Question

A researcher is interested in the relationship between oral contraceptive (OC) use and blood pressure in women. Suppose a sample of ten 35-to 39 year old nonpregnant, premenopausal OC users who work in a company and have a mean systolic blood pressure (SBP) of 132.86 mm Hg and sample standard deviation of 15.34 mm Hg are identified. A sample of 21 nonpregnant, premenopausal, non-OC users in the same age group are similarly identified who have mean SBP of 127. 44 mm Hg and a sample standard deviation of 18.23 mm Hg. A researcher assumes that the two groups are independent and normal will equal variances.

1. Using the p-value, does the underlying mean difference in blood pressure between the two groups differ from zero? Use the significance level α = 0.05.

2. Compute a 95% confidence interval for the true mean difference in systolic blood pressure (SBP) between 35 to 39 year old OC users and non-OC users. Interpret this result.

researcher assumes that the two groups are independent and normal will equal variances. 1. Using the p-value, does the underlying mean difference in blood pressure between the two groups differ from zero? Use the significance level a = 0.05. 2. Compute a 95% confidence interval for the true mean difference in systolic blood pressure (SBP) between 35 to 39 year old OC users and non-OC users. Interpret this result.

Answer 1

1.
Given that,
mean(x)=132.86
standard deviation , s.d1=15.34
number(n1)=10
y(mean)=127.44
standard deviation, s.d2 =18.23
number(n2)=21
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.045
since our test is two-tailed
reject Ho, if to < -2.045 OR if to > 2.045
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (9*235.316 + 20*332.333) / (31- 2 )
s^2 = 302.224
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=132.86-127.44/sqrt((302.224( 1 /10+ 1/21 ))
to=5.42/6.679
to=0.811
| to | =0.811
critical value
the value of |t α| with (n1+n2-2) i.e 29 d.f is 2.045
we got |to| = 0.811 & | t α | = 2.045
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 0.8115 ) = 0.4235
hence value of p0.05 < 0.4235,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.811
critical value: -2.045 , 2.045
decision: do not reject Ho
p-value: 0.4235
we do not have enough evidence to support the claim that underlying mean difference in blood pressure between the two groups.
2.
TRADITIONAL METHOD
given that,
mean(x)=132.86
standard deviation , s.d1=15.34
number(n1)=10
y(mean)=127.44
standard deviation, s.d2 =18.23
number(n2)=21
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (9*235.316 + 20*332.333) / (31- 2 )
s^2 = 302.224
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 302.224 * (1/10+1/21) )
=6.679
III.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and value of |t α| with (n1+n2-2) i.e 29 d.f is 2.045
margin of error = 2.045 * 6.679
= 13.659
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (132.86-127.44) ± 13.659 ]
= [-8.239 , 19.079]
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DIRECT METHOD
given that,
mean(x)=132.86
standard deviation , s.d1=15.34
sample size, n1=10
y(mean)=127.44
standard deviation, s.d2 =18.23
sample size,n2 =21
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 132.86-127.44) ± t a/2 * sqrt( 302.224 * (1/10+1/21) ]
= [ (5.42) ± 13.659 ]
= [-8.239 , 19.079]
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interpretations:
1. we are 95% sure that the interval [-8.239 , 19.079]contains the true population proportion
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 true population proportion