S2MA3 Assignment #3 Solutions

Statistics 2MA3 - Assignment #3 Solutions

2001-04-06

Due 2001-04-06 17:00

Full marks = 100

You can submit answers calculated with a pocket calculator or with R..

Q1 [15]

Considering the small sample size, the histogram is reasonably close to a normal distribution.

> mean(hosp$temp1)
[1] 98.308

Confidence interval for mean first temperature, assuming standard deviation = 1 degree: (97.92, 98.70)

> mean(hosp$temp1)-qnorm(.975)*(1/sqrt(25))
[1] 97.916
> mean(hosp$temp1)+qnorm(.975)*(1/sqrt(25))
[1] 98.7

Test the hypothesis that mean first temperature = 98.6, assuming standard deviation = 1 degree:

> z0 <- (mean(hosp$temp1)-98.6)/(1/sqrt(25))
> z0
[1] -1.46
> 2*pnorm(z0)
[1] 0.1442901

If standard deviation = 1, there is no evidence (p-value = 0.14) that mean first temperature is not 98.6.

Test the hypothesis that standard deviation = 1 degree:

> var(hosp$temp1)
[1] 0.4641
> chsq0 <- 24*var(hosp$temp1)/1
> chsq0
[1] 11.1384
> 2*pchisq(chsq0,24)
[1] 0.0240196

There is significant evidence (p-value = 0.024) that the standard deviation is not 1 degree.

Confidence interval for mean first temperature: (98.03, 98.59)

> mean(hosp$temp1)-qt(.975,24)*(sqrt(var(hosp$temp1))/sqrt(25))
[1] 98.0268
> mean(hosp$temp1)+qt(.975,24)*(sqrt(var(hosp$temp1))/sqrt(25))
[1] 98.5892

Test the hypothesis that mean first temperature = 98.6:

> t0 <- (mean(hosp$temp1)-98.6)/(sqrt(var(hosp$temp1))/sqrt(25))
> t0
[1] -2.143123
> 2*pt(t0,24)
[1] 0.04244861

or, using t.test()

> t.test(hosp$temp1,mu=98.6)
 
	 One Sample t-test 
 
data:  hosp$temp1 
t = -2.1431, df = 24, p-value = 0.04245 
alternative hypothesis: true mean is not equal to 98.6 
95 percent confidence interval:
 98.0268 98.5892 
sample estimates:
mean of x 
   98.308

There is significant evidence (p-value = 0.042) that the true mean is not 98.6.

The calculations assume independent, normal data. We can't test the assumption of independence. The histogram does not indicate non-normality. Since the hypothesis that standard deviation = 1 was rejected, the confidence intervals and tests that assumed standard deviation = 1 will be invalid. It is safe to assume that the other confidence intervals and tests are valid.

Q2 [2]

> (0.5*(qnorm(0.95)+qnorm(0.99))/(37-38))^2
[1] 3.94261

Rounding up, we conclude that at least n = 4 observations will be needed.

Q3 [2]

> (0.5*qnorm(0.975)/0.02)^2
[1] 2400.912

Rounding up, we conclude that at least n = 2401 observations will be needed. This calculation is based on the "worst case" where p = 0.5 and the variance of the estimate is maximized.

Q4 [2]

> qchisq(0.95,19)/qchisq(0.05,19)
[1] 2.979489
> qchisq(0.95,18)/qchisq(0.05,18)
[1] 3.074324

By trial and error, we see that at least 19 degrees of freedom are needed, that is, n = 19+1 = 20.

Q5 [8]

You have to "read between the lines"; if the hypothesis is that the mean body temperature is "unaffected" the alternative would seem to be 2-sided. However, these patients are fevered and the objective is obviously to reduce the temperature, so a right-tailed test could also be justified. But since the p-value is extremely small (p-value = 0.000023 2-sided, p-value = 0.000012 right-tailed) it doesn't affect the conclusion.

> pharma
   before after
1   102.4  99.6
2   103.2 100.1
3   101.9 100.2
4   103.0 101.1
5   101.2  99.8
6   100.7 100.2
7   102.5 101.0
8   103.1 100.1
9   102.8 100.7
10  102.3 101.1
11  101.9 101.3
12  101.4 100.2
 
> t.test(pharma$before,pharma$after,paired=T)
 
	 Paired t-test 
 
data:  pharma$before and pharma$after 
t = 6.9747, df = 11, p-value = 2.346e-05 
alternative hypothesis: true difference in means is not equal to 0 
95 percent confidence interval:
 1.197756 2.302244 
sample estimates:
mean of the differences 
                   1.75

We conclude that aspirin does affect mean body temperature (p-value = 0.000023) or does reduce mean body temperature (p-value = 0.000012). The assumptions are that the differences are independent and normal. We can't test independence. The sample is too small to test normality but the histogram of differences is consistent with the assumption of normality.

Q6 [3]

There are 12 positive differences out of 12 non-zero differences, so the exact p-value for the sign test is easily computed from the binomial distribution. The exact 2-sided p-value is 2*(0.5^12) = 0.000488. The result is very highly significant, so we reject the hypothesis.

> 2*(1-pbinom(11,12,0.5))
[1] 0.0004882812

The normal approximation to the binomial does not work very well in this example, but still gives a highly significant p-value.

> 2*(1-pnorm((12-6-0.5)/sqrt(12/4)))
[1] 0.001496164

Q7 [8]

> obstet
   bwt     group
1  6.9 treatment
2  7.6 treatment
3  7.3 treatment
4  7.6 treatment
5  6.8 treatment
6  7.2 treatment
7  8.0 treatment
8  5.5 treatment
9  5.8 treatment
10 7.3 treatment
11 8.2 treatment
12 6.9 treatment
13 6.8 treatment
14 5.7 treatment
15 8.6 treatment
16 6.4   control
17 6.7   control
18 5.4   control
19 8.2   control
20 5.3   control
21 6.6   control
22 5.8   control
23 5.7   control
24 6.2   control
25 7.1   control
26 7.0   control
27 6.9   control
28 5.6   control
29 4.2   control
30 6.8   control
 
> boxplot(split(obstet$bwt,obstet$group),ylab="Birth Weight (lb.)")

> bwtctl <- obstet$bwt[obstet$group=="control"]
> bwttrt <- obstet$bwt[obstet$group=="treatment"]
 
> t.test(bwtctl,bwttrt,var.equal=T)
 
	  Two Sample t-test 
 
data:  bwtctl and bwttrt 
t = -2.4136, df = 28, p-value = 0.02259 
alternative hypothesis: true difference in means is not equal to 0 
95 percent confidence interval:
 -1.5159418 -0.1240582 
sample estimates:
mean of x mean of y 
     6.26      7.08

The 2-sample t-test, assuming equal variances, is significant (p-value = 0.023), indicating that the means are not equal. Independence can't be tested. the sample sizes are too small to test normality but the box plots look reasonably symmetric so there is no reason to reject normality.

Assuming independence and normality, the assumption of equal variances (homoscedasticity) can be tested with an F-test.

> sqrt(var(bwtctl))
[1] 0.9605058
> sqrt(var(bwttrt))
[1] 0.8993649
> sp <- sqrt((14*var(bwtctl) + 14*var(bwttrt))/28)
> sp
[1] 0.9304377
 
> F0 <- var(bwtctl)/var(bwttrt)
> F0
[1] 1.140586
> 2*(1-pf(F0,14,14))
[1] 0.8090466

There is no evidence (p-value = 0.81) from these data that the variances of the control and treatment groups are unequal.

Q8 [4]

> anova(lm(bwt~group,data=obstet))
Analysis of Variance Table
 
Response: bwt
          Df  Sum Sq Mean Sq F value  Pr(>F)  
group      1  5.0430  5.0430  5.8252 0.02259 *
Residuals 28 24.2400  0.8657                  
---
Signif. codes:  0  `***'  0.001  `**'  0.01  `*'  0.05  `.'  0.1  ` '  1

The F statistic from the ANOVA is the square of the t statistic

> -2.4136^2
[1] -5.825465

The mean squared residual from the ANOVA is just the pooled variance estimate

> sp^2
[1] 0.8657143

Q9 [2]

The mean squared residual from the ANOVA, which is the same as the pooled variance estimate, is on 28 df, so the 95% confidence interval is (0.545, 1.584).

> c((sp^2)/(qchisq(0.975,28)/28), (sp^2)/(qchisq(0.025,28)/28))
[1] 0.5451995 1.5835002

Q10 [8]

> pulmon
   bronch lfg
1    20.8   A
2     4.1   A
3    30.0   A
4    24.7   A
5    13.8   A
6     7.5   B
7     7.5   B
8    11.9   B
9     4.5   B
10    3.1   B
11    8.0   B
12    4.7   B
13   28.1   B
14   10.3   B
15   10.0   B
16    5.1   B
17    2.2   B
18    9.2   C
19    2.0   C
20    2.5   C
21    6.1   C
22    7.5   C
 
> is.factor(pulmon$lfg)
[1] TRUE
 
> boxplot(split(pulmon$bronch,pulmon$lfg), ylab="Bronchial Reactivity to Sulphur Dioxide", xlab="Lung Function Group")

Lung function Group A, with FEV₁/FVC < 75%, has a much higher reactivity than group B (75-84%) and group C (> 84%). Group A appears to have a larger variance than the others and the distributions appear to be positively skewed.

Taking logs (NOT REQUIRED) seems to over-correct; the three variances are more nearly equal but the distributions appear to be negatively skewed. The square-root transformation (NOT REQUIRED and not shown) seems to be about right.

> boxplot(split(pulmon$bronch,pulmon$lfg), ylab="Bronchial Reactivity to Sulphur Dioxide", xlab="Lung Function Group", log="y")

> anova(lm(bronch~lfg,data=pulmon))
Analysis of Variance Table
 
Response: bronch
          Df Sum Sq Mean Sq F value  Pr(>F)  
lfg        2 503.55  251.77  4.9893 0.01813 *
Residuals 19 958.80   50.46                  
---
Signif. codes:  0  `***'  0.001  `**'  0.01  `*'  0.05  `.'  0.1  ` '  1

There is significant evidence (p-value = 0.018) that mean bronchial reactivity is not the same in each lung-function group.

> msr <- anova(lm(bronch~lfg,data=pulmon))$"Mean Sq"[[2]]
> msr
[1] 50.46329
> c(msr/(qchisq(0.975,19)/19), msr/(qchisq(0.025,19)/19))
[1]  29.18522 107.65180

The 95% confidence interval for the residual variance s² is (29.2, 107.7).

Q11 [10]

> anesthetics$subject<-as.factor(anesthetics$subject)
> anesthetics$anesthesia<-as.factor(anesthetics$anesthesia)
 
> anesthetics
   subject anesthesia epinephrine
1        1          1        0.28
2        1          1        0.36
3        1          2        0.30
4        1          2        0.88
5        1          3        1.07
6        1          3        1.53
7        2          1        0.51
8        2          1        0.32
9        2          2        0.39
10       2          2        0.39
11       2          3        1.35
12       2          3        0.49
 
> anova(lm(epinephrine~subject*anesthesia, data=anesthetics))
 
Analysis of Variance Table
 
Response: epinephrine
                   Df  Sum Sq Mean Sq F value  Pr(>F)  
subject             1 0.07841 0.07841  0.7074 0.43255  
anesthesia          2 1.26762 0.63381  5.7181 0.04075 *
subject:anesthesia  2 0.11502 0.05751  0.5188 0.61968  
Residuals           6 0.66505 0.11084                  
---
Signif. codes:  0  `***'  0.001  `**'  0.01  `*'  0.05  `.'  0.1  ` '  1

Assume that the observations are independent and normal, and that the variance of plasma epinephrine concentration is the same within each combination of Subject and Anesthetic.

There is no evidence (p-value = 0.62) of an interaction between Subject and Anesthesia. That is, both subjects responded to the three anesthesias in similar ways, so we can test the main effects. There is significant evidence (p-value = 0.041) of a difference between anesthesias, but no evidence (p-value = 0.43) of a difference between subjects.

> msr <- anova(lm(epinephrine~subject*anesthesia,data=anesthetics))$"Mean Sq"[[4]]
> msr
[1] 0.1108417
 
> c(msr/(qchisq(0.975,6)/6), msr/(qchisq(0.025,6)/6))
[1] 0.04602621 0.53748179

The 95% confidence interval for the residual variance s² is (0.0460, 0.5375).

If the study were to be done again, there should be more subjects. There should still be replication and a test of the interaction between Subject and Anesthesia because it is possible that with more subjects someone will respond differently to the different anesthesias.

Q12 [6]


> plot(hosp$age[-7],hosp$duration[-7])
> fith1 <- lm(duration~age,data=hosp[-7,])
> summary(fith1)
 
Call:
lm(formula = duration ~ age, data = hosp[-7, ])
 
Residuals:
    Min      1Q  Median      3Q     Max 
-5.4951 -2.0849  0.1927  1.8022  9.1326 
 
Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)  5.17552    1.63336   3.169  0.00445 **
age          0.06260    0.03629   1.725  0.09854 . 
---
Signif. codes:  0  `***'  0.001  `**'  0.01  `*'  0.05  `.'  0.1  ` '  1 
 
Residual standard error: 3.506 on 22 degrees of freedom
Multiple R-Squared: 0.1191,	Adjusted R-squared: 0.07911 
F-statistic: 2.976 on 1 and 22 degrees of freedom,	p-value: 0.09854 
 
> abline(fith1)
> anova(fith1)
Analysis of Variance Table
 
Response: duration
          Df  Sum Sq Mean Sq F value  Pr(>F)  
age        1  36.573  36.573  2.9758 0.09854 .
Residuals 22 270.385  12.290                  
---
Signif. codes:  0  `***'  0.001  `**'  0.01  `*'  0.05  `.'  0.1  ` '  1

Assume that the data are independent, that the true relationship is linear, and that the conditional distribution of duration, given age, is normal, with variance independent of age.

There is no evidence (p-value = 0.099) from these data that duration of stay depends on age. (Compare the corresponding analysis in Exercise #4 and note that removing the observation makes very little change to the results; the p-value is increased from 0.074 to 0.099.)

Q13 [12]

If you don't want to delete an extreme observation, taking logs can reduce its impact on the analysis. The intention here was that you would fit log-transformed data and not remove the 7th subject.

> hosp$logdur <- log(hosp$duration)
> dimnames(hosp)[[2]]
 [1] "id"       "duration" "age"      "sex"      "temp1"   
 [6] "wbc1"     "antib"    "bact"     "serv"     "logdur"  
> cor(hosp[,c(10,3,5,6)])
            logdur        age      temp1        wbc1
logdur  1.00000000  0.3515604  0.1846740 -0.01404079
age     0.35156041  1.0000000 -0.3816794 -0.36932443
temp1   0.18467398 -0.3816794  1.0000000  0.42354776
wbc1   -0.01404079 -0.3693244  0.4235478  1.00000000
 
> pairs(hosp[,c(10,3,5,6)])

Note that first white blood cell count is correlated with first temperature but not with log(duration), so if temperature is in the model then adding white blood cell count will not improve the fit significantly. Similarly, adding first temperature to the model after age model does not improve the fit significantly.

> fith2 <- lm(logdur~age+temp1+wbc1, data=hosp)
 
> anova(fith2)
Analysis of Variance Table
 
Response: logdur
          Df Sum Sq Mean Sq F value  Pr(>F)  
age        1 0.9599  0.9599  3.4275 0.07824 .
temp1      1 0.9243  0.9243  3.3002 0.08357 .
wbc1       1 0.0011  0.0011  0.0038 0.95116  
Residuals 21 5.8815  0.2801                  
---
Signif. codes:  0  `***'  0.001  `**'  0.01  `*'  0.05  `.'  0.1  ` '  1 
 
> plot(fitted(fith2),hosp$logdur)
> abline(0,1)
> plot(fitted(fith2),residuals(fith2))
> abline(h=0)

The ANOVA and the two plots indicate that a multiple linear regression model is reasonable, but does not give a significant improvement over the grand mean.

Since the instructions weren't clear, you could also analyze taking logs AND removing the 7th observation.

> cor(hosp[-7,c(10,3,5,6)])
            logdur        age       temp1        wbc1
logdur  1.00000000  0.2989780 -0.00494539 -0.02268732
age     0.29897801  1.0000000 -0.49544466 -0.37858725
temp1  -0.00494539 -0.4954447  1.00000000  0.45080402
wbc1   -0.02268732 -0.3785873  0.45080402  1.00000000
> pairs(hosp[-7,c(10,3,5,6)])

Note that first white blood cell count is correlated with first temperature but not with log(duration), so if temperature is in the model then adding white blood cell count will not improve the fit significantly. Similarly, adding first temperature to the model after age is in the model does not improve the fit significantly.

Removing the 7th subject as well as taking log(duration) makes age and first temperature even less significant.

> fith2a <- lm(logdur~age+temp1+wbc1, data=hosp[-7,])
 
> anova(fith2a)
Analysis of Variance Table
 
Response: logdur
          Df Sum Sq Mean Sq F value Pr(>F)
age        1 0.5084  0.5084  2.0286 0.1698
temp1      1 0.1545  0.1545  0.6166 0.4415
wbc1       1 0.0122  0.0122  0.0488 0.8274
Residuals 20 5.0123  0.2506               
 
> plot(fitted(fith2a),hosp$logdur[-7])
> abline(0,1)
> plot(fitted(fith2a),residuals(fith2a))
> abline(h=0)

We assume that the data are independent and normal and that the effects of age, first temperature and first white blood cell count are linear.

We conclude that the dependence of log(duration of stay) on age, first temperature and first white blood cell count is too weak to have any predictive power.

Q14 [4]

If we fit age after fitting first temperature the improvement in the fit is significant (p-value = 0.026) for predicting log(duration of stay) if the 7th observation is included, but (NOT REQUIRED) not if it is excluded.

> anova(lm(logdur~temp1+age+wbc1, data=hosp))
Analysis of Variance Table
 
Response: logdur
          Df Sum Sq Mean Sq F value Pr(>F)  
temp1      1 0.2649  0.2649  0.9458 0.3419  
age        1 1.6194  1.6194  5.7819 0.0255 *
wbc1       1 0.0011  0.0011  0.0038 0.9512  
Residuals 21 5.8815  0.2801                 
---
Signif. codes:  0  `***'  0.001  `**'  0.01  `*'  0.05  `.'  0.1  ` '  1 
 
> anova(lm(logdur~temp1+age+wbc1, data=hosp[-7,]))
Analysis of Variance Table
 
Response: logdur
          Df Sum Sq Mean Sq F value Pr(>F)
temp1      1 0.0001  0.0001  0.0006 0.9814
age        1 0.6628  0.6628  2.6446 0.1196
wbc1       1 0.0122  0.0122  0.0488 0.8274
Residuals 20 5.0123  0.2506

Q15 [10]

> cattle
   bodywt mcr bodywtf
1     110 235     110
2     110 198     110
3     110 173     110
4     230 174     230
5     230 149     230
6     230 124     230
7     360 115     360
8     360 130     360
9     360 102     360
10    360  95     360
11    505 122     505
12    505 112     505
13    505  98     505
14    505  96     505
 
> fitc1 <- lm(mcr~bodywt, data=cattle)
> plot(cattle$bodywt,cattle$mcr)
> abline(fitc1)
 
> anova(fitc1)
Analysis of Variance Table
 
Response: mcr
          Df  Sum Sq Mean Sq F value    Pr(>F)    
bodywt     1 16634.2 16634.2  27.567 0.0002043 ***
Residuals 12  7241.0   603.4                      
---
Signif. codes:  0  `***'  0.001  `**'  0.01  `*'  0.05  `.'  0.1  ` '  1 
 
> anova(lm(mcr~bodywtf, data=cattle))
Analysis of Variance Table
 
Response: mcr
          Df  Sum Sq Mean Sq F value    Pr(>F)    
bodywtf    3 19514.2  6504.7  14.916 0.0005067 ***
Residuals 10  4361.0   436.1                      
---
Signif. codes:  0  `***'  0.001  `**'  0.01  `*'  0.05  `.'  0.1  ` '  1 
 
> anova(lm(mcr~bodywt+bodywtf, data=cattle))
Analysis of Variance Table
 
Response: mcr
          Df  Sum Sq Mean Sq F value    Pr(>F)    
bodywt     1 16634.2 16634.2  38.143 0.0001047 ***
bodywtf    2  2880.0  1440.0   3.302 0.0792380 .  
Residuals 10  4361.0   436.1                      
---
Signif. codes:  0  `***'  0.001  `**'  0.01  `*'  0.05  `.'  0.1  ` '  1

Assume that the data are independent and normal and the variance in mcr does not depend on body weight.

Although the plot suggests that mcr levels out when body weight is greater than 350, the non-linearity is not significant (p-value = 0.079). Hence we can test the slope, which is very highly significant (p-value = 0.0001).

Interpolating on the fitted line, a body weight of 300 kg would predict a metabolic clearance rate of 142.1.

> predict(fitc1,data.frame(bodywt=300))
[1] 142.0674
 
> msrc1 <- anova(fitc1)$"Mean Sq"[[2]]
> msrc1
[1] 603.4178
> c(msrc1/(qchisq(0.975,12)/12), msrc1/(qchisq(0.025,12)/12))
[1]  310.2848 1644.2691
 
> msrc2 <- anova(lm(mcr~bodywtf,data=cattle))$"Mean Sq"[[2]]
> msrc2
[1] 436.1
> c(msrc2/(qchisq(0.975,10)/10), msrc2/(qchisq(0.025,10)/10))
[1]  212.9064 1343.0972

The 95% confidence interval for the residual variance is (310, 1644) if we assume a linear regression and use the MSR obtained after fitting the straight line. It is (213, 1343) if we do not assume a linear regression.

Q16 [4]

> library(ctest)
 
> std <- matrix(c(40,10,15,30,20,40,130,70,45),nrow=3)
> dimnames(std) <- list(c("Penicillin","Spect.low","Spect.high"),c("Spos","Sneg.Cpos","Sneg.Cneg"))
 
> std
           Spos Sneg.Cpos Sneg.Cneg
Penicillin   40        30       130
Spect.low    10        20        70
Spect.high   15        40        45
 
> chisq.test(std)$expected
            Spos Sneg.Cpos Sneg.Cneg
Penicillin 32.50      45.0    122.50
Spect.low  16.25      22.5     61.25
Spect.high 16.25      22.5     61.25
> chisq.test(std)
 
	 Pearson's Chi-squared test 
 
data:  std 
X-squared = 29.1401, df = 4, p-value = 7.322e-06

There is highly significant evidence (p-value = 0.0000007) that the response depends on which of the three treatments was used. In particular, penicillin gave more (- smear, + culture) results than expected under independence, while spectinomycin (high dose) gave more (- smear, - culture) than expected.

Statistics 2MA3 - Assignment #3 Solutions

2001-04-06

Due 2001-04-06 17:00

Full marks = 100

Q1 [15]

Considering the small sample size, the histogram is reasonably close to a normal distribution.

Confidence interval for mean first temperature, assuming standard deviation = 1 degree: (97.92, 98.70)

Test the hypothesis that mean first temperature = 98.6, assuming standard deviation = 1 degree:

If standard deviation = 1, there is no evidence (p-value = 0.14) that mean first temperature is not 98.6.

Test the hypothesis that standard deviation = 1 degree:

There is significant evidence (p-value = 0.024) that the standard deviation is not 1 degree.

Confidence interval for mean first temperature: (98.03, 98.59)

Test the hypothesis that mean first temperature = 98.6:

or, using t.test()

There is significant evidence (p-value = 0.042) that the true mean is not 98.6.

Q2 [2]

Rounding up, we conclude that at least n = 4 observations will be needed.

Q3 [2]

Rounding up, we conclude that at least n = 2401 observations will be needed. This calculation is based on the "worst case" where p = 0.5 and the variance of the estimate is maximized.

Q4 [2]

By trial and error, we see that at least 19 degrees of freedom are needed, that is, n = 19+1 = 20.

Q5 [8]

Q6 [3]

There are 12 positive differences out of 12 non-zero differences, so the exact p-value for the sign test is easily computed from the binomial distribution. The exact 2-sided p-value is 2*(0.5^12) = 0.000488. The result is very highly significant, so we reject the hypothesis.

The normal approximation to the binomial does not work very well in this example, but still gives a highly significant p-value.

Q7 [8]

The 2-sample t-test, assuming equal variances, is significant (p-value = 0.023), indicating that the means are not equal. Independence can't be tested. the sample sizes are too small to test normality but the box plots look reasonably symmetric so there is no reason to reject normality.

Assuming independence and normality, the assumption of equal variances (homoscedasticity) can be tested with an F-test.

There is no evidence (p-value = 0.81) from these data that the variances of the control and treatment groups are unequal.

Q8 [4]

The F statistic from the ANOVA is the square of the t statistic

The mean squared residual from the ANOVA is just the pooled variance estimate

Q9 [2]

The mean squared residual from the ANOVA, which is the same as the pooled variance estimate, is on 28 df, so the 95% confidence interval is (0.545, 1.584).

Q10 [8]

Lung function Group A, with FEV1/FVC < 75%, has a much higher reactivity than group B (75-84%) and group C (> 84%). Group A appears to have a larger variance than the others and the distributions appear to be positively skewed.

Taking logs (NOT REQUIRED) seems to over-correct; the three variances are more nearly equal but the distributions appear to be negatively skewed. The square-root transformation (NOT REQUIRED and not shown) seems to be about right.

There is significant evidence (p-value = 0.018) that mean bronchial reactivity is not the same in each lung-function group.

The 95% confidence interval for the residual variance s2 is (29.2, 107.7).

Q11 [10]

Assume that the observations are independent and normal, and that the variance of plasma epinephrine concentration is the same within each combination of Subject and Anesthetic.

The 95% confidence interval for the residual variance s2 is (0.0460, 0.5375).

If the study were to be done again, there should be more subjects. There should still be replication and a test of the interaction between Subject and Anesthesia because it is possible that with more subjects someone will respond differently to the different anesthesias.

Q12 [6]

Assume that the data are independent, that the true relationship is linear, and that the conditional distribution of duration, given age, is normal, with variance independent of age.

There is no evidence (p-value = 0.099) from these data that duration of stay depends on age. (Compare the corresponding analysis in Exercise #4 and note that removing the observation makes very little change to the results; the p-value is increased from 0.074 to 0.099.)

Q13 [12]

If you don't want to delete an extreme observation, taking logs can reduce its impact on the analysis. The intention here was that you would fit log-transformed data and not remove the 7th subject.

The ANOVA and the two plots indicate that a multiple linear regression model is reasonable, but does not give a significant improvement over the grand mean.

Since the instructions weren't clear, you could also analyze taking logs AND removing the 7th observation.

Removing the 7th subject as well as taking log(duration) makes age and first temperature even less significant.

We assume that the data are independent and normal and that the effects of age, first temperature and first white blood cell count are linear.

We conclude that the dependence of log(duration of stay) on age, first temperature and first white blood cell count is too weak to have any predictive power.

Q14 [4]

If we fit age after fitting first temperature the improvement in the fit is significant (p-value = 0.026) for predicting log(duration of stay) if the 7th observation is included, but (NOT REQUIRED) not if it is excluded.

Q15 [10]

Assume that the data are independent and normal and the variance in mcr does not depend on body weight.

Although the plot suggests that mcr levels out when body weight is greater than 350, the non-linearity is not significant (p-value = 0.079). Hence we can test the slope, which is very highly significant (p-value = 0.0001).

Interpolating on the fitted line, a body weight of 300 kg would predict a metabolic clearance rate of 142.1.

The 95% confidence interval for the residual variance is (310, 1644) if we assume a linear regression and use the MSR obtained after fitting the straight line. It is (213, 1343) if we do not assume a linear regression.

Q16 [4]

Statistics 2MA3

Lung function Group A, with FEV₁/FVC < 75%, has a much higher reactivity than group B (75-84%) and group C (> 84%). Group A appears to have a larger variance than the others and the distributions appear to be positively skewed.

The 95% confidence interval for the residual variance s² is (29.2, 107.7).

The 95% confidence interval for the residual variance s² is (0.0460, 0.5375).