Question

8 points Answer 005 Pound your answer A random sample of 45 fresh graduate has a mean starting camnings of $3250. Assume the population standard deviation is 675. Construct the confidence interval for the population means the nearest integer 2 27)

Answer 1

Solution :

c = 0.95, it means we need to construct 95% confidence interval.

The 95% confidence interval for population mean is given as follows :

$\large \left ( \bar{x}\pm Z_{\frac{0.05}{2}}\frac{\sigma}{\sqrt{n}} \right )$

Where, x̄ is sample mean, σ is population standard deviation and n is sample size and Z(0.05/2) is critical z-value to construct 95% confidence interval.

We have,  x̄ = $3250, σ = $675, n = 45

Using z-table we get, Z(0.05/2) = 1.96

Hence, 95% confidence interval for the true mean is,

$\large \left ( 3250 \pm 1.96\times\frac{675}{\sqrt{45}} \right )$

$\large =\left ( 3250 \pm 197.22 \right )$

$\large =\left ( 3250 - 197.22, 3250 + 197.22 \right )$

$\large =\left ( 3052.78, 3447.22 \right )$

On rounding to nearest integer we get,

$\large =\left ( 3053, 3447 \right )$

The 95% confidence interval for population mean is,

($3053, 3447).