Can someone guide me on how to solve this problem: A sample of 52 night students ages is obtained in order to estimate the mean age of night school students. sample mean = 25.9 years. the population variance is 23. Give a point estimate for the mean. Find the 95% confidence interval for the mean lower and upper and the 99% confidence interval

95% = mean ± 1.96 SEm

SEm = SD/√n

Variance = SD^2

99% = mean ± 2.575 SEm

To solve this problem, you will need to use the concepts of point estimation and confidence intervals. Let's break it down step by step:

1. Point estimate for the mean:
- In this case, the sample mean is given as 25.9 years. This is your point estimate for the mean age of night school students.

2. Finding the 95% confidence interval for the mean:
- To calculate the confidence interval, we need to determine the margin of error first. The margin of error depends on the sample size, the population standard deviation (or variance in this case), and the desired level of confidence.
- Since the variance is given as 23, we need to find the standard deviation. This can be done by taking the square root of the variance: √23 ≈ 4.796.
- The margin of error can then be calculated using the following formula: Margin of Error = Z * (σ / √n), where Z is the z-value corresponding to the desired level of confidence, σ is the standard deviation, and n is the sample size.
- For a 95% confidence interval, the z-value is approximately 1.96 (you can find this value in a standard normal distribution table).
- Plugging in the values, we get Margin of Error = 1.96 * (4.796 / √52) ≈ 1.34.
- Finally, the confidence interval can be calculated by subtracting and adding the margin of error to the point estimate: Lower Confidence Limit = Point Estimate - Margin of Error, Upper Confidence Limit = Point Estimate + Margin of Error.
- In this case, the 95% confidence interval for the mean is approximately 25.9 - 1.34 to 25.9 + 1.34, which can be simplified to 24.56 to 27.24 years.

3. Finding the 99% confidence interval for the mean:
- Here, we follow the same steps as above, but this time the z-value for the 99% confidence level is approximately 2.58.
- Using the formula, Margin of Error = 2.58 * (4.796 / √52) ≈ 2.23.
- The 99% confidence interval for the mean is approximately 25.9 - 2.23 to 25.9 + 2.23, which simplifies to 23.67 to 28.13 years.

To summarize:
- The point estimate for the mean age of night school students is 25.9 years.
- The 95% confidence interval for the mean is 24.56 to 27.24 years.
- The 99% confidence interval for the mean is 23.67 to 28.13 years.