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Step 1: State the Hypothesis Ha: # 100 hrs Ho:ĩ = 100 hrs Step 2: Level of Significance a= 0.05 (Two – tailed) Step 3: Test to be used Step 4: Calculations Z - test i = 100 hrs H- 980 hrs n= 100 bulbs a = 100 hrs 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 Since P(z = 2) = 0.0455 is within the acceptance region, we accept the nul hypothesis. Step 5: Conclusion | Therefore the sample mean is significantly different from 1000 hours

Step 1: State the Hypothesis Ha: # 100 hrs Ho:ĩ = 100 hrs Step 2: Level of Significance a= 0.05 (Two – tailed) Step 3: Test to be used Step 4: Calculations Z - test i = 100 hrs H- 980 hrs n= 100 bulbs a = 100 hrs 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 Since P(z = 2) = 0.0455 is within the acceptance region, we accept the nul hypothesis. Step 5: Conclusion | Therefore the sample mean is significantly different from 1000 hours

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.2: Expected Value And Variance Of Continuous Random Variables
Problem 10E
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please interpret and compute for the data using the 5-step solution

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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