For a random sample of 98 blue cars (group 1), 59 had been in an accident. For a
ID: 3323888 • Letter: F
Question
For a random sample of 98 blue cars (group 1), 59 had been in an accident. For a random sample of 129 white cars (group 2), 68 had been in an accident. Find an 80% confidence interval for the difference in the proportion of cars that had been in an accident, p1 - p2. (A) -0.0790 < p1 - p2 < 0.2288 (B) -0.0339 < p1 - p2 < 0.1837 (C) -0.0548 < p1 - p2 < 0.2046 (D) -0.0099 < p1 - p2 < 0.1597
The manager of an assembly process wants to determine if the number of defective products is independent of the day of the week. Find the critical value needed to test the claim that the row and column variables are independent at the 0.1 significance level.
Monday Tuesday Wednesday Thursday Friday
Total Nondefective 85 89 97 92 87 450
Defective 8 10 4 8 9 39
Total 93 99 101. 100 96 489
(A) 23.209 (B) 13.277 (C) 15.987 (D) 7.779
Find the p-value for a left-tailed test if the test statistic is t = -2.722 and n = 15. (A) 0.0083 (B) 0.0311 (C) 0.0108 (D) 0.0565 A police department reports the distribution of burglaries in a month for a particular neighborhood. Find the variance for the number of burglaries in a month.
X 0 1 2 3
P(X) 0.531 0.418 0.031 0.02
(A) 0.656 (B) 0.430 (C) 1.667 (D) 0.722
Explanation / Answer
1) p1 = 59/98 = 0.6020 and p2 = 68/129 = 0.5271
80% confidence interval : (0.6020-0.5271)+-1.28 ((0.6020)(1-0.6020)/98 + (0.5271)(1-0.5271)/129)0.5
= (-0.0099, 0.1597) option (d)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.