(a) The value of test statistic is 2.6795.
(b) The critical value is 2.1578 and p value is 0.0062.
(c) We conclude that if we have enough evidence to support the claim that first population has a large mean.
Given that,
mean(x)=12
standard deviation, sd(1)=3
number n(1)=20
mean y=10
standard deviation, sd(2)=1.5
number n(2)=21
null, H(0): μ1-μ2 = 0
Alternate, H(1): μ1-μ2 > 0
We use test statistic (t) = [tex](x-y)/\sqrt{(sd(1)^2/n(1))+(sd(2)^2/n(2))}[/tex]
t(0) =[tex]\frac{12-10}{\sqrt{9/20+2.25/21}}[/tex]
t(0) =2.6795
t(0) =2.6795
critical value
level of significance, alpha = 0.02
degrees of freedom (df) = [tex]\frac{sd(1) ^2/n(1)+sd(2) ^2/n(2))^2}{(s(1)^4/n(1)^2(n(1)-1) + (s(2)^4/n(2)^2(n(2)-1)}[/tex]
df = [tex]\frac{((3^2/20)+(1.5^2/21))^2}{(3^4/20^2(20-1))+(1.5^4/21^2(21-1))}[/tex]
df = 27.6364
df = 27(approx)
From standard normal table,right tailed t(a/2) =2.1578
Since our test is right-tailed,
Reject H(0), if t(0) > 2.1578
we got t(0) = 2.67946 and |t(a)| = 2.1578
make decision
Hence value of |t(0)| > | t(a)| and here we reject H(0)
p-value: right tail H(a): (p > 2.6795) = 0.0062
Hence value of p(0.02) > 0.0062, here we reject H(0)
H0: μ1-μ2 = 0
H1: μ1-μ2 > 0
test statistic: 2.6795
b. We have to find the t critical value for a significance level of 0.02 for an alternative hypothesis that the first population has a larger mean.
critical value: 2.1578
p value: 0.0062
c. We have to find the conclusion.
We have enough evidence to support the claim that first population has a large mean.
To learn more about null hypothesis link is here
brainly.com/question/28920252
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