Respuesta :
Answer:
[tex]p_v =P(t_{(59)}>t_{calc})<0.0001[/tex]
Since the p value is a very low value we have enough evidence to reject the null hypothesis in favor of the alternative hypothesis and we can conclude that the true mean for this case is significantly different from 98.6 F at any usual significance level used.
Step-by-step explanation:
Information given
[tex]\bar X=98.2[/tex] represent the sample mean
[tex]\sigma=0.62[/tex] represent the population standard deviation
[tex]n=60[/tex] sample size
[tex]\mu_o =98.6[/tex] represent the value that we want to test
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to verify if the true mean is equal to 98.6 F, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 98.6[/tex]
Alternative hypothesis:[tex]\mu \neq 98.6[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
The degrees of freedom, on this case:
[tex]df=n-1=60-1=59[/tex]
The p value would be given by:
[tex]p_v =P(t_{(59)}>t_{calc})<0.0001[/tex]
Since the p value is a very low value we have enough evidence to reject the null hypothesis in favor of the alternative hypothesis and we can conclude that the true mean for this case is significantly different from 98.6 F at any usual significance level used.
