{Exercise 10.17} a. State the null and alternative hypotheses. : mu1 - mu2 is le
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Question
{Exercise 10.17}
a. State the null and alternative hypotheses.
: mu1 - mu2 is less than 0- Select your answer -mu1 - mu2 is greater than 0mu1 - mu2 is greater than or equal to 0mu1 - mu2 is equal to 0mu1 - mu2 is not equal to 0mu1 - mu2 is less than 0mu1 - mu2 is less than or equal to 0Item 1
:mu1 - mu2 is greater than or equal to 0- Select your answer -mu1 - mu2 is greater than 0mu1 - mu2 is greater than or equal to 0mu1 - mu2 is equal to 0mu1 - mu2 is not equal to 0mu1 - mu2 is less than 0mu1 - mu2 is less than or equal to 0Item 2
b. Compute the value of the test statistic. (to 2 decimals)
c. What is the p-value? (to 4 decimals)
d. What is your conclusion?
Reject the null hypothesis- Select your answer -Reject the null hypothesisDo not reject the null hypothesisItem 6
Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u1 - u2 <= 0
Ha: u1 - u2 > 0 [ANSWER]
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b)
At level of significance = 0.05
As we can see, this is a right tailed test.
Calculating the means of each group,
X1 = 6.82
X2 = 6.25
Calculating the standard deviations of each group,
s1 = 0.64
s2 = 0.75
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 16
n2 = sample size of group 2 = 10
Thus, df = n1 + n2 - 2 = 24
Also, sD = 0.28609439
uD = hypothesized difference = 0
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = 1.992349448 [ANSWER]
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c)
Also, using p values,
p = 0.02891273 = 0.0289 [ANSWER]
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d)
As P < 0.05, WE REJECT THE NULL HYPOTHESIS.
Hence, there is significant evidence at 0.05 level that the consultant with more experience has the higher population mean service rating. [CONCLUSION]
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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!
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