Question

A random sample is drawn from a population of known standard deviation 22.1. Construct a 95% confidence interval for the population mean based on the information given (not all of the information given need be used).

a. n=121, x¯=82.4, s=21.9 n=121, x¯=82.4, s=21.9

b. n=81, x¯=82.4, s=21.9 n=81, x¯=82.4, s=21.9

Answer 1

The corresponding confidence interval is computed as shown below:

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

a) The critical value for $\alpha = 0.05, df=n-1=120$ degrees of freedom is $t_c=z_{1−α/2;n−1}=1.97993.$

$$CI=(82.4-1.97993\times\dfrac{21.9}{\sqrt{121}},$$

$$82.4+1.97993\times\dfrac{21.9}{\sqrt{121}})$$

$$=(78.458,86.342)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $78.458< \mu < 86.342,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(78.458,86.342).$

b) The critical value for $\alpha = 0.05, df=n-1=80$ degrees of freedom is $t_c=z_{1−α/2;n−1}=1.990063.$

$$CI=(82.4-1.990063\times\dfrac{21.9}{\sqrt{81}},$$

$$82.4+1.990063\times\dfrac{21.9}{\sqrt{81}})$$

$$=(77.558,87.242)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $77.558< \mu < 87.242,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(77.558,87.242).$