Question

In: Statistics and Probability

The national mean annual salary for a school administrator is $90,000 a year. A school official...

The national mean annual salary for a school administrator is $90,000 a year. A school official took a sample of 25 school administrators in the state of Ohio to learn about salaries in that state to see if they differed from the national average. Click on the datafile logo to reference the data. (a) Choose the hypotheses that can be used to determine whether the population mean annual administrator salary in Ohio differs from the national mean of $90,000. H0: Ha: (b) The sample data for 25 Ohio administrators is contained in the file named Administrator. What is the p value for your hypothesis test in part (a)? If required, round your answer to four decimal places. Do not round your intermediate calculations. 0.0213

Solutions

Expert Solution

Assumed data,
sample mean, x =89000
standard deviation, s =2030 because not given in the data,
Given that,
population mean(u)=90000
number (n)=25
null, Ho: μ=90000
alternate, H1: μ!=90000
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.064
since our test is two-tailed
reject Ho, if to < -2.064 OR if to > 2.064
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =89000-90000/(2030/sqrt(25))
to =-2.4631
| to | =2.4631
critical value
the value of |t α| with n-1 = 24 d.f is 2.064
we got |to| =2.4631 & | t α | =2.064
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.4631 ) = 0.0213
hence value of p0.05 > 0.0213,here we reject Ho
ANSWERS
---------------
a.
null, Ho: μ=90000
alternate, H1: μ!=90000
test statistic: -2.4631
critical value: -2.064 , 2.064
decision: reject Ho
b.
p-value: 0.0213
we have enough evidence to support the claim that whether the population mean annual administrator salary in Ohio differs from the national mean of $90,000.


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