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Wildcatozine Testing

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Introduction
This paper contains the report of the analysis carried out on a sample size of 30 to determine the extent to which Wildcatozine drug helps in the reduction of depression in people. Based on the data collected from the sample population, descriptive statistics, as well as the analysis of drug with the mean comparison, was done. The analysis involved the use of student t-test since a sample population of 30 was chosen. At 5% level of significance, the null hypothesis was then tested, and the overall results and discussion of the results were as presented below. Precisely, to test the effectiveness of the drug, the test was carried out at three levels: using descriptive statistics, using the z-score transformation statistics and …show more content…

It tells us how many standard deviations a given score is above the mean(González-Manteiga & Cao, 1993, p. 33). It is given as X=µ +zσ where x is a given score in the data set, σ is the standard deviation and z is the Z score. In the above data set, the Z-score associated with the sample mean is given by 5.63=7.00 +z(2.54) which imply that z-score is -0.54. From this value, we can determine the proportion of the sample that falls above and below the sample mean("Measures of Central Tendency, Dispersion, and Skewness," n.d., p. 35). In this case, the proportion that falls above the sample mean will contain score above 7 ( 5.63+0.54 = 6.17) while those that fall below the sample mean will be having a score less than 5 ( 5.63-0.54 = 5.09) and this gives 23.33% and 36.67% respectively.
Model 3: Hypothesis Testing
For this data set, the student t-test is best to be used in testing the hypothesis. Here, two tailed-sample tests would be used at a level of significance of 5%. This will help us to pick an alpha at 0.25 in each tail to test the direction, which the test statistic is significant(LAWRANCE, 1987, p. 280). The relationship in both directions is significantly tested by our hypothesis which will compare the mean of the given sample population to the given value x obtained by the t-test. We shall then use the obtained p-value to make a decision whether to accept or reject our null hypothesis. …show more content…

Therefore, there is sufficient evidence that there is a significant difference at the top 2.5% and the bottom 2.5% in the probability distribution of the given data set. Additionally, the resulting p-value at a level of significance of 0.05 is 2.76, which is greater than our hypothesized value. We, therefore, reject the null hypothesis and conclude that the sample mean is different from 5.63 as desired(McGowan & Vaughan, 2011, p. 64). These conclusions are based on the assumptions that no Type I or Type II errors were made in the course o hypothesis

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