HW9_DATA: Run Time (seconds): 15.045 15.073 15 15.027 14.995 14.991 14.916 15.05
ID: 3318156 • Letter: H
Question
HW9_DATA:
Run Time (seconds): 15.045 15.073 15 15.027 14.995 14.991 14.916 15.055 14.882 15.003 15.012 15.067 14.97 14.961 14.909 15.072 15.118 15.029 15.007 15.175 14.983 14.99 15.05 15.082 15.014 15.064 14.959 15.018 15 15.035 14.955 15.005 15.071 15.078 14.995 15.017 15.022 15.139 14.892 15.023 14.918 15.097 14.988 15.012 14.925 15.082 14.924 14.957 14.984 15.085 QUESTION 2: To test the computation time of a particular algorithm, an engineer runs this algorithm on 50 different computers. Each computer has the same hardware, and hence, the run time on each computer is assumed to be an IID random variable. The engineer would like to investigate whether there is evidence that the average run time exceeds 15.0 seconds. The 50 run times (in seconds) are given in the data file posted on Compass under the name "HW9_DATA" Find the sample variance of the measured run times. [NOTE: You will need to use sample variance as an estimate for population variance in the rest of this problem.] (a) (b) Suppose that the engineer decides that he will conclude that the average run time exceeds 15.0 seconds if the sample mean exceeds 15.01 seconds. Formulate a hypothesis test corresponding to this scenario. What is the value of for this hypothesis test? (c) Re-solve Part (b) if the engineer draws this conclusion if the sample mean exceeds 15.02 seconds. (d)Re-solve Part (b) if the engineer draws this conclusion if the sample mean exceeds 15.025 seconds (e) Consider the hypothesis test formulated in Part (b). What is the value of if the true mean is 15.005 seconds? (f) Re-solve part (e) if the true mean is 15.02 seconds. (g)Re-solve part (e) if the true mean is 15.03 seconds. (h) Compute the sample mean of the measured run times, and consider the hypothesis test formulated in Part (b). What is the observed significance level? INOTE: You may need to use software to compute some probabilities]Explanation / Answer
Solution:
Part a
The sample variance of the measured run times is given as below:
Sample variance = S2 = 0.00399
Sample standard deviation = S = 0.063189734
Sample mean = Xbar = 15.01342
Sample size = n = 50
Part b
Here, we have to use one sample t test for population mean. The null and alternative hypothesis for this test is given as below:
H0: µ = 15.0 versus Ha: µ > 15.0
This is a one tailed test. This is an upper tailed or right tailed test.
Test statistic formula is given as below:
t = (Xbar - µ) / [S/sqrt(n)]
We are given
Sample standard deviation = S = 0.063189734
Sample mean = Xbar = 15.01
Sample size = n = 50
Degrees of freedom = n – 1 = 49
t = (15.01 – 15)/[ 0.063189734/sqrt(50)]
t = 1.1190
P-value = 0.1343
(By using t-table)
Required value = 0.1343
Part c
Here, we have to use one sample t test for population mean. The null and alternative hypothesis for this test is given as below:
H0: µ = 15.0 versus Ha: µ > 15.0
This is a one tailed test. This is an upper tailed or right tailed test.
Test statistic formula is given as below:
t = (Xbar - µ) / [S/sqrt(n)]
We are given
Sample standard deviation = S = 0.063189734
Sample mean = Xbar = 15.02
Sample size = n = 50
Degrees of freedom = n – 1 = 49
t = (15.02 – 15)/[ 0.063189734/sqrt(50)]
t = 2.2380
P-value = 0.0149
(By using t-table)
Required value = 0.0149
Part d
Here, we have to use one sample t test for population mean. The null and alternative hypothesis for this test is given as below:
H0: µ = 15.0 versus Ha: µ > 15.0
This is a one tailed test. This is an upper tailed or right tailed test.
Test statistic formula is given as below:
t = (Xbar - µ) / [S/sqrt(n)]
We are given
Sample standard deviation = S = 0.063189734
Sample mean = Xbar = 15.025
Sample size = n = 50
Degrees of freedom = n – 1 = 49
t = (15.025 – 15)/[ 0.063189734/sqrt(50)]
t = 2.7976
P-value = 0.0037
(By using t-table)
Required value = 0.0037
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