Question

The fabled "Freshmen 15" is unstoppable. College students are expected to put on an additional 15 pounds on average throughout the first year of their studies. Maybe it’s the excellent cafeteria food, or the special ingredient in the mystery meatloaf. A group of 9 high school friends are trying to buck this trend by proving that they can go through their first year in college without gaining 15 pounds. They instead gained an average of 10 pounds with a standard deviation of 6. Can they safely celebrate their conquest over the "Freshmen 15"? (Assume α=0.05) So should the 9 friends celebrate?

Answer 1

`t=(10-15)/((6)/(√(9)))=-2.5`

`p_v =P(t_8<-2.5)=0.018`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is lower than 15 at 5% of significance. So they should celebrate.

Explanation:

Data given and notation

`\bar X=10` represent the sample mean

`s=6` represent the sample standard deviation

`n=9` sample size

`\mu_o =15` represent the value that we want to test

`\alpha=0.05` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is lower than 15, the system of hypothesis are :

Null hypothesis:
`\mu \geq 15`

Alternative hypothesis:
`\mu < 15`

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

`t=(10-15)/((6)/(√(9)))=-2.5`

P-value

We calculate the degrees of freedom given by:

`df = n-1 = 9-1=8`

Since is a one-side lower test the p value would given by:

`p_v =P(t_8<-2.5)=0.018`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is lower than 15 at 5% of significance. So they should celebrate.