Respuesta :
Answer:
a) Assuming 95% of confidence and [tex]\alpha=0.05[/tex] we can use the t distribution with 5 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(1-0.05,5)". And we got the critical value [tex]t_{\alpha/2}=2.015[/tex].
Since our calculates 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 highr than 3.5 hours.
b) [tex]p_v =P(t_{5}>0.204)=0.423[/tex] Â
If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly higher than 3.5 hours at 5% of significance. Â
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: Â
[tex]\bar X=3.507[/tex] represent the sample mean Â
[tex]s=0.084[/tex] represent the standard deviation for the sample
[tex]n=6[/tex] sample size Â
[tex]\mu_o =3.5[/tex] represent the value that we want to test Â
[tex]\alpha[/tex] represent the significance level for the hypothesis test. Â
t would represent the statistic (variable of interest) Â
[tex]p_v[/tex] represent the p value for the test (variable of interest) Â
State the null and alternative hypotheses to be tested Â
We need to conduct a hypothesis in order to determine if the mean producing a type of resin is higher than 3.5 hours, the system of hypothesis would be: Â
Null hypothesis:[tex]\mu \leq 3.5[/tex] Â
Alternative hypothesis:[tex]\mu > 3.5[/tex] Â
Compute the test statistic Â
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: Â
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1) Â
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". Â
We can replace in formula (1) the info given like this: Â
[tex]t=\frac{3.507-3.5}{\frac{0.084}{\sqrt{6}}}=0.204[/tex] Â
Now we need to find the degrees of freedom for the t distirbution given by:
[tex]df=n-1=6-1=5[/tex]
What do you conclude? Â
a. Use the critical value approach.
Assuming 95% of confidence and [tex]\alpha=0.05[/tex] we can use the t distribution with 5 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(1-0.05,5)". And we got the critical value [tex]t_{\alpha/2}=2.015[/tex].
Since our calculates 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 highr than 3.5 hours.
b. Use the p-value approach
Since is a one right tailed test the p value would be: Â
[tex]p_v =P(t_{5}>0.204)=0.423[/tex] Â
If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly higher than 3.5 hours at 5% of significance. Â
