Question

A random sample of size 150 is drawn from a large population. The population standard deviation is 2.3 and the sample mean is 21.3. Find a 99% confidence interval for the population mean μ . Round your answer to the nearest hundredth.

Answer 1

The formula for finding the confidence interval is:

`CI=\bar{x}\pm Z^{}(\frac{\sigma}{\sqrt[]{n}})`

mean=21.3

S.D=2.3

Sample size (n) =150

From theory, the critical value (Z) for a 99% confidence interval is 2.58

So Z=2.58

Substituting all these parameters into the confidence interval formula,

we will obtain:

`CI=21.3\pm(2.58*(\frac{2.3}{\sqrt[]{150}}))`
`\begin{gathered} =21.3\pm(2.58*0.18779) \\ =21.3\pm(0.4844982) \end{gathered}`
`\begin{gathered} =21.3+0.4844982\text{ to 21.3-0.4844982} \\ =21.7844982\text{ to 20.8155018} \\ =21.78\text{ to 20.82 (to the nearest hundredth)} \end{gathered}`

The final answer is 21.78 to 20.82 confidence interval.