STAT 211 EXAM 4 – FORM A SPRING 2004
The number of pounds of steam used per month by a chemical plant is thought to be related to the average ambient temperature in (°F) for that month. The past year’s usage and temperature are shown in the following table.
Month / Jan. / Feb. / Mar. / Apr. / May / June / July / Aug. / Sept. / Oct. / No. / Dec.Temp. / 21 / 24 / 32 / 47 / 50 / 59 / 68 / 74 / 62 / 50 / 41 / 30
Usage/1000 / 185.79 / 214.47 / 288.03 / 424.84 / 454.58 / 539.03 / 621.55 / 675.06 / 562.03 / 452.93 / 369.95 / 273.98
Using MINITAB, true average temperature that year (m temp) is tested being 50°F and the results as follows.
Test of m temp = 50 versus m temp ¹ 50
Variable n Mean StDev SE Mean
temp 12 46.50 17.34 5.01
Variable 95.0% CI T P
temp ( 35.48, 57.52) -0.70 0.499
Steam usage (y) is regressed on the average monthly temperature (x). The simple linear regression model, is fitted using MINITAB software. Residuals are checked at the end of the analysis and independent residuals are found to be normally distributed with the mean zero and the constant variance. The following are the results.
Predictor Coef SE Coef T P
Constant -5.452 1.620 -3.37 0.007
temp 9.19295 0.03281 280.17 0.000
S = 1.887 R-Sq = 100.0% R-Sq(adj) = 100.0%
Analysis of Variance
Source DF SS MS F P
Regression 1 279645 279645 78493.85 0.000
Residual Error 10 36 4
Total 11 279680
Unusual Observations
Obs temp usage Fit SE Fit Residual St Resid
12 30.0 273.980 270.336 0.768 3.644 2.11R
R denotes an observation with a large standardized residual
Use the given information to answer the following 8 questions.
- We claim that true average temperature that year should be 50°F. Do the data support this claim ?
(a) Yes (b) No
H0: m temp = 50 versus Ha: m temp ≠ 50 p-value=0.499 > 0.05 then fail to reject H0
- Which of the following is the test statistics for testing the true standard deviation of monthly average temperature being 18 °F ?
(a) 0.93 (b) 10.21 (c) 183.75 (d) 280.17 (e) 78493.85
H0: s temp = 18 versus Ha: s temp ≠ 18 test statistics: =
- If the true standard deviation of monthly average temperature is known to be 18 °F, Would the data support the true average temperature that year being 50°F?
(a) Yes (b) No
H0: m temp = 50 versus Ha: m temp ≠ 50 test statistics: < -1.96 then fail to reject H0
- What is the estimate of the expected steam usage when the average temperature is 55°F?
(a) we cannot answer this question because temperature 55 is not listed on the data
(b) we cannot answer this question because temperature 55 is not in the range of the temperature data
(c) 500.16
(d) 517.38
(e) 524.43
55 is in the range of the temperature data then estimate of the expected steam usage is -5.452+9.19295(55)
- What is the estimated change in expected steam usage when the monthly average temperature changes by 1°F?
(a) -5.452 (b) 9.193 (c) 17.34 (d) 46.50
It is the definition of the slope
- Suppose the monthly average temperature is 47°F, what is the corresponding residual on the fitted model?
(a) -1.7767 (b) -1.3213 (c) 0.5574 (d) 1.3213 (e) 1.7767
e=y-yhat=424.84-[-5.452+9.19295(47)]
- What proportion of total variability is accounted for by the simple linear regression model?
(a) 0% (b) 25% (c) 50% (d) 75% (e) 100%
Definition of R2
- We believe that the change in expected steam usage when the monthly average temperature changes by 1°F should be 10 pounds. Which of the following is the corresponding test statistics to test this?
(a) -24.60 (b) -3.37 (c) 10.21 (d) 280.17 (e) 78493.85
H0: b 1 = 10 versus Ha: b 1 ≠ 10 test statistics:
A paper on the Journal of the Association of Asphalt Paving Technologists describes an experiment to determine the effect of air voids on percentage retained strength of Asphalt. For purposes of the experiment, air voids are controlled at three levels: low (2-4%), medium (4-6%), and high (6-8%). The data are shown in the following table.
Air Voids Retained strength (%)
Low / 106 / 90 / 103 / 90 / 79 / 88 / 92 / 95Medium / 80 / 69 / 94 / 91 / 70 / 83 / 87 / 83
High / 78 / 80 / 62 / 69 / 76 / 85 / 69 / 85
The following are the analysis results from MINITAB. The true mean of percentage retained strength is shown by mlow, mmedium, mhigh for the levels of low, medium, and high, respectively.
Analysis of Variance for strength where the model is single factor ANOVA
Source DF SS MS F P
level 2 1230.3 615.1 8.30 0.002
Error 21 1555.8 74.1
Total 2786.0
Level N Mean StDev
low 8 92.875 8.560
medium 8 82.125 9.015
high 8 75.500 8.229
Pooled StDev = 8.607
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0200
Critical value = 3.56
Intervals for (column level mean) - (row level mean)
high low
low -28.21
-6.54
medium -17.46 -0.08
4.21 21.58
Test for equal true means (mlow =mmedium)in independent samples assuming unequal true variances
Estimate for difference: 10.75
95% CI for difference: (1.25, 20.25)
T-Test of difference = 0 (vs not =): T-Value = 2.45 P-Value = 0.029 DF = 13
Test for equal true means (mlow =mmedium)in independent samples assuming equal true variances
95% CI for difference: (1.32, 20.18)
T-Test of difference = 0 (vs not =): T-Value = 2.45 P-Value = 0.028 DF = 14
Both use Pooled StDev = 8.79
Test for equal true means (mlow =mhigh)in independent samples assuming unequal true variances
Estimate for difference: 17.38
95% CI for difference: (8.31, 26.44)
T-Test of difference = 0 (vs not =): T-Value = 4.14 P-Value = 0.001 DF = 13
Test for equal true means (mlow =mhigh)in independent samples assuming equal true variances
95% CI for difference: (8.37, 26.38)
T-Test of difference = 0 (vs not =): T-Value = 4.14 P-Value = 0.001 DF = 14
Both use Pooled StDev = 8.40
Test for equal true means(mmedium =mhigh)in independent samples assuming unequal true variances
Estimate for difference: 6.63
95% CI for difference: (-2.70, 15.95)
T-Test of difference = 0 (vs not =): T-Value = 1.54 P-Value = 0.149 DF = 13
Test for equal true means(mmedium =mhigh)in independent samples assuming equal true variances
95% CI for difference: (-2.63, 15.88)
T-Test of difference = 0 (vs not =): T-Value = 1.54 P-Value = 0.147 DF = 14
Both use Pooled StDev = 8.63
Testing equal variances for all levels
Bartlett's Test (normal distribution)
Test Statistic: 0.055
P-Value : 0.973
Levene's Test (any continuous distribution)
Test Statistic: 0.018
P-Value : 0.982
Use the given information to answer the following 8 questions.
- Do the different levels of air voids significantly affect mean percentage retained strength?
(a) Yes (b) No
H0: mlow =mmedium=mhigh
P-value=0.002 < 0.05 then reject H0
- Which of the following Tukey’s test suggest?
(a) The different levels of air voids do not affect mean percentage retained strength
(b) Low and medium levels have different true mean percentage retained strength
(c) Low and high levels have different true mean percentage retained strength
(d) Medium and high levels have different true mean percentage retained strength
(e) Exactly two of the above
- Is the constant variance assumption in analysis of variance satisfied for us to rely on the output MINITAB gives us?
(a) Yes. The p-value is 0.002
(b) No. The p-value is 0.002
(c) Yes. The p-value is 0.973 Bartlett’s test is used
(d) No. The p-value is 0.973
- Assuming that the constant variance assumption is satisfied, which of the following is the point estimate for the constant standard deviation of residuals?
(a) 8.6 (b) 24.8 (c) 74.1 (d) 615.1
Square root of MSE
- We would think the true means of percent retained strength for medium and high levels should not be different assuming equal true variances. Do the data support this?
(a) Yes (b) No
0 falls in the confidence interval or the p-value is larger than 0.05.
- We would think the true means of percent retained strength for low and high levels should not be different assuming equal true variances. We tested this using the given data. Based on our conclusion, which of the following we have made?
(a) Possible type II error (b) Possible type I error (c) Correct decision
Rejected equal means when the true means were equal
- If you were testing the equal variances of the percent retained strength for medium and high levels, which of the following would be the corresponding test statistics?
(a) 0.018 (b) 0.055 (c) 1.096 (d) 1.200
then the test statistics: F=
- I do not know if you noticed it, but the total degree of freedom is not written on the analysis of variance table. Which of the following is the degree of freedom for total?
(a) 20 (b) 21 (c) 22 (d) 23 (e) 24
The warranty for batteries of mobile phones is set at 200 operating hours, with proper charging procedures. A study of 5000 batteries is carried out and 15 stop operating prior to 200 hours. The following is the results from MINITAB software.
Test of p=0.002 vs p¹0.002 where p:true proportion of phones stop operating prior to 200 hours
Exact
Sample x n Sample p 95.0% CI for p P-Value
1 15 5000 0.003000 (0.001680, 0.004943) 0.150
Use the given information to answer the following 2 questions.
- The claim of less than 0.2% of the company’s batteries failing during the warranty period will be tested. Which of the following would be the corresponding null hypothesis?
(a) p£ 0.002 (b) p ³ 0.002 (c) p < 0.002 (d) p ³ 0.005 (e) p < 0.005
claim is under the Ha for the hypothesis to be valid
- Do these experimental results support the claim that less than 0.2% of the company’s batteries will fail during the warranty period?
(a) Yes (b) No
The P-value on the output is twice the area above positive test statistics where test statistics is positive. We have a lower tailed test then the P-value is 1-0.15/2=0.925 which cause failing to reject H0.
In a random sample of 200 College Station residents who drive a domestic car, 165 reported wearing their seat belt regularly, while another sample of 250 College Station residents, who drive a foreign car, revealed 198 who regularly wore their seat belt. The following is the results from MINITAB software.
Sample x n Sample p
1 165 200 0.825000
2 198 250 0.792000
Estimate for p(1) - p(2): 0.033
95% CI for p(1) - p(2): (-0.0398310, 0.105831)
Test for p(1) - p(2) = 0 (vs not = 0): Z = 0.89 P-Value = 0.375
Where p(1) and p(2) correspond to the proportion of domestic and foreign car drivers who wore their seat belt, respectively.
Use the given information to answer the following 4 questions.
- Which of the following is the point estimate for p(1)-p(2)?
(a) We do not have enough information to answer this question
(b) -0.033
(c) -0.375
(d) 0.033
(e) 0.375
- Are there any significant differences in seat belt usage between domestic and foreign car drivers?
(a) No (b) Yes
0 falls in the confidence interval or the p-value is larger than 0.05 where H0:p(1)-p(2)=0
- If all the numbers are doubled on the initial given data, does this effect your decision for the previous question?
(a) Yes (rejected H0 in the previous question and failing to reject H0 in this question or vice versa)
(b) No (rejected or failed to reject H0 in the previous question and rejecting or failing to reject H0 in this question)
Sample proportions stay the same.
- If p(1) < p(2), it means
(a) The proportion of domestic car drivers who wore their seat belts is the same as the foreign car drivers
(b) The proportion of domestic car drivers who wore their seat belts is more than the foreign car drivers
(c) The proportion of domestic car drivers who wore their seat belts is less than the foreign car drivers
- If the P-value is 0.0222 for the upper tailed test with the test statistics of z=2.01? Which of the following is the corresponding p-value for the two tailed test?
(a) 0.0111 (b) 0.0222 (c) 0.0333 (d) 0.0444
- I would like to test if the true correlation between the overall grade and the number of hours of studying is zero. Random sample is collected from 200 students and the Pearson’s correlation is found to be 0.975. Which of the following test statistics should be used for testing?
(a) 33.27 (b) 49.13 (c) 58.94 (d) 61.74 (e) 280.17
The test statistics is
- Which of the following is incorrect?
(a) If the P-value is very large in any test, you have to fail to reject H0
(b) Only if the value of the test statistics is above the critical value in two tailed test, you should reject H0
(c) If there is a less than sign under the alternative hypothesis, you have a lower tailed test
(d) The P-value is the area below the test statistics in the lower tailed test