A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, s2, is determined to be 19.8.

(a) Construct a 95% confidence interval for s2 if the sample size, n, is 10.
(b) Construct a 95% confidence interval for s2 if the sample size, n, is 25. How does increasing the sample size affect the width of the interval?
(c) Construct a 99% confidence interval for s2 if the sample size, n, is 10. Compare the results with those obtained in part (a). How does increasing the level of confidence affect the width of the confidence interval?

Given that a random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, s2, is determined to be 19.8

Std error of sample would be =[tex]\sqrt{\frac{s^2}{n} } =0.890[/tex] for n =25[tex](19.8-2.065*0.890,19.8+2.065*0.890)\\=(17.962, 21.638)[/tex]

Margin of error = critical value * std error

a) Here n <30. (95%)So we use t critical value for df 9 is used for confidence interval

t critical = 2.262

95% confidence interval =[tex](19.8-2.262*1.483,19.8+2.262*1.483)\\=(16.445, 23.155)[/tex][

b) Here n <30. (95%) So t critical for df 24 is used.

t critical = 2.065

Confidence interval = (17.962, 21.638)

c) For 99% t critical for df 9 is used.

t critical=3.250

Std error = 1.483

confidence interval = (14.980, 24.620)

a) whenever we increase level of confidence for the same sample size we find that confidence interval becomes wider.

Similarly for the same confidence level, if sample size is increases, confidence interval becomes narrower.