stem cell research has changed since 2002

Solutions

Expert Solution

p1cap = X1/N1 = 645/1500 = 0.43
p1cap = X2/N2 = 725/1500 = 0.4833
pcap = (X1 + X2)/(N1 + N2) = (645+725)/(1500+1500) = 0.4567

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 ≠ p2

Rejection Region
This is two tailed test, for α = 0.01
Critical value of z are -2.58 and 2.58.
Hence reject H0 if z < -2.58 or z > 2.58


Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.43-0.4833)/sqrt(0.4567*(1-0.4567)*(1/1500 + 1/1500))
z = -2.93

P-value Approach
P-value = 0.0034
As P-value < 0.01, reject the null hypothesis.


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