Question

suppose a study estimated the population mean for a variable of interest using a 99% confidence interval. If the width of the estimated confidence interval( the difference between the upper limit and the lower limit) is 600 and the sample size used in estimating the mean is 1000, what is the population standard deviation?

Answer 1

Answer:

Population standard deviation, $\sigma$ = 3683.063 .

Step-by-step explanation:

We are given that the width of the estimated confidence interval i.e. 99% is 600 and the sample size used in estimating the mean is 1000 which means ;   n = 1000 and width = 600

We know that Width of confidence interval = 2 * Margin of error

Margin of error = $Z_\frac{\alpha}{2} * \frac{\sigma}{\sqrt{n} }$ = 2.5758 * $\frac{\sigma}{\sqrt{1000} }$ {because at 1% significance level

                                                                          z table has value of 2.5758 .}

Therefore,  600 = 2 * 2.5758 * $\frac{\sigma}{\sqrt{1000} }$

      ⇒ $\sigma$ = $\frac{600 * \sqrt{1000} }{2 * 2.5758}$ = 3683.063 .

Hence, the Population standard deviation = 3683.063 .