Respuesta :
Answer:
A 98% confidence interval for the population mean of printouts per ink cartridge is [430.5, 489.5].
Step-by-step explanation:
We are given that a consumer advocate group selects a random sample of 20 ink cartridges and finds that the average number of printouts per ink cartridge is 460 with a standard deviation of 52.
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 average number of printouts per ink cartridge = 460
s = sample standard deviation = 52
n = sample of ink cartridges = 20
[tex]\mu[/tex] = population mean printouts per ink cartridge
Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.
So, 98% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.54 < [tex]t_1_9[/tex] < 2.54) = 0.98 {As the critical value of t at 19 degrees of
freedom are -2.54 & 2.54 with P = 1%}
P(-2.54 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.54) = 0.98
P( [tex]-2.54 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\bar X-\mu}[/tex] < [tex]2.54 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.98
P( [tex]\bar X-2.54 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.54 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.98
98% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.54 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.54 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]460-2.54 \times {\frac{52}{\sqrt{20} } }[/tex] , [tex]460+2.54 \times {\frac{52}{\sqrt{20} } }[/tex] ]
= [430.5, 489.5]
Therefore, a 98% confidence interval for the population mean of printouts per ink cartridge is [430.5, 489.5].
