Assume that you have a sample of n_1 = 5, with the sample mean X_1 = 50, and a s

Question

Assume that you have a sample of n_1 = 5, with the sample mean X_1 = 50, and a sample standard deviation of S_1 = 7, and you have an independent sample of n_2 = 6 from another population with a sample mean of X_2 = 39 and the sample standard deviation S_2 = 6. Assuming the population variances are equal, at the 0.01 level of significance, is there evidence that mu_1 > mu_2? Calculate the p-value. 1) None of the other answers 2) 0.0102 3) All of the other answers (except for "None of the other answers") 4) 3.9139 5) 2.8214 6) 2.8105

Explanation / Answer

Given that,
mean(x)=50
standard deviation , s.d1=7
number(n1)=5
y(mean)=39
standard deviation, s.d2 =6
number(n2)=6
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, a = 0.01
from standard normal table,right tailed t a/2 =2.821
since our test is right-tailed
reject Ho, if to > 2.821
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (4*49 + 5*36) / (11- 2 )
s^2 = 41.7778
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=50-39/sqrt((41.7778( 1 /5+ 1/6 ))
to=11/3.9139
to=2.8105
| to | =2.8105
critical value
the value of |t a| with (n1+n2-2) i.e 9 d.f is 2.821
we got |to| = 2.8105 & | t a | = 2.821
make decision
hence value of |to | < | t a | and here we do not reject Ho
p-value: right tail -ha : ( p > 2.8105 ) = 0.01018
hence value of p0.01 < 0.01018,here we do not reject Ho

ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2.8105
critical value: 2.821
decision: do not reject Ho
p-value: 0.01018 =0.0102

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