Conducting a Z-Test

Conducting a z-Test

Jennifer Vogel

Argosy University

Dr. Edith Nolan

1/10/2015

In the situation we are looking at there is a researcher that believes and even foresees that watching a film on institutionalization will convert the outlook of 36 students seem to have about mentally ill patients. So the researcher is pretty confident that the attitude of the students is going to transform after they watch a film that focuses on institutionalization. A sample of thirty six students are picked from class for this survey. The researcher provides an array of questions to evaluate the outlook after the presentation of the film centred on institutionalization. He goes ahead with another sample and presents them with the same array of questions to the students who did not watch the film.

Once the survey was completed it did show a deviation of the results within the two groups. There definitely were dissimilar results between the two groups. Accordingly, with what the researcher had claimed does align with his hypothesis. So, the sample size is n = 36

The mean score for the questions given to the students who watched the film is 70. So, we can say that x = 70. Now, for the people who did not see the film the mean score for the questions given to them was 75. So we can say that µ = 75

The standard deviation (σ) is 12 (Given)

For examining the claim, we use the hypothesis testing method where the significance level is set at 5%. We just need to test if the claim of the researcher is true or not.

In these methods, we will have a null hypothesis and one alternate hypothesis.

So, the null hypothesis is

Ho: μ=75

The reason behind this is that the null hypothesis is an indication that the claim of the researcher is true. Then, it would be true saying that the attitude of the students will change for the mentally ill patients if they watch a film on institutionalization.

We conclude the alternate hypothesis to be,

Ha: μ≠ 75

The reason behind this would be that...