Question

5.       You take a sample of 30 items and obtain a sample mean of 15 and a sample standard deviation of 5

1.       Construct a 95% confidence interval about the mean
2.       Construct a 99% confidence interval about the mean
3.       If I took 100 different samples of 30 items from a given population and obtained 100 different sample means and standard deviations and formed 100 90% confidence intervals, then about how many of the confidence intervals formed from these 100 samples would contain the true population mean?

Answer 1

a) DF = 30 - 1 = 29

At 95% confidence level, the critical value is t^* = 2.045

The 95% confidence interval is

$\dpi{120} \bar x$ +/- t^* * s/$\dpi{120} \sqrt n$

= 15 +/- 2.045 * 5/$\dpi{120} \sqrt {30}$

= 15 +/- 1.8668

= 13.1332, 16.8668

b) At 99% confidence level, the critical value is t^* = 2.045

The 99% confidence interval is

$\dpi{120} \bar x$ +/- t^* * s/$\dpi{120} \sqrt n$

= 15 +/- 2.045 * 5/$\dpi{120} \sqrt {30}$

= 15 +/- 2.756

= 12.244, 17.756

C) 90% of the confidence intervals formed from these 100 samples would contain the true population mean.

So total 90 of the confidence intervals formed from these 100 samples would contain the true population mean.