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Springdale Shopping Survey

Please see attached Excel sheet. Also below is the link to the free book for reference. (I would start there!) The question is on Page #308.

http://www.xn--klker-kva.hu/wp-content/uploads/2013/03/Introduction-to-Business-Statistics-7th-Edition-2008.pdf

1. Item C in the description of the data collection instrument lists variables 7, 8, and 9, which represent
the respondent's general attitude toward each of the three shopping areas. Each of these variables has
numerically equal distances between the possible responses, and for purposes of analysis they may be
considered to be of the interval scale of measurement.

a. Determine the point estimate, then construct the 95% confidence interval for µ7 = the average attitude
toward Springdale Mall. What is the maximum likely error in the point estimate of the population mean?

b. Repeat part (a) for µ8 and µ9, the average attitudes toward Downtown and West Mall, respectively.

2. Given the breakdown of responses for variable 26 (sex of respondent), determine the point estimate,
then construct the 95% confidence interval for π26 = the population proportion of males. What is the
maximum likely error in the point estimate of the population proportion?

3. Given the breakdown of responses for variable 28 (marital status of respondent), determine the point
estimate, then construct the 95% confidence interval for π28 = the population proportion in the "single
or other" category. What is the maximum likely error in the point estimate of the population proportion?

Attachments

Solution Preview

See the attachment.

a. Determine the point estimate, then construct the 95% confidence interval for 7 5 the average
attitude toward Springdale Mall. What is the maximum likely error in the point estimate of the population
mean?

Point estimate

Point estimate of Mean, u7 = 4.086
Point estimate of population standard deviation = 0.776
n = 150
z= z value corresponding to the level of confidence desired
Construction of the 95% confidence interval for population mean
The normal distribution for 95% confidence, z will be +- 1.96
So the 95% confidence interval for u is
mean - 1.96(0.776/sqrt(150)
4.086 - 1.96(0.776/12.24)
4.083 - 1.96 (0.06)
4.083 - 0.117= 3.966 OR 4.083 + 0.117 = 4.2
So the 95% confidence interval for u is between 3.966 and 4.2

The maximum standard error of the sampling distribution of the mean is std/ sqrt(150)
0.776 / 12.24 = 0.063

"b. Repeat part (a) for 8 and 9, the average attitudes toward Down ...

Solution Summary

The solution discusses the Springdale Shopping survey.

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