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.
(=.025, DF=7) =2.365 =.05 /2= .025 (two tailed test) So if the sample > 1.895
We reject H0
Sample = 2.366
Step 5. Since 2.66>2.365 we will reject the null hypothesis/HO
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
M1=38 m2= 35
S1=3 s2= 5
S2 = 19(3)2 + 29(5)2 = 896 = 18.667
(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
Cut off (=..05, DF= 48) = 1.677
Sample = 38-35 = 2.405
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.
Null H0: M1=M2=M3 (DON’T EXPECT DIFFERENCE)
H1: At least one of them...
Please join StudyMode to read the full document