The null and alternate hypotheses are: A random sample of 12 observations from o
ID: 3362224 • Letter: T
Question
The null and alternate hypotheses are: A random sample of 12 observations from one population revealed a sample mean of 21 and a sample deviation of 4.5. A random sample of 4 observations from another population revealed a sample mean of 22 and a sample standard deviation of 3.2 At the 05 significance level, is there a difference between the population means? (a) State the decision rule. (Negative amounts should be indicated by a minus sign. Round your answer to 3 decimal places.) Thedecision rule is to rejectH0 1 rx or tol (b) Compute the pooled estimate of the population variance (Round your answer to 3 decimal places.) Pooled estimate of the population variance (c) Compute the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) Test statistic (d) State your decision about the null hypothesis Ho (Click to select)Explanation / Answer
Given that,
mean(x)=21
standard deviation , s.d1=4.5
number(n1)=12
y(mean)=22
standard deviation, s.d2 =3.2
number(n2)=4
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.145
since our test is two-tailed
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (11*20.25 + 3*10.24) / (16- 2 )
s^2 = 18.105
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=21-22/sqrt((18.105( 1 /12+ 1/4 ))
to=-1/2.4566
to=-0.4071
| to | =0.4071
critical value
the value of |t | with (n1+n2-2) i.e 14 d.f is 2.145
we got |to| = 0.4071 & | t | = 2.145
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -0.4071 ) = 0.6897
hence value of p0.05 < 0.6897,here we do not reject Ho
ANSWERS
---------------
reject Ho, if to < -2.145 OR if to > 2.145
s^2 = 18.105
test statistic: -0.4071
decision: do not reject Ho
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