hypothesis test using the critical value method. Be sure to state the null and a
ID: 3151349 • Letter: H
Question
hypothesis test using the critical value method. Be sure to state the null and alternative hypotheses, identify the critical value, calculate the test statistic, compare the test statistic to the critical value, and state the conclusion.
The mean lasting time of 2 competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Wax#1 had a mean time of 3 months, while Wax#2's mean was 2.9 months. The population standard deviations are 0.33 and 0.36, respectively. Does the data indicate that Wax#1 lasts longer than Wax#2? Test at a 5% level of significance.
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.
n1 = sample size of group 1 = 20
n2 = sample size of group 2 = 20
Thus, df = n1 + n2 - 2 = 38
Now, the critical value for t is, at 0.05 right tailed test,
tcrit = 1.68595446 [ANSWER]
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c)
Calculating the means of each group,
X1 = 3
X2 = 2.9
Calculating the standard deviations of each group,
s1 = 0.33
s2 = 0.36
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
Also, sD = 0.109201648
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = 0.915737093 [TEST STATISTIC]
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d)
where uD = hypothesized difference = 0
As t < 1.686, WE FAIL TO REJECT THE NULL HYPOTHESIS.
Hence, there is no significant evidence that Wax 1 lasts longer than wax 2. [CONCLUSION]
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