Question

In a sample of 413 male college students the average number of energy drinks consumed per month was 2.49 with a standard deviation of 4.87 and in a sample of 382 female college students the average was 1.22 with a standard deviation of 3.24. Can you conclude that the mean number of energy drinks is greater for male students than for female students?

Answer 1

`t=\frac{(2.49-1.22)-0}{\sqrt{(4.87^2)/(413)+(1.22^2)/(382)}}}=5.129`

`p_v =P(t_(793)>5.129)=1.83x10^(-7)`

Comparing the p value with the significance level assumed is
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for males is significantly higher than the mean for female.

Explanation:

Data given and notation

`\bar X_(m)=2.49` represent the mean for the sample male

`\bar X_(f)=1.22` represent the mean for the sample female

`s_(m)=4.87` represent the sample standard deviation for the males

`s_(f)=3.24` represent the sample standard deviation for the females

`n_(m)=413` sample size for the group male

`n_(f)=382` sample size for the group female

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for males students is higher than the mean for female students, the system of hypothesis would be:

Null hypothesis:
`\mu_(m)-\mu_(f) \leq 0`

Alternative hypothesis:
`\mu_(m) - \mu_(f)> 0`

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

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

And the degrees of freedom are given by
`df=n_m +n_f -2=413+382-2=793`

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.

Calculate the statistic

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

`t=\frac{(2.49-1.22)-0}{\sqrt{(4.87^2)/(413)+(1.22^2)/(382)}}}=5.129`

What is the p-value?

Since is a right tailed test the p value would be:

`p_v =P(t_(793)>5.129)=1.83x10^(-7)`

Comparing the p value with the significance level assumed is
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for males is significantly higher than the mean for female.