A philanthropic organization sent free mailing labels and greeting cards to a ra

Question

A philanthropic organization sent free mailing labels and greeting cards to a random sample of 100,000 potential donors on their mailing list and received 4780 donations. They have had a contribution rate of 5% in past campaigns, but a staff member worries that the rate will be lower if they run this campaign as currently designed. Complete parts a through c below. a) What are the hypotheses? Which hypotheses below will test the staff member's worries? OA. Ho: p 0.05 HA: p= 0.05 OE. Ho:p=0.05 HA p 0.05 b) Are the assumptions and conditions for inference met? Which assumptions and conditions below are met? Select all that apply A. The randomization condition is met B. The success/failure condition is met C. The 10% condition is met D. The independence assumption is met c) Do you think the donation rate would drop? A. 0 B. ° C. 0 D. 0 E. The P-value is too high to conclude that the rate would be below 5% The P-value is too low to conclude that the rate would be below 5% The P-value is high enough to conclude that the rate would be below 5% The P-value is low enough to conclude that the rate would be below 5%. Inference is not possible since the assumptions and conditions are not met.

Explanation / Answer

Solution:-

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

Null hypothesis: P > 0.05

Alternative hypothesis: P < 0.05

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.000689

z = (p - P) /

z = - 3.19

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than - 3.19. We use the Normal Distribution Calculator to find P(z < - 3.19) = 0.0007

Thus, the P-value = 0.0007

Interpret results. Since the P-value (0.0007) is less than the significance level (0.05), we cannot accept the null hypothesis.

The p-value is low enough to conclude that the rate would be below 5%.

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