Respuesta :
Answer:
A 99.9% confidence interval for the population mean is [22.31, 59.91] .
Step-by-step explanation:
We are given the following sample values;
X = 45.3, 42.3, 53, 49, 15.2, 52.3, 45.6, 39.6, 39.4, 16.1, 54.4.
Firstly, the pivotal quantity for finding the confidence interval for the population 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 = [tex]\frac{\sum X}{n}[/tex] = 41.11
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X - \bar X)^{2} }{n-1} }[/tex] = 13.59
n = sample size = 11
Here for constructing a 99.9% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.
So, 99.9% confidence interval for the population mean, [tex]\mu[/tex] is;
P(-4.587 < [tex]t_1_0[/tex] < 4.587) = 0.999 {As the critical value of t at 10 degrees of
freedom are -4.587 & 4.587 with P = 0.05%}
P(-4.587 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 4.587) = 0.999
P( [tex]-4.587 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}{[/tex] < [tex]4.587 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.999
P( [tex]\bar X-4.587 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+4.587 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.999
99.9% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-4.587 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+4.587 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]41.11-4.587 \times {\frac{13.59}{\sqrt{11} } }[/tex] , [tex]41.11+4.587 \times {\frac{13.59}{\sqrt{11} } }[/tex] ]
= [22.31, 59.91]
Therefore, a 99.9% confidence interval for the population mean is [22.31, 59.91] .
