# Week Two Assignment Questions

Pages: 5 (1160 words) Published: July 25, 2013

Week Five Assignment Questions
Sharnez Lipscomb
PSY/315
July 18, 2013
Dennis Outcalt

Week Five Assignment Questions
Chapter 7
14.| | | Evolutionary theories often emphasize that humans have adapted to their physical environment. One such theory hypothesizes that people should spontaneously follow a 24-hour cycle of sleeping and waking—even if they are not exposed to the usual pattern of sunlight. To test this notion, eight paid volunteers were placed (individually) in a room in which there was no light from the outside and no clocks or other indications of time. They could turn the lights on and off as they wished. After a month in the room, each individual tended to develop a steady cycle. Their cycles at the end of the study were as follows: 25, 27, 25, 23, 24, 25, 26, and 25.Using the .05 level of significance, what should we conclude about the theory that 24 hours is the natural cycle? (That is, does the average cycle length under these conditions differ significantly from 24 hours?) (a) Use the steps of hypothesis testing. (b) Sketch the distributions involved, (c) Explain your answer to someone who has never taken a course in statistics. 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| | | | | | | |
| | | | | | | | | |
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|

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)
Sample > 1.895 Reject HO
Step 4: Sample = 2.366
Step 5: 2.66>2.365 reject the null hypothesis/HO
The average sleep cycle is not 24 hours according to the results; therefore we are allowed to reject the null hypothesis.

Chapter 8

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.a) Null hypothesis: µ1 = µ2

Alternate hypothesis: µ1 ≠ µ2
Pooled standard deviation = sqrt((19*3^2+29*5^2)/(20+30-2)) = 4.32 t-statistic = (38 – 30)/(4.32*sqrt(1/20+1/30)) = 2.4056
degree of freedom = 20+30-2 = 48
t-critical = 2.0106
As t-statistic is greater than critical t, we reject null hypothesis. Scores for two groups are not same. The experimental group had seemingly higher scores.  c) T-test for two sample independent means is same as using the t-test for single sample for the difference of two means. Only the standard error in this case is computed slightly differently as shown above using pooled standard deviation.Chapter 9 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, M = 4, S2 = 1.5; and for University Z,M = 6, S2 = 2.5. 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.Answer a)

Null hypothesis: Mean time for each college is same...