Question

The price of a champagne glass is changing randomly. Using data of previous 25 weeks (see Sheet ‘champagne’), at 95% confidence level find interval estimate for the expected price. What if the selling price of a champagne glass is changing randomly within this confidence interval? a. What is the sample mean? b. What is the sample standard deviation? c. What is the standard error of the mean? d. What is the margin of error? e. What is the lower limit of the confidence interval? f. What is the upper limit of the confidence interval? g. Is the confidence interval lying within the allowed interval of this objective coefficient? h. If the selling price of a champagne glass is changing randomly within this confidence interval, must the optimal production plan ( 400 3 ; 0; 0; 20 3 ) be revised?

selling price of champagne glasses

10
7,6
9,8
10,5
8,4
7,7
9,8
10,5
8,4
10,9
7,5
10,1
10,7
9,6
9,1
7,9
7,2
6,8
9,4
7,4
8,8
8,8
11,1
8,3
9

Available packing time   

290
369
387
335
302
299
396
337
318
352

Answer 1

Solution

Let X = price of a champagne glass.

Let x₁, x₂, ………, x₂₅ represent the given 25 values of X.

Back-up Theory

100(1 - α) % Confidence Interval for μ, when σ is not known is: Xbar ± (t_{n- 1, α /2})s/√n ……….. (1)where

Xbar = sample mean = (1/n) Σ(i = 1 to 25)x₁ …………………………………………………........ (2)

s = sample standard deviation = {1/(n - 1)}Σ(i = 1 to 25)(x₁ – Xbar)² …………………....…….. (3)

t_{n – 1, α /2} = upper (α /2)% point of t-distribution with (n - 1) degrees of freedom, and

n = sample size.

Part (a)

Vide (2), sample mean = Xbar = 9.5 Answer 1

Part (b)

Vide (3), sample standard deviation = s = 0.7071 Answer 2

Part (c)

Standard error of sample mean = s/√n = 0.1414 Answer 3

Part (d)

95% confidence interval => α = 5% or 0.05, given n = 25, df = 24 and hence

t_{n- 1, α /2} = upper 2.5% point of t-distribution with 24 degrees of freedom

= 2.7969[Using Excel Function: Statistical TINV]

Margin of error = (t_{n- 1, α /2})s/√n

= 0.3955   Answer 4

Part (e)

Vide (1), Lower Limit = Xbar - (t_{n- 1, α /2})s/√n

= 9.10 Answer 5

Part (f)

Vide (1), Upper Limit = Xbar + (t_{n- 1, α /2})s/√n

= 9.90 Answer 6

DONE