A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in the table. The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution.
Installation a b c d e f g h
Before 3 6 4 2 5 8 2 6
After 1 5 2 0 1 0 2 2
-What is the value of the appropriate test statistic (round to 2 decimal places)?
T almost equal to Answer
-What is the p-value associated with these sample results (round to 3 decimal places)?
p almost equal to Answer
-At the α = 0.01 level of significance is there sufficient evidence to conclude that software patch was effective in reducing system failures?
Select one:
a.
Yes
b.
No
To find the appropriate test statistic, we need to calculate the mean difference and the standard deviation of the differences.
Mean difference = sum of differences / n = (3-1 + 6-5 + 4-2 + 2-0 + 5-1 + 8-0 + 2-2 + 6-2) / 8 = 3.25
Standard deviation of differences = sqrt((sum of (differences - mean difference)^2) / (n-1)) = sqrt((2.75^2 + 0.25^2 + 1.25^2 + 2.25^2 + 3.25^2 + 4.25^2 + 0^2 + 4.25^2) / 7) ≈ 2.21
To calculate the test statistic (T), we use the formula T = (mean difference - population mean) / (standard deviation / sqrt(n)).
Population mean = 0 (since we are testing if the software patch reduced system failures)
n = number of pairs = 8
T = (3.25 - 0) / (2.21 / sqrt(8)) ≈ 3.69
Therefore, the value of the appropriate test statistic is approximately 3.69.
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To find the p-value, we need to calculate the probability of observing a test statistic (T) as extreme or more extreme than the calculated value, assuming the null hypothesis is true.
Let's use a two-tailed test.
Degrees of freedom = n - 1 = 8 - 1 = 7
The p-value can be found using a t-distribution table or a t-distribution calculator. Using a t-distribution calculator, we find that the p-value for T>3.69 with 7 degrees of freedom is approximately 0.005.
However, since this is a two-tailed test, we need to double the p-value.
p-value = 0.005 * 2 ≈ 0.01
Therefore, the p-value associated with these sample results is approximately 0.01.
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At the α = 0.01 level of significance, we compare the p-value to the significance level (α).
Since the p-value (0.01) is less than the significance level (0.01), we reject the null hypothesis.
Therefore, there is sufficient evidence to conclude that the software patch was effective in reducing system failures at the α = 0.01 level of significance.
The answer is:
a. Yes