A simple random sample of a size n is drawn from a population that is normally distributed. The sample mean, x is found to be 110, and the sample standard deviation, s, is found to be 10.
A- Construct a 95% confidence interval about u if the sample size, n, is 17
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (±.025) and its Z score.
95% = mean ± Z SEm
SEm = SD/√n
To construct a confidence interval for the population mean, we can use the formula:
Confidence interval = x ± (t * (s / √n))
Where:
- x is the sample mean
- t is the critical value from the t-distribution table for a given confidence level and degrees of freedom
- s is the sample standard deviation
- n is the sample size
In this case, we are given:
- x = 110
- s = 10
- n = 17
To find the critical value, we need to determine the degrees of freedom. Since we're using a t-distribution, the degrees of freedom are n - 1.
Degrees of freedom = 17 - 1 = 16
Next, we need to find the critical value for a 95% confidence level and 16 degrees of freedom. Looking this up on the t-distribution table (or using a statistical calculator), we find the critical value to be approximately 2.120.
Now, we can substitute the values into the confidence interval formula:
Confidence interval = 110 ± (2.120 * (10 / √17))
Calculating this, we get:
Confidence interval = 110 ± (2.120 * (10 / 4.123))
Simplifying further:
Confidence interval = 110 ± (2.120 * 2.420)
Confidence interval = 110 ± 5.139
Therefore, the 95% confidence interval for the population mean, μ, is approximately (104.861, 115.139).