The mean downtime for a computer has been 6.34 hours a week. A new supervisor ta
ID: 2926036 • Letter: T
Question
The mean downtime for a computer has been 6.34 hours a week. A new supervisor takes over and initiates new procedures in the computer lab. Now, for a sample of 23 weeks, the average downtime is 6.05 hours per week with a sample standard deviation of 1.82 hours. The new supervisor claims she deserves a raise for improving efficiency, in other words, that dowr LESS with her working for them.. For an alpha of 0.05, test whether the supervisor has ac reduced downtime (show your work). Report the appropriate null and alternative hypotheses. Is this a one tailed or a two tailed test?Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 6.34
Alternative hypothesis: < 6.34
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.3795
DF = n - 1 = 23 - 1
D.F = 22
t = (x - ) / SE
t = - 0.7642
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of - 0.764. We use the t Distribution Calculator to find P(t < - 7642) = 0.2265
Thus the P-value in this analysis is 0.2265
Interpret results. Since the P-value (0.2265) is greater than the significance level (0.05), we cannot reject the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that supervisor has actually reduced downtime.
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