Answer to Question #340287 in Statistics and Probability for dharyl

Question

the average grade of 30 students in their statistics and probability examination is 85 with a standard deviation of 5. find the interval for the true mean at 95% interval

Answer 1

The critical value for $\alpha = 0.05$ and $df = n-1 = 29$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} =2.04523.$

The corresponding confidence interval is computed as shown below:

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

$$=(85-2.04523\times\dfrac{5}{\sqrt{30}},85+2.04523\times\dfrac{5}{\sqrt{30}})$$

$$=(83.133, 86.867)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $83.133 < \mu < 86.867,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(83.133, 86.867).$