Respuesta :
Answer:
99% confidence interval for the mean time taken to execute the algorithm is (775.92 , 833.88).
Step-by-step explanation:
We are given that a random sample of 61 times are collected. The mean time in this sample is 804.9 seconds and the sample standard deviation is found to be 85.1.
Assuming data follows normal distribution.
So, firstly the pivotal quantity for 99% confidence interval for the population variance 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 time = 804.9 seconds
[tex]\mu[/tex] = population mean time
s = sample standard deviation = 85.1 seconds
n = sample size = 61
So, 99% confidence interval for population mean, [tex]\mu[/tex] is;
P(-2.66 < [tex]t_6_0[/tex] < 2.66) = 0.99 {As the t table at 60 degree of freedom
gives critical values of -2.66 & 2.66}
P(-2.66 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.66) = 0.99
P( [tex]-2.66 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.66 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
P( [tex]\bar X-2.66 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.66 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = ( [tex]\bar X-2.66 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X +2.66 \times {\frac{s}{\sqrt{n} } }[/tex] )
= ( [tex]804.9-2.66 \times {\frac{85.1}{\sqrt{61} } }[/tex] , [tex]804.9+2.66 \times {\frac{85.1}{\sqrt{61} } }[/tex] )
= (775.92 , 833.88)
Therefore, 99% confidence interval for the mean time taken to execute the algorithm is (775.92 , 833.88).
