Question 7.1: Kimberley Grocery Store Purchase Value Study A survey of a r......
Kimberley Grocery Store Purchase Value Study
A survey of a random sample of 300 grocery shoppers in Kimberley found that the mean value of their grocery purchases was R78. Assume that the population standard deviation of grocery purchase values is R21.
Find the 95% confidence limits for the average value of a grocery purchase by all grocery shoppers in Kimberley.
To construct confidence limits around a single sample mean, the following sample evidence is required:
- the sample mean, \bar{x} (= R78)
- the sample size, n (= 300)
In addition, the level of confidence and a measure of the sampling error is required.
- The sampling error is represented by the standard error of the sample means:
\sigma _{\bar{x}}=\frac{\sigma }{\sqrt{n} }=\frac{21}{\sqrt{300} }=1.2124 (Formula 6.1)
It requires that the population standard deviation, \sigma, is known.
- The 95% confidence level refers to z-limits that bound a symmetrical area of 95% around the mean (centre) of the standard normal distribution.
From the z-table (Appendix 1), a 95% confidence level corresponds to z-limits of ±1.96. Thus an area of 95% is found under the standard normal distribution between z = −1.96 and z = +1.96.
These z-limits represent the 95% confidence interval in z terms. To express the confidence limits in the same unit of measure of the random variable under study (i.e. value of grocery purchases), the z-limits must be transformed into \bar{x} -limits using the following z transformation formula:
z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n} } } (7.1)
When re-arranged to isolate μ:
\mu =\bar{x}\pm z\frac{\sigma }{\sqrt{n} }=78\pm1.96(1.2124)=78\pm2.376This gives a lower and an upper confidence limit about μ :
- lower 95% confidence limit of 78 – 2.376 = 75.624 (R75.62)
- upper 95% confidence limit of 78 + 2.376 = 80.376 (R80.38).
Management Interpretation
There is a 95% chance that the true mean value of all grocery purchases by grocery shoppers in Kimberley lies between R75.62 and R80.38.
Thus the confidence interval for a single population mean, μ, is given as:
\underset{(\text{lower limit}) \ \ \ \ \ \ \ \ \ (\text{upper limit})}{\bar{x}-z\frac{\sigma }{\sqrt{n} } \leq \mu \leq\bar{x}+z\frac{\sigma }{\sqrt{n}}} (7.2)
The term z\frac{\sigma }{\sqrt{n} } in the confidence interval formula is called the margin of error in estimation. It determines the width of a confidence interval and hence the precision of the interval estimate.