Statistics: Research Methods - Assignment Example

1. the advantages and disadvantages of graphical approaches as opposed to numerical approaches in descriptive statistics. (5 marks) The advantages of graphical approaches as opposed to numerical approaches in descriptive statistics are written below: Trend (trend-line) in data can easily be seen when using graphical approaches than when using numerical approaches. Graphical method reduces large amounts of data to manageable forms. Diagrams and graphs can easily be used to clarify a feature or to support interpretations. Information on graphs and diagrams can easily be understood by anyone. The disadvantages of graphical approaches as opposed to numerical approaches in descriptive statistics are that important information maybe lost, since they are usually a summerisation of the whole data set. 2. Explain briefly, using examples each of the following terms: (a) Mean, Median, Mode (b) Nominal, Ordinal, Interval scales (c) Normal Distribution, Positive, Negative Skew (3x5) = 15 marks a) Mean: According to Bell (2005: p. 207), mean or arithmetic mean is obtained by adding together each item (or value) and dividing by the total number of values. The formula for calculating mean is: The arithmetic mean (X) = total (sum) value of items/total number of items X = ∑x/n Where ∑x is the sum of all value of the items and n is the number of items For example if four pupils have £10, £20, £28 and £32, then the mean is: X = ∑x/n = £ (10 + 20 + 28 + 32)/4 = £ 90/4 = £22.50 Median: According to Burns & Grove (2001; p. 501) median is the score at the exact centre of the ungrouped frequency distribution. It is the value of a random variable, which is at the middle of an ordered data set. The median is a measure, which is not affected, by outliers and therefore it is a useful measure of central tendency when the distribution of random variable is severely skewed. Median computation depends on whether the data is grouped or ungrouped. Ungrouped data Method – where (n) is odd:- Arrange the observations in ascending order. Median is the value of the ((n+1)/2)th item Ungrouped data Method – where (n) is even:- Arrange the observations in ascending order. Identify the (n/2)th and ((n/2) + 1)th item Median is the average value of (n/2)th and ((n/2) + 1)th item For example; 12, 18, 23, 24, 27, 34, 28, 40, 41 and 20 Arranging this: 12, 18, 20, 23, 24, 27, 28, 34, 40, 41 Median = (24+ 27)/2 = 25.5 Grouped data method – the arithmetic method Median = [{(n + 1)/2} – (F + 1)]*(w/fm) + Lm Where, Lm is the lower limit of the median-class interval, w is the class-interval width, n is the sample size (total number of items), fm is frequency of the median class, and F is sum of all the class frequencies up to, but not including, the median class. Graphical methods: The median is the value associated with 50% cumulative relative frequency on the vertical axis. Mode: According to Lanoë (2002; p. 63), the mode is the score that occurs most frequently in a range of scores. It is the number that appears the most in a data set. For example, the mode of the figures 4, 8, 12, 12, 12, 12, 15, 15, and 18 is 12 because it is the most frequently occurring number. In a frequency distribution, the mode is the item with the highest frequency as shown in the table below: Size of component (cm) Number of makes of cars using the size 20 4 21 10 22 15 23 (Mode) 20 24 1 25 7 The mode in this example is 23 cm. Because it has the highest frequency (20), 23 cm is the modal size of the component. If the data are already grouped in a frequency distribution, then it is assumed that the mode is located in the class with the most items, that is, the class with the highest frequency. Mode = Lmo + {d1/(d1 + d2)}w Where, Lmo is lower limit of modal class, d1 is frequency of the modal class minus the frequency of the class directly below it, d2 is frequency of the modal class minus the frequency of the class directly above it, and w is width of the modal class interval. For example: Class in number of trips 0-3 4-7 8-11 12 and more Frequency 6 8 1 5 Modal class Lmo = 4, d1 = 8 - 6 = 2, d2 = 8 – 1 = 7 and w = 3 Mode = 4 + (2/9)3 = 4.67 b) Nominal scales: According to Riegelman (2000: p. 282) nominal scales are a method of classification and in this the function of numbers is the same as names, to label categories. Nominal scales are qualitative rather than quantitative. For example, in football say Liverpool or England team Steven Gerrald wears a jersey with number 8 at his back. This only allows viewers to identify him during the match. However, this does not mean that Steven Gerrald is better than Ashley Cole and John Terry who wear jerseys with numbers 3 and 5 respectively or is he worse than Wayne Rooney who puts on jersey number 9 or Michael Owen who puts on the jersey with number 10. Thus, it is pointless to calculate statistics based on these numbers because the numbers do not have value but are nominal, used as names. Data that fall under this category include gender, ethnicity, marital status, and so on. Ordinal Scale: Ordinal Scales are a measurement scale, which distinguishes events, based on the relative amounts of characteristics they posses. For example, different sized balls from the smallest to the largest. The differences between the measurements are not reflected in the rank. Ordinal scales are asymmetrical, meaning that a special relationship may hold between B and A. For instance, if A is greater than B, then B cannot be greater than A. Furthermore, ordinal scales do not have the property of transitivity, meaning that if A is greater than B and B is greater than C then A is greater than C. Interval scales: Interval scales are the measurement scale that can be applied when events can be distinguished from each other and ranked. The differences between measurements have meaning such as thermometer, which measures temperature. c) Normal Distribution : As explained by Parahoo (1997:p 239), Normal distribution is the most of the scores cluster around the mean, and the extreme scores are few and are more or less equally distributed above and below the mean. A normal distribution has the following features: 34 per cent of all scores fall between the mean and the 1 standard deviation on either side of the mean, median and mode are equal and are the dividing point between the left and the right side of the curve. 28 per cent (14 per cent on both sides) of all scores fall between 1 standard deviation and 2 standard deviation from the mean as shown in the diagram below. Source: Parahoo (1997: p. 350) Positive Skewness: A distribution curve that is not symmetrical is referred to as skewed or asymmetrical (distribution) curve and skewness is a measure of a distribution from symmetry. A probability distribution is said to be positively skewed if the mass of the distribution in concentrated on the left of the figure or if the largest portion of data is below the mean. In this case, the right tail will be the longest and the distribution is said to be right skewed. Take for instance, enrolment of the elderly people in nursing home scene to be positively skewed because the majority of these people died within a few months of enrolment. Below diagram shows positive skewness. Source: Polgar & Thomas (1995: p. 210) Negative Skewness: A distribution curve is considered to be negatively skewed if the largest portion of data is above the mean or whereby the mass of the probability distributions is concentrated on the right of the figure. In this case the left tail will be the longest and the distribution is said to be left skewed. For example, data on the occurrence of chronic illness by age in a population are negatively skewed because most chronic illness occurring in older age groups. See diagram below Source: Polgar & Thomas (1995: p. 210) 3. A teacher wishes to know whether in his/her class have more favourable attitudes toward E-learning than do than do the females. All students in the class are given a questionnaire about E-learning and the mean responses of the males and the females are compared. Briefly explain the type of statistics involved in this exercise. (5 marks) The type of statistics used here is hypothesis testing with two samples. Comparing the means of the sample population will tell the teacher the differences he/she would except when the null hypothesis is true. The teacher can plot these differences and will get the distribution of differences between sample means, as shown in the diagram below; Source: Hinton (1995: p. 79) The graph tends to be a normal distribution as it is a sampling distribution with a mean of the distribution zero. A z score needs a score, a mean and a standard deviation and can be calculated using the following formula:- Z = (X1 – X2) – μx1 – x2 σx1 – x2 Since the mean of the distribution is zero Z = (X1 – X2) – 0 = X1 – X2 σx1 – x2 σx1 – x2 Replacing σx1 – x2 in the formula with sx1 – x2 which is the standard error of the differences between two sample means above, Therefore, (X1 – X2) – 0 Sx1 – x2 Therefore, T = X1 – X2 Sx1 – x2 Sx1 – x2 is (an estimate of) how much we would except out means to differ by chance (when they come from the same distribution). X1 – X2 is the actual difference in means. Therefore (X1 – X2)/ Sx1 – x2 tells how much bigger our difference in means is relative to the difference expected by chance alone. The larger the ratio, the greater our confidence that the mean difference is not due to chance but due to 2 different population distributions. 4. A medical researcher is testing the effectiveness of s new drug for testing Alzheimer’s disease. Ten subjects with the disease are given the new drug and 10 are given a placebo. Improvement in symptoms is measured. Explain the roles of descriptive and inferential statistics in the analysis of the data? (5 marks) Roles of descriptive statistics in the analysis of the data would be to: Record and summarize the observed results (important in symptoms) Present the information (graphs, means, standard deviations etc) in such a way that inferences can easily be made. Roles of inferential statistics in the analysis of the data would be to: Form statistical hypothesis Define the null and the alternative/experimental hypothesis. Calculate the statistical test. Define the rejection or critical region. Draw conclusions from the test of the hypothesis and the data collected. 5. Explain the following terms: a) Standard deviation b) Correlation coefficient c) Standard error of the mean d) Type 1 error e) Type 2 error (2x5) = 10 marks a) Standard Deviation (σ): Standard deviation is the square root of the variance. Variance is the mean or average of the sum of squares of differences between the items and their mean. Standard deviation is important in the sense that it reverses the changes brought about by squaring and is obtained by using the following equation: Variance = σ2 Standard Deviation (σ) = √Variance Thus, standard deviation is a measure of the spread of a distribution around the mean and is the square root of sample variable (variance). b) Correlation coefficient: Correlation coefficient is a point of estimate of the strength of the association between two continuous variables usually referred to as Pearson’s Correlation Coefficient. The correlation coefficient has a range of possible values from -1 to +1. A correlation coefficient of zero shows no relationship between the dependent and independent variable. Positive correlation coefficient indicates that as the value of the independent variable increases, the value of the dependent variable also increases. On the other hand, negative correlation coefficient indicates that as the value of the independent variable increases, the value of the dependent variable decreases. In strictly saying the correlation value 1 indicates increasing linear relationship, -1 indicates decreasing linear relationship and some value in between in all other cases, indicate the degree of linear dependence between the variables. The correlation coefficient value of -1 or +1 (or closer values) indicates the stronger the correlation between the independent and dependent variables. c) Standard error of the mean: The standard error of the mean of a sample from a population is the standard deviation of the sampling distribution of the mean. The formula for estimating Standard error of the mean is: Where, is an estimate of the standard deviation σ of the population, and n is the size (number of items) of the sample. d) Type I Error: Rejecting a null hypothesis when it is true is called a Type I error. The probability of a Type 1 error occurring is symbolized by ά (alpha). The probability of making a Type 1 error is less or equal to 0.01. Thus, the smaller ά, the less the chance of making a Type 1 error. e) Type II Error: Accepting a null hypothesis when it is false is called a Type II error. The probability of a Type II error occurring is symbolized by β (beta). 6. Smokers attending a ‘rehabilitation clinic’ were randomly divided into three treatments groups. All had smoked 5 and 10 packets of cigarettes per day for at least five years. Each smoker was asked to keep a written record of the number of cigarettes smoked and time of day of smoking a cigarette during their rehabilitation period and for 6 months following the end of treatment. Group 1: 20 smokers discussed a plan of gradual reduction in smoking, individual with a counsellor. Each smoker met the counsellor once a week for six weeks. Group 2: 20 received hypnotherapy for 6 weeks. Group 3: 20 smokers were instructed about abstinence each week for 6 weeks with a psychologist. Six months after treatment those who had not smoked any cigarette on any day preceding month were regarded as a rehabilitation success, all other were rehabilitation failures. The null hypothesis was that there would de no differences between the treatment groups. The experimental hypothesis was that there would be differences in the success rate across the three groups. The hypothesis was not supported (a) What is meant by the term null hypothesis? (2 marks) (b) Explain briefly, why it is usual to state both the experimental and null hypothesis (3 marks) (c) Identify: (i) the independent variable (1 mark) (ii) the dependent variable (1 mark) (iii) the design of the study (1 mark) (iv) the most appropriate statistical test (2 marks) (Provide justification for you answer) a) Null Hypothesis: The null hypothesis is statistical hypothesis, which asserts or says that no change from the existing state or level has occurred. An example of this is that “There is no difference in the occurrence of a bruise at injection site low-dose heparin therapy when administered in three different subcutaneous sites (Burns and Grove). b) Stating both the experimental and null hypothesis is necessary because this gives a clear and precise statement of the problem, which would make it easier to identify the treatments whose effects are to be investigated, identify the experimental units to be used and to identify the responsed variable. The null alternative hypothesis rather gives an idea of what needs to be done in an experiment and answers the question under study/investigation. c) i. An independent variable according to Helewa and Walker (2000; p. 198) is “… a stimulus or activity that is manipulated varied by the researcher to create an effect on the dependent variable.” The independent variable is also referred to an intervention, treatment or experimental variable. c) ii. A dependant variable is the response, behaviour, or outcome that the researcher wants to predict or explain (Burns and Grove). The dependent variable can also be called an effect variable or criterion measure according to Kerlinger quoted in (Bell 2005). Here dependent variable is outcome of the study, which is a rehabilitation success. c) iii. The design of the study is to complete randomized design (CRD) c) iv. The most appropriate statistical test is to reject null hypothesis Ho in favour of the experimental hypothesis. If, F = Treatment mean sums of squares/Error mean sums of squares This is so because of the null and alternative hypothesis we are using (i.e. we are using a Type 2 model of a complete randomized design) 7. You read a study in which the result is just barely significant at the 0.05 level. You then look at the size of the sample. If the sample is very large (rather than very small), how should this affect your interpretation of: a) the probability that the null hypothesis is actually true and b) the practical importance of the result. (5 marks) a) If the sample size is very large, then I would assume that the probability that the null hypothesis is actually true to be high. b) The fact that n is very large (300) will make the conclusions drawn from the sample to be better estimates of the population parameters. 8. A data set is heavily skewed. Outline the possible ways in which you might handle this explaining your approach. (8 marks) Here it is given that data set is heavily skewed. It can be positively heavily skewed or negatively heavily skewed. The possible ways of handling this is explained below: Methods of percentiles: Split the sample in percentiles, where for instance, the 23rd percentile is that value below which 23% of the measurement lay. Supposed the data is positively skewed i.e. the frequency density has the form as shown below. Source: Polgar & Thomas (1995: p. 210) The above diagram shows that the data is positively skewed unlike when data is negatively skewed as shown below; Source: Polgar & Thomas (1995: p. 210) Instead, data should be symmetrical in a normal distribution as shown in the diagram below; Source: Parahoo (1997: p. 350) Now the cumulative frequencies, which lie between, say 21st and 22nd percentiles are assigned between the 21st and 22nd percentiles of the required distribution. This method alters the mean and the spread of data in the original data set and correct skewness. Histogram Interpretation: In a skewed distribution, the two halves of the histogram not appear as mirror image of one another (non-symmetric) as compared to normal distribution (symmetrical). To summarize this data set (distribution) by means of estimating a “typical value” brings certain philosophical complexity. However, if the data set is distributed normally then the value is unambiguous; it is well-defined center of distribution. Now since the data set are skewed, there will be no center. We can use several typical values such as mean, median and mode for the distribution. For severely skewed distribution; the mode may be at or near the left or right of the data and it will be not a good representative of the center of distribution. Similarly, the mean and median will be also not good representative. The most commonly reported typical value for skewed distribution is mean; the next is median and the least common is the mode. Now, as each of the three metrics (mean, median and mode) reflects a different aspect of “centerness”, therefore it is recommended that at least two (mean and median) or preferably all three should taken for summarizing and characterizing the data set. The recommended steps, in Histogram Interpretation for heavily skewed data set are: Step I. Quantitatively summarize the data by computing and reporting the sample mean, the sample median, and the sample mode. Step II. Determine the best-fit distribution (Weibull family or Gamma family , etc) Step III. Consider a normalizing transformation (Box-Cox transformation). 9. State 3 non parametric tests indicating their null hypotheses and briefly outlining the key features of each test. (12 marks) Non-parametric tests are often used in place of their parametric counter parts when certain assumptions about the underlining population are questionable. Non-parametric tests are often more powerful in detecting population differences when certain assumptions are not satisfied. All tests involving ranked data are non-parametric. Among the many available nonparametric tests are two tests that can be used in place of t-tests when the population distribution is so non-normal that the t-test is not robust. Wilcoxon developed both tests that use the median instead of the mean, and the calculations for both involve ranking the data. a) The Wilcoxon signed rank test (single samples): The Wilcoxon signed rank test is a test designed to test a hypothesis about the location (median) of a population distribution. The Wilcoxon signed rank test is a test, which is one of the most powerful of the non-parametric tests for comparing two populations. It is used to test the null hypothesis that two populations have identical distribution functions against the alternative hypothesis that two distribution functions differ only with respect to location (median) if at all. The Wilcoxon signed rank test is a nonparametric statistical procedure that evaluates an estimate of the population median with respect to some specific value. The Wilcoxon signed rank test often involves the use of matched pairs, e.g. before and after data, in which case it tests for median difference of zero. The brief outline of the Wilcoxon signed rank test is given in below table: Type of question answered: Is the population median significantly different from a specified value? Model or structure Independent variable A single numerical variable whose mean value is of interest Dependent variable None Equation model Corresponding parametric test One-sample t test Required assumptions Minimum sample size 5 Level of measurement Interval Distributional assumptions Any roughly symmetrical distribution Other assumptions The true population values are continuous. Source: Steven M. Kemp & Sid Kemp (2004: p. 309) b) The Wilcoxon rank sum test (two groups): The Wilcoxon rank sum test is a non-parametric statistical procedure that determines whether the difference between the medians of two groups is significant. It is a good replacement for the t test when the population distribution is non-normal. It works for ordinal data with less than five levels. The brief outline of The Wilcoxon rank sum test is given below: Type of question answered: Does the median of the dependent variable differ between two groups? Model or structure Independent variable A dichotomous variable designating group assignment. Usually zero for the control group and one for the experimental group. Dependent variable A numerical variable measuring some quantity predicted to be affected by the differing treatments/interventions applied to each group. Equation model Corresponding parametric test The t test (for independent groups). Required assumptions Minimum sample size 8 (4 per group) Level of measurement Ordinal Distributional assumptions None Source: Steven M. Kemp & Sid Kemp (2004: p. 310) c) The Kruskal–Wallis Test (Multi-Group Testing):The Kruskal–Wallis test is to one-factor ANOVA as the Wilcoxon rank sum test is to the two-group t test. The Kruskal–Wallis test is a non-parametric statistical procedure that determines whether the difference between the medians of several groups is significant. The brief outline of The Wilcoxon rank sum test is given below: Type of question answered: Are the medians of any of the groups unequal? Model or structure Independent variable A single categorical variable designating group assignment for k groups. Dependent variable A numerical variable measuring some quantity predicted to be affected by the differing treatments/interventions applied to each group. Equation model Corresponding parametric test One-factor ANOVA test Required assumptions Minimum sample size 5 per group Level of measurement Ordinal Distributional assumptions If the group populations are assumed to have the same shape and variability, the null hypothesis is that the medians are equal. If not, the null hypothesis is that there is no difference between the distributions of the groups. Other assumptions The true population values are continuous. Source: Steven M. Kemp & Sid Kemp (2004: p. 311) Steven M. Kemp & Sid Kemp (2004). Business Statistics Demystified, McGraw Hill Publication. (Chapter 14, p. 298-311) 10. A group of 450 subjects take part in a clinical trial which aims to determine whether a new treatment is effective or not. Data are drawn from the subjects before and after the treatment in order to see whether their condition improves. When data are generated an improved condition gives a higher score than ‘failure to improve’. It is decided to check any potential improvement by comparing the mean scores of the two sets of data. Outline how you would do this giving brief explanation of the steps involved. (20 marks)   Procedure to come out with the required results of this experiment is explained below: Step I: Determining the parameters of interest Let be the mean score before treatment - be the mean score after treatment S.d. = Standard deviations of the paired differences Step II: Stating the hypothesis to be tested Null hypothesis = There is no improvement. For example, lack of exercise does not cause constipation in elderly patients. Experimental hypothesis = There is an improvement. For instance, lack of exercise does cause constipation in elderly patients. Step III: Calculating the test statistic N = 450 patients is large sample data for both before and after treatment. To conduct this test, the difference between the two means is used in conjunction with the variation found in both samples (SD) and sample sizes to compute a t test statistic. The t test formula is Where, is the mean of the subjects before treatment, is the mean of the subjects after treatment, and S is the variance. The pooled standard error of the difference between two sample means is Where n1 is the size of the subjects before treatment, which is equal to n2, the size of the subjects after treatment which is 450. s1 is the variance of the subjects before treatment, and s2 is the variance of the subjects after treatment. Step IV: Rejection criteria Reject 4o if Ho Thus, if To, we conclude that there is an improvement (i.e. new treatment is effective). Read More