A study was done using a treatment group and a placebo group. The results are sh
ID: 3429355 • Letter: A
Question
A study was done using a treatment group and a placebo group. The results are shown in the table. 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 0.05 significance level for both parts.
Treatment
Placebo
?
?1
?2
n
32
38
x
2.38
2.65
s
0.92
0.55
a.?Test the claim that the two samples are from populations with the same mean.
What are the null and alternative hypotheses?
A. H0:?1<?2 H1: ?1??2
B. H0: ?1=?2 H1: ?1>?2
C. H0: ?1=?2 H1: ?1??2
D. H0: ?1??2 H1: ?1<?2
The test statistic, t, is_____
(Round to two decimal places as needed.)
The P-value is _____
(Round to three decimal places as needed.)
Treatment
Placebo
?
?1
?2
n
32
38
x
2.38
2.65
s
0.92
0.55
a.?Test the claim that the two samples are from populations with the same mean.
What are the null and alternative hypotheses?
A. H0:?1<?2 H1: ?1??2
B. H0: ?1=?2 H1: ?1>?2
C. H0: ?1=?2 H1: ?1??2
D. H0: ?1??2 H1: ?1<?2
The test statistic, t, is_____
(Round to two decimal places as needed.)
The P-value is _____
(Round to three decimal places as needed.)
Explanation / Answer
C. H0: mu1=mu2 H1: mu1 not equal to mu2
-----------------------------------------------------------------------------------------------------------------
t=(xbar1-xbar2)/sqrt(s1^2/n1+s2^2/n2)
=(2.38-2.65)/sqrt(0.92^2/32+0.55^2/38)
=-1.46
-----------------------------------------------------------------------------------------------------------------
It is a two-tailed test.
Th degree of freedom =n1+n2-2=32+38-2=68
So the p-value = 2*P(t with df=68 <-1.46) = 0.149 (from student t table)
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