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6166 UNIT 2 Section 2 Answers

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This problem will be worked in Minitab. The associated Minitab worksheet can be downloaded from here.
Do problem 6.37 on page 319. (This problem relates to comparing alloys for the manufacture of steel beams).
a. What are the populations of interest? In this case, the populations consist of load capacities (in tons) of beams that would be constructed using the old and new alloys respectively. Note that the populations are defined in terms of the responses (load capacities) and characteristics (alloy type) that we are interested in.
b. Place 99% confidence intervals on the mean load capacity of beams produced with both the new and old alloy. Basic statistics are computed in Minitab to produce the following table.
Descriptive Statistics: Load by Beam Variable Beam N Mean Median TrMean StDev Load new 10 28.850 28.800 29.075 2.510 old 10 23.370 23.200 23.412 1.218 Variable Beam SE Mean Minimum Maximum Q1 Q3 Load new 0.794 23.600 32.300 27.650 31.125 old 0.385 21.100 25.300 22.750 24.250
With the mean estimates and the standard errors of the means, we need only the appropriate tcritical values to construct the confidence intervals using the equation on page 234. The critical value is t(0.005,9)= 3.25 from Table 2 in the book. The confidence intervals are then computed as:
New Beam (28.85±(3.25)(0.794)) = (26.27,31.43)
Old Beam (23.370±(3.25)(0.385)) = (22.12,24.62)
We could also have obtained these confidence intervals by using the "unstacked" data columns in the worksheet and the onesample ttests. On the onesample ttest dialog there is the option to set the confidence interval to 99%. Remember to set the test to "twosided". The result is the following (in part):
Variable 99.0% CI T P Load_new ( 26.270, 31.430) 36.35 0.000 Load_old ( 22.117, 24.623) 60.65 0.000
c. Is the mean load capacity of the new alloy significantly greater than the mean load capacity of the currently used alloy? Use a = 0.01. Report the pvalue of your test. From Minitab we have for the separate variance two sample ttest the following:
TwoSample TTest and CI: Load, Beam Beam N Mean StDev SE Mean new 10 28.85 2.51 0.79 old 10 23.37 1.22 0.39 Difference = mu (new)  mu (old) Estimate for difference: 5.480 95% lower bound for difference: 3.917 TTest of difference = 0 (vs >): TValue = 6.21 PValue = 0.000 DF = 13
From this we would reject the null hypothesis that the two alloys produce beams with equal load capacities in favor of the onesided alternative that the new alloy is greater than the old alloy. The pvalue is < 0.001. We can redo this test assuming a common variance and using a pooled standard deviation estimate to get the following:
Twosample T for Load Difference = mu (new)  mu (old) Estimate for difference: 5.480 95% lower bound for difference: 3.950 TTest of difference = 0 (vs >): TValue = 6.21 PValue = 0.000 DF = 18 Both use Pooled StDev = 1.97
Note that the conclusions of the test do not change, nor is the pvalue any different, but the DF for the test is now 18 versus the 13 of the previous problem. The pooled standard deviation estimate is 1.97.
d. Do the condition required for the statistical techniques used in (b) and (c) appear to be satisfied? By this we wonder if the data look sufficiently normal (symmentric, unimodal). We could look at histograms of the data, but there is very little data here to do this. Box plots don't indicate much to worry about although the distribution do not look exactly symmetric (the median is not in the middle of the inter quartile box..
We could also look at a quantilequantile plot to determine if the data look sufficiently normal.
In this plot, if the data are truly normally distributed, they will fall mostly on the straight line. In both cases the data points fall pretty close to the straight line and hence we would conclude that the data are sufficiently normal and that the conditions required for the ttest are met.
e. The beams produced from the new alloy are more expensive thatn the beams produced from the currently used alloy. Thus, the new alloy will be used only if the mean load capacity is at least 5 tons greater than the mean load capacity of the currently used alloy. Based on this information, would you recommend that the company use the new alloy?
To check this, we need to test whether the difference between the two are sufficiently greater than 5 tons. In Minitab, we reperform the two sample ttest only now we set the significant difference at 5. The results are as follows:
Difference = mu (new)  mu (old) Estimate for difference: 5.480 95% lower bound for difference: 3.950 TTest of difference = 5 (vs >): TValue = 0.54 PValue = 0.297 DF = 18 Both use Pooled StDev = 1.97
Note that now the null hypothesis of the test is that the difference is 5 versus the alternative that it is greater than 5. The associated pvalue is 0.297, suggesting that we cannot reject the null hypothesis that the difference is actually 5 (or less). Since we do not reject the null hypothesis we cannot conclude that the difference in mean load capacity is greater than 5 tons, we would stick with the old alloy for this task.
Note that you may be a little confused with this result. Remember what we are really trying to do is show that the new alloy is 5 tons or greater load capacity than the old alloy. Our test says it is 5 OR LESS but NOT more than 5. Hence we cannot go with the new alloy since there is a high probability the load capacity is actually less than 5 tons and a very low probability that it is greater then 5.
The above computations were run using the full Minitab (on my portable).
The ttest output is from the STAT > BASIC STATISTICS > 2 SAMPLE T.
There is an OPTIONS button on the dialog that allows you put in the difference.
When I attempted to do this in the Student Edition of Minitab, there is no
OPTION button. I didn't notice this before. So, what to do. Well, we know
that subtracting a number from any random variable changes the mean, but does
not change the variance. So in the 'not Stacked' data, create a new "New_New"
column which is the old "NEW" data with 5 subtracted from it. Now
do the two sample ttest of the "NewNew" vesus the "OLD"
data to get a ttest that these two groups have the same means. If the original
null hypothesis was that the mean of the NEW beams was 5 tons greater than
that of the OLD beams, then the mean of the NEW5 group should be the same
as the mean of the OLD group. You should get the correct result.
For the stacked data, create a new column (call it DIFF) with 5 in the rows
for all NEW beam cases and 0 for all OLD beam cases. Now create a second "New_Load"
column with the old response column minus DIFF. Now do the twosample ttest
(stacked) using this "New_Load" as the response.
2. Using the data from problem 6.37, perform the Wilcoxon rank sum test. Do the ttest and Wilcoxon test give different results?
We will use Minitab. Note first that the Wilcoxon Rank Sum test in Minitab is the same as the MannWhitney test (whose name is actually the MannWhitneyWilcoxon Rank Sum test). Next we also note that the MannWhitney test does not use "stacked data" but used "unstacked data". Originally I entered the data as stacked data (as I would like you to get into the habit of doing), so I needed to use the Manip>Unstack Columns menu option to get the data into a stacked configuration. The worksheet I have provided to you has already done this. The results of running the MannWhitney test is given as:
MannWhitney Test and CI: Load_new, Load_old Load_new N = 10 Median = 28.800 Load_old N = 10 Median = 23.200 Point estimate for ETA1ETA2 is 5.700 95.5 Percent CI for ETA1ETA2 is (3.801,7.400) W = 151.0 Test of ETA1 = ETA2 vs ETA1 > ETA2 is significant at 0.0003 The test is significant at 0.0003 (adjusted for ties)
When this says the test is significant at the 0.0003 level we interpret this to mean the associated pvalue is 0.0003. With such a low pvalue we would reject the null hypothesis and conclude that the load capacity of the new alloy is greater than that of the old alloy. This result coincides with the ttest results.
3. What is the 95% confidence interval for the difference between the two means?
Minitab will compute the 95% CI for the difference between the two means if you are a little clever and/or observant. Use the two independent sample ttest and test the two sided alternative, Minitab will provide you with the 95% CI (equation 1 page 317) as given below.
Difference = mu (new)  mu (old) Estimate for difference: 5.480 95% CI for difference: (3.626, 7.334)
Doing the same thing with the MannWhitneyWilcoxon two independent sample rank sum test we get a 95% confidence interval for the difference in the medians as:
Point estimate for ETA1ETA2 is 5.700 95.5 Percent CI for ETA1ETA2 is (3.801,7.400)
4. How many beams would be needed for each group if we wanted to estimate the difference between the average loading of the two groups to be within plus or minus 0.5 tons with 95% confidence (Hint see page 314).Use the pooled variance estimate from the ttest as if it were the true variance.
Minitab doesn't do this computation directly so we will have to do this on our own. The pooled standard deviation estimate is given as: s^{2} =1.97. The difference of interest is E=0.5 and the P(type I error) =a = 0.05. Using equation 9 page 318 we have:
Thus we would need to test 120 beams of each type to be this precise in our estimate of the difference between the two beams.