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Chapter 9: Problem 587

Find a 95 per cent confidence interval for \(\mu\), the true mean of a normal population which has variance \(\sigma^{2}=100\). Consider a sample of size 25 with a mean of \(67.53\).

Short Answer

Expert verified
The 95% confidence interval for the true population mean (\(\mu\)) is approximately \((63.61, 71.45)\).

Step by step solution

01

Identify the given values

We are given: - The population variance: \(\sigma^2 = 100\) - Sample size: \(n = 25\) - Sample mean: \(\bar{x} = 67.53\) - Confidence level: 95%
02

Calculate the standard error

The standard error (SE) is the standard deviation of the sample mean. Since we know the population variance, we can calculate the standard error using the following formula: \(SE = \frac{\sigma}{\sqrt{n}}\) Here, \(\sigma\) is the population standard deviation, and can be calculated as the square root of the population variance: \(\sigma = \sqrt{\sigma^2} = \sqrt{100} = 10\) Now, we can find the standard error: \(SE = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2\)
03

Find the critical value from the Z-table

We need to find the critical value from the standard normal (Z) table corresponding to the 95% confidence level. Since the confidence interval is symmetric (i.e., the area under the curve is 95%), we need to find the Z-value that bounds 97.5% (95% in the middle plus 2.5% on the right side). From the standard normal table, the critical value corresponding to 0.975 (97.5%) is approximately 1.96.
04

Calculate the confidence interval

Now, we have everything we need to calculate the 95% confidence interval for the true population mean. The formula for the confidence interval is: \(\bar{x} \pm Z * SE\) Plugging in the values, we get: \(67.53 \pm 1.96 * 2 = 67.53 \pm 3.92\) This means the 95% confidence interval for the true population mean (\(\mu\)) is approximately: \((67.53 - 3.92, 67.53 + 3.92) = (63.61, 71.45)\) Therefore, we can be 95% confident that the true population mean (\(\mu\)) lies between 63.61 and 71.45.

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