
Respuesta :
Answer:
[tex]t_{calculated}=2.445[/tex]
[tex]p_v =P(t_{(41)}>2.445)=0.0094[/tex] Â
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis
Step-by-step explanation:
Data given and notation Â
[tex]\bar X[/tex] represent the sample mean
[tex]s[/tex] represent the sample standard deviation for the sample Â
[tex]n=42[/tex] sample size Â
[tex]\mu_o [/tex] represent the value that we want to test
[tex]\alpha=0.05[/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. Â
We need to conduct a hypothesis in order to check if the mean is lower than specified value, the system of hypothesis would be: Â
Null hypothesis:[tex]\mu \geq \mu_o[/tex] Â
Alternative hypothesis:[tex]\mu < \mu_o[/tex] Â
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: Â
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] Â (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
Calculate the statistic
The statistic for this case is given: Â
[tex]t_{calculated}=2.445[/tex]
P-value
The first step is calculate the degrees of freedom, on this case: Â
[tex]df=n-1=42-1=41[/tex] Â
Since is a one side test the p value would be: Â
[tex]p_v =P(t_{(41)}>2.445)=0.0094[/tex] Â
Conclusion Â
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis