Respuesta :
Answer:
Option D - [tex][0.12<\sigma<0.24][/tex]
Step-by-step explanation:
Given : The manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be s=0.16.
To find : The 95% confidence interval for the population standard deviation sigma?
Solution :
Number of sample n=19
The degree of freedom is Df=n-1=19-1=18
The standard deviation of the sample is s=0.16
Applying chi-square table to find critical value,
Upper critical value of [tex]\chi^2[/tex] is [tex]UC=\chi(\frac{0.05}{2},18) = 31.5264[/tex]
Lower critical value of [tex]\chi^2[/tex] is
[tex]LC=\chi(1-\frac{0.05}{2},18) = 8.2307[/tex]
Lower limit of the 95% confidence interval for the population variance
[tex]L=\frac{(df)\times (s^2)}{UC}[/tex]
[tex]L=\frac{18\times (0.16^2)}{31.5264}[/tex]
[tex]L=\frac{18\times0.0256}{31.5264}[/tex]
[tex]L=\frac{0.4608}{31.5264}[/tex]
[tex]L=0.0146[/tex]
Upper limit of the 95% confidence interval for the population variance
[tex]U=\frac{(df)\times(s^2)}{LC}[/tex]
[tex]U=\frac{18\times (0.16^2)}{8.2307}[/tex]
[tex]U=\frac{18\times0.0256}{8.2307}[/tex]
[tex]U=\frac{0.4608}{8.2307}[/tex]
[tex]U=0.0559[/tex]
So, The 95% confidence interval for the population variance is [0.0146, 0.0560]
Now, The 95% confidence interval for the population standard deviation is
[tex][\sqrt{0.0146}<\sigma<\sqrt{0.0560}][/tex]
[tex][0.1208<\sigma<0.2366][/tex]
or [tex][0.12<\sigma<0.24][/tex]
Therefore, Option D is correct.
The 95% confidence interval for the population standard deviation is [tex][0.12<\sigma<0.24][/tex]
