Answer to Question #273740 in Statistics and Probability for grace

Question

You are testing whether students at your school are overweight. You sample 25 students and measure how much more they weigh than the average weight for their height. The mean of your sample is 5 lbs, and the standard deviation is 5 lbs. 

Find a 99% confidence interval for the true population mean. (Round to THREE decimal places.)

Answer 1

The critical value for $\alpha = 0.01$ and $df = n-1 = 24$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.79694$

The corresponding confidence interval is computed as shown below:

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

$$=(5-2.79694\times\dfrac{5}{\sqrt{25}}, 5+2.79694\times\dfrac{5}{\sqrt{25}})$$

$$=(2.203, 7.797)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $2.203 < \mu < 7.797,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(2.203, 7.797).$