Respuesta :
Answer:
[tex] t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74[/tex]
The degrees of freedom are given by:
[tex] df =n-1= 25-1=24[/tex]
Now we can calculate the p value with the following probability:
[tex] p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}[/tex]
And for this case since the p value is lower compared to the significance level [tex]\alpha=0.05[/tex] we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05
Step-by-step explanation:
For this case we have the following info given:
[tex] \bar X = 39.5[/tex] represent the sample mean
[tex] s =6.6[/tex] represent the sample deviation
[tex]\mu = 30.6[/tex] represent the reference value to test.
[tex] n = 25[/tex] represent the sample size selected
The statistic for this case is given by:
[tex] t =\frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74[/tex]
The degrees of freedom are given by:
[tex] df =n-1= 25-1=24[/tex]
Now we can calculate the p value with the following probability:
[tex] p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}[/tex]
And for this case since the p value is lower compared to the significance level [tex]\alpha=0.05[/tex] we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05
