# Week 5 Practice Problems Topics: Statistical hypothesis testing, Statistics, Statistical significance, Null hypothesis, Arithmetic mean / Pages: 5 (1211 words) / Published: Mar 4th, 2014

Practice Problems
Ch. 7, Practice Problem: 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.
Answer
Null hypothesis H0: = 24 hours Alternative hypothesis: H1: ≠ 24 hours
Df = 7 Critical t value = ±2.36 Sample Mean = 25
Standard deviation = 1.195 The test statistic used is
P-value = 0.049867231

Since calculated p-value 0.049867231 is slightly less than 0.05(significance level) therefore we reject the null hypothesis.
Because there is enough evidence to support the claim that the average cycle length under experimental conditions is significantly different from 24 hours

Ch. 8, Practice Problem: 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