Respuesta :
Answer:
[tex]t=\frac{27.1-27.4}{\frac{7.3}{\sqrt{4934}}}=-2.887[/tex]
The degrees of freedom are given by:
[tex] df= n-1 = 4934-1= 4933[/tex]
Then the p value for this case calculated as:
[tex]p_v =P(t_{4933}<-2.887) =0.002[/tex]
Since the p value is a very lower value using any significance level for example 1% or 5% we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significanctly less than 27.4. So then is not anything wrong with the conclusion
Step-by-step explanation:
Information provided
[tex]\bar X=27.1[/tex] represent the sample mean
[tex]s=7.3[/tex] represent the sample standard deviation
[tex]n=4934[/tex] sample size
[tex]\mu_o =27.4[/tex] represent the value to test
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to verify if the mean age of full-time students did decline (less than 27.4), the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 27.4[/tex]
Alternative hypothesis:[tex]\mu < 27.4[/tex]
The statistic for this case is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info we got:
[tex]t=\frac{27.1-27.4}{\frac{7.3}{\sqrt{4934}}}=-2.887[/tex]
The degrees of freedom are given by:
[tex] df= n-1 = 4934-1= 4933[/tex]
Then the p value for this case calculated as:
[tex]p_v =P(t_{4933}<-2.887) =0.002[/tex]
Since the p value is a very lower value using any significance level for example 1% or 5% we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significanctly less than 27.4. So then is not anything wrong with the conclusion