Question

2) In a hospital, a random sample of 12 weeks was selected, and it was found that an average sample of 537 patients were treated in the emergency room each week. The standard deviation was 21.82. Assuming that the variable is normally distributed, find the 90% confidence interval of the true mean. 90% confidence interval:

Answer 1

solution

Degrees of freedom = df = n - 1 = 12 - 1 = 11

At 90% confidence level the t is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.1

$\alpha$ / 2 = 0.1 / 2 = 0.05

t$\alpha$ /2,df = t 0.05,11 = 1.796 ( using student t table)

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 1.796 * (21.82 / $\sqrt$ 12) = 11.3

The 90% confidence interval is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

537 - 11.3 < $\mu$ < 537 + 11.3

525.7 < $\mu$ < 548.3