Question

An article reports that in a sample of 413 male college students, the average number of energy drinks consumed per month was 2.48 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 3.23. Can you conclude that the mean number of energy drinks is greater for male students than for female students? Find the P-value and state a conclusion.

Answer 1

t= 0.367

pv=0.357

We don't have enough evidence to reject the null hypothesis, and we don't have an evidence to ocnclude that the mean number of energy drinks is greater for male students than for female students at 10% of significance.

Explanation:

1) Data given and notation

`\bar X_(M)=2.48` 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 sample Male

`s_(F)=3.23` represent the sample standard deviation for the sample Female

`n_(M)=413` sample size for the group Stick

`n_(F)=382` sample size for the group Liquid

t would represent the statistic

`p_v` represent the p value

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean number of energy drinks for male students is greater then for female students, the system of hypothesis would be:

Null Hypothesis:
`\mu_(M)\leq \mu_(F)`

Alternative Hypothesis:
`\mu_(M) > \mu_(F)`

If we analyze the size for the samples both are greater than 30, but we don't know the population deviations for male and female, 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)}{\sqrt{(s^2_(M))/(n_(M))+(s^2_(F))/(n_(F))}}` (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

We can replace in formula (1) like this:

`t=\frac{2.48-1.22}{\sqrt{(4.87^2)/(413)+(3.23^2)/(382)}}}=0.367`

4) Statistical decision

For this case we don't have a significance level provided
`\alpha`, but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

`df=n_(M)+n_(F)-2=413+382-2=793`, that would be approximately normal

Since is a one tailed test and if we look the alternative hypothesis, the the p value would be:

`p_v =P(t_((793))>0.367)=0.357`

So the p value is a very high value and using any significance level for example
`\alpha=0.05, 0,1,0.15` always
`p_v>\alpha` so we can conclude that we don't have enough evidence to reject the null hypothesis, and we don't have an evidence to ocnclude that the mean number of energy drinks is greater for male students than for female students at 10% of significance.