(1) In a sample of sixty four patients admitted to a hospital for pneumonia, the average stay was 8.4 days and the sample standard deviation was 7.2 days.
(a) Derive a 95% confidence interval for the mean hospital stay for such patients without assuming any distribution for the length of stay.
(b)
Suppose you are told that the hospital stay has approximately an
exponential distribution. How would you change your answer in
part (a). Recall that for an exponential distribution, 
.
Solution to Exercise 1.
(The solution is an MS Word document. You may have to right click
and choose "Save Target As ..." or something like that.)
(2) An ornithologist observes grackles' attempts at mating. In 75 such attempts, 21 are successful.
(a) What is your point estimate of the proportion of mating attempts which are successful?
(b) What is the estimated standard error of that estimate?
(c) Construct a 90% confidence interval for the true proportion of mating attempts which are successful.
(d) It has been claimed in the literature that fewer than 40% of the mating attempts are successful. Do these data offer strong evidence for that claim?
Solution to Exercise 2.
(The solution is an MS Word document. You may have to right click
and choose "Save Target As ..." or something like that.)
(3) Define the following hypothesis testing concepts:
Solution to Exercise 3.
(The solution is an MS Word document. You may have to right click
and choose "Save Target As ..." or something like that.)
(4) To estimate the weight gain in chickens from a supplement, the chickens are weighed before hand, put on the supplement regimen for two weeks and then weighed again. The data are entered into minitab, and two analyses are performed. First the data set:
before after 2.23 2.67 1.76 1.90 1.84 2.15 2.08 2.01 1.96 2.40
Now the analyses:
Two Sample T-Test and Confidence Interval
Two sample T for after vs before
N Mean StDev SE Mean
after 5 2.226 0.311 0.14
before 5 1.974 0.188 0.084
95% CI for mu after - mu before: ( -0.15, 0.649)
T-Test mu after = mu before (vs >): T = 1.55 P = 0.086 DF = 6
Paired T-Test and Confidence Interval
Paired T for after - before
N Mean StDev SE Mean
after 5 2.226 0.311 0.139
before 5 1.974 0.188 0.084
Difference 5 0.2520 0.2181 0.0975
95% CI for mean difference: (-0.0188, 0.5228)
T-Test of mean difference = 0 (vs > 0): T-Value = 2.58 P-Value = 0.031
(a) Which of these analyses is correct? Explain.
(b) What is a 95% confidence interval for the mean weight gain under this supplement regimen?
(c) Is the observed average weight gain significantly greater than 0?
Solution to Exercise 4.
(The solution is an MS Word document. You may have to right click
and choose "Save Target As ..." or something like that.)
(5) In a study of the effectiveness of an antidepressant drug, 800 subjects are randomly split into two groups of 400 each. The treatment group is given the drug and the control group is given a placebo. The number of subjects who show improvement in the treatment group is 205 and the number in the control group is 168. The researchers wish to know if these data show that the drug is statistically significantly better than the placebo. State the relevant null and alternative hypotheses, compute the P-value, and determine if the drug is significantly better than the placebo.
Solution to Exercise 5.
(The solution is an MS Word document. You may have to right click
and choose "Save Target As ..." or something like that.)
(6) In this exercise we investigate the minitab data set "tumours.mtw" which is apparently the measurements of tumour size vs. time (in days? ... weeks?) for some experimental animal. We also considered the natural logarithm and square root of the tumour size. The data are:
Time Size logsize sqrtsize 14 1.25 0.22314 1.11803 16 1.90 0.64185 1.37840 19 4.75 1.55814 2.17945 21 5.45 1.69562 2.33452 23 7.53 2.01890 2.74408 26 14.50 2.67415 3.80789 28 16.70 2.81541 4.08656 30 21.00 3.04452 4.58258 33 27.10 3.29953 5.20577 35 30.30 3.41115 5.50454 37 40.50 3.70130 6.36396 41 51.40 3.93964 7.16938The results of regression analysis considering each of the response variables. We also computed predicted values at Time = 50.
Regression Analysis
The regression equation is
Size = - 30.3 + 1.81 Time
Predictor Coef StDev T P
Constant -30.319 4.192 -7.23 0.000
Time 1.8149 0.1489 12.19 0.000
S = 4.246 R-Sq = 93.7% R-Sq(adj) = 93.1%
Analysis of Variance
Source DF SS MS F P
Regression 1 2677.6 2677.6 148.52 0.000
Residual Error 10 180.3 18.0
Total 11 2857.9
Unusual Observations
Obs Time Size Fit StDev Fit Residual St Resid
12 41.0 51.40 44.09 2.43 7.31 2.10R
R denotes an observation with a large standardized residual
Predicted Values
Fit StDev Fit 95.0% CI 95.0% PI
60.43 3.65 ( 52.29, 68.56) ( 47.95, 72.90) X
X denotes a row with X values away from the center
The plots produced as part of the above analysis:
scatterplot,
residuals vs. fitted values,
and Normal probability plot of the residuals.
=================================================================
Regression Analysis
The regression equation is
logsize = - 1.23 + 0.136 Time
Predictor Coef StDev T P
Constant -1.2326 0.2750 -4.48 0.001
Time 0.135650 0.009770 13.88 0.000
S = 0.2786 R-Sq = 95.1% R-Sq(adj) = 94.6%
Analysis of Variance
Source DF SS MS F P
Regression 1 14.958 14.958 192.78 0.000
Residual Error 10 0.776 0.078
Total 11 15.734
Predicted Values
Fit StDev Fit 95.0% CI 95.0% PI
5.5499 0.2394 ( 5.0164, 6.0833) ( 4.7314, 6.3683) X
X denotes a row with X values away from the center
The plots produced as part of the above analysis:
scatterplot,
residuals vs. fitted values,
and Normal probability plot of the residuals.
=================================================================
Regression Analysis
The regression equation is
sqrtsize = - 2.25 + 0.228 Time
Predictor Coef StDev T P
Constant -2.2548 0.1572 -14.34 0.000
Time 0.227656 0.005586 40.75 0.000
S = 0.1593 R-Sq = 99.4% R-Sq(adj) = 99.3%
Analysis of Variance
Source DF SS MS F P
Regression 1 42.131 42.131 1660.77 0.000
Residual Error 10 0.254 0.025
Total 11 42.385
Predicted Values
Fit StDev Fit 95.0% CI 95.0% PI
9.1280 0.1369 ( 8.8229, 9.4330) ( 8.6600, 9.5960) X
X denotes a row with X values away from the center
The plots produced as part of the above analysis:
scatterplot,
residuals vs. fitted values,
and Normal probability plot of the residuals.
(a) Which regression fit do you think is best? Justify your answer.
(b) Based on the best regression model in (a), give your best prediction of the tumour size at Time = 50 with a 95% prediction interval. Don't forget to convert back to the original units if you use one of the transformed responses logsize or sqrtsize.
Solution to Exercise 6. (The solution is an MS Word document. You may have to right click and choose "Save Target As ..." or something like that.)
(7) The minitab data set "winde.mtw" has quality ratings of a number of wines from three different regions of France. Is there a significant difference in the mean quality rating among the regions? The minitab analysis follows. Plots appear here and here. Describe the results of this analysis as completely as possible. Discuss the assumptions behind the analysis and whether or not there is any evidence concerning the validity of the assumptions.
One-way Analysis of Variance Analysis of Variance for Quality Source DF SS MS F P Region 2 94.62 47.31 27.52 0.000 Error 35 60.17 1.72 Total 37 154.79 Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -+---------+---------+---------+----- 1 17 11.976 1.328 (---*---) 2 9 10.444 1.515 (----*-----) 3 12 14.583 1.112 (----*----) -+---------+---------+---------+----- Pooled StDev = 1.311 9.6 11.2 12.8 14.4 Tukey's pairwise comparisons Family error rate = 0.0500 Individual error rate = 0.0196 Critical value = 3.46 Intervals for (column level mean) - (row level mean) 1 2 2 0.210 2.854 3 -3.816 -5.553 -1.397 -2.724
Solution to Exercise 7. (The solution is an MS Word document. You may have to right click and choose "Save Target As ..." or something like that.)
(8) Exercise 14.07, p. 624.
Solution to Exercise 8. (The solution is an MS Word document. You may have to right click and choose "Save Target As ..." or something like that.)
