6. Professional employees who work for large corporations often content that the

Question

6. Professional employees who work for large corporations often content that the mean salary paid by a company differs by location in Ontario. To test that claim, data were collected on financial analysts working for a large corporation at locations in Toronto and in Ottawa. The study found the following results: Locations Toronto Ottawa n1 = 25 T1 -$38, 348 2-$36,782 1-S2, 336 2-S2, 258 Assume that the populations are normal (a) Formulate the null and the alternative hypotheses to test whether the average salary in (b) Test the hypotheses from part (a) with -5% by computing the p-Value. Is there (c) Compute a 95% confidence interval for the difference between the average salary in Toronto is higher than in Ottawa. enough evidence that the average salary in Toronto is higher than in Ottawa? Toronto and Ottawa. Does the confidence interval indicate that the salary in Toronto is higher than in Ottawa on average?

Explanation / Answer

Given that,
mean(x)=38348
standard deviation , 1 =2336
number(n1)=25
y(mean)=36782
standard deviation, 2 =2258
number(n2)=18
null, Ho: u1 = u2
alternate, H1: 1 > u2
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=38348-36782/sqrt((5456896/25)+(5098564/18))
zo =2.21
| zo | =2.21
critical value
the value of |z | at los 0.05% is 1.645
we got |zo | =2.211 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail -Ha : ( p > 2.21 ) = 0.01351
hence value of p0.05 > 0.01351,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 > u2
test statistic: 2.21
critical value: 1.645
decision: reject Ho
p-value: 0.01351
evidence that avaerge salary in toranto is higher tha in ottawwa

PART C.
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 38348-36782) ±Z a/2 * Sqrt( 5456896/25+5098564/18)]
= [ (1566) ± Z a/2 * Sqrt( 501529.3956) ]
= [ (1566) ± 1.96 * Sqrt( 501529.3956) ]
= [177.9527 , 2954.0473]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [177.9527 , 2954.0473] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true mean
difference is zero

yes we have reason that avaerge salary in toranto is higher tha in ottawwa

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.