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

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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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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