Question

The prices of CD players are normally distributed, with a standard deviation of $12. A random sample of 16
CD player prices is drawn. A confidence interval for the mean price shows a $9.30 margin of error. What was
the level of confidence?

Answer 1

Given that, standard deviation $(\sigma)$ = $12

sample size ( n ) = 16

Margin of error ( E ) = $9.30

where,

$E = Z_{\alpha/2} * \frac { \sigma}{\sqrt { n }}$

$=>9.30= Z_{\alpha/2} * \frac { 12}{\sqrt { 16}}$

$=>9.30= Z_{\alpha/2} * \frac { 12}{4}$

$=>9.30= 3 * Z_{\alpha/2}$

$=> Z_{\alpha/2} = \frac { 9.30}{3}$

$=> Z_{\alpha/2} = 3.10$

Using standard normal z-table we get,

P(Z < 3.10) = 0.9990

Therefore, we get, $\alpha/2 = 1- 0.9990 = 0.001$

$\alpha = 2 * 0.001 = 0.002$

Hence, level of confidence ( c ) = $1-\alpha = 1- 0.002 = 0.998$

Level of confidence =  99.8%