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)

χ^{2} (0.02/2, 50 - 1 ) = 74.9195

χ^{2} (1 - 0.02/2, 50 - 1) ) = 28.9406

Lower Limit = (( 50-1 ) 47.61 / χ^{2} (0.02/2) ) =
31.1386

Upper Limit = (( 50-1 ) 47.61 / χ^{2} (0.02/2) ) =
80.6096

98% Confidence interval is ( 31.1386 , 80.6096 )

**( 31.1386 < σ ^{2} < 80.6096 )
( 5.5802 < σ < 8.9783 )**

Assumptions :- Samples selected should be from normal population.

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.

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