Answer to Question #190309 in Statistics and Probability for fizzah

Question

Arterial blood gas analyses performed on a sample of 15 physically active adult males yielded the following resting values: 75, 80, 80, 74, 84, 78, 89, 72, 83, 76, 75, 87, 78, 79, 88 Compute the 90 and 95 percent confidence interval for the mean of the population

Answer 1

Solution:

Mean$=\mu=\bar{x}=\dfrac{\Sigma x_i}{n}$ $=\dfrac{75+80+ 80+ 74+ 84+ 78+ 89+ 72+83+ 76+ 75+ 87+ 78+ 79+ 88}{15}$

$=\dfrac{1198}{15}=79.867$

Variance$=\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}$

$=\dfrac{(75-\dfrac{1198}{15})^2+(80-\dfrac{1198}{15})^2+...+(88-\dfrac{1198}{15})^2}{15}$

$=\dfrac{393.733}{15}=26.2488$

Then, standard deviation, $\sigma=\sqrt{26.2488} =5.123$

Now, confidence interval, CI$=(\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}})$

For 90% CI, $z=1.645$

So, 90% CI$=(79.867\pm1.645\times \dfrac{5.123}{\sqrt{15}})=(77.791,82.142)$

For 95% CI, $z=1.96$

So, 95% CI$=(79.867\pm1.96\times \dfrac{5.123}{\sqrt{15}})=(77.374,82.559)$