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Chapter 4: Problem 1

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

Expert verified
The 90% confidence interval for \(\mu\) is \( [\bar{X} - 1.645*\sigma_{\bar{x}}, \bar{X} + 1.645*\sigma_{\bar{x}}] \). The 95% confidence interval for \(\mu\) is \( [\bar{X} - 1.96*\sigma_{\bar{x}}, \bar{X} + 1.96*\sigma_{\bar{x}}] \). The 99% confidence interval for \(\mu\) is \( [\bar{X} - 3.3*\sigma_{\bar{x}}, \bar{X} + 3.3*\sigma_{\bar{x}}] \). The increase in the length of the confidence intervals as the confidence increases is consistent with the statistical principle 'greater confidence requires a wider interval'.

Step by step solution

01

- Identifying the parameters

First, identify the values given in the problem. We have a sample size (\(n\)) of 20, a sample mean (\(\bar{X}\)) of 81.2, and a sample variance (\(\sigma^{2}\)) of 26.5.
02

- Find standard deviation

Calculate the sample standard deviation. Standard deviation is square root of variance, so in this case, \(\sigma = \sqrt{26.5}\)
03

- Find the standard error

Calculate the standard error. It's the standard deviation divided by the square root of the sample size, so \(\sigma_{\bar{x}} = \sigma / \sqrt{n}\)
04

- Find the Z-value

Find the appropriate values of the Z-distribution for a 90%, 95%, and 99% confidence interval from the standard normal distribution table. The values are roughly 1.645, 1.96, and 2.576 respectively
05

- Calculate Confidence Intervals

Finally, find the confidence intervals. We calculate them following the formula: \(\bar{X} \pm (Z * \sigma_{\bar{x}})\), where Z is the value from step 4, \(\bar{X}\) is the sample mean and \(\sigma_{\bar{x}}\) is the standard error

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