Please answer the following questions on confidence intervals: 1. A study is run
ID: 3061370 • Letter: P
Question
Please answer the following questions on confidence intervals: 1. A study is run to estimate the mean total cholesterol level in children 2-6 years of age. A sample of 9 participants is selected and their total cholesterol levels (in mg/dL) are measured as follows. marks) 185 225 240 196 175 180 194 147 223 a) Please generate a 99% confidence interval for the true mean total cholesterol levels in children. b) The proportion of children with cholesterol levels higher than 200 mg/dl, is 3 out of 9 in this sample. Please generate a 99% conf dence interval for the proportion of children with higher than 200 mg dl total cholesterol levels 2. A elinical trial is conducted to compare an experimental medication designed to er blood pressure to a placebo, and involves a total of 200 patients. Patients are enrolled and randomized to receive either the experimental medication or the placebo. The data shown below are data collected at the end of the study after 6 weeks on the assigned treatment. 6 marks) low Experimental n 100 Placebo (n 100 120.2 (15.4 Mean (Std. Dev) Systolie Blood Pressure % Hypertensive %with Sidel:neets 31.4 (18.9) 22% 8% 6% a) Please, generate a 95% confidence interval for the difference in mean systolic blood pressures between groups b) Please, generate a 95% confidence interval for the difference in proportions of patients with hypertension between groupsExplanation / Answer
Question 1:
a) let X : total cholesterol level in children.
(1-alpha)% confidence interval for the true mean total cholesterol level in children is
( xbar - (s/sqrt(n) * tn-1,alpha/2 , xbar + (s/sqrt(n) * tn-1,alpha/2)
( Since population standard deviation is unknown, we use one sample t-test)
n = 9 alpha =level of significance = 0.01
tn-1,alpha/2 = t8,0.005 = 2.896
xbar = sample mean = 1765/9 = 196.1111
s2 = sample variance = sum ( x-xbar) ^2 / (n-1) = 6728.88889/8 = 745.6543
s = sample s.d. = sqrt(745.6543)=27.3432
99% confidence interval for true mean.
( 196.1111- ( 27.3432/sqrt(9) * 2.896) , 196.1111 + ( 27.3432/sqrt(9) * 2.896))
= (169.7157, 222.5064)
b) p = propertion of children with higher than 200 mg/dl total cholesterol levels = 3/9 = 0.3333
q = 1 -p = 0.6667
pq/n = ( 0.3333*0.6667) / 9 = 0.02469
sqrt(pq/n) = 0.1571
alpha = level of signifiance = 0.01
Zalpha/2 = 2.58
99% Confidence interval for propertion of children with higher than 200 mg/dl total cholesterol levels
( p- sqrt(pq/n) * Zalpha/2 , p +sqrt(pq/n) * Zalpha/2)
= ( 0.3333 - ( 2.58 * 0.1571) , 0.3333 +( 2.58 *0.1571))
= (-0.0720 , 0.7386)
Queston 2.
a) From the information
n1= 100 ,n2 =100
xbar = 120.2 , ybar = 131.4
s1 = 15.4, s2 = 18.9
95% confidence interval for the difference in mean systolic blood pressure between groups.
( (xbar-ybar) - ( Zalpha/2) * sigmahat , (xbar-ybar) + ( Zalpha/2) * sigmahat)
Since sample size is lagre
sigmahat = sqrt (( n1* s12 + n2*s22) / (n1+n2))
sigmahat= sqrt(100 (15.42+18.92)/200) = 17.2390
alpha= level of significance = 0.05
Zalpha/2 = 1.96
95% confidence interval for the difference in mean systolic blood pressure between groups.
= ( ( 120.2-131.4) - ( 1.96 *17.2390) , ( 120.2-131.4) - ( 1.96 *17.2390) )
= ( -44.9884, 22.5884)
b) p1= 0.14, p2 = 0.22
95% confidence interval for the difference in propertion of patients with hypertension between groups
( (p1-p2) - sqrt ( phat *qhat ( 1/n1 + 1/n2) ) * Zalpha/2 , (p1-p2) - sqrt ( phat *qhat ( 1/n1 + 1/n2) ) * Zalpha/2)
phat = ( n1 *p1 + n2 *p2 ) / (n1+n2)
= ( 100 *0.14 +100*0.22) /(200)
= 0.18
qhat = 1-phat = 0.82
sqrt ( phat *qhat ( 1/n1 + 1/n2) ) = sqrt ( 0.18 *0.82 ( 1/100 +1/100) = 0.05433
Zalpha/2 = 1.96
95% confidence interval for the difference in propertion of patients with hypertension between groups
= ( (0.14-0.22) - ( 1.96 * 0.05433) , (0.14-0.22) +( 1.96 * 0.05433))
= ( -0.1864, 0.02648)
X (X-xabr) ^2 185 123.45679 225 834.567901 240 1926.23457 196 0.01234568 175 445.679012 180 259.567901 194 4.45679012 147 2411.90123 223 723.012346 1765 6728.88889