A local church parish wants to raise money to add to their campus. In a sample f

Question

A local church parish wants to raise money to add to their campus. In a sample from a previous fund raising campaign, the parish found that of the 120 people in the sample they contacted, that 85 actually contributed money. Of those that contributed money, the average contribution was exist800 with a standard deviation of exist250. In the recent fundraising campaign, a sample of 85 people revealed that 52 have contributed money with an average contribution of exist950 and a standard deviation of exist300. When testing (at the 5% level of significance) whether the proportion of people who are contributing has increased, what is the null and alternative hypotheses? (please write out the null and alternative hypotheses below AND on your scratch work... failure to do BOTH will result in 0 points)

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 >= P2
Alternative hypothesis: P1 < P2 (Claim, th proportion of people who are contributing has increased.)

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the proportion of women catching cold (p1) is sufficiently smaller than the proportion of men catching cold (p2).

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate 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) = [(85/120 * 120) + (52/85 * 85)] / (120 + 85) = 137/205 = 0.668

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = sqrt [ 0.668 * 0.332 * ( 1/120 + 1/85 ) ] = 0.06676

z = (p1 - p2) / SE = (85/120 - 52/85)/0.06676 = 1.4465

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than 1.4465. We use the Normal Distribution Calculator to find P(z < 1.4465)

.The P-Value is 0.074019.
The result is not significant at p < 0.05.

Interpret results. Since the P-value (0.074) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Conclusion. Fail to reject the null hypothesis. We have insufficient evidence to prove the claim that the proportion of people who are contributing has increased.

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