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:

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

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:

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

4 0

1 year ago

Find the square root of 68.1 to the nearest hundredth

Ira Lisetskai [31]

I think that is is 8.3, but I could be incorrect!

8 0

3 years ago

Read 2 more answers