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Step 1: State the Hypothesis Ha: ĩ # 100 hrs Ho:ĩ = 100 hrs Step 2: Level of Significance a =0.05 (Two – tailed) Z - test f = 100 hrs H- 980 hrs n= 100 bulbs a = 100 hrs Step 3: Test to be used Step 4: Calculations 100 – 980 100/V100 From the normal probability distribution table P(z = 2) = 0.0455 Rejection Region: P(Z)S 0.025 or P(Z) 20.975 Acceptance Region: 0.025 < P(Z) < 0.0975 Step 5: Conclusion Since P(z = 2) = 0.0455 is within the acce ptance region, we accept the nul hypothesis. | Therefore the sample mean is significantly different from 1000 hours

Step 1: State the Hypothesis Ha: ĩ # 100 hrs Ho:ĩ = 100 hrs Step 2: Level of Significance a =0.05 (Two – tailed) Z - test f = 100 hrs H- 980 hrs n= 100 bulbs a = 100 hrs Step 3: Test to be used Step 4: Calculations 100 – 980 100/V100 From the normal probability distribution table P(z = 2) = 0.0455 Rejection Region: P(Z)S 0.025 or P(Z) 20.975 Acceptance Region: 0.025 < P(Z) < 0.0975 Step 5: Conclusion Since P(z = 2) = 0.0455 is within the acce ptance region, we accept the nul hypothesis. | Therefore the sample mean is significantly different from 1000 hours

MATLAB: An Introduction with Applications
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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interpret and compute for the table using the 5-step solution provided

a) Sex & Sleep hours /weeknights Levene's Test for Equality of Variances t df Sig. (2-tailed) Sleephours/ Equal variances assumed weeknights Equal variances not assumed -.913 669 .362 -.911 657.048 .363
Transcribed Image Text:a) Sex & Sleep hours /weeknights Levene's Test for Equality of Variances t df Sig. (2-tailed) Sleephours/ Equal variances assumed weeknights Equal variances not assumed -.913 669 .362 -.911 657.048 .363
Step 1: State the Hypothesis Ha: # 100 hrs Ho:i = 100 hrs Step 2: Level of Significance a = 0.05 (Two – tailed) Z- test f = 100 hrs H= 980 hrs n= 100 bulbs o = 100 hrs Step 3: Test to be used Step 4: Calculations 100 – 980 2 100/V100 From the normal probability distribution table P(z = 2) = 0.0455 Rejection Region: P(Z)s 0.025 or P(Z) 2 0.975 Acceptance Region: 0.025 < P(Z) < 0.0975 Step 5: Conclusion Since P(z = 2) = 0.0455 is within the acceptance region, we accept the nul hypothesis. | Therefore the sample mean is significantly different from 1000 hours
Transcribed Image Text:Step 1: State the Hypothesis Ha: # 100 hrs Ho:i = 100 hrs Step 2: Level of Significance a = 0.05 (Two – tailed) Z- test f = 100 hrs H= 980 hrs n= 100 bulbs o = 100 hrs Step 3: Test to be used Step 4: Calculations 100 – 980 2 100/V100 From the normal probability distribution table P(z = 2) = 0.0455 Rejection Region: P(Z)s 0.025 or P(Z) 2 0.975 Acceptance Region: 0.025 < P(Z) < 0.0975 Step 5: Conclusion Since P(z = 2) = 0.0455 is within the acceptance region, we accept the nul hypothesis. | Therefore the sample mean is significantly different from 1000 hours
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