Answer to Question #329446 in Statistics and Probability for Moon

Question

a random sample of size n 16 is drawn from a normally distributed population. the sample mean is 98 with known standard deviation of 5. Construct a 98% confidence interval for the population mean.

Answer 1

Let n = 16 - sample size, $\mu=98$ - mean, $\sigma=5$ - standard deviation.

t-value for n - 1 = 15 degrees of freedom and 98% confidence is 2.602.

Confidence interval:

$\mu=\mu_0\pm t\frac{\sigma}{\sqrt{n}}=98\pm 2.602\frac{5}{\sqrt{16}}=\\ =98\pm 3.2525\approx98\pm3.3$