Answer to Question #227979 in Statistics and Probability for NDUME

Question

Suppose that a survey was commissioned by the Ministry of Finance in order to determine the income of local vendors as part of their project aimed at investigating the feasibility of the presumptive tax. A sample of 26 local vendors was selected and their sample mean monthly income and variance was found to be N$ 7 600 and 1 250 (N$ squared), respectively. a) What is the 90% confidence interval estimate for the population mean monthly income of the local vendors? 

B). Provide an interpretation of the answer in a

c) What can be done to improve the precision of the confidence interval estimate in a)? Discuss. 

Answer 1

a)

Confidence interval for the general average.

(x^- - t_кр*(s/n^1/2) ; x^- + t_кр*(s/n^1/2))

where x^- = mean and s = standard deviation

Since n ≤ 30, we determine the value of tkp from the Student's distribution table

According to the Student's table, we find:

T_table (n-1;α/2) = (26;0.05) = 2.385

(7600 - 16.537;7600 + 16.537) = (7583.46;7616.54)

With a probability of 0.90, it can be argued that the average value for a larger sample will not go beyond the found interval.

b) With 90% confidence, we can say that the average monthly income of suppliers lies in the range from 7583.46 to 7616.54. If we conduct a similar study on samples of the same size many times, independently of

each other, 90% of the confidence intervals will include the true value of the average monthly income of suppliers.

с) The number of local suppliers may affect the estimation of the confidence interval