Respuesta :
Answer:
[tex]t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28[/tex]
The degrees of freedom are given by
[tex]df=n-1=17-1=16[/tex]
The p value would be given by this probability:
[tex]p_v =P(t_{(16)}<-4.28)=0.000287[/tex]
Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3
Step-by-step explanation:
Information given
[tex]\bar X=8.19 cm^3[/tex] represent the sample mean
[tex]s=0.8 cm^3[/tex] represent the sample deviation
[tex]n=17[/tex] sample size
[tex]\mu_o =9.02[/tex] represent the value to verify
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis to check
We want to check if the true mean is less than the normal value of 9.02 cm^3, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 9.02[/tex]
Alternative hypothesis:[tex]\mu < 9.02[/tex]
The statistic for this case 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{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28[/tex]
The degrees of freedom are given by
[tex]df=n-1=17-1=16[/tex]
The p value would be given by this probability:
[tex]p_v =P(t_{(16)}<-4.28)=0.000287[/tex]
Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3