In a sample of 55 business trips taken by employees in the HR department, a comp
ID: 3153994 • Letter: I
Question
In a sample of 55 business trips taken by employees in the HR department, a company finds that the average amount spent for the trips was $1,413 with a standard deviation of $443. In a sample of 81 trips taken by the employees in the sales department is $1,617 with a standard deviation of $534. When testing the hypothesis (at the 5% level of significance) that the average amount spent on trips taken by the sales department are higher than those taken by the HR department what is the test statistic? (please round your answer to 2 decimal places)
Explanation / Answer
Formulating the null and alternative hypotheses,
Ho: u1 - u2 >= 0
Ha: u1 - u2 < 0
At level of significance = 0.05
As we can see, this is a left tailed test.
Calculating the means of each group,
X1 = 1413
X2 = 1617
Calculating the standard deviations of each group,
s1 = 443
s2 = 534
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 55
n2 = sample size of group 2 = 81
Also, sD = 84.19387199
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = -2.422979193 [ANSWER]
[If you are looking for a positive t value, then disregard the negative sign.
The answer would be t = 2.42. [ANSWER] ]
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