Respuesta :
Answer:
[tex]t=\frac{370.4-300}{\frac{300.1}{\sqrt{30}}}=1.285[/tex]
[tex]p_v =P(z>1.285)=0.104[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X=370.4[/tex] represent the sample mean
[tex]s=300.1[/tex] represent the sample standard deviation for the sample
[tex]n=30[/tex] sample size
[tex]\mu_o =300[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
z 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 more than 300, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 300[/tex]
Alternative hypothesis:[tex]\mu > 300[/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".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{370.4-300}{\frac{300.1}{\sqrt{30}}}=1.285[/tex]
P-value
The degrees of freedom are given by:
[tex]df = n-1=30-1=29[/tex]
Since is a right tailed test the p value would be:
[tex]p_v =P(t_{29}>1.285)=0.104[/tex]
