Respuesta :
Answer:
[tex]t=\frac{203.8-221.4}{\frac{21.6}{\sqrt{100}}}=-8.148[/tex]
The degrees of freedom are given by:
[tex]df=n-1=100-1=99[/tex]
and the p value would be given by:
[tex]p_v =P(t_{(99)}<-8.148) \approx 0[/tex]
For this case 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 less than 221.4 the value of study.
Step-by-step explanation:
Information given
[tex]\bar X=203.8[/tex] represent the sample mean
[tex]s=21.6[/tex] represent the sample standard deviation
[tex]n=100[/tex] sample size
[tex]\mu_o =221.4[/tex] represent the value to verify
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test
Hypothesis to test
We want to verify if the true mean is lower than 221.4, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 221.4[/tex]
Alternative hypothesis:[tex]\mu < 221.4[/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{203.8-221.4}{\frac{21.6}{\sqrt{100}}}=-8.148[/tex]
The degrees of freedom are given by:
[tex]df=n-1=100-1=99[/tex]
and the p value would be given by:
[tex]p_v =P(t_{(99)}<-8.148) \approx 0[/tex]
For this case 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 less than 221.4 the value of study.