Question

Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standarddeviation s = 2.40.

Answer 1

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

`\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}`

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

`CI(\mu)=\lbrack x-Z_{1-(\alpha)/(2)}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-(\alpha)/(2)}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack`

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

`CI(\mu)=\lbrack30.0-Z_(0.99)\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_(0.99)\cdot\frac{2.40}{\sqrt[]{28}}\rbrack`

Where (from tables):

`Z_(0.99)=2.33`

Finally, the interval at 98% confidence level is:

`CI(\mu)=\lbrack28.94,31.06\rbrack`