Question

A magazine reported the mean annual household income of its readers to be $150,000 with a standard deviation of $30,000. A recent sample of 80 households from among those subscribing to this magazine found a mean of $139,155. What is the upper bound of a 95% confidence interval for the mean income (correct to no decimal places)?

Answer 1

SOLUTION:

From given data,

A magazine reported the mean annual household income of its readers to be $150,000 with a standard deviation of $30,000. A recent sample of 80 households from among those subscribing to this magazine found a mean of $139,155. What is the upper bound of a 95% confidence interval for the mean income (correct to no decimal places)?

Given

= $150,000,

Standard deviation = = $30,000,

= $139,155,

n = 80 and

95% confidence interval

Confidence interval is 95%

95% = 95/100 = 0.95

= 1 - Confidence interval = 1-0.95 = 0.05

/2 = 0.05 / 2

= 0.025

Z_/2 = Z_0.025 = 1.96

The Upper Bound of a 95% confidence interval

= + Z_/2 *( /sqrt{n})

= 139155 + 1.96 * (30000/sqrt(80))

=139155 + 6574.0398

= 145729.0398

  145,729