The production manager at Invacare, a manufacturer of wheelchairs, wants to comp
ID: 3125421 • Letter: T
Question
The production manager at Invacare, a manufacturer of wheelchairs, wants to compare the number of defective wheelchairs produced on the day shift with the number on the afternoon shift. A random sample of the production from 6 day shifts and 8 afternoon shifts revealed the following number of defects: At a significance level of alpha = .03, is there a statistically significant difference in the mean number of defects per shift? Do a complete and appropriate hypothesis test using the 5-step procedure and p-value approach discussed in class. Step 1: State the null and alternative hypotheses. Step 2: State the decision rule for rejecting the null hypothesis. Step 3: State the p-value of the hypothesis test. Step 4: Evaluate the null hypothesis. Step 5: State the practical conclusion(s) to be drawn from the hypothesis test, in the context of the problem, in plain English. Construct a 99% confidence interval estimate of the true difference in the average number of defects per shift. Make sure that your interval estimate is in the correct format, as discussed in class. Interpret the practical meaning of the resulting interval estimate, in plain English (i.e., in the context of the problem).Explanation / Answer
A)
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0 [STEP 1]
At level of significance = 0.03
As we can see, this is a two tailed test.
As
df = n1 + n2 - 2 = 12
Now, the critical value for t is
tcrit = +/- 2.460700166
Hence, reject Ho if |t| > 2.4607. [STEP 2]
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Calculating the means of each group,
X1 = 10.5
X2 = 7.5
Calculating the standard deviations of each group,
s1 = 2.449489743
s2 = 1.048808848
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 8
n2 = sample size of group 2 = 6
Thus, df = n1 + n2 - 2 = 12
Also, sD = 0.966091783
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = 3.105295017
where uD = hypothesized difference = 0
Thus, the P value is
p = 0.009100036 [STEP 3, P VALUE]
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As P < 0.03, WE REJECT THE NULL HYPOTHESIS. [STEP 4]
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Hence, there is no significant evidence at 0.03 level that the mean number of defects for day shift and afternoon shift differ. [STEP 5]
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b)
For the 0.99 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.005
t(alpha/2) = 3.054539589
lower bound = [X1 - X2] - t(alpha/2) * sD = 0.049034402
upper bound = [X1 - X2] + t(alpha/2) * sD = 5.950965598
Thus, the confidence interval is
( 0.049034402 , 5.950965598 ) [ANSWER]
As this whole interval is positive, then we are 99% confident that there is a difference between the mean number of defects between the day shift and afternoon shift.
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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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