Part: 2/4Part 3 of 4H: H=1.3 versus H₁: μ = 1.3Since the 95% confidence interval does not contain Ho, then Ho will notPart: 3/4Part 4 of 4H₁: H=2.7 versus H₁: μ*2.7Since the 95% confidence interval containsHo then Ho willbe rejected at the 0.05 level.be rejected at the 0.05 level.XXŚŚola A 95% confidence interval for is computed to be (1.55, 4.10). For each of the following hypotheses, state whether Ho will be rejected at 0.05 level.Part: 0/4Part 1 of 4H₁ μ=2 versus H₁: μ#2Since the 95% confidence interval does not contain Ho, then Ho will notPart: 1/4Part 2 of 4H₁ μ = 4 versus H₁: μ#4Since the 95% confidence interval containsPart: 2/4Ho then Ho willbe rejected at the 0.05 level.be rejected at the 0.05 level.XX35da园

Given: The 95% confidence interval is (1.55, 4.10)

Answered: A 95% confidence interval for His…

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 95% two-sided confidence interval for μ which has been calculated using R turns out to be (0, 1). A 90% two-sided confidence interval based on the same data will contain the value 0.9 (True, cannot tell, or False)
n=10 b = 0.50 sb = 0.02 α=0.05 find the confidence interval for β.
Construct a 90% confidence interval for u, - Ho with the sample statistics for mean cholesterol content of a hamburger from two fast food chains and confidence interval construction formula below. Assume the populations are approximately normal with unequal variances. Stats X = 122 mg, s, = 3.72 mg, n, = 12 X2 = 51 mg, s, =2.17 mg, n2 = 14 %3D %3D Confidence interval when (X1 – X2) -t variances are not equal
We have created a 95% confidence interval for μ with the result (10, 15). What decision will we make if we test H0: μ =16 versus H1: μ ≠ 16 at a = 0.10? Select one: a. Reject H0 in favor of H1. b. We cannot tell what our decision will be from the information given. c. Fail to reject H0 in favor of H1. d. Accept H0 in favor of H1.
5. Consider the following estimated equation that is used to study the effect of skipping class on colege GPA: COLGPA- 1.39+412 HSGPA+015 ACT - 083 skipped (011) (.026) (33) (.094) n= 141, R2= 234 (1) Using the standard normal approximation, find the 95% confidence interval for puA (2) Based on the 95% confidence interval you obtained, can you reject the hypothesis H BOPA = 0.4 against the two-sided alternative at the 5% level? (3) Can you reject the hypothesis H: PPA = 0.1 at the 5% level? (4) Can you reject the hypothesis H,: P 0.8 at the 5% level?
5. Consider the following estimated equation that is used to study the effect of skipping class on college GPA: colGPA= 1.39+ .412 hsGPA+ .015 ACT- .083 skipped (.011) (33) (.094) (.026) n= 141, R= 234 (1) Using the standard normal approximation, find the 95% confidence interval for GPA (2) Based on the 95% confidence interval you obtained, can you reject the hypothesis H: PisGPA 0.4against the two-sided alternative at the 5% level? (3) Can you reject the hypothesis H BGPA 0.1 at the 5% level? (4) Can you reject the hypothesis Ho PGPA 0.8 at the 5% level?
Give a 99.5% confidence interval, for μ1−μ2μ1-μ2 given the following information. n1=60, x1=2.67, s1=0.92n2=35, x2=2.46, s2=0.5
Construct a 99% confidence interval for μ1−μ2 with the sample statistics for mean cholesterol content of a hamburger from two fast food chains and confidence interval construction formula below. Assume the populations are approximately normal with unequal variances. Stats x1=55 mg, s1=3.94 mg, n1=17
Give a 99.9% confidence interval, for μ1−μ2μ1-μ2 given the following information. n1=50n1=50, ¯x1=2.13x¯1=2.13, s1=0.98s1=0.98n2=40n2=40, ¯x2=1.65x¯2=1.65, s2=0.86s2=0.86 Rounded to 2 decimal places.
Construct a 90% confidence interval for u, - H2 with the sample statistics for mean cholesterol content of a hamburger from two fast food chains and confidence interval construction formula below. Assume the populations are approximately normal with unequal variances. X2 = 78 mg, sɔ = 2.02 mg, nɔ = 16 Stats X1 = 91 mg, s1 = 3.67 mg, n1 = 10 %3D s s3 Confidence interval when (X1 -X2)-t variances are not equal
Construct the 90 % confidence interval for the difference −μ1μ2 when =x1479.30 , =x2323.56 , =s145.04 , =s222.40 , =n19 , and =n217 . Use tables to find the critical value and round the answers to at least two decimal places. A 90% confidence interval for the difference in the population means is
You would like to explore the relationship between income level and self-rated happiness among city dwellers, and data from your first pilot study of 20 participants produces a confidence interval for Bi of (-0.73, -0.15). What can you conclude? Nothing; your pilot study is too underpowered to detect a true correlation O A linear relationship between income level and self-rated happiness appears to be statistically unlikely O Your method for measuring happiness is probably unreliable O There is preliminary evidence for a negative correlation between income level and self-rated happiness

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