Question

A coffee shop claims that its fresh-brewed drinks have a mean caffeine content of 80 milligrams per 5 ounces. A city health agency believes that the coffee shop’s fresh- brewed drinks have higher caffeine content. To test this claim the health agency takes a random sample of 100 five-ounce servings and found the average mean caffeine content of the sample was 87 milligrams with standard deviation of 25 milligrams. Does this provide enough evidence at the 1% significance level to claim that the coffee shop’s fresh- brewed drinks have higher caffeine content?

Answer 1

`t=(87-80)/((25)/(√(100)))=2.8`

`p_v =P(t_((99))>2.8)=0.00307`

If we compare the p value and the significance level given
`\alpha=0.01` 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 significantly higher than 80 mg pr 5 ounces at 1% of signficance.

Explanation:

1) Data given and notation

`\bar X=87` represent the mean content for the sample

`s=25` represent the sample standard deviation

`n=100` sample size

`\mu_o =68` represent the value that we want to test

`\alpha=0.01` represent the significance level for the hypothesis test.

t 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 actually higher than 80, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 80`

Alternative hypothesis:
`\mu > 80`

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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=(87-80)/((25)/(√(100)))=2.8`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=100-1=99`

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

`p_v =P(t_((99))>2.8)=0.00307`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.01` 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 significantly higher than 80 mg pr 5 ounces at 1% of signficance.