Question

Suppose that you used a 9-point rating scale and that you wanted to compare men who had an annual income over $50,000 (Group 1) with men who had an annual income less than or equal to $50,000 (Group 2) on their liking of a new product. If you studied 40 men in Group 1 and they have a mean of 7 and a standard deviation of 2.5, while the 35 men in Group 2 have a mean of 5 and a standard deviation of 1.4, what is the approximate value of t using the t-test?

Answer 1

`t=\frac{7-5}{\sqrt{(2.5^2)/(40)+(1.4^2)/(35)}}=4.34`

Explanation:

1) Data given and notation

`\bar X_(1)=7` represent the mean for group 1

`\bar X_(2)=5` represent the mean for group 2

`s_(1)=2.5` represent the sample standard deviation for the sample 1

`s_(2)=1.4` represent the sample standard deviation for the sample 2

`n_(1)=40` sample size for the group 1

`n_(2)=35` sample size for the group 2

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 are equal, the system of hypothesis would be:

H0:
`\mu_(1)=\mu_(2)`

H1:
`\mu_(1) \\eq \mu_(2)`

If we analyze the size for the samples both are higher than 30, but we don't know the population deviation, so for this case is better apply a t test to compare means, and the statistic is given by:

`t=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}` (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.

In order to calculate the mean and the sample deviation we can use the following formulas:

3) Calculate the statistic

We have all in order to replace in formula (1) like this:

`t=\frac{7-5}{\sqrt{(2.5^2)/(40)+(1.4^2)/(35)}}=4.34`