Question

A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standard deviation of 3.72 days. Do these data provide sufficient evidence that ACME employees use more than 15 vacation days on average each year? (a) Write appropriate hypotheses for this test. (b) At the 5% level of significance, what would the conclusion be if the P-value for the test is 0.0392? (c) Explain the meaning of a Type I error in the context of the test

Answer 1

a) Null hypothesis:
`\mu \leq 15`

Alternative hypothesis:
`\mu > 15`

b)
`df=n-1=25-1=24`

For this case the p value is given
`p_v = 0.0392`

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 higher than 15 at 5% of signficance.

c) Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Explanation:

Data given and notation

`\bar X=18.3` represent the sample mean

`s=3.72` represent the sample standard deviation

`n=25` sample size

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

`\alpha=0.05` 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)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 15`

Alternative hypothesis:
`\mu > 15`

If we analyze the size for the sample is < 30 and 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".

Part b: P-value and conclusion

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

`df=n-1=25-1=24`

For this case the p value is given
`p_v = 0.0392`

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 higher than 15 at 5% of signficance.

Part c

Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.