# Psy 315 Final

**Topics:**Statistics, Statistical hypothesis testing, Standard deviation

**Pages:**5 (1051 words)

**Published:**May 9, 2011

Part A.

Step 1.

Ho=M=24(null)

H1=M24(alternative)

Step 2.

X| | |

25| 0| 0|

27| 2| 4|

25| 0| 0|

23| -2| 4|

24| -1| 1|

25| 0| 0|

26| 1| 1|

25| 0| 0|

x= 200 S2=10/7= 1.42857

S2M=S2/N=10÷7/8=0.17857

SM=Sm2= 0.422577

Step 3.

(=.025, DF=7) =2.365 =.05 /2= .025 (two tailed test) So if the sample > 1.895

We reject H0

Step 4.

Sample = 2.366

Step 5. Since 2.66>2.365 we will reject the null hypothesis/HO

Part B.

.

The results show that the average sleep cycle is not 24 hours. This allowed us to reject the null hypothesis. We used the data to determine if the average person’s sleep cycle was 24 hours. After examining the information we see that is not the case.

18. Twenty students randomly assigned to an experimental group receive an instructional program; 30 in a control group do not. After 6 months, both groups are tested on their knowledge. The experimental group has a mean of 38 on the test (with an estimated population standard deviation of 3); the control group has a mean of 35 (with an estimated population standard deviation of 5). Using the .05 level, what should the experimenter conclude? (a) Use the steps of hypothesis testing, (b) sketch the distributions involved, and (c) explain your answer to someone who is familiar with the t test for a single sample but not with the t test for independent means.

Exp Control a=.05

N1=20 n2=30

M1=38 m2= 35

S1=3 s2= 5

Step 1.

Ho:M1=M2

H1:M1>M2

Step 2.

S2 = 19(3)2 + 29(5)2 = 896 = 18.667

30+20-2 48

(SM1) 2 = 18.667/20 (SM2)2 = 18.667/30

(SM1)2 = .9333(SM2)2 = .62222

S2 Difference = S2M1 + S2M2 = 1.5556

DF= 20+30-2=48 Sdiff= 1.5556 = 1.247

Step 3

Cut off (=..05, DF= 48) = 1.677

Step 4

Sample = 38-35 = 2.405

1.247

Step 5

Since 2.405 > 1.677 Reject H0

This exprerimanet shows that the on average the experiment group scored better than the controlled group. This is show that the treatment was successful.

17. Do students at various universities differ in how sociable they are? Twenty-five students were randomly selected from each of three universities in a region and were asked to report on the amount of time they spent socializing each day with other students. The result for University X was a mean of 5 hours and an estimated population variance of 2 hours; for University Y, ; and for University Z, . What should you conclude? Use the .05 level. (a) Use the steps of hypothesis testing, (b) figure the effect size for the study; and (c) explain your answers to parts (a) and (b) to someone who has never had a course in statistics.

College XCollegeYCollege Z

M=5 S2=2M=4 S2=1.5M=6 S2=2.5

N=25N=25N=25

Step 1

Null H0: M1=M2=M3 (DON’T EXPECT DIFFERENCE)

H1: At least one of them...

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