Question

Im having a hard time wirh this question please explain the solution in details

Problem 5 (7 marks) A survey of 25 retail stores revealed that the average price of a DVD was $375 with a standard deviation of $20. a) What is the 95% confidence interval to estimate the true cost of the DVD? (4 marks) b) What sample size would be needed to estimate the true average price of a DVD with an error of +3$ and a 99% confidence? (3 marks)

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 25- 1 ) = 2.064
375 ± t(0.05/2, 25 -1) * 20/√(25)
Lower Limit = 375 - t(0.05/2, 25 -1) 20/√(25)
Lower Limit = 366.744
Upper Limit = 375 + t(0.05/2, 25 -1) 20/√(25)
Upper Limit = 383.256
95% Confidence interval is ( 366.744 , 383.256 )

Part b)

Sample size can be calculated by below formula
n = (( Z(α/2) * σ) / e )²
n = (( Z(0.01/2) * 20 ) / 3 )²
Critical value Z(α/2) = Z(0.01/2) = 2.5758
n = (( 2.5758 * 20 ) / 3 )²
n = 295
Required sample size at 99% confidence is 295