Course evaluation ratings for four college instructors are shown in the following table. Use a = .05 and test for a significant difference among the rating for these instructors. What is your conclusion?
|
Instructor |
|||
|
Black |
Jennings |
Swanson |
Wilson |
|
88 |
87 |
88 |
80 |
|
80 |
78 |
76 |
85 |
|
79 |
82 |
68 |
56 |
|
68 |
85 |
82 |
71 |
|
96 |
99 |
85 |
89 |
|
69 |
99 |
82 |
87 |
|
85 |
84 |
||
|
94 |
83 |
||
|
81 |
|||
Delegate your assignment to our experts and they will do the rest.
Solution:
| Anova: Single Factor | ||||||
| SUMMARY | ||||||
|
Groups |
Count |
Sum |
Average |
Variance |
||
| Column 1 |
6 |
480 |
80 |
117.2 |
||
| Column 2 |
8 |
709 |
88.625 |
61.41071 |
||
| Column 3 |
9 |
729 |
81 |
34.25 |
||
| Column 4 |
6 |
468 |
78 |
157.6 |
||
| ANOVA | ||||||
Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |
| Between Groups |
477.9181 |
3 |
159.306 |
1.916694 |
0.152703 |
2.991241 |
| Within Groups |
2077.875 |
25 |
83.115 |
|||
| Total |
2555.793 |
28 |
||||
Hypothesis:
From the Anova: Single Factor output,
Here, the F-observed value is less than F-critical. Thus, we fail to reject the null hypothesis.
Conclusion:
There is not sufficient evidence to support the claim that there is a significant difference.
In a quality control test of parts manufactured at Dabco Corporation, an engineer sampled parts produced on the first, second, and third shifts. The research study was designed to determine if the population proportion of good parts was the same for all three shifts. Sample data follow.
Using a .05 level of significance, conduct a hypothesis test to determine if the population proportion of good parts is the same for all three shifts. What is the p-value and what is your conclusion?
|
Quality |
Production Shift |
|||
|
First |
Second |
Third |
Total | |
| Good |
285 |
368 |
176 |
829 |
| Defective |
15 |
32 |
24 |
71 |
| Total |
300 |
400 |
200 |
900 |
Hypothesis Testing:
|
Quality |
Production Shift |
||
|
First |
Second |
Third |
|
| Good |
285 |
368 |
176 |
| Defective |
15 |
32 |
24 |
Solution:
Hypothesis testing:
To test these hypotheses, we need to convert the variables to expected frequencies using the equation shown below.
With regard to box 1,1,
Thus, the expected value at 1,1 is 276.3. This process is repeated for each row. This was done using Excel and the results are shown in the table below:
|
Quality |
Production Shift |
|||
|
First |
Second |
Third |
Total |
|
| Good |
276.3333 |
368.4444 |
184.2222 |
829.0 |
| Defective |
23.6667 |
31.5556 |
15.7778 |
71.0 |
| Total |
300.0 |
400.0 |
200.0 |
900.0 |
The next step is to convert the table to chi-squared. The equation shown below can be used to do the conversion.
Repeat the process for each row. This process was done using Excel and the table below shows the results.
|
Quality |
Production Shift |
|||
|
First |
Second |
Third |
Total |
|
| Good |
0.2718 |
0.0005 |
0.3670 |
0.6393 |
| Defective |
3.1737 |
0.0063 |
4.2848 |
7.4648 |
|
8.1041 |
||||
From the chi-square table,
Therefore, reject the null hypothesis.
Conclusion:
The population proportion of good parts is the same for all three shifts.
If the conclusion is that the population proportions are not all equal, use a multiple comparison procedure to determine how the shifts differ in terms of quality. What shift or shifts need to improve the quality of parts produced?
The table below shows the proportions of the samples.
To compare each point, we need calculate the critical value for each point using the equation shown below:
This was done using Excel, and the results are shown below:
| Comparison |
pi |
pj |
Difference |
ni |
nj |
Critical Value |
Sig Diff > CV |
| 1 vs 2 |
0.95 |
0.92 |
0.03 |
300 |
400 |
0.0453 |
No |
| 1 vs 3 |
0.95 |
0.88 |
0.07 |
300 |
200 |
0.0641 |
Yes |
| 2 vs 3 |
0.92 |
0.88 |
0.04 |
400 |
200 |
0.0653 |
No |
Based on the comparisons, statistically first and second shift are not alike. Statistically first and second shift are like third shift.
The Consumer Reports Restaurant Customer Satisfaction Survey is based upon 148,599 visits to full-service restaurant chains ( Consumer Reports website). One of the variables in the study is meal price, the average amount paid per person for dinner and drinks, minus the tip. Suppose the reporter for the Sun Coast Times thought that it would be of interest to her readers to conduct a similar study for restaurants located on the Grand Strand section in Myrtle Beach, South Carolina. The reporter selected a sample of 8 seafood restaurants, 8 Italian restaurants, and 8 steakhouses. The following data show the meal prices ($) obtained for the 24 restaurants sampled. Use a = .05 to test whether there is a significant difference among the mean meal price for the three types of restaurants.
|
Italian |
Seafood |
Steakhouse |
|
$12 |
$16 |
$24 |
|
13 |
18 |
19 |
|
15 |
17 |
23 |
|
17 |
26 |
25 |
|
18 |
23 |
21 |
|
20 |
15 |
22 |
|
17 |
19 |
27 |
|
24 |
18 |
31 |
Solution:
Descriptive Statistics
|
Italian |
Seafood |
Steakhouse |
|
|
$12 |
$16 |
$24 |
|
|
13 |
18 |
19 |
|
|
15 |
17 |
23 |
|
|
17 |
26 |
25 |
|
|
18 |
23 |
21 |
|
|
20 |
15 |
22 |
|
|
17 |
19 |
27 |
|
|
24 |
18 |
31 |
|
| Mean |
$17 |
$19 |
$24 |
| STDEV |
3.8545 |
3.7033 |
3.7417 |
| Mean Average |
$20 |
ANOVA Analysis
| Anova: Single Factor | ||||||
| SUMMARY | ||||||
|
Groups |
Count |
Sum |
Average |
Variance |
||
| Column 1 |
8 |
136 |
17 |
14.85714 |
||
| Column 2 |
8 |
152 |
19 |
13.71429 |
||
| Column 3 |
8 |
192 |
24 |
14 |
||
| ANOVA | ||||||
Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |
| Between Groups |
208 |
2 |
104 |
7.328859 |
0.003852 |
3.4668 |
| Within Groups |
298 |
21 |
14.19048 |
|||
| Total |
506 |
23 |
||||
Hypothesis Testing:
From the ANOVA: Single Factor output,
Here, the p value is less than the alpha value. Thus, the null hypothesis is rejected.
Conclusion:
There is a sufficient evidence to support the claim that there is a significant difference among the mean meal price for the three types of restaurants.
The Scholastic Aptitude Test (SAT) contains three areas: critical reading, mathematics, and writing. Each area is scored on an 800-point scale. A sample of SAT scores for six students follows.
Using a .05 level of significance, do students perform differently on the three areas of the SAT?
| Anova: Single Factor | ||||||
| SUMMARY | ||||||
|
Groups |
Count |
Sum |
Average |
Variance |
||
| Column 1 |
6 |
3012 |
502 |
4686 |
||
| Column 2 |
6 |
3090 |
515 |
3161.2 |
||
| Column 3 |
6 |
2964 |
494 |
5042.8 |
||
| ANOVA | ||||||
Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |
| Between Groups |
1348 |
2 |
674 |
0.156866 |
0.856203 |
3.68232 |
| Within Groups |
64450 |
15 |
4296.667 |
|||
| Total |
65798 |
17 |
||||
Hypothesis Testing:
From the ANOVA: Single Factor output,
Here, the p value is greater than the alpha value
Since the p value is greater than the alpha value, we accept the null hypothesis and reject the alternative hypothesis.
Conclusion: There is no sufficient evidence to support the claim that the means are different among the three proportions of the SAT.
Which area of the test seems to give the students the most trouble? Explain.
|
Student |
Critical Reading |
Mathematics |
Writing |
|
1 |
526 |
534 |
530 |
|
2 |
594 |
590 |
586 |
|
3 |
465 |
464 |
445 |
|
4 |
561 |
566 |
553 |
|
5 |
436 |
478 |
430 |
|
6 |
430 |
458 |
420 |
| Mean |
502 |
515 |
494 |
Out of the three areas, the Writing area of the test seems to be the area of the SAT that gives students the most trouble. This is because its mean score is the least.
|
Student |
Critical Reading |
Mathematics |
Writing |
|
1 |
526 |
534 |
530 |
|
2 |
594 |
590 |
586 |
|
3 |
465 |
464 |
445 |
|
4 |
561 |
566 |
553 |
|
5 |
436 |
478 |
430 |
|
6 |
430 |
458 |
420 |
