Respuesta :
Answer:
A 98% confidence interval for the mean assembly time is [21.34, 26.49] .
Step-by-step explanation:
We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.
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 time = 23.92 minutes
s = sample standard deviation = 6.72 minutes
n = sample of times = 40
[tex]\mu[/tex] = population mean assembly time
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, a 98% confidence interval for the population mean, [tex]\mu[/tex] is;
P(-2.426 < [tex]t_3_9[/tex] < 2.426) = 0.98 {As the critical value of z at 1% level
of significance are -2.426 & 2.426}
P(-2.426 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.426) = 0.98
P( [tex]-2.426 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.426 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.98
P( [tex]\bar X-2.426 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.426 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.98
98% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.426 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.426 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]23.92-2.426 \times {\frac{6.72}{\sqrt{40} } }[/tex] , [tex]23.92+2.426 \times {\frac{6.72}{\sqrt{40} } }[/tex] ]
= [21.34, 26.49]
Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .
