Question

A consumer affairs investigator records the repair cost for 4 randomly selected TVs. A sample mean of $91.78 and standard deviation of $23.13 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

= ( $72.756, $110.804)

Therefore, the 90% confidence interval (a,b) = ( $72.756, $110.804)

Critical value at 90% confidence = 1.645

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 = $91.78

Standard deviation r = $23.13

Number of samples n = 4

Confidence interval = 90%

Using the z table;

z(α=0.05) = 1.645

Critical value at 90% confidence = 1.645

Substituting the values we have;

$91.78+/-1.645($23.13/√4)

$91.78+/-1.645($11.565)

$91.78+/-$19.024425

$91.78+/-$19.024

= ( $72.756, $110.804)

Therefore, the 90% confidence interval (a,b) = ( $72.756, $110.804)