Significance Level (a): 0.01 Null & Alternative Hypothesis: H0:μN=μDH1:μN>μD Data Sample (S1): 55 Sample mean (X1): 3.44 Standard deviation (SD1): 0.06 Sample (S2): 25 Sample mean (X2): 3.41 Standard deviation (SD2): 0.03. Critical value (2 decimal places): I used the InvT function on the calculator to solve for this, but it says it the wrong answer. I also used the T.INV function on excel, but it was still wrong. This is the information I inputted into both functions a = 0.01 & df = 78 (I added the samples, then subtracted by two 25+55 = 80-2 = 78) Can you help me figure out how to solve for the critical value.
Significance Level (a): 0.01 Null & Alternative Hypothesis: H0:μN=μDH1:μN>μD Data Sample (S1): 55 Sample mean (X1): 3.44 Standard deviation (SD1): 0.06 Sample (S2): 25 Sample mean (X2): 3.41 Standard deviation (SD2): 0.03. Critical value (2 decimal places): I used the InvT function on the calculator to solve for this, but it says it the wrong answer. I also used the T.INV function on excel, but it was still wrong. This is the information I inputted into both functions a = 0.01 & df = 78 (I added the samples, then subtracted by two 25+55 = 80-2 = 78) Can you help me figure out how to solve for the critical value.
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
Topic Video
Question
- Significance Level (a): 0.01
- Null & Alternative Hypothesis:
- H0:μN=μD
H1:μN>μD
- H0:μN=μD
- Data
- Sample (S1): 55
- Sample mean (X1): 3.44
- Standard deviation (SD1): 0.06
- Sample (S2): 25
- Sample mean (X2): 3.41
- Standard deviation (SD2): 0.03.
Critical value (2 decimal places): I used the InvT function on the calculator to solve for this, but it says it the wrong answer. I also used the T.INV function on excel, but it was still wrong. This is the information I inputted into both
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