Question

In: Statistics and Probability

Suppose the population variance is unknown and the sample standard deviation is 1.92. Compute the correct...

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.

Solutions

Expert Solution

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 )

orchestra answered 2 years ago
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