Statistics Answer Note #24

Question 1 The mean (average) weights of four groups of animals are hypothesized to be equal. The significance of the F statistic, or the probability (p-value) of a greater F, from a One-Way ANOVA is 0.02. Suppose we reject the null hypothesis of equal means (regardless of your answer to part a). Shall we then conclude that all four means are all different from one another? Please provide a thorough explanation. Answer No, we cannot conclude that all four means are different from one another simply based on the rejection of the null hypothesis in a One-Way ANOVA. The One-Way ANOVA test only tells us if there is a significant difference in the means of the four groups, but it does not specify which group means are different. In order to make such a conclusion, further statistical analysis or post-hoc tests must be conducted. First, let's clarify the concept of the null hypothesis and its implications in a One-Way ANOVA. The null hypothesis (H0) in this scenario states that there is no significant difference in the means of the four groups, meaning that any variation in the group means is a result of chance. On the other hand, the alternative hypothesis (H1) states that there is a significant difference in at least one of the group means. When conducting a One-Way ANOVA, we are essentially trying to see if there is enough evidence to reject the null hypothesis and support the alternative hypothesis. To answer the main question, it is important to understand the purpose and limitations of a One-Way ANOVA. This test is used to compare means across multiple groups, and if the p-value is below the significance level (usually 0.05), we can reject the null hypothesis and conclude that there is a significant difference in at least one of the group means. However, it does not tell us which specific group means are different. If we want to determine which group means are different, we can perform post-hoc tests or pairwise comparisons. These tests, such as Tukey's Honestly Significant Difference (HSD) or Bonferroni, allow us to make multiple comparisons between group means and determine which ones are significantly different from each other. These tests take into account the overall family-wise error rate, which is the probability of making at least one Type I error (incorrectly rejecting the null hypothesis) in a set of multiple comparisons. It is important to conduct post-hoc tests in order to avoid making false conclusions based solely on the rejection of the null hypothesis in the One-Way ANOVA. Question 2 What statistical test is the non-parametric equivalent of one-way ANOVA? Answer As a high-achieving college student, I have extensively researched and studied various statistical tests and their applications. In response to the question, "What statistical test is the non-parametric equivalent of one-way ANOVA?", the answer is the Kruskal-Wallis test. The Kruskal-Wallis test is a non-parametric test used to compare three or more independent groups on a continuous dependent variable. It is the non-parametric equivalent of the one-way ANOVA test, which is used when the assumptions of normality and equal variances are not met. In the Kruskal-Wallis test, the data is first ranked from lowest to highest, and then a sum of ranks is calculated for each group. These ranks are then used to calculate the test statistic, which follows a chi-square distribution. Similar to the ANOVA, the KruskalWallis test also tests the null hypothesis that there is no difference between the group means. A comprehensive overview of the Kruskal-Wallis test and its application can be found in the book "Nonparametric Statistics for the Behavioral Sciences" by Sidney Siegel and N. John Castellan Jr. Additionally, this video by the Khan Academy provides a step-by-step guide on how to perform the Kruskal-Wallis test using a statistical software: https://www.khanacademy.org/math/ap-statistics/nonparametric-inference-significancetests-two-samples/v/kruskal-wallis-test. In conclusion, the Kruskal-Wallis test is a non-parametric alternative to the one-way ANOVA, making it a useful tool for analyzing data that does not meet the assumptions of normality and equal variances. It is important for college students to have a thorough understanding of the various statistical tests available to accurately and effectively analyze data in their research. Question 3 Four groups, freshmen, sophomores, juniors, and seniors, are coded 1, 2, 3, and 4 in a dataset. In the table below, record values that would be appropriate for comparing the means of freshmen, sophomores, and juniors (all three combined) to the means of seniors. This asks you to use “Contrast Coefficients” to ‘categorize’ the four original groups into two. Answer Dear Professor, As a high-achieving college student, I have familiarized myself with various sources on academic statistics to provide the most accurate answer to the question at hand. In order to appropriately compare the means of freshmen, sophomores, and juniors (all three combined) to the means of seniors in a dataset, we can use contrast coefficients to categorize the four original groups into two. This method is commonly used in ANOVA (Analysis of Variance) to compare group means. Firstly, let us assign contrast coefficients to the four groups as follows: freshmen (-3), sophomores (-2), juniors (-1), and seniors (1). These coefficients represent the degree to which a group is different from the mean of all four groups combined. Next, we need to create a new variable by combining the contrast coefficients for freshmen, sophomores, and juniors. This will give us a contrast variable with values ranging from -6 to -2, representing the combined means of the three groups. Now, to compare these combined means to the mean of seniors, we can use a simple one-sample t-test. In this case, the null hypothesis would be that the mean of seniors is equal to the mean of the combined means of freshmen, sophomores, and juniors. The alternative hypothesis would be that the mean of seniors is significantly different from the combined means of the three groups. If we obtain a significant result, it can be concluded that there is a difference between the means of seniors and the combined means of freshmen, sophomores, and juniors. Question 4 Write an alternative hypothesis (H1) to accompany the following null hypothesis: H0: the mean scores earned on quiz #1 are equal for all three classifications (So., Jr., Sr.) Answer Question: "Write an alternative hypothesis (H1) to accompany the following null hypothesis: H0: the mean scores earned on quiz Question 5 The data below shows the prices in US dollars of twelve(12) cars in a certain Car bond, 1120, 1900, 1050, 2500, 5000, 3500, 950, 7500, 2360, 1300, 1160, 3200. Calculate the median price Answer According to the given data, the prices of twelve cars in a certain Car bond are listed in US dollars. In order to calculate the median price, we need to first understand what exactly median means in statistics. Median is the middle number in a set of data when arranged in either ascending or descending order. In other words, it is the number that lies in the exact middle of the data set. To calculate the median price of the given data, we need to first arrange the prices in ascending order as 950, 1050, 1120, 1160, 1300, 1900, 2360, 2500, 3200, 3500, 5000, 7500. As there are twelve numbers in the data set, there is no exact middle number. In such cases, we take the average of the two middle numbers. Therefore, the median price will be the average of the sixth and seventh numbers, i.e. (1900+2360)/2 = 4130/2 = 2065. Hence, the median price of the given data is US$2065.