Respuesta :
Answer:
[tex]t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2[/tex]
The degrees of freedom are:
[tex] df = n--1=16-1=15[/tex]
Now since we are conducting a right tailed test the p value can be calculated as follows:
[tex]p_v =P(t_{15}>2)=0.0320[/tex]
Since the p value is lower than the significance level of [tex]\alpha=0.05[/tex] we have enough evidence to conclude that the true mean for the ages is significantly higher than 24
Step-by-step explanation:
Information provided
[tex]\bar X=25[/tex] represent the mean for the ages
[tex]s=2[/tex] represent the sample standard deviation
[tex]n=16[/tex] sample size
[tex]\mu_o =24[/tex] represent the value to check
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to verify
We want to check if the average age of all the students at the university is significantly more than 24, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 24[/tex]
Alternative hypothesis:[tex]\mu > 24[/tex]
Since we don't know the population deviation the statistic would be given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the data given we got:
[tex]t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2[/tex]
The degrees of freedom are:
[tex] df = n-1=16-1=15[/tex]
Now since we are conducting a right tailed test the p value can be calculated as follows:
[tex]p_v =P(t_{15}>2)=0.0320[/tex]
Since the p value is lower than the significance level of [tex]\alpha=0.05[/tex] we have enough evidence to conclude that the true mean for the ages is significantly higher than 24
