Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample? mean, is found to be 112?, and the sample standard? deviation, s, is found to be 10.

**?(a)** Construct a 95?% confidence interval about
mu? if the sample? size, n, is 18.

**?(b)** Construct a 95?% confidence interval about
mu? if the sample? size, n, is 11.

**?(c)** Construct a 70?% confidence interval about
mu? if the sample? size, n, is 18.

**?(d)** Could we have computed the confidence
intervals in parts? (a)-(c) if the population had not been
normally? distributed?

Answer #1

**Solution:**

= 112

= 10

**( a )**

n = 18

z = 1.96

95% confidence interval = z ( / )

= 112 1.96 ( 10/ )

= 112 4.6197

= **( 107.3803 , 116.6197 )**

**( b )**

n = 11

z = 1.96

95% confidence interval = z ( / )

= 112 1.96 ( 11 / )

= 112 5.0817

= **( 106.9183 , 117.0817 )**

**( c )**

n = 18

z = 1.04

95% confidence interval = z ( / )

= 112 1.04 ( 10/ )

= 112 2.4513

= **( 109.5487 , 114.4513 )**

**( d )**

**No, the confidence interval computed in parts (a)-(c),
if the popolation had not been normally distributed**

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