Question

A recent journal article reports that the average life expectancy for all women who live in poverty plausibly lies between 68.4 and 75.2 at approximate 95% confidence. Their data comes from a sample of 70 women. (Use 1 decimal place.)

a. Give the value of x for the sample in the study.

b. Give the value of the margin of error for the sample in this study.

Answer 1

Solution :

The 95% confidence interval = ( 68.4, 75.2 )

a)   = (Lower confidence interval + Upper confidence interval ) / 2

Sample mean = = (68.4 + 75.2) / 2

Sample mean = = 71.8

b) Margin of error = E = Upper confidence interval -

Margin of error = E = 75.2 - 71.8

Margin of error = E = 3.4