Please help with this business STATS problem, and explain/show work. Thank you!!
ID: 3367195 • Letter: P
Question
Please help with this business STATS problem, and explain/show work. Thank you!!?
To study the effectiveness of sending registration reminders to citizens, two groups of citizens were randomly selected. In one group of 611 citizens, who are potential voters, NO registration reminders were sent, and the number of citizens in this group who registered to vote was 296. In the other group of the same size, 611 citizens, registration reminders were sent, and the number of citizens in this group who registered to vote was 344 the proportion of registered voters who received reminders, and the proportion of registered voters who did not receive reminders respectively. At Let P1 , p2 be significance level 0.01, is this good evidence showing that the proportion of registered voters who did NOT receive reminders was less than that of registered voters who received reminders? A) State the Null and the Alternate Hypotheses Ho: p1 = p2 (dp = 0) Hi: Pi P2 (dp P2 (dp > 0) Ho: p1 = P2 (dp-0) Ho: pi=P2(dp = 0) Hi: Pi > p2 (dp > 0) B) Is this a (left-tail, right-tail, one-tail, two-tail) test? Oright-tail one-tail left-tail two-tail C) Using 4 decimal places for pi and p2, find the value of dp *. Note: dp must be in 4 decimal places.Explanation / Answer
Solution:
A. Null Hypothesis (Ho): p1 = p2
Alternative Hypothesis (Ha): p1 < p2
B. Left-tailed
C. p1’ = x1/n1 = 296/611 = 0.4845
p2’ = x2/n2 = 344/611 = 0.5630
dp’ = p1’ – p2’
dp’ = 0.4845 – 0.5630
dp’ = -0.0786
D. p’c = (x1 +x2)/(n1 + n2)
p’c = (296 + 344)/(611 + 611)
p’c = 0.5237
E. ?dp’ = ?p’c (1 – p’c) (1/n1 + 1/n2)
?dp’ = ?0.5237*(1-0.5237)(1/611+1/611)
?dp’ = 0.0286
F.
Z = dp’/?dp’
Z = -0.0786/0.0286
Z = -2.75
G. Using Z-tables, the p-value is
P [Z < -2.75] = 0.0030
H. P-value < 0.01, Yes
I. Zc = 1.960
dp’ ± 0.0560