® Save & End Certify Lesson: 12.3a Comparing Two Means (Larg... CASEY CARROLL Qu
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® Save & End Certify Lesson: 12.3a Comparing Two Means (Larg... CASEY CARROLL Question 2 of 4, Step 3 of 4 4/16 Correct A systems analyst tests a new algorithm designed to work faster than the currently used algorithm. Each algorithm is applied to a group of 49 sample problems. The new algorithm completes the sample problems with a mean time of 18.71 hours. The current algorithm completes the sample problems with a mean time of 20.83 hours. The standard of si is found to be 5.184 hours for the new algorithm, and 5.810 hours for the current algorithm. Conduct a hypothesis test at the 0.01 level of the darn that the new algorithm has a lower mean completion time than the current algorithm. Let be the true mean completion time for the new algorithm and p, be the true mean completion time for the current algorithm. Step 3 of 4: Determine the decision rule for rejecting the null hypothesisH Round the numerical portion of your answer to two decimal places Tables Keypad Reject Ho if zExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1> 2
Alternative hypothesis: 1 < 2
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 1.112
DF = 96
t = [ (x1 - x2) - d ] / SE
t = - 1.91
tcritical = - 2.628
Reject the H0, if t-value is less than - 2.628.
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of - 1.91. We use the t Distribution Calculator to find P(t < - 1.91).
Therefore, the P-value in this analysis is 0.0296.
Interpret results. Since the P-value (0.0296) is greater than the significance level (0.01), we cannot reject the null hypothesis.
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