Respuesta :
Answer:
95% confidence interval for the true population mean for the amount of soda served is [12.37 , 14.23].
Step-by-step explanation:
We are given that quality control specialist for a restaurant chain takes a random sample of size 13 to check the amount of soda served in the 16 oz. serving size.
The sample mean is 13.30 with a sample standard deviation of 1.54.
Firstly, the pivotal quantity for 95% confidence interval for the true 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 = 13.30
s = sample standard deviation = 1.54
n = sample size = 13
[tex]\mu[/tex] = true population mean
Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.179 < [tex]t_1_2[/tex] < 2.179) = 0.95 {As the critical value of t at 12 degree of
freedom are -2.179 & 2.179 with P = 2.5%}
P(-2.179 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.179) = 0.95
P( [tex]-2.179 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.179 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.179 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.179 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.179 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.179 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]13.30-2.179 \times {\frac{1.54}{\sqrt{13} } }[/tex] , [tex]13.30+2.179 \times {\frac{1.54}{\sqrt{13} } }[/tex] ]
= [12.37 , 14.23]
Therefore, 95% confidence interval for the true population mean for the amount of soda served is [12.37 , 14.23].
