Respuesta :
Answer:
[tex]t=\frac{2567-2564}{\frac{11}{\sqrt{21}}}=1.250[/tex]
The degrees of freedom are given by:
[tex]df=n-1=21-1=20[/tex]
the p value for this case would be given by:
[tex]p_v =2*P(t_{(20)}>1.250)=0.2113[/tex]
For this case we see that the p value is higher than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 2564 mm
Step-by-step explanation:
Information given
[tex]\bar X=2567[/tex] represent the mean height for the sample
[tex]s=\sqrt{121}= 11[/tex] represent the sample standard deviation
[tex]n=21[/tex] sample size
[tex]\mu_o =2564[/tex] represent the value that we want to test
[tex]\alpha=0.1[/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 check if the true mean is equal to 2564 mm, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 2564[/tex]
Alternative hypothesis:[tex]\mu \neq 2564[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing we got:
[tex]t=\frac{2567-2564}{\frac{11}{\sqrt{21}}}=1.250[/tex]
The degrees of freedom are given by:
[tex]df=n-1=21-1=20[/tex]
the p value for this case would be given by:
[tex]p_v =2*P(t_{(20)}>1.250)=0.2113[/tex]
For this case we see that the p value is higher than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 2564 mm
