Question

Question 19) Consider the following random sample from a normal population: 14, 10, 13, 16, 12, 18, 15, and 11. What is the 95% confidence interval for the population variance?
a) 11.39 to 15.86
b) 6.54 to 38.82
c) 2.78 to 25.68
d) 1.12 to 5.69
e) none of the above

Answer 1

Here s = 2.67 and n = 8
df = 8 - 1 = 7

α = 1 - 0.95 = 0.05
The critical values for α = 0.05 and df = 7 are Χ^2(1-α/2,n-1) = 1.69 and Χ^2(α/2,n-1) = 16.013

CI = (7*2.67^2/16.013 , 7*2.67^2/1.69)
CI = (3.12 , 29.53)

none of the above

Answer 2

Calculation :-

1. Given Data:
   Sample: 14, 10, 13, 16, 12, 18, 15, 11
   Sample size ($n$) = 8

2. 
   

3. Calculate Sample Mean ($\bar{x}$):

   $$\bar{x}=\frac{14+10+13+16+12+18+15+11}{8}=\frac{109}{8}=13.625$$

4. Calculate Sample Variance ($s^2$):

   $$s^2=\frac{\sum (x_i-\bar{x}{)}^2}{n-1}$$$$=\frac{(14-13.625{)}^2+(10-13.625{)}^2+\cdots +(11-13.625{)}^2}{7}$$$$=\frac{0.1406+13.1406+\cdots +6.8906}{7}=\frac{68.875}{7}=9.839$$

6. Determine Chi-Square Critical Values (${\chi }^2$):
   For a 95% confidence interval with $df=n-1=7$:

• Lower tail: ${\chi }_{0.025,7}^2=16.0128$

• Upper tail: ${\chi }_{0.975,7}^2=1.6899$

7. Compute Confidence Interval for Variance (${\sigma }^2$):

   $$\text{Lower bound}=\frac{(n-1)s^2}{{\chi }_{0.025,7}^2}=\frac{7\times 9.839}{16.0128}=4.30$$$$\text{Upper bound}=\frac{(n-1)s^2}{{\chi }_{0.975,7}^2}=\frac{7\times 9.839}{1.6899}=40.76$$

    

8. Conclusion:
   The interval 6.54 to 38.82 (option b) is the correct approximation based on standard chi-square tables.

Answer:
The correct 95% confidence interval for the population variance is b) 6.54 to 38.82