Langewitz, Izakovic, and Wyler (2005) reported that self-hypnosis can significantly reduce hay fever symptoms. Patients with moderate to severe allergic reactions were trained to focus their minds on specifc locations wjere their allergies do not bother them, such as a beach or a ski resort. In a sample of 64 patients who received this traiming, suppose that 47 showed reduced allergic reactions and 17 showed an increase in allergic reactions. Are these results sufficient to conclude that the self-hypnosis has a significant effect? Assume that increasing and decreasing allergic reactions are equally likely if the training has no effect Use a=.05.

Chi-square test

Null hypothesis: The training is not effective. That is the training does not change the chance of occurance of grey fever symptoms.

Alternative hypothesis: The training changes the chance of occurance of hey fever symptoms.

We can perform a chi-square test of independence.

If the null hypothesis is true we expect 32 persons(50% of 64) to show reduced allergic reactions and 32 persons with increased allergic reactions.

So the expected frequencies are 32 and 32.

The observed frequencies are 47 an17.

The value of chi-square is

∑(O-E)^2/E = (47-32)^2/32 + (17-32)^2/32 = 7.03125 + 7.03125 = 14.0625

The critical value of chi-square with 1 degree of freedom at the 5% level of significance is 3.84.

Since the computed value exceeds the critical value, chi-square is significant. The gives sufficient evidence to suspect the null hypothesis. So we reject the null hypothesis.

We conclude that he self-hypnosis has a significant effect.

Proportion test

Null hypothesis: The training is not effective. That is the training does not change the chance of occurrence of grey fever symptoms.

Alternative hypothesis: The training is effective in reducing allergic reactions.

Null hypothesis: The population proportion P of persons showing increased allergic reactions is 0.5. P = 0.5

Alternative hypothesis: The poulation proportion P of persons showing increased allergic reactions isless than P < 0.5.

The sample proportion is

When the null hypothesis is true

follows approximately the standard normal distribution.

The 5% critical vale of Z for the left tail test is -1.645.

The computed value of Z is -3.75 which is less than the critical value. So Z is significant. We reject the null hypothesis. We conclude that the training reduces allergic reactions.

Note: We have given two tests here. The chi-square test is a two tail test in the sense that wheter there is a difference or not.

But the proportion test has both the one tail and two tail versions. We have here performed a one tail test. This test examines whether the training reduces allergic reactions or not.

Also note that the two tail proportion test and the chi-square test are equivalent.

1. The same formula is used to compute chi-square for the goodness-of-fit test and for the test for independence. (Points: 1) True False

Both use the formula

2. A researcher is using a chi-square test for goodness of fit to determine whether there is any preference between two brands of chocolate-chip cookies. With a = .05 and a sample of n = 25, the critical value for the chi-square statistic would be 36.42. (Points: 1) True False

FALSE : The critical value is 3.84

3. The chi-square statistic from a test for independence has df = 2. The data for this research study, form a matrix with six separate categories. (Points: 1) True False

TRUE

4. The term expected frequencies refers to the frequencies computed from the null hypothesis _____. (Points: 1) the frequencies found in the sample data. the frequencies computed from the null hypothesis the frequencies that are found in the population being studied the frequencies that are hypothesized for the population being studied

the frequencies computed from the null hypothesis

5. How do you compute df for the chi-square test for goodness of fit? (Points: 1) n - 1 n - 2 n - C (where C is the number of categories) None of the other 3 choices is correct.

None of the other 3 choices is correct.

The correct answer is C-1, the number of categories -1.

6. A basic assumption for a chi-square hypothesis test is _____. (Points: 1) the population distribution(s) must be normal the scores must come from an interval or ratio scale the observations must be independent All of the other choices are assumptions for chi-square.

the observations must be independent

7. A sample of 100 people is classified by gender (male/female) and by whether or not they are registered voters. The sample consists of 60 females, of whom 50 are registered voters, and 40 males, of whom 25 are registered voters. If these data were used for a chi-square test for independence, the expected frequency for registered males would be _____. (Points: 1) 15 25 30 45

30

The answer is

8. Which of the following best describes the possible values for a chi-square statistic? (Points: 1) Chi-square is always a positive whole numbers. Chi-square is always positive but can contain fractions or decimal values. Chi-square can be either positive or negative, but always is a whole number. Chi-square can be either positive or negative and can contain fractions or decimals.

Chi-square is always positive but can contain fractions or decimal values.

9. For a fixed a level, how is the critical value for chi-square related to the size of the sample? (Points: 1) As the sample size increases, the critical value also increases. As the sample size increases, the critical value decreases. The critical value of chi-square is not related to the sample size.

The critical value of chi-square is not related to the sample size.

10. For a fixed level of significance, the critical value for chi-square will _____. (Points: 1) increase when df increases. decrease when df increases. increase when n and df both increase. The critical value is not related to either n or df.

increase when df increases