A survey is being planned by the marketing research department of a manufacturer

Question

A survey is being planned by the marketing research department of a manufacturer of car audio systems. The purpose of the research is to assess consumer perceptions of the company’s products. Management would like to estimate the average overall amount spent on all stereo items (household, portable, etc.) per year, the average amount spent on accessories, and the average amount spent on car security systems per year. They would like to obtain a 95% confidence interval in their results. Furthermore, the magnitude of error is not to exceed +/- $10 for all estimates. The research department noted that while some households spent nothing on overall stereo expenditures per year, others might spend as much as $3000 per year. Also, some customers may spend nothing on accessories while others may pay as much as $1200 per year. Finally, some customers may spend nothing on car security systems while others may spend as much as $600 per year.

(a)    How large a sample would you recommend if each of the three variables were considered separately (overall expenditures, accessories, security system)? Show your calculations for the 3 sample sizes.

(b)   What sample size would you recommend overall given that the average stereo expenditures is most important? Why?

(c)    Now, assume that the survey eventually included 900 respondents and indicated that the average stereo expenditures was $300, and the standard deviation was $90. Estimate a 95% confidence interval for the population mean.

Explanation / Answer

Answer to part a)

Z = 1.96

Margin of error E = 10

.

Now for normal distribution: Maximum value - minimum value = 6 times the standard deviation

Thus standard deviation = rnage of expenses / 6

Standard deviation for stereo expenditures is : 3000/6 = 500

Sample size = (1.96*500/10)^2 = 9604

.

For Accessories, standard deviation = 1200/6 = 200

Sample size = (1.96 *200 /10) = 1536.64 ~ 1537

.

For security systems , Standard deviation = 600/6 = 100

Sample size = (1.96*100/10) = 384.16 ~ 385

.

Answer to part b)

In case if the average stereo expenses are more important then the sample size must be 9604

.

Answer to part c)

n = 900

x bar = 300

s = 90

z = 1.96

.

The formula of confidence interval is:

x bar - z* s/ sqrt(n) , xbar + z * s/ sqrt(n)

.

On plugging the values we get

300 - 1.96 *90 /sqrt(900) , 300 + 1.96 *90 /sqrt(900)

294.12 , 305.88

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