Hypothesis Testing - Summary
Hypothesis Testing - Summary
Hypothesis testing involves four steps: (1) develop evidence supporting.ii judgment; (2) conceptualize the null hypothesis; (3) determine the ^probability of obtaining the evidence if the null hypothesis were true; and (4) calculate the probability—that is, the p-value—of the hypothesis test. There are several points worth remembering about hypothesis testing:
1. Hypothesis testing is a screening test. If the evidence does not pass this test, it may not be worth much attention. If it does pass this test, then it might at least be worth further analysis.
2. Hypothesis testing really measures the sample size. A large sample nearly always will yield "statistically significant" results, and a small enough sample probably will not be statistically significant. Thus, the test really does no more than provide a measure of sample size.
3. Hypothesis testing does not establish whether the null hypothesis is true or false: it only quantifies how persuasive the evidence is against it. Alow p-value indicates impressive evidence, and a .hjLgh^zyalue indicates the evidence is not impressive!
A cross-tabulation has associated with it a dii-square test of the relationship between the two variables. The null hypothesis is that the two variables are statistically independent, that is, knowledge of one provides no information about the other. The resulting p-value represents the probability under the null hypothesis of getting a chi-square statistic as large or larger than the one observed. If the p-value is low, then the evidence is rather impressive that the independence assumption is not true. If it is high, then the evidence is very weak that the independence assumption is not true.
The difference between means is a commonly used statistic in data analysis, particularly when experimentation is involved. The appropriate null hypothesis in this context is that there is no difference between means in the population. The p-level then represents the probability that the observed differences between means could have been that large if, in fact, the null hypothesis that all means are the same were true. The test is based on the F-statistic and the accompanying F-distribution. The extended ANOVA table provides the ability to conduct several difference-between-means tests at the same time.
b. Interpret the term Ex = 17.8
c. Calculate the x2 value.
d. Is the x2 value significant? At what level? What exactly is the
null hypothesis?
e. This data set proves that the usage of the library differs by
age. True or false? Why?
5. Determine the p-value and the significance level of a hypothesis test on the table presented in problem 4 of Chapter 13. What is the null hypothesis? Interpret the p-value.
6. An experiment was conducted to determine which of three advertisements to use in introducing a new personal computer. A total of 120 people who were thinking of buying a personal computer was split randomly into three groups of 40. Each of the groups was shown a different advertisement and each person was asked his or her likelihood of buying the advertised brand. A scale of 1 (very unlikely) to 7 (very likely) was used. The results showed that the average likelihood of purchase was
Advertisement A 5.5 Advertisement B 5.8 Advertisement C 5.2
The ANOVA table was as follows:
Source of Variation SS d. f. MSS F-ratio p-value
Due to advertisements Unexplained Total 12 234 246 2117 119 6.0 2.0
a. What is the appropriate null hypothesis? The alternate
hypothesis?
b. What is the F-ratio? The p-value?
c. Is the result significant at the 0.10 level? The 0.05 level? The
0.01 level?
d. Are there any differences among the impacts of the three
advertisements ?
7. Using problem 6, assume that each of the three groups of respondents had been divided into two groups: younger (under 30) and older (over 30). The revised ANOVA table was as follows:
Source of Variation SS d. f. MSS F-ratio p-value
Due to advertisements 12 2 6.0
Due to age 24 1 24.0
Unexplained 210 116 1.81
Total 246 119
a. What are the F-ratio and p-value associated with the hypoth-
esis test that there is no advertisement effect? Why is it
different than that in problem 6? Notice that the total SS and
the advertisement SS have not changed.
b. Test the hypothesis that there is no age effect.
8. Recall Case 13-1, The American Conservatory Theater. Calculating from Table 13-5, the F-statistics for the difference between means for the five variables are 27.9, 27.7, 2.4, 14.3, and 2.4. The number of degrees of freedom for the numerator is 2 (three minus one). The total degrees of freedom are 823. Why? Thus the degrees of freedom for the denominator are 821. Determine the significance level to be associated with each variable. What is the null hypothesis? Calculating from Table 13-6 of the same case, the chi-square value is 23.9. What is the significance level of the appropriate hypothesis test? What is the appropriate null hypothesis?
Hypothesis testing involves four steps: (1) develop evidence supporting.ii judgment; (2) conceptualize the null hypothesis; (3) determine the ^probability of obtaining the evidence if the null hypothesis were true; and (4) calculate the probability—that is, the p-value—of the hypothesis test. There are several points worth remembering about hypothesis testing:
1. Hypothesis testing is a screening test. If the evidence does not pass this test, it may not be worth much attention. If it does pass this test, then it might at least be worth further analysis.
2. Hypothesis testing really measures the sample size. A large sample nearly always will yield "statistically significant" results, and a small enough sample probably will not be statistically significant. Thus, the test really does no more than provide a measure of sample size.
3. Hypothesis testing does not establish whether the null hypothesis is true or false: it only quantifies how persuasive the evidence is against it. Alow p-value indicates impressive evidence, and a .hjLgh^zyalue indicates the evidence is not impressive!
A cross-tabulation has associated with it a dii-square test of the relationship between the two variables. The null hypothesis is that the two variables are statistically independent, that is, knowledge of one provides no information about the other. The resulting p-value represents the probability under the null hypothesis of getting a chi-square statistic as large or larger than the one observed. If the p-value is low, then the evidence is rather impressive that the independence assumption is not true. If it is high, then the evidence is very weak that the independence assumption is not true.
The difference between means is a commonly used statistic in data analysis, particularly when experimentation is involved. The appropriate null hypothesis in this context is that there is no difference between means in the population. The p-level then represents the probability that the observed differences between means could have been that large if, in fact, the null hypothesis that all means are the same were true. The test is based on the F-statistic and the accompanying F-distribution. The extended ANOVA table provides the ability to conduct several difference-between-means tests at the same time.
b. Interpret the term Ex = 17.8
c. Calculate the x2 value.
d. Is the x2 value significant? At what level? What exactly is the
null hypothesis?
e. This data set proves that the usage of the library differs by
age. True or false? Why?
5. Determine the p-value and the significance level of a hypothesis test on the table presented in problem 4 of Chapter 13. What is the null hypothesis? Interpret the p-value.
6. An experiment was conducted to determine which of three advertisements to use in introducing a new personal computer. A total of 120 people who were thinking of buying a personal computer was split randomly into three groups of 40. Each of the groups was shown a different advertisement and each person was asked his or her likelihood of buying the advertised brand. A scale of 1 (very unlikely) to 7 (very likely) was used. The results showed that the average likelihood of purchase was
Advertisement A 5.5 Advertisement B 5.8 Advertisement C 5.2
The ANOVA table was as follows:
Source of Variation SS d. f. MSS F-ratio p-value
Due to advertisements Unexplained Total 12 234 246 2117 119 6.0 2.0
a. What is the appropriate null hypothesis? The alternate
hypothesis?
b. What is the F-ratio? The p-value?
c. Is the result significant at the 0.10 level? The 0.05 level? The
0.01 level?
d. Are there any differences among the impacts of the three
advertisements ?
7. Using problem 6, assume that each of the three groups of respondents had been divided into two groups: younger (under 30) and older (over 30). The revised ANOVA table was as follows:
Source of Variation SS d. f. MSS F-ratio p-value
Due to advertisements 12 2 6.0
Due to age 24 1 24.0
Unexplained 210 116 1.81
Total 246 119
a. What are the F-ratio and p-value associated with the hypoth-
esis test that there is no advertisement effect? Why is it
different than that in problem 6? Notice that the total SS and
the advertisement SS have not changed.
b. Test the hypothesis that there is no age effect.
8. Recall Case 13-1, The American Conservatory Theater. Calculating from Table 13-5, the F-statistics for the difference between means for the five variables are 27.9, 27.7, 2.4, 14.3, and 2.4. The number of degrees of freedom for the numerator is 2 (three minus one). The total degrees of freedom are 823. Why? Thus the degrees of freedom for the denominator are 821. Determine the significance level to be associated with each variable. What is the null hypothesis? Calculating from Table 13-6 of the same case, the chi-square value is 23.9. What is the significance level of the appropriate hypothesis test? What is the appropriate null hypothesis?
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