Question

Construct a 99% confidence interval to estimate the population mean using the data below x53 G 12 n 41 With 99% confidence, when n 41 the population mean is between a lower limit of (Round to two decimal places as needed.) and an upper limit of

Answer 1

solution:

$\bar X$ = 53, $\sigma$ = 12, n = 41

as the standard deviation of the population is known and sample size is greater than 30 so we use z table

confidence level = 99% or 0.99

$\alpha$ = 1 - 0.99 = 0.01

confidence interval = $\bar X \pm z_{\alpha/2}*\frac{\sigma}{\sqrt{n}}$

critical value of $z_{\alpha/2} =z _{0.005} =$ 2.58

marginal error = $z_{\alpha/2} *\frac{\sigma}{\sqrt{n}} = 2.58*\frac{12}{\sqrt{41}}=4.84$

lower boundary = 53 - 4.84 = 48.16

upper boundary = 53 + 4.84 = 57.84

with 99% confidence, when n = 41 the population mean is between a lower limit of 48.16 and an upper limit of 57.84