QUESTION 11 A sample of 5000 c
QUESTION 11

A sample of 5000 college students were asked if they support thelegalization of marijuana and if they support the legalization ofabortion (results are presented below). Using StatCrunch (you don’tneed to open a dataset from the textbook), calculate the confidenceinterval for percentage of people more likely to support legalizingmarijuana than legalizing abortion. List the lower confidenceinterval below ____________ round to 2 decimal places.
Support the Legalization of Abortion Support the Legalization of Marijuana Yes No Totals Yes 2000 2000 4000 No 200 800 1000 Totals 2200 2800 5000

List the upper confidence interval from question _______ – roundto 2 decimal places.

Now perform a hypothesis test on the data presented in question11. You are testing the null hypothesis that there is no differencein the proportion of those that believe in legalizing marijuana andthe proportion of those that believe in legalizing abortion. Youralpha or plevel is set at .05. List the z statistic below – roundto 2 decimal places.____________
What is the pvalue in the test you just ran? _________ Round to2 decimal places.
Assuming that the plevel or alpha level was set at .05, basedon the pvalue you calculated, would you reject or fail to rejectthe null hypothesis that there is no difference in the proportionof people who support the legalization of marijuana and theproportion that support the legalization of abortion?
Reject 

Fail to reject 
Answer:
Solution:
The confidence interval for percentage of people morelikely to support legalizing marijuana than legalizing abortion isC.I = (0.34 , 0.38).
C.I = 0.36 + 0.019032
C.I = (0.341 , 0.379)
State the hypotheses. The first step is tostate the null hypothesis and an alternative hypothesis.
Null hypothesis: P_{1} = P_{2}Alternative hypothesis: P_{1} P_{2}
Note that these hypotheses constitute a twotailed test. Thenull hypothesis will be rejected if the proportion from population1 is too big or if it is too small.
Formulate an analysis plan. For this analysis,the significance level is 0.05. The test method is a twoproportionztest.
Analyze sample data. Using sample data, wecalculate the pooled sample proportion (p) and the standard error(SE). Using those measures, we compute the zscore test statistic(z).
p = (p_{1} * n_{1} + p_{2} *n_{2}) / (n_{1} + n_{2})p = 0.62SE = sqrt{ p * ( 1 – p ) * [ (1/n_{1}) + (1/n_{2})] }SE = 0.00971z = (p_{1} – p_{2}) / SE
z = 37.08
where p_{1} is the sample proportion in sample 1, wherep_{2} is the sample proportion in sample 2, n_{1}is the size of sample 1, and n_{2} is the size of sample2.
Since we have a twotailed test, the Pvalue is the probabilitythat the zscore is less than 37.08 or greater than 37.08.
Thus, the Pvalue = 0.00.
Interpret results. Since the Pvalue (0.00) is less thanthe significance level (0.05), we reject the nullhypothesis.
Reject H_{0}, we reject the null hypothesis thatthere is no difference in the proportion of people who support thelegalization of marijuana and the proportion that support thelegalization of abortion.