Consider the following hypothesis
\(H_0:\mu= 315\) and \(H_a: \mu \ne 315\)
A sample of 60 provided a sample mean of 324.6. The standard deviation (\(\sigma\)) is 14 and level of significance \(\alpha = 0.05\). Then the confidence interval is:
[Given: \(Z_{\alpha/2}\displaystyle\frac{14}{\sqrt{60}} = 3.54\)]
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (321.06, 328.14)
With \(\sigma\) known, the confidence interval for the population mean at level \(1 - \alpha = 0.95\) is the sample mean plus or minus the margin of error:
\( \displaystyle \bar{x} \pm Z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \)
\( = 324.6 \pm 3.54 \)
\( = (324.6 - 3.54,\ 324.6 + 3.54) \)
\( = (321.06,\ 328.14) \)