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 0.01 significance level for both parts. Male BMI Female BMI μ μ1 μ2 n 45 45 x 27.3958 24.7599 s 7.837628 4.750044 a. Test the claim that males and females have the same mean body mass index (BMI). What are the null and alternative hypotheses? A. H0: μ1 = μ2 H1: μ1 > μ2 B. H0: μ1 ≠ μ2 H1: μ1 < μ2 C. H0: μ1 = μ2 H1: μ1 ≠ μ2 D. H0: μ1 ≥ μ2 H1: μ1 < μ2 The test statistic, t, is---------- . (Round to two decimal places as needed.) The P-value is---------- . (Round to three decimal places as needed.) State the conclusion for the test. A. Fail to reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI. B. Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI. C. Fail to reject the null hypothesis. There is not sufficient evidence to warrant the rejection of the claim that men and women have the same mean BMI. D. Reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI. b. Construct a confidence interval suitable for testing the claim that males and females have the same mean BMI. ---------< μ1 − μ2 <----------(Round to three decimal places as needed.) Does the confidence interval support the conclusion of the test? (1)---------- because the confidence interval contains (2)----------- (1) Yes, No, (2) only positive values. only negative values. zero.

Answer 1

a.
Given that,
mean(x)=27.3958
standard deviation , s.d1=7.837628
number(n1)=45
y(mean)=24.7599
standard deviation, s.d2 =4.750044
number(n2)=45
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =2.692
since our test is two-tailed
reject Ho, if to < -2.692 OR if to > 2.692
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =27.3958-24.7599/sqrt((61.42841/45)+(22.56292/45))
to =1.93
| to | =1.93
critical value
the value of |t α| with min (n1-1, n2-1) i.e 44 d.f is 2.692
we got |to| = 1.92938 & | t α | = 2.692
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 != 1.9294 ) = 0.06
hence value of p0.01 < 0.06,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.93
critical value: -2.692 , 2.692
decision: do not reject Ho
p-value: 0.06
we do not have enough evidence to support the claim that men and women have the same mean BMI.
b.
TRADITIONAL METHOD
given that,
mean(x)=27.3958
standard deviation , s.d1=7.837628
number(n1)=45
y(mean)=24.7599
standard deviation, s.d2 =4.750044
number(n2)=45
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((61.428/45)+(22.563/45))
= 1.366
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.01
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 44 d.f is 2.692
margin of error = 2.692 * 1.366
= 3.678
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (27.3958-24.7599) ± 3.678 ]
= [-1.042 , 6.314]
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DIRECT METHOD
given that,
mean(x)=27.3958
standard deviation , s.d1=7.837628
sample size, n1=45
y(mean)=24.7599
standard deviation, s.d2 =4.750044
sample size,n2 =45
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 = [( 27.3958-24.7599) ± t a/2 * sqrt((61.428/45)+(22.563/45)]
= [ (2.636) ± t a/2 * 1.366]
= [-1.042 , 6.314]
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interpretations:
1. we are 99% sure that the interval [-1.042 , 6.314] 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
yes,
confidence interval support the conclusion of the test because the confidence interval contains positive and negative values.