Answer to Question #206289 in Statistics and Probability for Tano

Question

    A sample of 60 Grade 11 student’s ages was obtained to estimate the mean age of all Grade 11 students. X=16.7 years and the population variance is 16.

1. Find the 90% confidence interval for ?

1. Find the 95% confidence interval for ?

1. What conclusions can you make based on each estimate

Answer 1

1.1

The critical value for $\alpha=0.1$ is $z_c=z_{1-\alpha/2}=1.6449.$

The corresponding confidence interval is computed as shown below:

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

$$=(16.7-1.6449\times\dfrac{4}{\sqrt{60}}, 16.7+1.6449\times\dfrac{4}{\sqrt{60}})$$

$$\approx(15.8506, 17.5494)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $15.8506<\mu<17.5494,$ which indicates that we are 90% confident that the true population mean $\mu$

is contained by the interval $(15.8506, 17.5494).$

1.2

The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$

The corresponding confidence interval is computed as shown below:

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

$$=(16.7-1.96\times\dfrac{4}{\sqrt{60}}, 16.7+1.96\times\dfrac{4}{\sqrt{60}})$$

$$\approx(15.6879, 17.7121)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $15.6879<\mu<17.7121,$ which indicates that we are 95% confident that the true population mean $\mu$

is contained by the interval $(15.6879, 17.7121).$

1.3

The width of the confidence interval will be larger when the confidence level is higher (because you can have greater confidence when you are less precise).