Question

A random sample of 100 is taken from a population. The mean and standard deviation are respectively 76 and 12. Obtain a 99% confidence interval for the mean of the population.

Hint:

To calculate a confidence interval for a population mean, use the sample mean, standard deviation, sample size, and the Z-value for the desired confidence level.

Answer 1

We are asked to find the 99% confidence interval for the population mean based on a sample. The formula for the confidence interval for the population mean when the sample size is large (n>30) is: \[ \text{Confidence Interval} = \bar{X} \pm Z \cdot \frac{\sigma}{\sqrt{n}} \] Where: \(\bar{X} = 76\) (sample mean) \(\sigma = 12\) (population standard deviation) \(n = 100\) (sample size) \(Z\) is the Z-value corresponding to a 99% confidence level. Step 1: Find the Z-value.
For a 99% confidence level, the Z-value is approximately 2.576 (from standard Z-tables).
Step 2: Calculate the standard error.
The standard error is given by: \[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{100}} = \frac{12}{10} = 1.2 \]
Step 3: Calculate the margin of error.
The margin of error is given by: \[ \text{Margin of Error} = Z \cdot \text{Standard Error} = 2.576 \cdot 1.2 = 3.0912 \]
Step 4: Calculate the confidence interval.
Now, we can calculate the confidence interval: \[ \text{Confidence Interval} = 76 \pm 3.0912 \] Thus, the 99% confidence interval for the mean is: \[ (76 - 3.0912, 76 + 3.0912) = (72.91, 79.09) \] Therefore, the 99% confidence interval for the population mean is approximately \( \boxed{(72.91, 79.09)} \).