Respuesta :
Answer:
[tex]\chi^2 =\frac{60-1}{64} 36 =33.19[/tex]
The degrees of freedom are:
[tex]df= n-1= 60-1= 59[/tex]
And the p value would be given by:
[tex]p_v =2*P(\chi^2 >33.19)=0.0053[/tex]
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 deviation is different from 8 days.
Step-by-step explanation:
Data given
[tex]n=60[/tex] represent the sample size
[tex]\alpha=0.01[/tex] represent the confidence level
[tex]s^2 =6^2 =36 [/tex] represent the sample variance obtained
[tex]\sigma^2_0 =8^=64[/tex] represent the value that we want to test
Hypothesis to test
We want to verify if the true deviation is equal to 8 days o not, so the system of hypothesis would be:
Null Hypothesis: [tex]\sigma^2 = 64[/tex]
Alternative hypothesis: [tex]\sigma^2 \neq 64[/tex]
The statistic is given by:
[tex]\chi^2 =\frac{n-1}{\sigma^2_0} s^2[/tex]
Replacing we got:
[tex]\chi^2 =\frac{60-1}{64} 36 =33.19[/tex]
The degrees of freedom are:
[tex]df= n-1= 60-1= 59[/tex]
And the p value would be given by:
[tex]p_v =2*P(\chi^2 >33.19)=0.0053[/tex]
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 deviation is different from 8 days.
