Respuesta :
Answer:
[tex] t = 2.24[/tex]
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=23-1=22[/tex]
Since is a one side right tailed test the p value would be:
[tex]p_v =P(t_{(22)}>2.24)=0.01776[/tex]
And for this case we can conclude that:
[tex] 0.01 < p_v < 0.025[/tex]
And we will reject the null hypothesis at [tex] \alpha=0.025[/tex] since [tex] p_v < \alpha[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X[/tex] represent the mean height for the sample
[tex]s[/tex] represent the sample standard deviation
[tex]n=23[/tex] sample size
[tex]\mu_o =15[/tex] represent the value that we want to 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.
We need to conduct a hypothesis in order to check if the mean is higher than 15, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 15[/tex]
Alternative hypothesis:[tex]\mu > 15[/tex]
If we analyze the size for the sample is > 30 but 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".
Calculate the statistic
For this case the statistic is given:
[tex] t = 2.24[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=23-1=22[/tex]
Since is a one side right tailed test the p value would be:
[tex]p_v =P(t_{(22)}>2.24)=0.01776[/tex]
And for this case we can conclude that:
[tex] 0.01 < p_v < 0.025[/tex]
And we will reject the null hypothesis at [tex] \alpha=0.025[/tex] since [tex] p_v < \alpha[/tex]