Question

The 90% confidence interval estimate for a population standard deviation when a sample variance of 50 is obtained from a sample of 15 items is

Answer 1

Answer:

$Z_{score}=\frac{x-\mu}{\sigma}$

$\mu=15\\\\ \sigma=\sqrt{Variance}=\sqrt{50}=7.07$

$Z_{90 percent}=1.645\\\\1.65=\frac{x- 15}{7.07}\\\\ 1.65 \times 7.07=x-15\\\\11.6655=x-15\\\\x=11.67 +15\\\\x=26.67$

Total Population = 26 (approx)

So, sample of 15 items is chosen from Population of 26.

Answer 2

90% confidence interval estimate for a population standard deviation is [5.44 , 10.32].

Step-by-step explanation:

We are given that a survey of the age of people denied promotion was conducted. In a random sample of 23 people the average age was 47.0 with a sample standard deviation of 7.2.

Assume this sample comes from a population that is normally distributed.

Firstly, the Pivotal quantity for 90% confidence interval for the population standard deviation is given by;

                           P.Q. =  $\frac{(n-1)s^{2} }{\sigma^{2} }$  ~ $\chi^{2}__n_-_1$

where, $s^{2}$ = sample variance = 50

            n = sample of items = 15

            $\sigma$ = population standard deviation

Here for constructing 90% confidence interval we have used One-sample chi-square test statistics.

So, 90% confidence interval for the population standard deviation, $\sigma$ is ;

P(6.57 < $\chi^{2}__1_4$ < 23.68) = 0.90  {As the critical value of chi at 14 degree

                                                     of freedom are 6.57 & 23.68}  

P(6.57 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 23.68) = 0.90

P( $\frac{ 6.57}{(n-1)s^{2}}$ < $\frac{1}{\sigma^{2} }$ < $\frac{ 23.68 }{(n-1)s^{2}}$ ) = 0.90

P( $\frac{(n-1)s^{2}}{23.68}$ < $\sigma^{2}$ < $\frac{(n-1)s^{2}}{6.57}$ ) = 0.90

90% confidence interval for $\sigma^{2}$ = [ $\frac{(n-1)s^{2}}{23.68}$ , $\frac{(n-1)s^{2}}{6.57}$ ]

                                                   = [ $\frac{14 \times 50}{23.68}$ , $\frac{14 \times 50}{6.57}$ ]

                                                   = [29.56 , 106.54]

90% confidence interval for $\sigma$  = [ $\sqrt{29.56} , \sqrt{106.54}$ ]

                                                   = [5.44 , 10.32]

     

Therefore, 90% confidence interval estimate for a population standard deviation is [5.44 , 10.32].