Question

From a random sample of 58 businesses, it is found that the mean time the owner spends on administrative issues each week is 21.69 with a population standard deviation of 3.23. What is the 95% confidence interval for the amount of time spent on administrative issues?

Answer 1

Answer: (20.86, 22.52)

Explanation:

Formula to find the confidence interval for population mean :-

`\overline{x}\pm z^*(\sigma)/(√(n))`

, where
`\overline{x}` = sample mean.

z*= critical z-value

n= sample size.

`\sigma` = Population standard deviation.

By considering the given question , we have

`\overline{x}= 21.69`

`\sigma=3.23`

n= 58

Using z-table, the critical z-value for 95% confidence = z* = 1.96

Then, 95% confidence interval for the amount of time spent on administrative issues will be :

`21.69\pm (1.96)(3.23)/(√(58))`

`=21.69\pm (1.96)(1.7)/(7.61577)`

`=21.69\pm (1.96)(0.223221)`

`\approx21.69\pm0.83`

`=(21.69-0.83,\ 21.69+0.83)=(20.86,\ 22.52)`

Hence, the 95% confidence interval for the amount of time spent on administrative issues = (20.86, 22.52)