Question

Trucks traveling on a certain highway have to stop and be weighed at a weigh station. The federal government drew a random sample of 150 trucks and measured their weights. The sample mean was 45,000 pounds. Assume the population standard deviation is 8,000 pounds.

Calculate a 99% confidence interval for the true unknown mean weight of all trucks on the highway. Round to one decimal place.

What is the margin of error for this estimate? Round to one decimal place.

What sample size would be required to obtain a margin of error of 1,000 pounds (still with 99% confidence)? Round up to the nearest whole number.

Answer 1

99% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.576
45000 ± Z (0.01/2 ) * 8000/√(150)
Lower Limit = 45000 - Z(0.01/2) 8000/√(150)
Lower Limit = 43317.4
Upper Limit = 45000 + Z(0.01/2) 8000/√(150)
Upper Limit = 46682.6
99% Confidence interval is ( 43317.4 , 46682.6 )

Margin of error = Z( α /2) σ / √ ( n )

= 2.576 * 8000/√(150)

= 1682.6

Sample size = (Z(α/2) * σ / E)2

= (2.5758 * 8000 / 1000)2

= 424.6

Sample size = 425 (Rounded up to nearest integer)