Question

The Corona beer company produces bottles that hold 12 ounces of liquid. Periodically, the company gets complaints that their bottles are not holding enough liquid. To test this claim, the bottling company randomly samples 64 bottles and finds the average amount of liquid held by the bottles is 11.9155 ounces with a standard deviation of 0.40 ounce. Suppose the p-value of this test is 0.0455. State the proper conclusion.

Answer 1

`df=n-1=64-1=63`

Since is a one side test the p value would be:

`p_v =P(t_((63))<t_(calc))=0.0455`

If we compare the p value and the significance level most commonly used
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significantly lower than 12 at 5% of signficance. But be careful at 1% of significance the
`p_v >\alpha` and on this case we FAIL to reject the null hypothesis.

Explanation:

Data given and notation

`\bar X=11.9155` represent the sample mean

`s=0.4` represent the sample standard deviation

`n=64` sample size

`\mu_o =12`represent the value that we want to test

`\alpha` 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 true mean is less than 12 ounces, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 12`

Alternative hypothesis:
`\mu < 12`

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 = t_(calc)`

P-value

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

`df=n-1=64-1=63`

Since is a one side test the p value would be:

`p_v =P(t_((63))<t_(calc))=0.0455`

Conclusion

If we compare the p value and the significance level most commonly used
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significantly lower than 12 at 5% of signficance. But be careful at 1% of significance the
`p_v >\alpha` and on this case we FAIL to reject the null hypothesis.