Assume that both populations are normally distributed a) Test whether 1 > 2 at t

Question

Assume that both populations are normally distributed a) Test whether 1 > 2 at the = 0.10 level of significance for the given sample data b) Construct a 90% confidence interval about 1-u2 Sample 2 39 11.1 Sample 1 47.9 8.3 Click the icon to view the Student t-distribution table a) Perform a hypothesis test. Determine the null and alternative hypotheses. Determine the test statistio. t(Round to two decimal places as needed.) Determine the critical value(s). Select the correct choice below and fill in the answer box(es) within your choice. Round to three decimal places as needed.) A. The critical value is B. The lower critical value is . The upper critical value is Click to select your answer(s).

Explanation / Answer

Given that,
mean(x)=47.9
standard deviation , s.d1=8.3
number(n1)=27
y(mean)=39
standard deviation, s.d2 =11.1
number(n2)=22
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.1
from standard normal table,right tailed t /2 =1.323
since our test is right-tailed
reject Ho, if to > 1.323
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =47.9-39/sqrt((68.89/27)+(123.21/22))
to =3.12
| to | =3.12
critical value
the value of |t | with min (n1-1, n2-1) i.e 21 d.f is 1.323
we got |to| = 3.11716 & | t | = 1.323
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 3.1172 ) = 0.00261
hence value of p0.1 > 0.00261,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 > u2
b.
test statistic: 3.12
c.
critical value: 1.323
d.
decision: reject Ho, outside
p-value: 0.00261

PART B.
given that,
mean(x)=47.9
standard deviation , s.d1=8.3
sample size, n1=27
y(mean)=39
standard deviation, s.d2 =11.1
sample size,n2 =22
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 = [( 47.9-39) ± t a/2 * sqrt((68.89/27)+(123.21/22)]
= [ (8.9) ± t a/2 * 2.855]
= [3.99 , 13.81]
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interpretations:
1. we are 90% sure that the interval [3.986 , 13.814] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion

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