Respuesta :
Answer:
[tex]t=\frac{19.9-19}{\frac{4.5}{\sqrt{45}}}=1.342[/tex]
The degrees of freedom are given by:
[tex] df=n-1= 45-1=44[/tex]
And the p value would be;
[tex]p_v =P(t_{44}>1.342)=0.0932[/tex]
Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 19 mph
Step-by-step explanation:
Information given
[tex]\bar X=19.9[/tex] represent the sample mean
[tex]s=4.5[/tex] represent the sample standard deviation
[tex]n=45[/tex] sample size
[tex]\mu_o =19[/tex] represent the value to verify
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to verify if the true mean is higher than 19, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 19[/tex]
Alternative hypothesis:[tex]\mu > 19[/tex]
The statistic would be given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]t=\frac{19.9-19}{\frac{4.5}{\sqrt{45}}}=1.342[/tex]
The degrees of freedom are given by:
[tex] df=n-1= 45-1=44[/tex]
And the p value would be;
[tex]p_v =P(t_{44}>1.342)=0.0932[/tex]
Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 19 mph