# A group of 62 randomly selected students have a mean score of 28.3 on a standardized placement test. The population SD for all students taking the test is sigma = 3.4. What is the 90 percent confidence interval for the mean score?

Jul 13, 2016

Population mean is likely to fall between $29.005$ and $27.595$

#### Explanation:

Given -

$\overline{x} = 28.3$
$n = 62$
$\sigma = 3.4$
$z$ score for 90% confidence interval $1.64$
$S E = \frac{\sigma}{\sqrt{n}} = \frac{3.4}{\sqrt{62}} = \frac{3.4}{7.87} = 0.43$

90% confidence interval is defined by the formula

$\mu = \overline{x} + \left(S E \times z\right)$ --------upper limit
$\mu = 28.3 + \left(0.43 \times 1.64\right) = 28.3 + 0.705 = 29.005$

$\mu = \overline{x} - \left(S E \times z\right)$ --------Lower limit
$\mu = 28.3 - \left(0.43 \times 1.64\right) = 28.3 - 0.705 = 27.595$

Sample means are normally distributed around the population mean $\mu$

An interval is developed as $29.005$ and $27.595$. There is 90% chance for the population mean to fall in this interval.