Question

A company is interested in estimating , the mean number of days of sick leave taken by its employees. The firm's statistician randomly selects 100 personnel files and notes the number of sick days taken by each employee. The sample mean is 12.2 days and the sample standard deviation is 10 days. Calculate a 93% confidence interval for , the mean number of days of sick leave. Assume the population standard deviation is 10.

Answer 1

Answer:

93% confidence interval for the mean number of days of sick leave is [10.39 days , 14.01 days].

Step-by-step explanation:

We are given that the firm's statistician randomly selects 100 personnel files and notes the number of sick days taken by each employee. The sample mean is 12.2 days and the sample standard deviation is 10 days.

Assume the population standard deviation is 10.

Firstly, the pivotal quantity for 93% confidence interval for the population mean is given by;

                          P.Q. = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\bar X$ = sample mean = 12.2 days

            $\sigma$ = population standard deviation = 10 days

            n = sample of files = 100

            $\mu$ = population mean

Here for constructing 93% confidence interval we have used One-sample z  test statistics because we know about population standard deviation.

So, 93% confidence interval for the population mean, $\mu$ is ;

P(-1.8121 < N(0,1) < 1.8121) = 0.93  {As the critical value of z at 3.5%

                                           level of significance are -1.8121 & 1.8121}  

P(-1.8121 < $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.8121) = 0.93

P( $-1.8121 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.8121 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.93

P( $\bar X-1.8121 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.8121 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.93

93% confidence interval for $\mu$ = [ $\bar X-1.8121 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X +1.8121 \times {\frac{\sigma}{\sqrt{n} } }$ ]

                                       = [ $12.2-1.8121 \times {\frac{10}{\sqrt{100} } }$ , $12.2+1.8121 \times {\frac{10}{\sqrt{100} } }$ ]

                                       = [10.39 days , 14.01 days]

Therefore, 93% confidence interval for the mean number of days of sick leave is [10.39 days , 14.01 days].