statistics Suppose in a random sample of 400 registered nurs

Suppose in a random sample of 400 registered nurses in Texas, it was found that the average salary was .V-S67.S60. Estimate that aver-! if salary i.e. n, of all registered nurses in Texas, by a 95 percentage confidence interval. Assume the standard deviation in the population is sigma= $6240.

Solution

For 95% confidence interval ,

z = 1.96

Margin of Error ME = z * SD / sqrt(n) = (1.96 * 6240) / sqrt(400) = 611.52

Confidence Interval:

( Mean - ME , Mean + ME )

= ( 67580 - 611.52 , 67580 + 611.52)

= (66968.48 , 68191.52) Answer