An english teacher has been teaching a sixth grade composition class for many years. he has the feeling that over the past several years, the writing ability of students has changed. a national test of proficiency in composition was administered 5 years ago. the resulting distribution of scores was normally shaped, had a mean of 85 and a standard deviation of 10.9. in order to test his feeling, he gives his present class of 43 students the same proficiency test. the resulting mean is 80 and the standard deviation is 8.7. using the z test with α= 0.052 tail, the appropriate critical value

We have here as as , then the values are.

s=8.7

For the resulting of the test by the teacher he had,

With all of this dates to make the comparition we use the formula for Z values, that is

We know moreover that = 0.052.

To find we need to find

Searching in the table of Normal Distribution for Z, and making the lecture we find that z_{critic} is, 1.95.

256π units²   (answer b)

step-by-step explanation:

the formula for the surface area of a sphere of radius r

is             a = 4πr²

and so, if r = 8 units, then the surface area is:

a = 4π(8 units)² = 256π units²   (answer b)

1. d.90

2. e. none of these (80)

3. a. 0

4. b. 6

5. e. none of these (10)

6. a. 5