I think if I knew the equation I could get it
Construct a 99% confidence interval for the population mean given the following sample statistics: sample size is 45, sample mean is 10.5, and standard deviation is 2.14.
All help is extremely appreciated!!!
To construct a confidence interval for the population mean, you'll need to use the formula:
Confidence Interval = sample mean ± (critical value * standard error)
The critical value is based on the desired confidence level, sample size, and the assumption of a normal population distribution. For a 99% confidence level, the critical value can be obtained from a normal distribution table or a statistical software.
The standard error, denoted as SE, is calculated as:
SE = standard deviation / √(sample size)
Given the sample statistics you provided:
Sample size (n) = 45
Sample mean (x̄) = 10.5
Standard deviation (σ) = 2.14
First, calculate the standard error:
SE = 2.14 / √(45) ≈ 0.319
Next, determine the critical value corresponding to a 99% confidence level. This value can be found using statistical software or a critical value table. For a normal distribution, with 99% confidence level, the critical value is approximately 2.61.
Lastly, you can calculate the confidence interval:
Confidence Interval = 10.5 ± (2.61 * 0.319)
= [10.5 - (2.61 * 0.319), 10.5 + (2.61 * 0.319)]
= [10.5 - 0.833, 10.5 + 0.833]
= [9.667, 11.333]
Therefore, the 99% confidence interval for the population mean is [9.667, 11.333].
Note: It's important to note that constructing a confidence interval assumes a random and representative sample from the population, as well as the assumption that the population follows a normal distribution.