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
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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.
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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
So, we should reject the null hypothesis H0. At a 0.05 level of significance level, we conclude that there is a significant difference between the average height for females and the average height for the males.
Topics Distribution of the sample mean. Central Limit Theorem. Confidence intervals for a population mean. Confidence intervals for a population proportion. Sample size for a given confidence level and margin of error (proportions). Poll articles. Hypotheses tests for a mean, and differences in means (independent and paired samples). Sample size and power of a test. Type I and Type II errors. You will be given a table of normal probabilities. You may wish to be familiar with the follow formulae and their application.
1. Why is a z score a standard score? Why can standard scores be used to compare scores from different distributions? It is a scores relationship to the mean indicating whether it is above or below the mean. It does this by converting scores to z score. Yes – keep going – just a bit more is needed.2 out of 3 pts
Mean would be the most appropriate measure of central tendency to describe this data. This is because the mean is the average of all scores in the data set. If Dr. Williams were to graph the data into a bell shaped distribution, then the mean would be in the center where most of the scores are located. The mean is calculated using all information of the data set, and is the best score to use if you want to predict an individual score.
We can use z-table instead of t-table because the standard deviation comes from the population.
The t-test is a parametric analysis technique used to determine significant differences between the scores obtained from two groups. The t-test uses the standard deviation to estimate the standard error of the sampling distribution and examines the differences between the means of the two groups. Since the t-test is considered fairly easy to calculate, researchers often use it in determining differences between two groups. When interpreting the results of t-tests, the larger the calculated t ratio, in absolute value, the greater the difference between the two groups. The significance of a t ratio can be determined by comparison with the critical values in a
Based on the given sample of student test scores of 50, 60, 74, 83, 83, 90, 90, 92, and 95 after rearranging them from least to greatest. As the mean is based on the average of sum, the average of this sample is 79.67 or 80. The mode refers to numbers that appear the most in a sequence and in this case 83 and 90 both appear twice. Range calculates the difference between the largest and smallest number, which are 95 and 50 which have a difference of 45. The variance is the difference between the sum of squares divided by the sample size, which is the number in the sample minus one (Hansen & Myers, 2012), meaning it takes each number of the set and subtracts
At one school, the average amount of time that tenth-graders spend watching television each week is 18.4 hours. The principal introduces a campaign to encourage the students to watch less television. One year later, the principal wants to perform a hypothesis test to determine whether the average amount of time spent watching television per week has decreased. Formulate the null and alternative hypotheses for the study described.
The null hypothesis was that the female and male shoe sizes have an equal mean while the alternative hypothesis was that female and male shoe sizes do not have an equal mean. With the degrees of freedom being 33, the t-statistic is -8.27. The probability that -8.27 is ≤-1.69 is 7.5×10-10 for the one-tailed test. Also, the probability that -8.27 is ≤ ±2.03. is 1.5×10-9 for the two-tailed test. Due to both probabilities being under the alpha value of 0.05, the null hypothesis is rejected, and the alternative hypothesis is accepted at the 95% confidence level.
In order to know whether the evidence of research studies are accurate, one must be able to have a fundamental understanding in statistical analyses to determine if such descriptions and findings within manuscripts and articles are presented correctly and explicitly (Sullivan, 2012). Proper use of statistics begins with the understanding of both descriptive and inferential statistics. Correct organization and description of data characteristics from the population sample being studied leads the researcher to identify a hypothesis and formulate inferences about such characteristics. It is with inferential statistics that researchers conduct appropriate tests of significance and determine whether to accept or reject the identified null
We conduct an independent sample t-test using Excel, and obtain the following output (see sheet T-TEST)
| Based on explicit knowledge and this can be easy and fast to capture and analyse.Results can be generalised to larger populationsCan be repeated – therefore good test re-test reliability and validityStatistical analyses and interpretation are
With a P-value of 0.00, we have a strong level of significance. No additional information is needed to ensure that the data given is accurate.
This paper will examine a data analysis and application for an independent t test comparing the mean GPAs of a sample of male and female students. It will pose a research question that the data will set out to answer. It will provide a null hypothesis and an alternative hypothesis, and will provide an analysis showing why the null hypothesis should be accepted or rejected in favor of the alternative hypothesis.
The objective of this chapter is to describe the procedures used in the analysis of the data and present the main findings. It also presents the different tests performed to help choose the appropriate model for the study. The chapter concludes by providing thorough statistical interpretation of the findings.