Respuesta :
Answer:
a) Reject H0 if [tex]t_{cal}>1.796[/tex]
b) [tex]t=\frac{5.417-5}{\frac{1.165}{\sqrt{12}}}=1.2404[/tex]
c-1) FALSE. Since our calculated value < critical value. We fail to reject the null hypothesis, and we can say that at 5% of significance we don't have enough evidence to conclude that the true mean is higher than 5 days.
c-2) FALSE (see explanation below)
c-3) Yes (see explanation below)
Step-by-step explanation:
Data given and notation
We can calculate the sample mean and standard deviation with these formulas:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
The results obtained are given below:
[tex]\bar X=5.417[/tex] represent the sample mean
[tex]s=1.165[/tex] represent the 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[/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)
State the null and alternative hypotheses to be tested
We need to conduct a hypothesis in order to determine if the mean is higher than 5 days, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 5[/tex]
Alternative hypothesis:[tex]\mu > 5[/tex]
Now we need to find the degrees of freedom for the t distirbution given by:
[tex]df=n-1=12-1=11[/tex]
What do you conclude?
a. Use the critical value approach.
Assuming 95% of confidence and [tex]\alpha=0.05[/tex] we can use the t distribution with 11 degrees of freedom in order to calculate a critical value that accumulates 0.05 of the area on the right tail of the distribution. We can use excel and the code to do this is given by: "=T.INV(1-0.05,11)". And we got the critical value [tex]t_{\alpha/2}=1.796[/tex].
The rejection zone would be on this case
Reject H0 if [tex]t_{cal}>1.796[/tex]
b) Compute the test statistic
We don't know the population deviation, so for this case 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".
We can replace in formula (1) the info given like this:
[tex]t=\frac{5.417-5}{\frac{1.165}{\sqrt{12}}}=1.2404[/tex]
c-1) The null hypothesis should be rejected.
Since our calculated value < critical value. We fail to reject the null hypothesis, and we can say that at 5% of significance we don't have enough evidence to conclude that the true mean is higher than 5 days.
c-2) The average repair time is longer than 5 days.
FALSE, we FAIL to reject the null hypothesis that the mean is lower or equal then 5 days, so then the alternative hypothesis that the mean is higher than 5 days not makes sense.
c-3) At[tex] \alpha = .05[/tex] is the goal being met?
Since the goal is given by:
"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". On this case Yes, since we fail to reject the null hypothesis.
