A given poll states 56 of sixth graders read at or above the

A given poll states \"56% of sixth graders read at or above the sixth-grade level.\"
If there is a stated margin of error of +/-1.8%, what is the 95% confidence interval for this statistic?
Express your answers rounded to the nearest tenth of a percent!

Lower value = %

Upper value = %

Solution

A given poll states \"56% of sixth graders read at or above the sixth-grade level.\" Thus, p=0.56.

Margin of error= +/-1.8%

The 95% confidence interval is

P’ +/- 0.018

=(0.56 +/- 0.018)

=(0.542,0.578)

Lower value 54.2%

Upper value 57.8%