An accounting firm is trying to decide between it training conducted in-house and the use of the third party consultants. to get some preliminary cost data, each type of training implemented at two of the firm’s offices located in different cities. the table below shows the average annual training cost per employee at each location. are the mean costs significantly different? write appropriate hypotheses, find the test statistic and the p-value and draw a conclusion at the 0.05 level of significance.

Comparing the p value with the significance level provided we see that so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average of the two groups.  

Step-by-step explanation:

Assuming the following info

IT Training     N   mean Stdev

In-house       210 $490 $32

Consultants  180 $500 $48

1) Data given and notation

represent the mean for the sample of In-house

represent the mean for the sample Consutants

represent the population standard deviation for the sample In-house

represent the population standard deviation for the sample Consultants

sample size for the group In-house

sample size for the group Consultants

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null hypothesis:

Alternative hypothesis:

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

(1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

With the info given we can replace in formula (1) like this:

4) Statistical decision

First we need to calculate the degrees of freedom given by:

Comparing the p value with the significance level provided we see that so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average of the two groups.  

median- 36f

mode - none

mean- 33.7 ~ 34f