A recent survey of 236 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 90% confidence interval for the actual mean value of all purchases in convenient stores in Kimberly. Give the lower limit.

Answer 1

To construct a confidence interval for the mean value of all purchases in convenient stores in Kimberly, we can use the formula:

Confidence Interval = sample mean ± (critical value * (sample standard deviation / √sample size))

First, let's find the critical value for a 90% confidence interval. The critical value can be found using a z-table or a calculator. For a 90% confidence level, the critical value is 1.645.

Now we can calculate the confidence interval:

Confidence Interval = 48.45 ± (1.645 * (18.58 / √236))

Calculating the values inside the brackets:

(1.645 * (18.58 / √236)) ≈ 3.43

So the confidence interval is:

48.45 ± 3.43

The lower limit of the confidence interval is:

48.45 - 3.43 = 45.02

Therefore, the lower limit of the confidence interval for the actual mean value of all purchases in convenient stores in Kimberly is R 45.02.