An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:
420 425 427 427 432 433 434 437 439 446 447 448 453 454 465 469
Suppose the sample is from a normal population.
(a) Calculate a 95% confidence interval for the population mean, and interpret it.
(b) Calculate a 95% upper confidence bound for the population mean, and interpret it.

Respuesta :

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

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 mean = [tex]\frac{\sum X}{n}[/tex] = 441

            s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex]  = 14.34

            n = sample size = 16

            [tex]\mu[/tex] = population mean

Here for constructing a 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(-2.131 < [tex]t_1_5[/tex] < 2.131) = 0.95  {As the critical value of t at 15 degrees of

                                             freedom are -2.131 & 2.131 with P = 2.5%}  

P(-2.131 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.131) = 0.95

P( [tex]-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

P( [tex]\bar X-2.131 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -2.131 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X +2.131 \times {\frac{s}{\sqrt{n} } }[/tex] ]

                                      = [ [tex]441-2.131 \times {\frac{14.34}{\sqrt{16} } }[/tex] , [tex]441+2.131 \times {\frac{14.34}{\sqrt{16} } }[/tex] ]

                                      = [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.