Question

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.

Answer 1

Answer:

$t=\frac{500-490}{\sqrt{\frac{48^2}{180}+\frac{32^2}{210}}}}=2.379$  

$p_v =2*P(t_{388}>2.379)=0.0178$

Comparing the p value with the significance level provided $\alpha=0.05$ we see that $p_v<\alpha$ 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

$\bar X_{I}=490$ represent the mean for the sample of In-house

$\bar X_{C}=500$ represent the mean for the sample Consutants

$s_{I}=32$ represent the population standard deviation for the sample In-house

$s_{C}=48$ represent the population standard deviation for the sample Consultants

$n_{I}=210$ sample size for the group In-house

$n_{C}=180$ 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:$\mu_{I}=\mu_{C}$

Alternative hypothesis:$\mu_{I} \neq \mu_{C}$

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:

$t=\frac{\bar X_{C}-\bar X_{I}}{\sqrt{\frac{s^2_{C}}{n_{C}}+\frac{s^2_{I}}{n_{I}}}}$ (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:

$t=\frac{500-490}{\sqrt{\frac{48^2}{180}+\frac{32^2}{210}}}}=2.379$  

4) Statistical decision

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

$df=n_{C}+n_{I}-2=210+180-2=388[/tex[

The sample is large enough to assume that the distribution is also normal.

Since is a bilateral test the p value would be:

[tex]p_v =2*P(t_{388}>2.379)=0.0178$

Comparing the p value with the significance level provided $\alpha=0.05$ we see that $p_v<\alpha$ 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.