Question

A random sample of size 10 from a normal population gives the values,
64, 72, 65, 70, 68, 71, 65, 62, 66, 67.
If it is known that the standard error of the sample mean is √0.64 , find the 95% confidence limits for the population mean. Also find the population variance.

Answer 1

Answer on Question #46487 – Math – Statistics and Probability

Problem.

A random sample of size 10 from a normal population gives the values, 64, 72, 65, 70, 68, 71, 65, 62, 66, 67.

If it is known that the standard error of the sample mean is $\sqrt{0.64}$, find the 95% confidence limits for the population mean. Also find the population variance.

Solution:

The mean of this sample equals $\bar{x} = \frac{64 + 72 + 65 + 70 + 68 + 71 + 65 + 62 + 66 + 67}{10} = 67$.

For 95% confidence interval $z^{*} = 1.96$.

The confidence limits for the population mean are equal to $\bar{x} \pm z^{*} \cdot \mathrm{SE}$ ($\mathrm{SE} = \sqrt{0.64} = 0.8$) or 65.432 and 68.568.

$$\mathrm{SE} = \frac{\sqrt{\mathrm{Var}}}{\sqrt{n}},$$

so the population variance equals $\mathrm{Var} = n\mathrm{SE}^2 = 6.4$.

Answer: confidence limits for the population mean are 65.432 and 68.568; the population variance equals $\mathrm{Var} = 6.4$.

www.AssignmentExpert.com