Consider the question of whether taking vitamin C supplements helps to prevent c
ID: 3045533 • Letter: C
Question
Consider the question of whether taking vitamin C supplements helps to prevent colds. Starting in July, researchers assign 2000 volunteers to take a placebo pill with no vitamin C in it, and 2000 other people to take 500 mg of vitamin C. At the end of the main cold season (autumn and early winter), they survey the volunteers as to whether or not they had a cold that year. The results look like this:
Number of people who
got a cold
Number of people who
did not get a cold
Vitamin C
1200
800
No vitamin C
1250
750
a. Calculate the fraction of people with colds in each group.
b. Consulting the examples on page 94 for hints, calculate the 95% confidence
interval for the fraction of people getting colds in each group.
c. Comparing the confidence intervals for the two groups, what conclusion
do you reach about whether vitamin C helped prevent colds? Explain your
logic.
Number of people who
got a cold
Number of people who
did not get a cold
Vitamin C
1200
800
No vitamin C
1250
750
Explanation / Answer
a. Calculate the fraction of people with colds in each group.
people with placebo pill with vitamin C = 1200/2000 = 0.6
people with placebo pill with no vitamin C = 1250/2000 = 0.625
b. Consulting the examples on page 94 for hints, calculate the 95% confidence
interval for the fraction of people getting colds in each group.
with Vitamin C
given that,
possibile chances (x)=1200
sample size(n)=2000
success rate ( p )= x/n = 0.6
CI = confidence interval
confidence interval = [ 0.6 ± 1.96 * Sqrt ( (0.6*0.4) /2000) ) ]
= [0.6 - 1.96 * Sqrt ( (0.6*0.4) /2000) , 0.6 + 1.96 * Sqrt ( (0.6*0.4) /2000) ]
= [0.5785 , 0.6215]
------------------------------
with no Vitamin C
given that,
possibile chances (x)=1250
sample size(n)=2000
success rate ( p )= x/n = 0.625
CI = confidence interval
confidence interval = [ 0.625 ± 1.96 * Sqrt ( (0.625*0.375) /2000) ) ]
= [0.625 - 1.96 * Sqrt ( (0.625*0.375) /2000) , 0.625 + 1.96 * Sqrt ( (0.625*0.375) /2000) ]
= [0.6038 , 0.6462]
c. Comparing the confidence intervals for the two groups, what conclusion
do you reach about whether vitamin C helped prevent colds? Explain your
logic.
TRADITIONAL METHOD
given that,
sample one, x1 =1200, n1 =2000, p1= x1/n1=0.6
sample two, x2 =1250, n2 =2000, p2= x2/n2=0.625
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.6*0.4/2000) +(0.625 * 0.375/2000))
=0.0154
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0154
=0.0302
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.6-0.625) ±0.0302]
= [ -0.0552 , 0.0052]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =1200, n1 =2000, p1= x1/n1=0.6
sample two, x2 =1250, n2 =2000, p2= x2/n2=0.625
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.6-0.625) ± 1.96 * 0.0154]
= [ -0.0552 , 0.0052 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ -0.0552 , 0.0052] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean P1-P2
we have evidence that vitamin C helped the group
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