Question

We wish to obtain a 90% confidence interval for the standard deviation of a normally distributed random variable. To accomplish this we obtain a simple random sample of 16 elements from the population on which the random variable is defined. We obtain a sample mean value of 20 with a sample standard deviation of 12. Give the 90% confidence interval (to the nearest integer) for the standard deviation of the random variable. a) 83 to 307 b) 9 to 18 c) 91 to 270 d) 15 to 25 e) 20 to 34

Answer 1

Answer: d) 15 to 25

Step-by-step explanation:

Given : Sample size : n= 16

Degree of freedom = df =n-1 = 15

Sample mean : $\overline{x}=20$

sample standard deviation : $s= 12$

Significance level : $\alpha= 1-0.90=0.10$

Since population standard deviation is unavailable , so the confidence interval for the population mean is given by:-

$\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}$

Using t-distribution table , we have

Critical value = $t_{\alpha/2, df}=t_{0.05 , 15}=1.7530$

90% confidence interval for the mean value will be :

$20\pm (1.7530)\dfrac{12}{\sqrt{16}}$

$20\pm (1.7530)\dfrac{12}{4}$

$20\pm (1.7530)(3)$

$20\pm (5.259)$

$(20-5.259,\ 20+5.259)$

$(14.741,\ 25.259)\approx(15,\ 25 )$[Round to the nearest integer]

Hence, the 90% confidence interval (to the nearest integer) for the standard deviation of the random variable.= 15 to 25.