Question

Suppose the population variance is unknown and the sample standard deviation is 1.92. Compute the correct quantile for each confidence interval: (Use 3 decimal places)

(a) A 95% confidence interval from a sample of size 10.

(c) A 80% confidence interval from a sample of size 6.

Answer 1

Solution :

Given that,

(a)

s = 1.92

Point estimate = s2 = 3.6864

2R = 2/2,df = 19.023

2L = 21 - /2,df = 2.700

The 95% confidence interval for 2 is,

(n - 1)s2 / 2/2 < 2 < (n - 1)s2 / 21 - /2

(9) (3.6864) / 19.023 < 2 < (9) (3.6864) / 2.700

1.744 < 2 < 12.286

(1.744 , 12.286)

(c)

s = 1.92

Point estimate = s2 = 3.6864

2R = 2/2,df = 9.236

2L = 21 - /2,df = 1.610

The 80% confidence interval for 2 is,

(n - 1)s2 / 2/2 < 2 < (n - 1)s2 / 21 - /2

(5) (3.6864) / 9.236 < 2 < (5) (3.6864) / 1.610

1.996 < 2 < 11.446  

(1.996 , 11.446 )