Question

Find the mean, mod, median, and standard deviation of the following data. And Based on these results, check whether the value of 10 is usual? 5, 6, 7, 8, 9,8,7,8 _________________________________________________________________________ Pre-Employment Drug Screening Results are shown in the following Table: Positive Test Result Negative Test Result Subject Uses Drugs 8 (True Positive) 2 (False Negative) Subject is not a Drug User 10 (False Positive) 180 (True Negative) If 1 of the 200 test subjects is randomly selected, find the probability that the subject had a positive test result, given that the subject actually uses drugs. That is, find (positive test result subject uses drugs). If 1 of the 200 test subjects is randomly selected, find the probability that the subject actually uses drugs, given that he or she had a positive test result. That is, find ( subject uses drugs positive test result ). _______________________________________________________________________ This is observation from previous years about the impact of students working while they are enrolled in classes, due to students too much work, they are spending less time on their classes. First, the observer need to find out, on average, how many hours a week students are working. They know from previous studies that the standard deviation of this variable is about 5 hours. A survey of 200 students provides a sample mean of 7.10 hours worked. What is a 95% confidence interval based on this sample?

Answer 1

(1)

(i)

Mean = 58/8 = 7.25

(ii)

Mode = 8

because 8 occurs maximum number of times

(iii)

Arranging numbers in ascending order, we get:

5,6,7,7,8,8,8,9

n = 8

Median = (8 + 1)/2th item = Average of 4th & 5th items = (7 + 8)/2 = 7.5

So,

Median = 7.5

(iv)

From the given data, the following Table is calculated:

x	(x - )	(x - )2
5	- 2.25	5.0625
6	- 1.25	1.5625
7	- 0.25	0.0625
8	0.75	0.5625
9	1.75	3.0625
8	0.75	0.5625
7	- 0.25	0.0625
8	0.75	0.5625
Total =		11.5

Standard Deviation (s) is given by:

(v)

Since 10 is greater than , we conclude:

10 is not usual.

(2)

From the given data, the following Table is calculated:

	Positive Test Result	Negative Test Result	Total
Subject uses drug	8 (True Positive)	2	10
Subject is not a drug user	10	180 (True Negative)	190
Total	18	182	200

(a)
P(Positive Test Result/ Subject uses drug) = P(Positive Test Result AND Subject uses drug)/ P(Subject uses drug)

                                                             = 8/10

                                                           = 0.8

So,

Answer is:

0.8

(b)
P(Subject uses drug/ Positive Test Result) = P(Subject uses drug AND Positive Test Result )/ P( Positive Test Result )

                                                             = 8/18

                                                           = 0.4444

So,

Answer is:

0.4444

(3)

SE = /

= 5/

= 0.3536

= 0.05

From Table, critical values of Z = 1.96

Confidence Interval:

7.10 (1.96 X 0.3536)

= 7.10 0.6930

= ( 6.4070,7.7930)

Confidence Interval:

6.4070 <   <   7.7930