To study the effectiveness of sending registration reminders to citizens, two gr

Question

To study the effectiveness of sending registration reminders to citizens, two groups of citizens were randomly selected. In one group of 630 citizens, who are potential voters, NO registration reminders were sent, and the number of citizens in this group who registered to vote was 295. In the other group of the same size, 630 citizens, registration reminders were sent, and the number of citizens in this group who registered to vote was 359.

Let p1,p2p1,p2 be the proportion of registered voters who received reminders, and the proportion of registered voters who did not receive reminders respectively. At significance level 0.005, is this good evidence showing that the proportion of registered voters who received reminders was greater than that of registered voters who did NOT receive reminders?

I) Give a 94% confidence interval for the difference in the proportions of registered voters who received reminders and who did not.
zc=____???? (3 decimal places)

dˆp± _____ ???? (4 decimal places)

Explanation / Answer

Solution:-

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

Null hypothesis: P1> P2
Alternative hypothesis: P1 < P2

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. 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)

p = 0.51905

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.02815
z = (p1 - p2) / SE

z = - 3.609

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 -

Thus, the P-value = 0.0002

Interpret results. Since the P-value (0.0002) is less than the significance level (0.01), we have to reject the null hypothesis.

From the above test we have sufficient evidence showing that the proportion of registered voters who received reminders was greater than that of registered voters who did NOT receive reminders.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.