Question

Problem 3 (CLO 1, 2, 3, 4) A sample of elementary school children from Abu Dhabi participated for two years in a nutrition program organized by their school. At the end of the second year, the heights of the students were measured in meters as 1.01, .95, 1.03, 1.04, 0.97, 0.97, 0.99, 1.01, and 1.03. Assume the population of elementary school children heights is normally distributed with mean µ = 1.04, but ? unknown. 1) What is the mean (average) student height in the sample? 2) What is the variance of the student heights in the sample? 3) What is the standard deviation of the student heights in the sample? 4) Given that µ = 1.04, what is the probability of observing a sample mean height greater than or equal to the value of your computed sample mean in question (1) ? 5) What is the 99 percent confidence interval for the mean student height from question (1)? Give an interpretation for this interval? 6) What is the 95 percent confidence interval for this mean student height? Give an interpretation for this interval?

Answer 1

Following table shows the calculations:

	X	(X-mean)^2
	1.01	0.0001
	0.95	0.0025
	1.03	0.0009
	1.04	0.0016
	0.97	0.0009
	0.97	0.0009
	0.99	0.0001
	1.01	0.0001
	1.03	0.0009
Total	9	0.008

1)

The mean (average) student height in the sample is

2)

The variance of the student heights in the sample is

3)

The standard deviation of the student heights in the sample is

4)

The sampling distriution of sample mean will follow t distribution with df=9-1=8.

The t-score is

The probability of observing a sample mean height greater than or equal to the value of your computed sample mean in question (1) is

5)

The critical value of t for 99% confidence interval is 3.355. The required confidence interval is

6)

The critical value of t for 95% confidence interval is 2.306. The required confidence interval is