Given in the table are the BMI statistics for random samples of men and women. A

Question

Given in the table are the BMI statistics for random samples of men and women. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a .01 significance level for both parts.

what are the null and alternative hypotheses?
test statistic, t?
p value?
state the conclusion?
confidence interval?

_

x

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x

Explanation / Answer

a.
Given that,
mean(x)=27.3955
standard deviation , s.d1=7.764582
number(n1)=40
y(mean)=25.3183
standard deviation, s.d2 =5.170436
number(n2)=40
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.708
since our test is two-tailed
reject Ho, if to < -2.708 OR if to > 2.708
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =27.3955-25.3183/sqrt((60.28873/40)+(26.73341/40))
to =1.408
| to | =1.408
critical value
the value of |t | with min (n1-1, n2-1) i.e 39 d.f is 2.708
we got |to| = 1.40829 & | t | = 2.708
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.4083 ) = 0.167
hence value of p0.01 < 0.167,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.408
critical value: -2.708 , 2.708
decision: do not reject Ho
p-value: 0.167
we do not have enough evidence to support the claim

b.
TRADITIONAL METHOD
given that,
mean(x)=27.3955
standard deviation , s.d1=7.764582
number(n1)=40
y(mean)=25.3183
standard deviation, s.d2 =5.170436
number(n2)=40
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((60.289/40)+(26.733/40))
= 1.475
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 39 d.f is 2.708
margin of error = 2.708 * 1.475
= 3.994
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (27.3955-25.3183) ± 3.994 ]
= [-1.917 , 6.071]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=27.3955
standard deviation , s.d1=7.764582
sample size, n1=40
y(mean)=25.3183
standard deviation, s.d2 =5.170436
sample size,n2 =40
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 27.3955-25.3183) ± t a/2 * sqrt((60.289/40)+(26.733/40)]
= [ (2.077) ± t a/2 * 1.475]
= [-1.917 , 6.071]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-1.917 , 6.071] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion

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