Question

A 95% confidence interval for a population mean, μ, is given as (18.985, 21.015). This confidence interval is based on a simple random sample of 36 observations. Calculate the sample mean and standard deviation. Assume that all conditions necessary for inference are satisfied. Use the t-distribution in any calculations.

Answer 1

Answer:

standard deviation of the sample =  s = 3.04

Step-by-step explanation:

given,

μ, is given as (18.985, 21.015)

sample mean = $\dfrac{(18.958+21.015)}{2}$

                      =19.9865

margin of error = 19.9865 - 18.958 =1.0285

The degree of freedom = n -1 = 36 - 1 = 35

a = 0.05,  t(0.025, df = 35) = 2.03 (from student t table)

margin error = $\dfrac{t\times s}{\sqrt{n}}= 1.0285$

      = $\dfrac{2.03 \times s}{\sqrt{36}}= 1.0285$

      = $\dfrac{2.03 \times s}{6}= 1.0285$

      = $s = \dfrac{1.0285\times 6}{2.03}$

      = s = 3.04

standard deviation of the sample =  s = 3.04