Respuesta :
Answer:
95% confidence interval for the mean length of sentencing for this crime is [47.53 months , 54.47 months].
Step-by-step explanation:
We are given that in a random sample of 41 criminals convicted of a certain crime, it was determined that the mean length of sentencing was 51 months, with a standard deviation of 11 months.
Firstly, the pivotal quantity for 95% 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 length of sentencing = 51 months
[tex]s[/tex] = sample standard deviation = 11 months
n = sample of criminals = 41
[tex]\mu[/tex] = population mean length of sentencing
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(-2.021 < [tex]t_4_0[/tex] < 2.021) = 0.95 {As the critical value of t at 40 degree of
freedom are -2.021 & 2.021 with P = 2.5%}
P(-2.021 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.021) = 0.95
P( [tex]-2.021 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]-2.021 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.021 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.021 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.021 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.021 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]51-2.021 \times {\frac{11}{\sqrt{41} } }[/tex] , [tex]51+2.021 \times {\frac{11}{\sqrt{41} } }[/tex] ]
= [47.53 , 54.47]
Therefore, 95% confidence interval for the mean length of sentencing for this crime is [47.53 months , 54.47 months].
