Answer :
Answer:
a) Null hypothesis:[tex]\mu \leq 76[/tex]
Alternative hypothesis:[tex]\mu > 76[/tex]
b) [tex]t=\frac{77.5-76}{\frac{3.3}{\sqrt{29}}}=2.45[/tex]
c) [tex]p_v =P(t_{(28)}>2.45)=0.0104[/tex]
d) Fail to reject H0. There is not enough evidence to support the claim.
Step-by-step explanation:
Data given and notation
[tex]\bar X=77.5[/tex] represent the sample mean
[tex]s=3.3[/tex] represent the sample standard deviation
[tex]n=29[/tex] sample size
[tex]\mu_o =76[/tex] represent the value that we want to test
[tex]\alpha=0.01[/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 population mean is higher than 76, the system of hypothesis are :
Null hypothesis:[tex]\mu \leq 76[/tex]
Alternative hypothesis:[tex]\mu > 76[/tex]
Since we don't know the population deviation, 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".
b) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{77.5-76}{\frac{3.3}{\sqrt{29}}}=2.45[/tex]
c) P-value
We need to calculate the degrees of freedom first given by:
[tex]df=n-1=29-1=28[/tex]
Since is a one-side right tailed test the p value would given by:
[tex]p_v =P(t_{(28)}>2.45)=0.0104[/tex]
We can use the following code in excel to find the answer: "=1-T.DIST(2.45,28,TRUE)"
d) Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
The best option on this case:
Fail to reject H0. There is not enough evidence to support the claim.