Answer to Question #229025 in Statistics and Probability for nat

Question

3.1 The average dividend yield of a random sample of 25 JSE-listed companies this year was found to be 14.5%, with a sample standard deviation of 3.4%. Assume that dividend yields are normally distributed. 3.1.1 Calculate, with 90% confidence, the actual mean dividend yield of all JSE-listed companies this year. Interpret the finding. (6) 3.1.2 Calculate, with 95% confidence, the actual mean dividend yield of all JSE-listed companies this year. Compare the interval with the one calculated in 3.1.1. (12)

Answer 1

3.1.1

We are given that the sample mean = 0.145 and the sample standard deviation, s = 0.034 for sample size n=25

The Z value corresponding to 90% CI = 1.645

The 90% CI = mean ± Z*s/$\sqrt{\smash[b]{n}}$

The lower bound = 0.145 - 1.645*0.034/$\sqrt{\smash[b]{25}}$  = 0.1338

The upper bound = 0.145 + 1.645*0.034/$\sqrt{\smash[b]{25}}$  = 0.1562

The 90% CI = (0.1338, 0.1562)

This means that we are 90% confident that the actual mean dividend yield of all JSE-listed companies this year will be contained in the interval from 13.38% to 15.62%

3.1.2

We are given that the sample mean = 0.145 and the sample standard deviation, s = 0.034 for sample size n=25

The Z value corresponding to 95% CI = 1.96

The 95% CI = mean ± Z*s/$\sqrt{\smash[b]{n}}$

The lower bound = 0.145 - 1.96* 0.034/$\sqrt{\smash[b]{25}}$ = 0.1317

The upper bound = 0.145 + 1.96*0.034/$\sqrt{\smash[b]{25}}$  = 0.1583

The 95% CI = (0.1317, 0.1583)

This means that we are 95% confident that the actual mean dividend yield of all JSE-listed companies this year will be contained in the interval from 13.17% to 15.83%

We find that the 95% confidence level has a wider interval as compared to the 90% confidence interval. This may imply that the more the confidence level, the wider the interval.