Question

A normal population with unknown variance is believe to have a mean 20. Is one likely to obtain random sample of size 9 from this population that has a mean X=24 and a standard deviation of s=4.1?If not ,what conclusion would you draw?site:socratic.org

Answer 1

We need to construct the $95\%$ confidence interval for the population mean $\mu.$

The critical value for $\alpha = 0.05$ and $df = n-1 = 8$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} =2.306002.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(24-2.306\times\dfrac{4.1}{\sqrt{8}},24+2.306\times\dfrac{4.1}{\sqrt{8}})$$

$$=(20.848, 27.152)$$

Therefore, based on the data provided, the $95\%$ confidence interval for the population mean is $20.848<\mu< 27.152,$ which indicates that we are $95\%$ confident that the true population mean $\mu$ is contained by the interval $(20.848, 27.152).$

A value of $20$ does not lie in the 95% confidence interval.

We need to construct the $99\%$ confidence interval for the population mean $\mu.$

The critical value for $\alpha = 0.01$ and $df = n-1 = 8$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} =3.355361.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(24-3.355361\times\dfrac{4.1}{\sqrt{8}},24+3.355361\times\dfrac{4.1}{\sqrt{8}})$$

$$=(19.414, 28.587)$$

Therefore, based on the data provided, the $99\%$ confidence interval for the population mean is $19.414<\mu< 28.587,$ which indicates that we are $95\%$ confident that the true population mean $\mu$ is contained by the interval $(19.414, 28.587).$

A value of $20$ lies in the 99% confidence interval.