Question

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought in to the time repairs are completed. A random sample of 12 repair records showed the following repair times (in days): 9, 2, 5, 1, 5, 4, 7, 5, 11, 3, 7, 2. Click here for the Excel Data File (a) H0: μ ≤ 5 days versus H1: μ > 5 days. At α = .05, choose the right option. Reject H0 if tcalc > 1.796 Reject H0 if tcalc < 1.796 a bCalculate the Test statistic.

(c-1) The null hypothesis should be rejected.

TRUE
FALSE
(c-2) The average repair time is longer than 5 days.

TRUE
FALSE
(c-3) At \alpha = .05 is the goal being met?

Yes
No

Answer 1

c-1) The null hypothesis should be rejected.

FALSE

c-2) The average repair time is longer than 5 days.

FALSE

c-3) At \alpha = .05 is the goal being met?

YES we fail to reject the null hypothesis

Explanation:

Data given and notation

`\bar X=5.08` represent the mean for the sample

`s=2.999` represent the sample standard deviation for the sample

`n=12` sample size

`\mu_o =5` 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is actually higher than 5, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 5`

Alternative hypothesis:
`\mu > 5`

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=(5.083-5)/((2.999)/(√(12)))=0.0958`

P-value

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

`df=n-1=12-1=11`

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

`p_v =P(t_((11))>0.0958)=0.4627`

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 FAIL reject the null hypothesis, so we can conclude that the population mean is not significantly higher than 5.

c-1) The null hypothesis should be rejected.

FALSE

c-2) The average repair time is longer than 5 days.

FALSE

c-3) At \alpha = .05 is the goal being met?

YES we fail to reject the null hypothesis