Question

A company that produces and markets video game systems wishes to assess its customers' level of satisfaction with a relatively new model, the XYZ-Box. In the six months since the introduction of the model, the company has received 73,219 warranty registrations from purchasers. The company will select a random sample of 65 of these registrations and will conduct telephone interviews with the purchasers. Specifically, each purchaser will be asked to state his or her level of agreement with each of the seven statements listed on the survey instrument given in the following table.. Here, the level of agreement for each statement is measured on a 7-point Likert scale. Purchaser satisfaction will be measured by adding the purchaser’s responses to the seven statements. It follows that for each consumer the minimum composite score possible is 7 and the maximum is 49. Furthermore, experience has shown that a purchaser of a video game system is “very satisfied” if his or her composite score is at least 42.

The Video Game Satisfaction Survey Instrument
	Strongly						Strongly
Statement	Disagree						Agree
The game console of the XYZ-Box is well designed.	1	2	3	4	5	6	7
The game controller of the XYZ-Box is easy to handle.	1	2	3	4	5	6	7
The XYZ-Box has high-quality graphics capabilities.	1	2	3	4	5	6	7
The XYZ-Box has high-quality audio capabilities.	1	2	3	4	5	6	7
The XYZ-Box serves as a complete entertainment center.	1	2	3	4	5	6	7
There is a large selection of XYZ-Box games to choose from.	1	2	3	4	5	6	7
I am totally satisfied with my XYZ-Box game system.	1	2	3	4	5	6	7

Suppose that when the 65 customers are interviewed, their composite scores are as given in the following table.

Composite Scores for the Video Game Satisfaction Rating Case
22	34	24	37	33
44	29	33	25	41
20	29	40	26	20
24	28	38	38	25
32	45	32	30	28
20	23	22	45
27	35	22	44
33	27	45	42
24	37	41	26
43	28	41	37
41	26	20	28
29	33	33	41
39	30	21	42
30	26	42	37
38	23	39	23

Using the data, estimate limits between which most of the 73,219 composite scores would fall. Also, estimate the proportion of the 73,219 composite scores that would be at least 42. (Round your proportion of scores answer to 2 decimal places.)

Answer 1

Solution

Back-Up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ²,

then, Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .………......................………..............................…(1)

Probability values for the Standard Normal Variable, Z, can be directly read off from Standard Normal Tables ..........(2a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) ........................................…(2b)

Empirical rule, also known as 68 – 95 – 99.7 percent rule: applicable to Normal and

symmetric (bell-shaped) distributions

Mean ± 2 Standard Deviations holds 95% of the observations ….................................……............................……….(2c)

In statistical analysis, ‘most’ is taken to be 95% .......................................................................................................... (3)

Now To Work Out The Solution,

Let X = composite score.

Since the composite score is a sum of so many other scores, X can be assumed to follow Normal distribution.

From the given data, using Excel Function: Statistical AVERAGE   and STDEV,

Mean = 32

Standard deviation = 7.74.

Thus, X ~ N(32, 7.74) .................................................................................................................................................... (4)

Part (a)

If t₁ and t₂ be the limits between which most of the 73,219 composite scores would fall, then vide (3), we should have:

P(t₁ ≤ X ≤ t2) = 0.95

=> vide (2c) and (4), t₁ = 32 – (2 x 7.74) = 16.52 and t₂ = 32 + (2 x 7.74) = 57.49.

Thus, the required limits are: [16.52, 57.49] Answer 1

Part (b)

Proportion of the 73,219 composite scores that would be at least 42

= P(X ≤ 42)

= P[Z ≤ {42 - 32)/7.74}] [vide (1) and (4)]

= P(Z ≤ 1.292)

= 0.9018 [vide (2b)] Answer 2

DONE