Normal Distribution Is Used to Find the

¶ … normal distribution is used to find the probability of the sample mean in this single sample z-test (we know the standard deviation of the population). Z-scores are measures of relative location. Suppose we are interested in whether enrichment sessions with parents and special tutors influence I.Q.s of children. We might start by learning what the baseline I.Q. scores are for children who participate in the I.Q. enrichment sessions. Standardized instruments for measuring I.Q. were used to assess the children.

We know that the mean is 100 and the standard deviation is 16 on these standardized tests. After testing our sample of fifty students, we observed an average score of 102. The z-score for the sample is a measure of the relative location of the observation in the data set: How does the average I.Q score of the students in our observed sample compare with the average I.Q. score in the population? Standard error of the mean is: 50 = 16/SOR (50) = 2.26.

The population mean (something that is usually not known, but in this case assumes the convention of IQ measures) is 100. The standard errors between the population mean and the sample mean is (102 -- 100) / 2.2.6 = 2 / 2.26 = .8849. On the normal curve, the percentage of area from the negative infinity tail to .8849 = 81.2%. Thus, the average intelligence of the students in the observed sample is 81.2% higher than all other samples of the 50 students. Moreover, and of greater interest to us, our sample students scored 31.2% higher than the average score of the overall population.

Research Question - t test for independent samples We use a simple sample case for this question. We select a simple random sample of workers in which each worker uses an approach to completing a job we call Approach A. We then select a second independent simple random sample of workers in which each worker used an approach to completing the job that we call Approach B. We want to know the difference between the means of the two groups of workers.

We will use a procedure based on the t distribution with n1 + n2 -- 3 degree of freedom. We assume that both populations have normal probability distribution and that the variances of the populations are equal. We want to know the relative efficiency of using the two approaches to complete the work. We assume that Approach B (?1) will be more efficient than Approach A (?2). Thus, Ha: 1 - ?2 > 0. Approach B. is more efficient than Approach A. And Ho: 1 - ?2 < or = 0. There is no difference between Approach A and Approach B.

Research Question - t test for dependent samples We use a matched sample design (dependent samples) for this question. We select a simple random sample of workers. Each worker first uses one Approach A and then the other Approach B. The order of the two approaches is randomly assigned to the workers, such that some workers perform Approach A first and some workers perform Approach B. first. In this manner, each worker provides a pair of data values, one for each approach.

We will use a procedure based on the t distribution with n1 + n2 -- 3 degree of freedom. We assume that both populations have normal probability distribution and that the variances of the populations are equal. In the matched sample design, the variation between the workers is eliminated as a source of sampling error. Is there a mean difference between the work completion rates of the two approaches is the means of the difference between the values for the population of workers.

Thus, the null hypothesis is that there is no difference between the means of the difference for the values of the two groups. However, we reject.