PSY 555 Homework 5

3.8.  Diagnostically meaningful cutoff:

Z Score      Area above Z

1.2800       .1003

1.2817       .1000

1.2900       .0985

z=

-1.2817=

111.549=

The diagnostically meaningful cutoff would be 112.

3.9(a).

Next year's salary raises:

Z=

-1.2817=

X=\$2512.68

So the most productive 10% of the faculty will have a raise equal to or greater than \$2512.68.

3.9(b).

z=

-1.645=

X=\$1342

The 5% of the faculty who haven't done anything useful in years will receive no more than \$1342 each and probably don't deserve that much.

3.10(a).

3.10(b).

z==

A count this high (or higher) would occur by chance only .5% of the time.  Thus, there is definite reason to suspect that this count has been fabricated by the student.

3.11. Transforming scores on a diagnostic test for language problems:

X1=original scores          m1=48     s1=7

X2=transformed scores       m2=80     s2=10

s2=s1/C

10=7/C

C=.7

Therefore, to transform the original standard deviation from 7 to 10, we need to divide the original scores by .7.  However, dividing the original scores by .7 divides their mean by .7.

We want to raise the mean to 80.80-68.57=11.43.  Therefore, we need to add 11.43 to each score.

X2= X1/.07+11.43 [This formula summarizes the whole process].

3.13. October 1981 GRES, all people taking exam:

z=

z=

p(larger portion)=.81

A GRE score of 600 would correspond to the 81st percentile.

3.14. For the data in Exercise 3.13:

z=

.6754=

X=573.987

Z Score      P

.6700        .7486

.6745        .7500

.6800        .7517

A GRE score of (.6745*126+489)=574 would correspond to the 75th percentile.

3.15. October 1981 GRE, all seniors and nonenrolled college graduates.

z=                z=

z=          .6745=

p=.785                 X=586.591

3.18. Diagnostically meaningful cutoff for the Behavior Problem scores:

z=

.6745=

X=cutoff=70.54

Z Score      P

2.050        .9798

2.054        .9800

2.060        .9803

1.    A.  We assume that many variables are normally distributed in the population, so the

normal distribution is a good approximation of the population distribution.

B.  Assumption of normality allows us to use tools to draw inferences about variable values.

C.  If we drew the distribution of the sample means of many different samples, it would theoretically be a normal distribution (i.e., the sampling distribution of the means would be a normal distribution).

D.  Most statistical tests that are commonly used assume a normal distribution.

(See p. 76 in the book for elaboration)

2(a). z=

z=.25.0987

z=.50.1915

.1915-.0987=.0928

9.28% of the distribution will fall between z-scores of .25 and .50.

2(b). z=

z=-1.5.4332

z=.5.1915

.4332+.1915=.6247

62.47% of the distribution will fall between z-scores of -1.5 and .5.

2(c). z=

z=1.5.4332

43.32% of the distribution will fall between z-scores of 0 and 1.5.

2(d). z=

z=.50.1915

z=1.75.4599

.4599-.1915=.2684

26.84% of the distribution will fall between z-scores of .50 and 1.75.

3(a).

The value of X is 38.92.

3(b).

The value of  is 19.05.

3(c). z=

The value of  is 18.31.

4.

z=

z=

z=-.5.1915+.5=.6915

The probability/likelihood that John will be late to his 11:30 class if he leaves his house at 11:10 is 69.15%.

5.

The cutoff scores for assigning students to the three math courses should be 81.305 and 92.695.