Question

A random sample is drawn from a population of known standard deviation 11.3. Construct 90%confidence interval for the population mean based on the information given (not all of the information given need be used).

a. n=36, x¯=105.2, s=11.2n=36, x¯=105.2, s=11.2

b. n=100, x¯=105.2, s=11.2n=100, x¯=105.2, s=11.2

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 $\alpha = 0.10, df=n-1=35$ degrees of freedom is $t_c=z_{1−α/2;n−1}=1.689572.$

$$CI=(105.2-1.689572\times\dfrac{11.2}{\sqrt{36}},$$

$$105.2+1.689572\times\dfrac{11.2}{\sqrt{36}})$$

$$=(102.046,108.354)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $102.046< \mu < 108.354,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(102.046,108.354).$

b) The critical value for $\alpha = 0.10, df=n-1=99$ degrees of freedom is $t_c=z_{1−α/2;n−1}= 1.660391.$

$$CI=(105.2- 1.660391\times\dfrac{11.2}{\sqrt{100}},$$

$$105.2+ 1.660391\times\dfrac{11.2}{\sqrt{100}})$$

$$=(103.340,107.060)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $103.340< \mu < 107.060,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(103.340,107.060).$