Question

The sick days of employees every two years in a company are normally distributed with a population standard deviation of 7 days and an unknown population mean. If a random sample of 20 employees is taken and results in a sample mean of 21 days, find a 95% confidence interval for the population mean.

Answer 1

Explanation:

Let
`$X_1, X_2, X_3,$` . . . ,Xn be the random sample of n employee's sick days. It is given that the random samples follows the Normal distribution along with standard deviation of 7 days. Let

`X_i\sim N(\mu ,7)`

`\bar{X} =(1)/(n)\Sigma X_i\sim N(\mu,(7)/(\sqrt n))`

or
`Z=\frac{\bar{X}-\mu}{7 /\sqrt n} \sim N(0,1)`

So,

`P(-Z_(\alpha /2) \leq Z\leq Z_(\alpha /2) ) = 1- \alpha`

`P(-Z_(\alpha /2) \leq \frac{\bar{X}- \mu}{7 / \sqrt n}\leq Z_(\alpha /2) ) = 1- \alpha`

`P(\bar{X}-(7)/(\sqrt n) Z_(\alpha / 2) \leq \mu \leq \bar{X}+(7)/(\sqrt n) Z_(\alpha / 2) ) = 1- \alpha`

Therefore, the confidence interval of the population mean for α = 0.05 is

=
`P(\bar{X}-(7)/(\sqrt n) Z_(\alpha / 2) , \bar{X}+(7)/(\sqrt n) Z_(\alpha / 2) )`

=
`P(21-(7)/(\sqrt 20) Z_(0.05 / 2) , 21+(7)/(\sqrt 20) Z_(0.05 / 2) )`

= (17.93, 24.07)