Question

A study was conducted of the effects of a special class designed to aid students with verbal skills. Each child was given a verbal skills test twice, both before and after completing a 4- week period in the class. Let Y = score on exam at time 2 - score on exam at time 1. Hence, if the population mean µ for Y is equal to 0, the class has no effect, on the average. For the four children in the study, the observed values of Y are 8-5=3, 10-3=7, 5-2=3, and 7-4=3 (e.g. for the first child, the scores were 5 on exam 1 and 8 on exam 2, so Y = 8-5=3). It is planned to test the null hypothesis of no effect against the alternative hypothesis that the effect is positive, based on the following results from a computer software package: Variable Number of Cases Mean Std. Dev. Std. Error Y 4 4.000 2.000 1.000 a) Set up the null and alternative hypotheses. b) Calculate the test statistic. c) Make a decision, using α = .05. Interpret. d) True or false? When we make a decision using α = .05, this means that if the special class is truly beneficial, there is only a 5% chance that we will conclude that it is not beneficial. Justify

Answer 1

Given Data

Special Class	Special Claas	Difference (Y)
8	5	3
10	3	7
5	2	3
7	4	3

Here, we are to test,

We use the test statistic:


We reject H0 at 1% level of significance, iff


Now,

Now,


Hence, we observe that,

Hence, we reject H0 at 5% level of significance, and conclude that the special class was truly beneficial.

When we make a decision using α = .05, this means that if the special class is truly beneficial, there is only a 5% chance that we will conclude that it is not beneficial.
This is a true statement. α signifies the probability of Type I error, that is, the probability of True Rejection of Null is 5% probable.


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