# QUESTION 11 A sample of 5000 c

QUESTION 11

1. 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
1. List the upper confidence interval from question _______ – roundto 2 decimal places.

2. 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 p-level is set at .05. List the z statistic below – roundto 2 decimal places.____________

What is the p-value in the test you just ran? _________ Round to2 decimal places.

Assuming that the p-level or alpha level was set at .05, basedon the p-value 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

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: P1 = P2Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed 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 two-proportionz-test.

Analyze sample data. Using sample data, wecalculate the pooled sample proportion (p) and the standard error(SE). Using those measures, we compute the z-score test statistic(z).

p = (p1 * n1 + p2 *n2) / (n1 + n2)p = 0.62SE = sqrt{ p * ( 1 – p ) * [ (1/n1) + (1/n2)] }SE = 0.00971z = (p1 – p2) / SE

z = 37.08

where p1 is the sample proportion in sample 1, wherep2 is the sample proportion in sample 2, n1is the size of sample 1, and n2 is the size of sample2.

Since we have a two-tailed test, the P-value is the probabilitythat the z-score is less than -37.08 or greater than 37.08.

Thus, the P-value = 0.00.

Interpret results. Since the P-value (0.00) is less thanthe significance level (0.05), we reject the nullhypothesis.

Reject H0, 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.

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