Respuesta :
Answer:
A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].
Step-by-step explanation:
We are given the weights, in the ounces, of a sample of 12 boxes below;
Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.
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 weight = [tex]\frac{\sum X}{n}[/tex] = 21.88 ounces
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex] = 0.201 ounces
n = sample of boxes = 12
[tex]\mu[/tex] = population mean weight
Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.
So, 90% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-1.796 < [tex]t_1_1[/tex] < 1.796) = 0.90 {As the critical value of t at 11 degrees of
freedom are -1.796 & 1.796 with P = 5%}
P(-1.796 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.796) = 0.90
P( [tex]-1.796 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.796 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90
P( [tex]\bar X-1.796 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.796 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90
90% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.796 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+1.796 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }[/tex] , [tex]21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }[/tex] ]
= [21.78, 21.98]
Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].
