(a Find a 99%% confidence interva for u, - u,- (Round your answers to two decimal places.) lower limit -2.32 upper limit 3.58 (b) Find a 99% confidence interval for u, - Hz. (Round your answers to two decimal places.) lower limit -1.41 upper limit 4.27 (c) Find a 99% confidence interval for l, - H3: (Round your answers to two decimal places.) lower limit -2.25 upper limit 3.85 (d) Comment on the meaning of each of the confidence intervals found in parts (a), (b), and (c). At the 99% confidence level, what can you say about the average differences in influence on self-esteem between competence and social acceptance? between competence and attractiveness? between social acceptance and

(a Find a 99%% confidence interva for u, - u,- (Round your answers to two decimal places.) lower limit -2.32 upper limit 3.58 (b) Find a 99% confidence interval for u, - Hz. (Round your answers to two decimal places.) lower limit -1.41 upper limit 4.27 (c) Find a 99% confidence interval for l, - H3: (Round your answers to two decimal places.) lower limit -2.25 upper limit 3.85 (d) Comment on the meaning of each of the confidence intervals found in parts (a), (b), and (c). At the 99% confidence level, what can you say about the average differences in influence on self-esteem between competence and social acceptance? between competence and attractiveness? between social acceptance and

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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d, e and f
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
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