3. A random sample of 200 school managers were administered a developed Leadership Skills Test. The sample mean and the standard deviation were 78 and 12 respectively. In the standardization of the test, the mean was 73 and the standard deviation was 8. Test for significance difference using a = .05 utilizing the p- value method. Ir R Step 1 Step 2 Step 3 Compute for Its value (z or t) =. Step 4 Critical Value and draw the rejection and acceptance regions Step 5 Decision and conclusion Hypotheses H: = 78 H.: Determine the level of confidence Ha: Test statistic (z or t) = CH a = DE KE
3. A random sample of 200 school managers were administered a developed Leadership Skills Test. The sample mean and the standard deviation were 78 and 12 respectively. In the standardization of the test, the mean was 73 and the standard deviation was 8. Test for significance difference using a = .05 utilizing the p- value method. Ir R Step 1 Step 2 Step 3 Compute for Its value (z or t) =. Step 4 Critical Value and draw the rejection and acceptance regions Step 5 Decision and conclusion Hypotheses H: = 78 H.: Determine the level of confidence Ha: Test statistic (z or t) = CH a = DE KE
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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Transcribed Image Text:3. A random sample of 200 school managers were administered a developed Leadership Skills Test. The
sample mean and the standard deviation were 78 and 12 respectively. In the standardization of the test, the
mean was 73 and the standard deviation was 8. Test for significance difference using a = .05 utilizing the p-
value method.
Ir
R
Step 1
Step 2 Determine the level of confidence
Step 3 Compute for its value (z or t) =
Step 4 Critical Value and draw the rejection and acceptance regions
Step 5 Decision and conclusion
Hypotheses
Ha:
Test statistic (z or t) =
CH
a =
DE
KE
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