Respuesta :
Answer:
(a) 80% confidence interval for the population mean is [109.24 , 116.76].
(b) 80% confidence interval for the population mean is [109.86 , 116.14].
(c) 98% confidence interval for the population mean is [105.56 , 120.44].
(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.
Step-by-step explanation:
We are given that a simple random sample of size n is drawn from a population that is normally distributed.
The sample mean is found to be 113 and the sample standard deviation is found to be 10.
(a) The sample size given is n = 13.
Firstly, the pivotal quantity for 80% 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 = 113
s = sample standard deviation = 10
n = sample size = 13
[tex]\mu[/tex] = population mean
Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
So, 80% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-1.356 < [tex]t_1_2[/tex] < 1.356) = 0.80 {As the critical value of t at 12 degree
of freedom are -1.356 & 1.356 with P = 10%}
P(-1.356 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.356) = 0.80
P( [tex]-1.356 \times }{\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.356 \times }{\frac{s}{\sqrt{n} } }[/tex] ) = 0.80
P( [tex]\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }[/tex] ) = 0.80
80% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }[/tex]]
= [ [tex]113-1.356 \times }{\frac{10}{\sqrt{13} } }[/tex] , [tex]113+1.356 \times }{\frac{10}{\sqrt{13} } }[/tex] ]
= [109.24 , 116.76]
Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].
(b) Now, the sample size has been changed to 18, i.e; n = 18.
So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.
80% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }[/tex]]
= [ [tex]113-1.333 \times }{\frac{10}{\sqrt{18} } }[/tex] , [tex]113+1.333 \times }{\frac{10}{\sqrt{18} } }[/tex] ]
= [109.86 , 116.14]
Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].
(c) Now, we have to construct 98% confidence interval with sample size, n = 13.
So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.
98% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }[/tex]]
= [ [tex]113-2.681 \times }{\frac{10}{\sqrt{13} } }[/tex] , [tex]113+2.681 \times }{\frac{10}{\sqrt{13} } }[/tex] ]
= [105.56 , 120.44]
Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].
(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.
