OOThe necessary value for Za/2 was determined to be 1.960. Recall the given information.Sample 1Sample 2n₂ = 300P₂ = 0.41Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places.P₁-P2-²a/2P₂(1-P₂)7₂n₁ = 400P₁ = 0.56lower bound == 0.56 0.41 - 1.960upper bound =P₁(1-P₁)n1P₁-P₂ + ²a/2V= 0.56 0.41 +=Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places.P₁(1-P₁) +P₂(1-P₂)n₁7₂)√++1.9600.56(1 0.56) 0.41(10.41)300✓4000.56(1 0.56)400+DELL+0.41(10.41)300Therefore, a 95% confidence interval for the difference in the population proportions (rounding to four decimal places) is from a lowerto an upper bound ofbound ofSi74°F Sunny^ C40)

The data is as follows: Sample 1 Sample 2 n1=400 n2=300 p¯1=0.56 p¯2=0.41 zα2=1.96

Answered: The necessary value for Za/2 was…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A random sample of 90 observations produced a mean x= 26.1 and a standard deviation s= 2.6. a. Find a 95% confidence interval for µ. b. Find a 90% confidence interval for p. c. Find a 99% confidence interval for µ. a. The 95% confidence interval is (OD- (Use integers or decimals for any numbers in the expression. Round to two decimal places as needed.)
A simple random sample of size n = 15 is drawn from a population that is normally distributed. The sample mean is found to be x= 68 and the sample standard deviation is found to be s = 11. Construct a 95% confidence interval about the population mean. The lower bound is The upper bound is (Round to two decimal places as needed.)
A simple random sample of size n = 40 is drawn from a population. The sample mean is found to be x = 121.2 and the sample standard deviation is found to be s = 12.6. Construct a 99% confidence interval for the population mean. The lower bound is (Round to two decimal places as needed.) The upper bound is . (Round to two decimal places as needed.)
Answer 1 2 3
Determine the t-score for a 95% confidence interval for a population mean from a sample of size 6. t = [three decimal accuracy]
A simple random sample of size n= 12 is drawn from a population that is normally distributed. The sample mean is found to be x= 63 and the sample standard deviation is found to be s = 13. Construct a 95% confidence interval about the population mean. The lower bound is The upper bound is (Round to two decimal places as needed.)
Step 3 The necessary value for Za/2 was determined to be 1.645. Recall the given information. Sample 1 n1 = 400 P1 = 0.56 P2 = 0.41 Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places. P₁(1-P₁) P₂(1-P₂) 1 72 lower bound = P₁-P₂-²a/2 = 0.560.41 Sample 2 n2 = 300 = 0.0559 = 0.56 0.41 = 02441 X Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places. P₁(1-P₁) P₂(1-P₂) upper bound = P₁ P₂ + ²a/2√ n1 n2 X + ✔ - 1.645, + 0.56(1 0.56) 0.41(10.41) 300 +1.645. 400 0.56(1 0.56) + 400 + 0.41(1 0.41) 300
6.2.13 Use the given confidence interval to find the margin of error and the sample mean. (14.2,24.0) The sample mean is (Type an integer or a decimal.)
The necessary value for z a/2 was determined to be 1.960. Recall the given information. Sample 1 Sample 2 n1 = 400 = 300 P1 = 0.56 P2 = 0.39 Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places. P;(1 - P1) P2(1 - P2) P1- P2 - Za/2V n1 lower bound = + n2 0.56(1 – 0.56) 0.39(1 – 0.39) + = 0.56 – 0.39 – 400 300 Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places. P(1 - P1) P2(1 – P2) P1 - P2 + Za/2V upper bound + n1 n2 = 0.56 – 0.39 + 0.56(1 – 0.56) 0.39(1 – 0.39) + 400 300 Therefore, a 95% confidence interval for the difference in the population proportions (rounding to four decimal places) is from a lower bound of to an upper bound of
6.2.9-T Construct the indicated confidence interval for the population mean u using the t-distribution. Assume the population is normally distributed. c = 0.95, x= 13.4, s= 3.0, n= 8 The 95% confidence interval using a t-distribution is (Round to one decimal place as needed.)

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