Could someone work one of these out so I can use as example.
A sample of 59 night-school students' ages is obtained in order to estimate the mean age of night-school students. x = 25 years. The population variance is 20.
(a) Give a point estimate for ì. (Give your answer correct to one decimal place.)
(b) Find the 95% confidence interval for ì. (Give your answer correct to two decimal places.)
Lower Limit
Upper Limit
(c) Find the 99% confidence interval for ì. (Give your answer correct to two decimal places.)
Lower Limit
Upper Limit
To work out these questions, you need to use the formulas for point estimate and confidence intervals.
(a) Point Estimate for μ:
The point estimate for the population mean (μ) is calculated by taking the sample mean (x̄) as the best estimate. In this case, the sample mean (x̄) is given as 25 years. So, the point estimate for μ is 25 years.
(b) 95% Confidence Interval for μ:
To calculate the confidence interval, you need to use the formula:
Confidence Interval = x̄ ± (Z * σ / √n)
Where:
x̄ = sample mean (25 years)
Z = Z-score corresponding to the desired confidence level (95% confidence level has a Z-score of 1.96)
σ = population standard deviation (in this case, the square root of the variance, which is √20)
n = sample size (59 students)
Now, let's plug in the values into the formula:
Confidence Interval = 25 ± (1.96 * √20 / √59)
Performing the calculations, you will get:
Confidence Interval = 25 ± 2.87
So, the 95% confidence interval for μ is (22.13, 27.87).
(c) 99% Confidence Interval for μ:
Using the same formula as above, but now with a Z-score of 2.58 (corresponding to a 99% confidence level), you can calculate the 99% confidence interval for μ.
Confidence Interval = 25 ± (2.58 * √20 / √59)
Performing the calculations, you will get:
Confidence Interval = 25 ± 3.52
Therefore, the 99% confidence interval for μ is (21.48, 28.52).