Question

1) Use the given confidence interval to find the margin of error and the sample mean. ​(12.8​,19.8​)

2) In a random sample of four microwave​ ovens, the mean repair cost was ​$60.00 and the standard deviation was ​$12.00. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean ?. What is the margin of error of ? Interpret the results.

Answer 1

1)

Margin of error = Width of confidence interval / 2 = (19.8 - 12.8) / 2 = 3.5

2)

df = n - 1 = 4 - 1 = 3

From T table, critical value at 0.01 significance level with 3 df = 5.841

99% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

60 - 5.841 * 12 / sqrt(4) < < 60 + 5.841 * 12 / sqrt(4)

24.95 < < 95.05

99% CI is ( 24.95 , 95.05 )

margin of error =  t * S / sqrt(n)

= 5.841 * 12 / sqrt(4)

= 35.05

Interpretation = We are 99% confident that true mean repair cost is between $24.95 and $95.05