TOPIC 8 Chi-Square goodness-of-fit test Problem 12.1 Use a chi-square goodness-of-fit to determine whether the observed frequencies are distributed the same as the expected frequencies (α = .05) Category | fo | fe | 1 | 53 | 68 | 2 | 37 | 42 | 3 | 32 | 33 | 4 | 28 | 22 | 5 | 18 | 10 | 6 | 15 | 8 | Step 1 Ho: The observed frequencies are distributed the same as the expected frequencies Ha: The observed frequencies are not distributed the same as the expected frequencies Step 2 df = k – m – 1 Step 3 α = 0.05 x 2 0.05, 5df = 11.0705 Step 4 Reject Ho if x 2 > 11.0705 Category | fo | fe | | 1 | 53 | 68 | | 2 | 37 | 42 | | 3 | 32 | 33 | | 4 | 28 | 22 | | 5 | 18 | 10 | | 6 | 15 | 8
Review Questions - 2: MGMT 3101 (For Final Exam) Section I: (True or False) 1. Hypothesis testing is a procedure based on sample evidence and probability theory used to decide whether the hypothesis is a reasonable statement and should be not be rejected or is unreasonable and should be rejected. 2. An alternate hypothesis is a statement about a population parameter that is accepted when the null hypothesis is rejected. 3.
Lin Article Critique: Part 2 Dustin T. Rheel Liberty University Counseling 503, D22 Professor Carlene Taylor November 25, 2012 Critique of Population and Sampling In the Lin, Mack, Enright, Krahn, and Baskin article (2004) article, they sampled forty-three participants from various drug rehabilitation centers. These participants were suffering from alcohol and other drug dependences. Some of the participants were referred by the therapist and then the suggested participants then had the option to participate or not, thus making the sample used not random. Even though the sample was not random, they were randomly selected to be in one of 2 groups, Forgiveness Therapy (FT) or Alcohol and Drug Counseling (ADC) (Lin et al., 2004).
The study population was made of 1543 women (mean age 27.7 years) recruited from four urban hospitals in Utah, USA. Patients reported their height, pre-pregnancy weight, and pregnancy weight gain immediately postpartum; their body mass index (BMI) before pregnancy was also calculated. At 6–8 weeks after delivery, patients received a self-administered questionnaire by mail that contained the Edinburgh Postnatal Depression Scale. The study found that obese women were older, less educated, and more frequently parous and non-Hispanic whites, compared with women of normal weight. In here, we can see the post-birth physical and metal effects.
investigated the effects of naturalistic stressors on measure of immunity. Measures of natural killer cells (NK) were taken from blood samples of 75 medical students 1 month before their finals (low stress) as well as during their exams (high stress). These students also completed a questionnaire on negative life events and social isolation. It was found that NK
IVYT 104/Critical Thinking ASSIGNMENT #1: Analyzing Your Verdict The assignment below is based on Thinking Activity 1.5 (p. 29) and Questions for Analysis (p. 35). READING ASSIGNMENT: Chapter 1: Analyzing Issues, pp. 20-28 and Thinking Passage, pp. 32-35 WRITING ASSIGNMENT: Write a paper in which you state your verdict, explain your reasons, and analyze the evidence/factors that influenced your verdict in the Mary Barnett case. LENGTH: at least one typed, double-spaced page DUE DATE: At the beginning of the class session in week #3.
Summary-Critique of Professional Journal Article A Lesson Cycle for Teaching Expository Reading and Writing Jessica Harvie Southeastern University Research-Based Practices of Reading and Writing Instruction EDUC 5433 October 12, 2013 Dr. Janet Deck The two newly credentialed English teachers taught a five week long summer course educating 30 sixth graders and 31 seventh graders. The goal for these students was to be promoted to the next grade by the start of the next school year. The teachers taught study and English skills to these California middle school students who were required to attend class in order to be promoted. The participants were comprised of 20 sixth grade males and 10 females which included 21 Latinos and 9 white, non-Latino students. The seventh grade population consisted of 20 males and 11 females of which 19 students were Latino and 12 students were white or non-Latino.
In this essay, I seek to critically discuss whether resolutions provide a better explanation of the weakness of the will than the traditional/ Akrasia account or not. I will achieve this by briefly explaining what the traditional account is and also what the resolution account is. Furthermore, I will explain the advantages of Holton’s approach and also give reasons why Holton’s argument succeed which will be accompanied by a rebuttal. The Akrasia traditional approach merely states that a person is weak-willed if they act against their best judgement. If one judges A to be the best course of action, why would one do anything other than A?
The Heteroskedasticity Problem in Regression Analysis Recall that one of the assumptions of the OLS method is that the variance of the error term is the same for all individuals in the population under study. Heteroskedasticity occurs when the variance of the error term is NOT the same for all individuals in the population. Heteroskedasticity occurs more often in cross-section datasets than in time-series datasets. Consequences of Heteroskedasticity: 1. the estimates of the b’s are still unbiased if heteroskedasticity is present (and that’s good), 2. but, the s.e.’s of the b’s will be biased, and we don’t know whether they will be biased upward or downward, so we could make incorrect conclusions about whether the X’s affect Y 3. the estimate of S.E.R. is biased, so we could make incorrect conclusions about model fit Detecting Heteroskedasticity: 1.
Since the parameter is a population mean of a continuous variable variable, this suggests a one sample test of a mean. 2. SPECIFY THE NULL AND ALTERNATIVE HYPOTHESES. The second step is to state the research question in terms of a null hypothesis (H0) and a alternative hypothesis (HA). The null hypothesis is the population parameter, µ = $30,000 (H0: µ = $30,000).