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

To obtain a 90% confidence interval for the standard deviation of a normally distributed random variable with a sample size of 16, sample mean of 20, and sample standard deviation of 12, use the chi-square distribution to calculate the lower and upper bounds. The 90% confidence interval is approximately 86 to 283.

Step-by-step explanation:

To obtain a 90% confidence interval for the standard deviation of a normally distributed random variable, we can use the chi-square distribution. Given a simple random sample of 16 elements with a sample mean of 20 and a sample standard deviation of 12, we can calculate the lower and upper bounds of the confidence interval.

Step 1: Calculate the chi-square values for the lower and upper bounds using the following formulas:

Lower bound: (n-1)s² / X², where n is the sample size, s is the sample standard deviation, and X² is the chi-square value for a 90% confidence level with (n-1) degrees of freedom.

Upper bound: (n-1)s² / X², where n is the sample size, s is the sample standard deviation, and X² is the chi-square value for a 10% significance level with (n-1) degrees of freedom.

Substituting the values into the formulas, we get:

Lower bound: (15)(144) / 24.996 = 86.437

Upper bound: (15)(144) / 7.633 = 283.368

Rounding to the nearest integer, the 90% confidence interval for the standard deviation of the random variable is approximately 86 to 283.

Answer 2

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)(s)/(√(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)(12)/(√(16))`

`20\pm (1.7530)(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.