Answer to Question #214089 in Statistics and Probability for Madimetja

Question

A random sample of 16 values of x, obtained from a N(u, Q²)distribution, had a sample mean of 4:2 and a sample variance of 3: Find the 95% confidence interval for u

Answer 1

The critical value for $\alpha=0.05$ and $df=n-1=16-1=15$ degrees of freedom is $t_c=2.131449.$

The corresponding confidence interval is computed as shown below:

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

$$=(4.2-2.131449\times\dfrac{ \sqrt{3}}{\sqrt{16}}, 4.2+2.131449\times\dfrac{\sqrt{3}}{\sqrt{16}})$$

$$=(3.277, 5.123)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $3.277<u<5.123,$ which indicates that we are 95% confident that the true population mean $u$ is contained by the interval $(3.277, 5.123).$