Review Questions 2 – for Final Exam - Solutions

  1. Peripheral neuropathy is a complication of uncontrolled diabetes. The number of cases of peripheral neuropathy among a control group of 350 diabetic patients was 122. Among a group of 115 patients who were taking an oral agent to prevent hyperglycemia, there were 35 cases of peripheral neuropathy. Is the proportion of patients with peripheral neuropathy comparable in both groups? Perform the test at a = 0.05.

Solution: For the control group

For the group taking an oral agent

The test statistic is

We will reject H0: p1 = p2 in favour of HA: p1 ≠ p2

if z < -z0.025 (-1.960) or z < z0.025 (1.960).

Hence H0 is accepted.

Construct 95% confidence intervals for the proportion of patients with peripheral neuropathy in the medication group and the proportion of patients in the control group.

Solution: 95% confidence limits for a proportion

For the control group: or 0.299 to 0.398

For the medicated group: or 0.220 to 0.388

Construct a 95% and a 99% confidence intervals for difference in the proportion of patients with peripheral neuropathy between the medicated group and the control group.

Solution: (1 – a)% confidence limits for a difference in proportions

95% limits: or -0.054 to 0.142

99% limits: or -0.084 to 0.173

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Review Questions 2 – for Final Exam - Solutions

  1. Researchers studied the association between birth mothers' smoking habits and the birth weights of their babies. Group 1 consisted of nonsmokers. Group 2 comprised smokers who smoked less than one pack of cigarettes per day. Group 3 smoked more than one but fewer than two packs per day. Group 4 smoked more than two packs per day.

Table: Birth Weights of Infants (n = 11 in Each Group) by Mother's Smoking Status
Group 1 / Group 2 / Group 3 / Group 4
3510 / 3444 / 2608 / 2232
3174 / 3111 / 2555 / 2331
3580 / 2890 / 3100 / 2200
3232 / 3002 / 1775 / 2121
3884 / 2995 / 2985 / 2001
3982 / 3101 / 2479 / 1566
4055 / 3400 / 2901 / 1676
3459 / 3764 / 2778 / 1783
3998 / 2997 / 2099 / 2002
3852 / 3031 / 2500 / 2118
3421 / 3120 / 2322 / 1882

a.  For each group

i.  Compute the summary statistics

1.  mean

2.  median

3.  standard deviation

4.  IQR

Group 1 / Group 2 / Group 3 / Group 4
mean / 3649.73 / 3168.64 / 2554.73 / 1992.00
median / 3580 / 3101 / 2555 / 2002
std. dev / 317.12 / 260.39 / 391.61 / 241.99
IQR / 561 / 403 / 579 / 417

ii.  Construct

1.  A stem-leaf diagram

  1. A box-whisker plot

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Review Questions 2 – for Final Exam

iii.  Compute 95% confidence limits for the mean of the group.

Solution: (1 – a)100 % confidence limits for the mean:

with d.f. = n – 1 = 10, t0.025 = 2.228 for 10 d.f.

iv.  Compute 95% confidence limits for the standard deviation of the group.

Solution: (1 – a)100 % confidence limits for the standard deviation:

with d.f. = n – 1 = 10, for 10 d.f.

b.  Carry out the Analysis of Variance F-test to determine if there is any significant difference in birth weight between the four groups. (Use a = 0.05 and a = 0.01)

Solution: The computations:

G = 125016, N = 44, ,

Thus SSBetween =

Thus SSWithin =

The Anova table

  1. A psychiatric epidemiology study collected information on the anxiety and depression levels of 11 subjects. The results of the investigation are presented in the following Table:

TABLE 12.13. Anxiety and Depression Scores of 11 Subjects
Subject ID / Anxiety Score / Depression Score
1 / 24 / 14
2 / 9 / 5
3 / 25 / 16
4 / 26 / 17
5 / 35 / 22
6 / 17 / 8
7 / 49 / 37
8 / 39 / 41
9 / 8 / 6
10 / 34 / 28
11 / 28 / 33
  1. Plot a scattergram of the data

  1. Compute Pearson’s correlation coefficient. r = 0.8937
  2. Compute Spearman’s Rank correlation coefficient. rS = 0.945
  3. Use Pearson’s correlation coefficient to test independence between the Anxiety score and the Depression score (Use a =0.05 and a = 0.01).

Solution:

test statistic ,

t0.025 = 2.262, t0.005 = 3.250 for 9 d.f.

Thus H0 is rejected for both a =0.05 and a = 0.01.

  1. Fit the least squares line for predicting the Depression score from the Anxiety score. (i.e. Estimate the slope and the intercept.)

Solution

n = 11, Sx = 294, Sx2 = 9398 , Sy = 227, Sy2 = 6273, Sxy = 7465.

S x x = 1540.18, Sy y = 1588.55 , Sxy = 1397.91

,

  1. Estimate the slope and the intercept of the least squares line with a 95% confidence interval.

Solution

Confidence Limits for the slope 0.564 to 1.251

Confidence Limits for the intercept -13.664 to 6.420

  1. Test (using a = 0.05) to determine if the slope of the regression line is zero.

Solution:

Test statistic = 5.976, t0.025 = 2,262 for 9 d.f., hence H0 is rejected.

  1. Using a 95% prediction interval predict the Depression score when the Anxiety score is 10, 20, 30, 40 and 50. Repeat these calculations using 99% prediction intervals.

The solution to these questions will be available a later today

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