Question

The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters.

Construct a 98% confidence interval for:

a) the mean height

b) the standard deviation of heights of all college students. State assumptions.

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.02 /2, 50- 1 ) = 2.405 ( Critical value from t table )
174.5 ± t(0.02/2, 50 -1) * 6.9/√(50)
Lower Limit = 174.5 - t(0.02/2, 50 -1) 6.9/√(50)
Lower Limit = 172.1532
Upper Limit = 174.5 + t(0.02/2, 50 -1) 6.9/√(50)
Upper Limit = 176.8468
98% Confidence interval is ( 172.1532 , 176.8468 )

Part b)

χ² (0.02/2, 50 - 1 ) = 74.9195
χ² (1 - 0.02/2, 50 - 1) ) = 28.9406
Lower Limit = (( 50-1 ) 47.61 / χ² (0.02/2) ) = 31.1386
Upper Limit = (( 50-1 ) 47.61 / χ² (0.02/2) ) = 80.6096
98% Confidence interval is ( 31.1386 , 80.6096 )
( 31.1386 < σ² < 80.6096 )
( 5.5802 < σ < 8.9783 )

Assumptions :- Samples selected should be from normal population.