Question

A chocolate company markets a 500g box of chocolates to large retail stores for Valentine’s Day. This year, they project the mean order quantity will be greater than last year. Last year, the mean order quantity was 500 boxes per large retail store. This year, a random sample of 4 large retail stores gave a sample mean order quantity of 490 boxes. Assume a population standard deviation of 50 boxes, and that the mean order quantity follows a Normal distribution.

Conduct a hypothesis test using the critical value method to determine whether this year’s population mean order quantity is more than 10% lower than last year’s. HINT: What is meant by “more than 10% lower”? For example, it might be 10.8885% lower, or 14.68% lower, etc. What does that mean in terms of “number of boxes”?

Answer 1

Solution

‘Last year, the mean order quantity was 500 boxes per large retail store.’ And ‘this year’s population mean order quantity is more than 10% lower than last year’s.’

=> this year’s population mean order quantity > 500 – (500 x 0.1) i.e., > 450.

Testing

Let X = order quantity (boxes).

Given X ~ N(µ, σ), where σ = 50 and µ is unknown.

Claim: This year’s population mean order quantity is more than 10% lower than last year’s.

Hypotheses:

Null H₀: µ = µ₀ = 450   Vs Alternative H_A: µ > 450

Test statistic:

Z = (√n)(Xbar - µ₀)/σ,

where

n = sample size;

Xbar = sample average;

σ = known population standard deviation.

Summary of Excel Calculations is given below:

n	4
µ0	450
σ	50
Xbar	490
Zcal	1.6
Given α	0.05
Zcrit	1.644854

Distribution, Significance level, α, Critical Value,

Under H₀, Z ~ N(0, 1)

Critical value = upper α% point of N(0, 1).

Using Excel Functions : Statistical NORMSINV, Zcrit is found to be as shown in the above table.

Decision:

Since Zcal < Zcrit, H₀ is accepted.

Conclusion:

There is not sufficient evidence to support the claim and hence we conclude that

This year’s population mean order quantity is NOT more than 10% lower than last year’s. Answer

DONE