Golubitsky & Guillemin, an Ontario accounting firm, is trying to decide between IT training conducted in-house and the use of third-party consultants. To get some preliminary cost data, each type of training was implemented at two of the firm's offices located in Kingston and Windsor. The table below shows the average annual training cost per employee at each location. Are the mean costs significantly different? Note, that as the sample sizes are large, the decision makers assumed that the sample standard deviations for both samples most likely are approximately equal to the population standard deviations and thereforez

z

-distribution can be used for hypothesis testing. Use4%level of significance.

IT TrainingSample Mean

x

x

Population Std. Dev.

Sample Size(1) In-House49745117(2) Consultants5186898

Hint: Use the test statistic formula:z

=

(

x

1

x

2

)

(

1

2

)

2

1

n

1

+

2

2

n

2

z=(x1-x2)-(1-2)12n1+22n2

, (Note:

1

2

1-2

= 0)

(a) State the null and alternative hypotheses, and identify which one is the claim.

H

0

H0

:Select an answer

-

- x

x - x

x -

?

<

=

>

H

1

H1

:Select an answer

-

x -

- x

x - x

?

<

>

=

Which one is the claim:

• H
• 1
• H1
• H
• 0
• H0

For parts (b), (c) round your answers to 3 decimal places.

(b) What is the criticalz

z

-value?

(c) What is the test statistic (from the formula)?

(d) Is the null hypothesis rejected?

• No
• Yes

(e) Is the claim supported?

• At 4% significance level, there is not sufficient sample evidence to support the claim. There is no significant difference between mean costs of two types of training.
• At 4% significance level, there is sufficient sample evidence to support the claim. There is a significant difference between mean costs of two types of training.