Question

The king of a large castle would like to estimate the average age of knights currently enrolled. Previous studies show a standard deviation of 2 years in knight ages.

a. If in a random sample of 25 knights at the castle, the average age x was found to be 23.2 years old, find a 95% confidence interval for the average age of all knights at the castle.

b. If the king wanted a 95% confidence interval for the average knight age that was no more than 1 year in width, how large a sample of knights would be needed?

Answer #1

a)

95% Confidence Interval :-

X̅ ± Z( α /2) σ / √ ( n )

Z(α/2) = Z (0.05 /2) = 1.96

23.2 ± Z (0.05/2 ) * 2/√(25)

Lower Limit = 23.2 - Z(0.05/2) 2/√(25)

Lower Limit = 22.416

Upper Limit = 23.2 + Z(0.05/2) 2/√(25)

Upper Limit = 23.984

95% Confidence interval is **( 22.416 , 23.984
)**

b)

Margin of error E = Width of confidence interval / 2 = 1 / 2 = 0.5

Sample size = (Z/2 * / E)2

= ( 1.96 * 2 / 0.5)2

= 61.5

**Sample size = 62** (Rounded up to nearest
integer)

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