Question

A recent survey of 81 customers who frequently visit convenient stores in Kimberly found that the mean value of purchases was R 48.45, with the standard deviation of R18.58. Assume that the values are normally distributed. Construct a 95% confidence interval for the actual mean value of all purchases in convenient stores in Kimberly. Give the lower limit.

Answer 1

To construct the confidence interval, we can use the formula:

Confidence interval = (mean - z * (standard deviation / √n), mean + z * (standard deviation / √n))

where:
mean = R 48.45 (mean value of purchases)
standard deviation = R 18.58 (standard deviation of purchases)
n = 81 (number of customers surveyed)
z = the critical value corresponding to the level of confidence (in this case, 95%)

The critical value for a 95% confidence interval (two-tailed) is approximately 1.96.

Substituting the given values into the formula:

Confidence interval = (48.45 - 1.96 * (18.58 / √81), 48.45 + 1.96 * (18.58 / √81))

Simplifying and calculating:

Confidence interval = (48.45 - 1.96 * 2.055, 48.45 + 1.96 * 2.055)
Confidence interval = (48.45 - 4.02, 48.45 + 4.02)
Confidence interval = (44.43, 52.47)

The lower limit of the 95% confidence interval for the actual mean value of all purchases in convenient stores in Kimberly is R 44.43.