Respuesta :
Answer:
95% confidence interval for the true mean cholesterol content of all such eggs is [173.82 , 196.18].
Step-by-step explanation:
We are given that a laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185 milligrams with s = 17.6 milligrams.
Assuming the population has a normal distribution.
Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;
P.Q. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean amount of cholesterol = 185 milligrams
s = sample standard deviation = 17.6 milligrams
n = sample of chicken eggs = 12
[tex]\mu[/tex] = true mean
Here for constructing 95% confidence interval we have used t statistics because we don't know about population standard deviation.
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.201 < [tex]t_1_1[/tex] < 2.201) = 0.95 {As the critical value of t at 11 degree of
freedom are -2.201 & 2.201 with P = 2.5%}
P(-2.201 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.201) = 0.95
P( [tex]-2.201 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.201 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.201 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.201 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.201 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.201 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]185-2.201 \times {\frac{17.6}{\sqrt{12} } }[/tex] , [tex]185+2.201 \times {\frac{17.6}{\sqrt{12} } }[/tex] ]
= [173.82 , 196.18]
Therefore, 95% confidence interval for the true mean cholesterol content of all such eggs is [173.82 , 196.18].
