Question

There are 350 Families lives in the small town of America. A poll of 50 families revealed the mean annual church contribution is $550 with the standard deviation of $ 75.Construct the 95% confidence interval for the mean annual contribution

Answer 1

$550+/-$20.79

= ( $529.21, $570.79)

Therefore, the 95% confidence interval (a,b) = ( $529.21, $570.79)

Explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $550

Standard deviation r = $75

Number of samples n = 50

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

$550+/-1.96($75/√50)

$550+/-1.96($10.60660171779)

$550+/-$20.7889393668

$550+/-$20.79

= ( $529.21, $570.79)

Therefore, the 95% confidence interval (a,b) = ( $529.21, $570.79)