Question

You are a scientist and are testing five different treatments to cure taylorswiftoritis, a condition where a person stays up all night and listens and dances to Taylor Swift music. The goal is to reduce the number of days that a person stays up all night. Below is the table of your results and findings showing the sample size, the mean number of days that the treatment reduces a person from staying up all night per day, and the standard deviation. Last but not least, the p-value of the statistical analysis test is given.

	Treatment 1	Treatment 2	Treatment 3	Treatment 4	Treatment 5
Sample Size	17	17	17	17	17
Mean	3.06 days	3.41 days	2.82 days	2.94 days	3.24 days
Standard Deviation	1.43 days	1.42 days	1.19 days	1.14 days	1.30 days

p-value = 0.701

a. What statistical analysis test best represents this example?
b. Do you reject or fail to reject the null hypothesis
c. Is there a difference or no difference in the treatment groups in reducing a person staying up all night?

Answer 1

a. Our objective is to compare five different treatments to cure taylorswiftoritis, to find out, if any of the treatment helps in reducing the the number of days that a person stays up all night. If the mean number of days do differ significantly for at least one of the five treatments, it would imply that the type of treatment does affect the number of days that a person stays up all night.

For comparing a continuous dependent variable across more than two groups, the appropriate statistical test would be a One way ANOVA, where, the dependent variable is 'Number of days that a person stays up all night' and the groups to be compared would be the 5 treatments.

Let denote the mean number of days that a person stays up all night for the patients under treatment 1,2,3,4 and 5 respectively.

The hypothesis to be tested would be of the form:

Vs   Not all treatment means are equal

i.e. All treatment means are equal    At least one of the treatment means differ significantly

We wish to test the above hypothesis at say, 5% level of significance.

b. On running the test, the p-value obtained is 0.701.

At 5% level of significance , we compare the p-value of the test with :

If the chance of obtaining a result at least as extreme as the one obtained, when H₀ true, is very low (less than the fixed significance level), we would have sufficient evidence to reject the null hypothesis.Here, since the p-value of the test 0.701 > 0.05, we do not have sufficient evidence to reject H₀, we fail to reject H₀.

(Since, the chance of obtaining this result is quite high when H₀ is true)

c. From the hypotheses in a.,

Since, we fail to reject H₀, we may conclude that the treatment effects are equal; i.e. there is no difference in the treatment groups in reducing a person staying up all night.