Answer to Question #105680 in Statistics and Probability for suresh

Question

Suppose we create a 99% confidence interval for a sample size of 20 units with mean 200 standard deviation 30, what are the upper and lower limits of the interval?

A> 186.85 and 213.15
B> 184.37 and 215.63
C> 184.24 and 215.76
D> 182.69 and 217.30
E> None of the above

Answer 1

We need to construct the 99% confidence interval for the population mean $\mu$ with known population standard deviation $\sigma.$

Given that $\bar{X}=200, \sigma=30, n=20, \alpha=0.01.$

The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha/2}=2.576\approx2.58$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-z_c\cdot{\sigma \over \sqrt{n}}, \bar{X}+z_c\cdot{\sigma \over \sqrt{n}})\approx$$

$$\approx(200-2.58\cdot{30 \over \sqrt{20}}, 200+2.58\cdot{30 \over \sqrt{20}})\approx$$

$$\approx(182.69,\ 217.30)$$

D) $182.69$ and $217.30$

We need to construct the 99% confidence interval for the population mean $\mu$ with unknown population standard deviation $(\sigma)$ for which reason the sample standard deviation $(s)$ is used instead.

Given that $\bar{X}=200,s=30, n=20, \alpha=0.01.$

The critical value for $\alpha=0.01, df=20-1=19$ is $t_c=2.861.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-t_c\cdot{s \over \sqrt{n}}, \bar{X}+t_c\cdot{s \over \sqrt{n}})\approx$$

$$\approx(200-2.861\cdot{30 \over \sqrt{20}}, 200+2.861\cdot{30 \over \sqrt{20}})\approx$$

$$\approx(180.81,\ 219.19)$$

E) None of the above