The solution to problem 2 d) . Page 3 of 3 Problem 2 A study was performed to as
ID: 3313162 • Letter: T
Question
The solution to problem 2d).
Page 3 of 3 Problem 2 A study was performed to assess the effect of metformin treatment during pregnancy (in women with PCOS) on offspring growth. Each woman either received metformin treatment or placebo treatment and the treatments were assigned randomly. One of the growth variables considered was the BMI z-score at 48 months. count, mean and standard deviation for the BMI z-score for the metformin and placebo groups are given in the table below: Metformin treatnt-81.49 Placebo treatment np 790.031 CountMean Standard deviation Pu=1.185 SP 1.057 Assume that BMI z-scores are normally distributed. a) What test can be used to test whether metformin treatment affects the variance of BMI z-scores? b) Perform the proposed test in a) at the 5% significance level. It is given that Fao, 0.64 and Feraums = 1.50. Let d' and be the population variances for the metformin and placebo groups, respectively. c) Use the probability statement P( to find a 95% confidence interval for the ratio of variances / test can be used to test whether metformin treatment affecets the mean of BMI d) What e) Perform the proposed test in d) at the 5% significance level. It is given that ti f) Find a 95% confidence interval for the rnean metformin effect. z-scores? 1 58.0975 = 1.98.Explanation / Answer
(d) Here the t - test hypothesis testing for two independent samples with equal variance would be most proper to the test whether metforin treatment affects the mean of BMI z - scores.
(e) Here pooled standard deviation sp = sqrt [{(n1 -1)s12 + (n2 -1)s22 }/(n1 + n2 -2)]
sp = sqrt [(80 * 1.1852 + 78 * 1.0572)/ 158] = 1.1236
t = (xm - xp)/ [sp * sqrt (1/n1 + 1/n2)]
t = (0.491 - 0.031)/ [1.1236 * sqrt (1/81 + 1/79)]
t = 0.46 / 0.1777
t = 2.5886
here t158,0.975 = 1.98 < t (2.5886)
so we shall reject the null hypothesis and can conclude that tthe metaformin treatment affects the mean of BMI z - scores.
(f) 95% confidenceinterval for the mean metformin effect.
95% CI = xm +- t80, 0.975 seM
t80, 0.975 = 1.99 and seM = sM / sqrt(N) = 1.185/ sqrt(81) = 0.13167
Here,
95% CI = xm +- t80, 0.975 seM
= 0.491 - 1.99 * 0.13167
= ( 0.229, 0.753)