Solutions to S2MA3 Test #2

1998-03-31 (PM)

The stem-leaf plot wasn't specifically asked for, but it is the quickest way to sort the data for the box plots, and it is better than the box plots for assessing normality. Plotted side-by-side, the stem-leaf plots are as good as the box plots for comparing the two groups.

LHD                     Normal
Stem: hundreds          Stem: hundreds
 
2 85                    2 45,70,90
3 10,25,75              3 00,00,10,40,50,60
4 50,50,60,95           4 05
5 
6 15
7 60
 
x_bar1 =   452.5        x_bar2 =  317.0
s12    = 21495.8333     s22    = 2217.777
s1     =   146.6146     s2     =   47.093
 
U Outliers     760      U Outliers      
UH+1.5*HS      750      UH+1.5*HS     440
U Inlier       615      U Inlier      405
UH             495      UH            350
Median         450      Median        305
LH             325      LH            290
L Inlier       285      L Inlier      245
LH-1.5*HS       70      LH-1.5*HS      70
L Outliers              L Outliers 
HS             170      HS             60

It is easy to re-draw the box plots on a log scale, but the log transformation doesn't help to equalize the variances in this example.

F-TEST FOR EQUALITY OF VARIANCES (TWO-SIDED TEST)

F₀ = (21495.8333) / (2217.7777) = 9.6925; Reference Distribution is F(9,9).

The F tables have F(8,9) and F(12,9) but not F(9,9). The quick way to get the P-value here is to see that F_8,9,0.995 = 6.69 and F_12,9,0.995 = 6.23. Since F₀ = 9.69 is bigger than both of these, we know that the area to the right of F₀ under the F(9,9) density is less than 0.005, hence the two=sided P < 2 (0.005) = 0.01.

If we really wanted to, we could use interpolation to get

F_9,9,0.995 = 6.69 + 0.25 (6.23 - 6.69) = 6.58

and interpolate between F_8,9,0.999 = 10.37 and F_12,9,0.999 = 9.57 to get

F_9,9,0.999 = 10.37 + 0.25 (9.57 - 10.37) = 10.17;

then, since 6.58 < F₀ < 10.17 we have that the two-sided P-value lies between 0.01 and 0.002.

Either way, there is evidence (P < 0.01) that the variances are not equal, so we should not assume that s₁² = s₂².

t-TEST FOR EQUALITY OF MEANS

Unequal-variances t (page 272):

t₀ = 2.7825, d' = 10.8; since t_11,0.99 < t_10,0.99 = 2.764, we find P < 2 (0.01) = 0.02.

Equal-variances t (page 259):

s_p² = 108.88, on 18 degrees of freedom, and t₀ = 2.7825; t_18,0.99 = 2.552.

Hence P < 2 (0.01) = 0.02

Assumptions:

• Groups independent (probably OK, unless THW was used to diagnose LHD);
• Data independent within each group (no way to check this);
• Distribution is normal within each group (can't check with small samples, but looks OK by stem-leaf plots and box plots);
• Variances equal (only required for the equal-variances t-test; this assumption was tested by the F test and found false, but it did not prevent the equal-variances t-test from reaching the same conclusion as the unequal-variances test).

Conclusions:

• There is evidence (P < 0.002) that variance of THW is different in LHD and normal subjects; it is more variable in the LHD group.
• There is evidence (P < 0.02) that the mean THW is different in LHD and normal subjects; the mean THW is higher in the LHD group.

Since for Bin(6, 0.5) we have f(0) = f(6) = 0.0156, the two-sided test that is as close as possible to 5% is to reject when x = 0 or x = 6; the size of the test is then a = 2 (0.0156) = 0.0313.

Power = g(p₁) = P(X = 0 | p₁ ) + P(X = 6 | p₁) = (1 - p₁)⁶ + p₁⁶

The power curve shows that with n = 6, we do not have a good chance of detecting when p <> 0.5 unless p is about 0.1 or smaller, or about 0.9 or larger.

Q3

If it is true that the treatment is no different from the control, and this hypothesis is tested independently many times, then 5% of all the results will be Type I errors. That is, 5% of all the studies will lead to the false conclusion that the treatment is different from the control. But these will be only studies that get published, so anyone doing a literature search will find only evidence that the treatment is significantly different from the control.

If all the tests are two-sided, then on average half of the published results will show the treatment is better than the control and half will show that it is worse, so there is a chance that a thorough literature search will suggest that the treatment is not different from the control.

Q4

OR_hat = {(75)(73)}/{(53)(44)} = 2.34777 log(OR_hat) = 0.85347

Var_hat(log(OR_hat)) = 1/75 + 1/53 + 1/44 + 1/73 = 0.06863

z_0.995 = 2.576

99% CI for log(OR) = 0.85347 ± (2.576)sqrt(0.06863)

= 0.85347 ± 0.67483

= (0.17864, 1.52830)

99% CI for OR = (exp(0.17864), exp(1.52830)) = (1.1956, 4.6103)

Hence we are 99% confident that the odds of remission are between 20% and 361% higher with the experimental treatment than with the control treatment.

MARKING SCHEME

Q1

• At least one graph (box plots, stem-leaf plots, histograms) that compares the two groups; box plot is preferred. The two groups must appear side-by-side on the same scale. The graphs must be accurate. [4]
• F-test to compare variances [2]
• t-test; since they reach the same conclusion, either the equal-variances or unequal-variances forms may be used. [3]
• Assumptions with discussion [3]
• Conclusions [1]

Q2

• Binomial graph; must be accurate [2]
• Rejection region [2]
• Size [2]
• Derive power [4]
• Plot power; graph must be accurate [3]

Q3

• Discussion [5]

Q4

• Confidence interval [5]