Treatment Placebo placebo group. The results are s hown in the table Assume that

Question

Treatment Placebo placebo group. The results are s hown in the table Assume that the two samples are 1 2 er group and a p "e independent deviations are equal ole ag dom samples selected from normalily distitbuted populations, and do not assume that the population standard a 0 10 significance level for both parts 26 36 | 2.37 | 2.67 S 0.98 0 0.62 a. Test the claim that the two samples are from populations with the same mean What are the null and alternative hypotheses? 41112 2 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. O A. Reject the null hypothesis. There is sufficient evidence to warrant rejectio n of the claim that the two samples are from populations with the same mean. O B. Falil to reject the nul hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with thes mean. O c. Fai toreject the nul hypothesis. There is sumicient evidence to warant rejection of the claim that the two O D. Reject the nul hypothesis. There is not sumicient evidence to warrant rejection of the claim that the two samples are from populations with n of the claim that the two samples are from populations with the same mea b. Construct a confidence interval suitable for testing the claim that the two samples are from populations with the same mean as needed.) (Round to three decimal places

Explanation / Answer

Given that,
mean(x)=2.37
standard deviation , s.d1=0.98
number(n1)=26
y(mean)=2.67
standard deviation, s.d2 =0.62
number(n2)=36
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.1
from standard normal table, two tailed t /2 =1.708
since our test is two-tailed
reject Ho, if to < -1.708 OR if to > 1.708
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =2.37-2.67/sqrt((0.9604/26)+(0.3844/36))
to =-1.3748
| to | =1.3748
critical value
the value of |t | with min (n1-1, n2-1) i.e 25 d.f is 1.708
we got |to| = 1.37481 & | t | = 1.708
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.3748 ) = 0.181
hence value of p0.1 < 0.181,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.37
critical value: -1.708 , 1.708
decision: do not reject Ho
p-value: 0.181
option C
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 = [( 2.37-2.67) ± t a/2 * sqrt((0.96/26)+(0.384/36)]
= [ (-0.3) ± t a/2 * 0.218]
= [-0.673 , 0.073]
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interpretations:
1. we are 90% sure that the interval [-0.673 , 0.073] 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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