Respuesta :
Answer:
Hypothesis
Null hypothesis:[tex]\mu = 12[/tex]
Alternative hypothesis:[tex]\mu \neq 12[/tex]
Statistic
[tex]t=\frac{11.82-12}{\frac{0.38}{\sqrt{36}}}=-2.84[/tex]
[tex]df=n-1=36-1=35[/tex]
P value
The p value for this case would be given by:
[tex]p_v =2*P(t_{(35)}<-2.84)=0.0075[/tex]
Conclusion
Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 12 oz.
Step-by-step explanation:
Information given
[tex]\bar X=11.82[/tex] represent the sample mean
[tex]s=0.38[/tex] represent the sample standard deviation
[tex]n=36[/tex] sample size
[tex]\mu_o =12[/tex] represent the value to verify
[tex]\alpha=0.01[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value for the test
Hypothesis to verify
We want to cehck if the true mean is 12 or not, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 12[/tex]
Alternative hypothesis:[tex]\mu \neq 12[/tex]
Statistic
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]t=\frac{11.82-12}{\frac{0.38}{\sqrt{36}}}=-2.84[/tex]
The degrees of freedom are given by:
[tex]df=n-1=36-1=35[/tex]
P value
The p value for this case would be given by:
[tex]p_v =2*P(t_{(35)}<-2.84)=0.0075[/tex]
Conclusion
Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 12 oz.
