Respuesta :
Answer:
Null hypothesis:[tex]\mu \geq 36[/tex]
Alternative hypothesis:[tex]\mu < 36[/tex]
[tex]p_v =P(t_{21}<-2.402)=0.0128[/tex]
Since the [tex]p_v< \alpha[/tex] we can reject the null hypothesis at 2.5 % of significance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=33.9[/tex] represent the sample mean
[tex]s=4.1[/tex] represent the standard deviation for the sample
[tex]n=22[/tex] sample size
[tex]\mu_o =36[/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 true mean less than 36, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 36[/tex]
Alternative hypothesis:[tex]\mu < 36[/tex]
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{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402[/tex]
Now we need to find the degrees of freedom for the t distirbution given by:
[tex]df=n-1=22-1=21[/tex]
What do you conclude? Use the p-value approach
Since is a one left tailed test the p value would be:
[tex]p_v =P(t_{21}<-2.402)=0.0128[/tex]
Since the [tex]p_v< \alpha[/tex] we can reject the null hypothesis at 2.5 % of significance.
