Question

Let the observed value of the mean \(\bar{X}\) and of the sample variance of a random sample of size 20 from a distribution that is \(N\left(\mu, \sigma^{2}\right)\) be \(81.2\) and \(26.5\), respectively. Find respectively \(90 \%, 95 \%\) and \(99 \%\) confidence intervals for \(\mu .\) Note how the lengths of the confidence intervals increase as the confidence increases.

Short answer

The 90%, 95% and 99% confidence intervals for \(\mu\) are approximately (78.53, 83.86), (78.00, 84.40), and (77.13, 85.27) respectively. The lengths of these intervals increase with the level of confidence.

Step-by-step solution

Determine Z-scores

First, determine the Z-scores for 90%, 95%, and 99% confidence levels by looking at a standard normal distribution table or using a calculator. The Z-scores for these levels are approximately 1.645, 1.96 and 2.576, respectively.

Calculate Standard Error

Next, calculate the standard error by dividing the square root of the sample variance by the square root of the sample size, \(SE = \sqrt{s^2/n} = \sqrt{26.5/20}\).

Compute Confidence Intervals

Then, plug the sample mean, the Z-scores, and the standard error in the confidence interval formula \((\bar{X}- Z*SE, \bar{X} + Z*SE)\) to obtain the confidence intervals for \(\mu\) at each confidence level. Calculate for a 90% CI, a 95% CI, and a 99% CI.

Discuss Results

Discuss the increase in the lengths of the confidence intervals as the level of confidence increases. This represents that as we try to be more and more certain about catching the true parameter, we need to allow our net (i.e. our interval) to be more wide.