Respuesta :
Answer:
a) Reject H0 if tcalc > 1.7960
b) [tex]t=\frac{5.42-5}{\frac{1.16}{\sqrt{12}}}=1.239[/tex]
c-1) ii. FALSE
c-2) ii.FALSE
c-3)i. TRUE
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=5.42[/tex] represent the mean time for the sample
[tex]s=1.16[/tex] represent the sample standard deviation for the sample
[tex]n=12[/tex] sample size
[tex]\mu_o =5[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
a) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less than 5 days, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 5[/tex]
Alternative hypothesis:[tex]\mu > 5[/tex]
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:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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".
Rejection zone
On this case we need a critical value that accumulates 0.05 of the area on the right tail. The degrees of freedom are given by 11. And we can use the following excel code to find the critical value : "T.INV(1-0.95,11)" and the critical value would be given by [tex]t_{\alpha/2}=1.795[/tex].
And the rejection zone is given by:
Reject H0 if tcalc > 1.7960
b) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{5.42-5}{\frac{1.16}{\sqrt{12}}}=1.239[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=12-1=11[/tex]
Since is a one side test the p value would be:
[tex]p_v =P(t_{(11)}>1.239)=0.121[/tex]
c-1. The null hypothesis should be rejected.
ii. FALSE
c-2. The average repair time is longer than 5 days.
ii. FALSE
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, and the true mean is not significantly higher than 5.
c-3 At α = .05 is the goal being met?
i. TRUE
We fail to reject the null hypothesis so then the goal is met.
