Question

The scores on an examination in economics are approximately normally distributed with mean 500 and an unknown standard deviation. The following is a random sample of scores from this examination. 422, 437, 472, 502, 534 Send data to Excel Find a 90% confidence interval for the population standard deviation. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)

What is the lower limit of the 90% confidence interval?

What is the upper limit of the 90% confidence interval?

Answer #1

Here sample mean = = 473.4

sample standard deviation = s = 45.99

as standard deviation is unknown, we will use t distribution to find critical value of t

Degree of freedom= Df = n -1 = 5-1 = 4

tcritical = TINV(0.10, 4) = 2.132

Standard error = se = s/sqrt(n) = 45.99/sqrt(5) = 20.566

lower limit of the 90% confidence interval =
- tcritical * se = 473.4 - 20.566 * 2.132 =
**429.55**

Upper limit of the 90% confidence interval =
+ tcritical * se = 473.4 + 20.566 * 2.132 =
**517.25**

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