Statistics Assignment Seven This paper will use inferential statistics to test two different research questions. There will be two different types of statistical test that will be used. The first statistical test will be a Chi-Square for independence and the second test will be a Person r test. Both test will have the seven hypotheses testing steps explained, descriptive discussion of the variables, possible errors and critique of research methods and implications for research and nursing practice will be provided. Research Scenario A: Question 1: Step 1 selection of test statistics Null hypothesis The null hypothesis is a statement that explains there is no difference between two variables (Salkind, 2013). Using the research question, a …show more content…
This test is selected as the research question is looking for the association between is NIofH Hypertension Category which is the independent variables and ordinal level of measurement and engagement in weekly moderate activity which is the dependent variable and is nominal level of measurement (Loiselle et al., 2011). There are two different samples and the researcher is interested in if there is an association between the two samples. A Chi-Square for independence test looks at the association between two samples that are both of a categorical level of measurement (Salkind, 2013). Lastly, the statistics assignment seven outline provided a cross tabulation SPSS output of frequencies and a Chi-Square test output. The SPSS output has four columns that consist of the independent variable NIofH Hypertension Category and two rows that consist of the dependent variable engagement in weekly moderate activity (Salkind, 2013). A Chi-Square test for independence contains two variables that are categorical, the independent variable in the column and the dependent variable in the row, within the cells are frequency of responses (Salkind, 2013). The goal of a Chi-Square for independence is to test if there is a relationship or association between two categorical variables (Salkind, …show more content…
At a level of significance of 0.05 the researcher is accepting a possible type I error 5 cases out of a 100 and 95 cases out of 100 a true null hypothesis (Loiselle et al., 2011). Step 3: Selection of one-tailed or two-tailed test The selection for a one-tailed or two-tailed test is not required for a Chi-Square for independence test (Salkind, 2013). Chi-Square test has both variables that are of a categorical level of measurement therefore, there is no direction (Salkind, 2013). Step 4: Excel output of test statistic The output for the Chi-Square for independence test is provided in the Statistic assignment seven outline and obtained from an SSPS output. Please refer to Statistics assignment seven to see the SPSS output for the Chi-Square tables. Step 5: Degrees of freedom The degrees of freedom (df) for a Chi-Square for independence is calculated as the number of rows subtracted by one multiplied by the number of columns subtracted by one (Salkind, 2013). From the SSPS output in assignment seven outline there are two rows and 4 columns. df = (Rows-1) (Coloums-1) df= (2-1)(4-1) df= (1)(3) df= 3 Therefore, the degree of freedom (df) is 3. Step 6: Critical
Provide 4 answers. Tell me why the correct answer is correct and why the incorrect answers are incorrect. 1. This occurs when a researcher fails to reject a null hypothesis that is
No 524 20.8965 12.46425 .54450 Yes 436 15.8239 10.13655 .48545 Independent Samples Test
Data was recorded every thirty seconds for ten minutes. In order to see if the results were significant or not the Chi-Square Distribution test was
Which of the following is a characteristic of a good hypothesis? A.Original idea not based observation B.Raises further questions C.Should be testable D.Contains more than two variables. 43. Which of the following gland regulate the function of other glands in the endocrine
Table 4.4 The Reliability of the Instrument Constructs Items Corrected Item-total Correlation Alpha if Item Deleted Cronbach's Alpha Value of the Constructs Parents' Attitude toward FB A1 .890 .920 .942 A2 .775 .940 A3 .853 .927 A4 .848 .928 A5 .855
In the following paragraphs, the grand theory of Jean Watson will be explored for its usefulness in practice. We will explore how the theory is congruent with current nursing standards and nursing interventions. Next, we will study if her theory has been tested empirically, if it is supported by research and if it is accurate. We will explore if there is evidence that her theory has been used by nursing educators, researchers, and nursing administrators. Then we will study how her theory is relevant socially and cross-culturally.
-.334 Std. Error of Kurtosis .468 .468 The table shows that the in the data set, the GPA variable is a left skewed distribution of -.053 and a left kurtosis of -.811. The final variable indicates that the left skewed distribution stands at -.334 and a left kurtosis of -.334.
Null hypothesis has no effect. A hypothesis gathers evidence against the null hypothesis, so the null can be rejected. The hypothesis is when you have an idea and you justify it that is right and gather evidence 6. Describe the differences between dependent and independent variables. a.
Use your results in Data Table 2 to support your answer.
1. CONCEPT 1-EVIDENCE-BASED PRACTICE IN NURSING This concept was deducted from module 2 with the topic “Concept of Evidence-based practice”. Evidenced-based nursing is a way of making decision and providing nursing care that is based on clinical issues and combine it with the most current, relevant research that is available on that issue. Evidence based nursing utilize the most up to date method of providing care, which have been proven via assessing high quality studies and statistically with significant research findings.
Analysis of Variance (ANOVA) is a statistical method used to test differences between two or more means. Though it is called "Analysis of Variance" but actually it is "Analysis of Means." ANOVA was developed by Ronald Fisher in 1918 and is the extension of the t and the z test. Before the use of ANOVA, the t-test and z-test were commonly used. But the problem with the T-test is that it cannot be applied for more than two groups.
For tests about one mean , when the population standard deviation s is known, we use s when calculating the test statistic. (for population standard deviation), , (for sample standard deviation), the equations above are called the z scores. Therefore in our case we use . Hence for bottle 19, , = 0.346. Using Moore’s table the p-value = 1, and significant level = 0.1.
1. Student details: 1.1 Name: Vaghela Deepikaben Maganbhai 1.2 Student ID:1525258 2. The programme of research 2.1 Title: To evaluate customer satisfaction in restaurant industry in India. 2.2 Research Objectives: • To explore the relationship exist among these factors, employee performance, food quality, price, physical environment and customer satisfaction with the help of literature review.
Why we use Probability Distribution: Some uses of probability distribution are as follows: Scenario Analysis Probability distributions can be used to create scenario analyses. A scenario analysis uses probability distributions to create several, theoretically distinct possibilities for the outcome of a particular course of action or future event.