A personnel director in a particular state claims that the mean annual income is

Question

A personnel director in a particular state claims that the mean annual income is the same in one of the state's counties (County A) as it is in another county (County B). In County A, a random sample of 15 residents has a mean annual income of $41, 300 and a standard deviation of $9000. In County B, a random sample of 9 residents has a mean annual income of $38, 100 and a standard deviation of $5700 At alpha = 0.01. answer parts (a) through (e). Assume the population variances are not equal. Assume the samples are random and independent, and the populations are normally distributed.

Explanation / Answer

Given that,
mean(x)=41300
standard deviation , s.d1=9000
number(n1)=15
y(mean)=38100
standard deviation, s.d2 =5700
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =3.355
since our test is two-tailed
reject Ho, if to < -3.355 OR if to > 3.355
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =41300-38100/sqrt((81000000/15)+(32490000/9))
to =1.066
| to | =1.066
critical value
the value of |t | with min (n1-1, n2-1) i.e 8 d.f is 3.355
we got |to| = 1.06607 & | t | = 3.355
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.0661 ) = 0.318
hence value of p0.01 < 0.318,here we do not reject Ho

ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.066
critical value: -3.355 , 3.355
decision: do not reject Ho
p-value: 0.318

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