91 98 105 94 79 68 84 C -value = (Round to three decimal places as needed.) - Select a value of a, the probability of Type I error. Interpret this value in the words of the problem. OA. There would still be sufficient evidence to reject the null hypothesis if a > 0.206. OB. There would still be sufficient evidence to reject the null hypothesis if a > 0.001 OC. There would still be sufficient evidence to reject the null hypothesis if a <0.206. OD. There would still be sufficient evidence to reject the null hypothesis if a = 0.001 nterpret this value in the words of the problem. OA. A Type I error would be to conclude that the true mean of the trap spacing measurements is 88.4 when, in fact, the mean is equal to 95. OB. A Type I error would be to conclude that the true mean of the trap spacing measurements is not 95 when, in fact, the mean is equal to 95. OC. A Type I error would be to conclude that the true mean of the trap spacing measurements is 95 when, in fact, the mean is not equal to 95. OD. A Type I error would be to conclude that the true mean of the trap spacing measurements is 95 when, in fact, the mean is equal to 88.4.
91 98 105 94 79 68 84 C -value = (Round to three decimal places as needed.) - Select a value of a, the probability of Type I error. Interpret this value in the words of the problem. OA. There would still be sufficient evidence to reject the null hypothesis if a > 0.206. OB. There would still be sufficient evidence to reject the null hypothesis if a > 0.001 OC. There would still be sufficient evidence to reject the null hypothesis if a <0.206. OD. There would still be sufficient evidence to reject the null hypothesis if a = 0.001 nterpret this value in the words of the problem. OA. A Type I error would be to conclude that the true mean of the trap spacing measurements is 88.4 when, in fact, the mean is equal to 95. OB. A Type I error would be to conclude that the true mean of the trap spacing measurements is not 95 when, in fact, the mean is equal to 95. OC. A Type I error would be to conclude that the true mean of the trap spacing measurements is 95 when, in fact, the mean is not equal to 95. OD. A Type I error would be to conclude that the true mean of the trap spacing measurements is 95 when, in fact, the mean is equal to 88.4.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question

Transcribed Image Text:Trap spacing measurements (in meters) for a sample of seven teams of red spiny lobster fisherman are reproduced in the accompanying table. Let u represent the average of the trap spacing
measurements for the population of red spiny lobster fisherman. The mean and the standard deviation of the sample measurements are x = 88.4 meters and s = 12.4 meters, respectively. Suppose
you want to determine if the true value of μ differs from 95 meters. Complete parts a through h below.
91 98 105 94 79 68 84
p-value= (Round to three decimal places as needed.).
e. Select a value of a, the probability of Type I error. Interpret this value in the words of the problem.
OA. There would still be sufficient evidence to reject the null hypothesis if a > 0.206.
OB. There would still be sufficient evidence to reject the null hypothesis if a > 0.001
OC. There would still be sufficient evidence to reject the null hypothesis if a < 0.206.
OD. There would still be sufficient evidence to reject the null hypothesis if α = 0.001
Interpret this value in the words of the problem.
OA. A Type I error would be to conclude that the true mean of the trap spacing measurements is 88.4 when, in fact, the mean is equal to 95.
OB. A Type I error would be to conclude that the true mean of the trap spacing measurements is not 95 when, in fact, the mean is equal to 95.
OC. A Type I error would be to conclude that the true mean of the trap spacing measurements is 95 when, in fact, the mean is not equal to 95.
OD. A Type I error would be to conclude that the true mean of the trap spacing measurements is 95 when, in fact, the mean is equal to 88.4.
Expert Solution

Step 1: Introduction
Given that information
Sample mean = barx=88.4
Sample standard Deviation=s=12.4
Sample size=n =7
Null and alternative hypothesis
Ho:mu=95 and Ha:mu ≠ 95
This is two tailed test
Test statistic t =(x-mu)/s/√n
=(88.4-95)/12.4/√7
= -1.408(approximately)
Step by step
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