Bio Statistics 1746 | Assignment

Running head: BIOSTATISTICS 1746
BIOSTATISTICS 1746
Name of the student
Name of the university

2BIOSTATISTICS 1746

3BIOSTATISTICS 1746
Table of Contents
Q1...............................................................................................................................................3
Q2...............................................................................................................................................4
Q3...............................................................................................................................................4
Q4...............................................................................................................................................5
Q5...............................................................................................................................................5
Q6...............................................................................................................................................6
Q7...............................................................................................................................................7
Q8...............................................................................................................................................8
Q9...............................................................................................................................................9
References:...............................................................................................................................10

4BIOSTATISTICS 1746
Q1.
Range IQR
Drug Sample
size
Mean Standard
deviation
Minimum P25 Median P75 Maximum (Max-
Min)
P75-
P25
0 18 66 10.9 46.6 58.3 65.7 72 91.2 44.6 13.7
1 23 49.5 8.1 28.5 44.9 47.9 56.1 63 34.5 11.2
Code:
by drug, sort : summarize painintensityscoremm, detail
99% 63 63 Kurtosis 3.167746
95% 60.6 60.6 Skewness -.4108626
90% 59.2 59.2 Variance 65.41364
75% 56.1 58.6
Largest Std. Dev. 8.08787
50% 47.9 Mean 49.5
25% 44.9 41.5 Sum of Wgt. 23
10% 41 41 Obs 23
5% 40.6 40.6
1% 28.5 28.5
Percentiles Smallest
painintensityscoremm
-> drug = 1
99% 91.2 91.2 Kurtosis 2.8825
95% 91.2 79.2 Skewness .3049078
90% 79.2 75.3 Variance 120.0706
75% 72 74.7
Largest Std. Dev. 10.95767
50% 65.75 Mean 66
25% 58.3 55.4 Sum of Wgt. 18
10% 52.6 53.5 Obs 18
5% 46.6 52.6
1% 46.6 46.6
Percentiles Smallest
painintensityscoremm
-> drug = 0

5BIOSTATISTICS 1746
Q2.
The subgroup of a normally distributed sample is also normally distributed.
Q3.
Estimated mean difference between pain intensity score between the new drug and the
placebo is 16.5 and the associated 95% CI for the difference is 10.20 to 22.80.
This means that there is a 95 % chance that the mean difference between pain intensity scores
for the patients who were given the placebo and the new drug lies between 10.20 to 22.80 in
the overall sample.
The appropriate test to be carried out for investigating the difference is the two group mean
comparison test which is done in stata. The test proceeds by considering the null hypothesis
that there is no mean difference between the two groups whereas the alternative hypothesis
states that a difference exists.
The t stat for the test was found to be 5.3 with associated probability value much less than
0.05 and it suggests that the difference exists between the mean of the two groups.
Code:
. ttest painintensityscoremm, by(drug) unequal

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Pr(T < t) = 1.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 0.0000
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Ho: diff = 0 Satterthwaite's degrees of freedom = 30.3266
diff = mean(0) - mean(1) t = 5.3492
diff 16.5 3.084584 10.20328 22.79672
combined 41 56.7439 1.948946 12.47934 52.80494 60.68287
1 23 49.5 1.686437 8.08787 46.00254 52.99746
0 18 66 2.582748 10.95767 60.55088 71.44912
Group Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]
Two-sample t test with unequal variances
Q4.
The confidence interval indicates that the difference in effectiveness between the new drug
and the placebo lies between 10.20 and 22.80.
Q5.
Proportion of the participants in the new drug group who experienced side effects: 50 %
Proportion of participants in the placebo group who experienced side effects: 30.43 %
60.98 39.02 100.00
Total 25 16 41
69.57 30.43 100.00
1 16 7 23
50.00 50.00 100.00
0 9 9 18
drug 0 1 Total
side_effects
Code: tabulate drug sideeffects, row

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Yes it is possible to calculate the confidence intervals of the proportions of participants who
experienced side effects in each drug group using the normal distribution because the overall
population can be assumed to be approximately normal.
Q6.
The difference in proportion of participants with a 95% CI that experienced side effects
between the new drug group and the placebo group is 0.196.
95% ci for the difference in population proportion between the two groups lies between -0.10
and .49.
This means that there is a 95% chance that the difference in the proportion of participants that
experienced side effects in the two groups lies between -0.10 and 0.49.
Pr(Z < z) = 0.8988 Pr(|Z| < |z|) = 0.2025 Pr(Z > z) = 0.1012
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Ho: diff = 0
diff = prop(0) - prop(1) z = 1.2745
under Ho: .1535104 1.27 0.202
diff .1956522 .1519675 -.1021986 .493503
1 .3043478 .0959439 .1163013 .4923944
0 .5 .1178511 .269016 .730984
Variable Mean Std. Err. z P>|z| [95% Conf. Interval]
1: Number of obs = 23
Two-sample test of proportions 0: Number of obs = 18
. prtest side_effects, by(drug)

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Q7.
A two sample t test is done to test the claim that mean log systolic blood pressure is different
between participants with healthy bmi ( coded as 1) and those with bmi in the
overweight/obese range ( coded 0) in examination period coded as Period = 1.
The null hypothesis is that there is no difference between the means of the two groups. And
the alternative hypothesis is that there is a difference between the two groups.
The df for the test is 4413. And the test statistic is t = 18.56.
The p value comes out to be much less than 0.05 and the null hypothesis can be rejected to
accept the claim that the log systolic blood pressure is different for the two groups.
Pr(T < t) = 1.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 0.0000
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Ho: diff = 0 degrees of freedom = 4413
diff = mean(0) - mean(1) t = 18.5600
diff .0860778 .0046378 .0769854 .0951702
combined 4415 4.87646 .002396 .1592063 4.871762 4.881157
1 1993 4.829239 .0033602 .1500081 4.822649 4.835829
0 2422 4.915317 .0031708 .1560489 4.909099 4.921535
Group Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]
Two-sample t test with equal variances
-> period = 1
Code: by period, sort : ttest logsys, by(healthybmi)

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Q8.
Proportion of participants in period 1 who are categorised as atrisk : 29.82 %
63.28 36.72 100.00
Total 7,357 4,270 11,627
55.84 44.16 100.00
3 1,822 1,441 3,263
61.65 38.35 100.00
2 2,423 1,507 3,930
70.18 29.82 100.00
1 3,112 1,322 4,434
period 0 1 Total
atrisk
Code: tabulate period atrisk, row
Q9.
(b)
A two sample t test is done to test if there is a difference between mean systolic blood
pressure between period 1 and period 3.
The null hypothesis is that there is no difference between the mean systolic blood pressure of
the two groups. And the alternative hypothesis states that there is a difference.
The df for the test is 7037.05 and test statistic is t = -14.6473.
The p value comes out to be much less than 0.05 and therefore the null hypothesis can be
rejected and the alternative hypothesis is accepted.

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Pr(T < t) = 0.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 1.0000
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Ho: diff = 0 Satterthwaite's degrees of freedom = 7037.05
diff = mean(period1logsys) - mean(period3logsys) t = -14.6473
diff -.0538188 .0036743 -.0610215 -.046616
combined 7696 4.899322 .0018415 .1615477 4.895712 4.902932
p~3log~s 3262 4.93033 .0027858 .1591056 4.924868 4.935792
p~1log~s 4434 4.876511 .0023959 .1595356 4.871814 4.881208
Variable Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]
Two-sample t test with unequal variances
Code: ttest period1logsys == period3logsys, unpaired unequal

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References:
Acock, A.C., 2016. A gentle introduction to Stata. Stata press.
Kohler, U. and Kreuter, F., 2014. Data analysis using Stata. Stata press.