Respuesta :
Answer:
[tex]t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.771[/tex]
The degrees of freedom are given by:
[tex]df=n-1=8-1=7[/tex]
And the p value is given by this probability:
[tex]p_v =P(t_{(7)}<-3.771)=0.00348[/tex]
Since the p value is lower than the significance level provided of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 16 ounces
Step-by-step explanation:
Information provided
[tex]\bar X=15.6[/tex] represent the sample mean
[tex]s=0.3[/tex] represent the sample standard deviation
[tex]n=8[/tex] sample size
[tex]\mu_o =16[/tex] represent the value to verify
[tex]\alpha=0.1[/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 amount of coffee in a container is less than 16 ounces when filled, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 16[/tex]
Alternative hypothesis:[tex]\mu < 16[/tex]
The statistic for this case would be given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.771[/tex]
The degrees of freedom are given by:
[tex]df=n-1=8-1=7[/tex]
And the p value is given by this probability:
[tex]p_v =P(t_{(7)}<-3.771)=0.00348[/tex]
Since the p value is lower than the significance level provided of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 16 ounces
