Respuesta :
Answer:
[tex]t=\frac{183-160}{\frac{12}{\sqrt{25}}}=9.58[/tex]
The degrees of freedom are given by:
[tex]df=n-1=25-1=24[/tex]
And the p value would be:
[tex]p_v =P(t_{(24)}>9.58)\approx 0[/tex]
Since the p value is very low at any significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 160
Step-by-step explanation:
Information provided
[tex]\bar X=183[/tex] represent the sample mean
[tex]s=12[/tex] represent the sample standard deviation
[tex]n=25[/tex] sample size
[tex]\mu_o =160[/tex] represent the value to test
[tex]\alpha=0.05[/tex] represent the significance level
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 greater than 160, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 160[/tex]
Alternative hypothesis:[tex]\mu > 160[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info we got:
[tex]t=\frac{183-160}{\frac{12}{\sqrt{25}}}=9.58[/tex]
The degrees of freedom are given by:
[tex]df=n-1=25-1=24[/tex]
And the p value would be:
[tex]p_v =P(t_{(24)}>9.58)\approx 0[/tex]
Since the p value is very low at any significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 160