Question

The average cost of living for a family of four in twelve different cities was found to be $65,351 with a standard deviation of $7,711. What is a 90% confidence interval for the true mean

Answer 1

Answer:

$65,351+/-$3,661.73

= ( $61,689.27, $69,012.73)

Therefore, the 90% confidence interval (a,b) = ( $61,689.27, $69,012.73)

Step-by-step 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 = $65,351

Standard deviation r = $7,711

Number of samples n = 12

Confidence interval = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

$65,351+/-1.645($7,711/√12)

$65,351+/-1.645($2,225.973962860)

$65,351+/-$3661.727168905

$65,351+/-$3,661.73

= ( $61,689.27, $69,012.73)

Therefore, the 90% confidence interval (a,b) = ( $61,689.27, $69,012.73)