Hypothesis Testing - Article Example

Hypothesis Testing Table of Content List of Figures Figure 1: Scatter Plot 20 Figure 2: Residual Histogram 21 Figure 3: Normal P-P Plot of Regression Standardized Residual 22 Figure 4 23 List of Tables Table 1: Independent Samples Test 5 Table 2: Descriptives 9 Table 3: Test of Homogeneity of Variances 9 Table 4: ANOVA 10 Table 5: Ranks 11 Table 6: Test Statisticsa 11 Table 7: Ranks 13 Table 8: Test Statisticsa,b 14 Table 9: preexisting hypertension * alive discharge Crosstabulation 16 Table 10: Chi-Square Tests 17 Table 11: Directional Measures 17 Table 12: Symmetric Measures 18 Table 13: Coefficientsa 24 Table 14: ANOVAb 25 Table 15: Model Summaryb 25 [Writer’s Name] [Instructor’s Name] [Course] [Date] Hypothesis Testing 1) Independent Samples t-test Assumptions Assumptions of the Independent Sample t Test: 1. The variances of the dependent variable in the two populations are equal. 2. The dependent variable is normally distributed within each population. 3. The data are independent (scores of one participant are not related systematically to scores of the others). Evaluation of the Assumptions SPSS automatically test assumption 1 with the Levene test for equal variances. Assumption 2 could be tested with the Explore command, to see that the dependent variables are at least approximately normally distributed for each gender. However, the t test is quite robust to violations of this assumption; therefore this assumption needs not to be tested. Assumption 3 probably is met because the genders are not matched or related pairs and there is no reason to believe that one person’s score might have influenced another person’s. Hypothesis Statement H0: There is no difference in the mean length of stay in hospital for males vs. females. H1: There is a difference in the mean length of stay in hospital for males vs. females. Analysis The procedure produces two tests of the difference between the two groups. One test assumes that the variances of the two groups are equal. The Levene statistic tests this assumption. In this example, the significance value of the statistic is 0.505 (See Sig. under Levene's Test for Equality of Variances). Because this value is greater than 0.05, we can assume that the groups have equal variances and ignore the second test. The t column in Table 1 presents the observed t statistic for each sample, calculated as the ratio of the difference between sample means divided by the standard error of the difference. The column labeled Sig. (2-tailed) displays a probability from the t distribution with 82 degrees of freedom. The value listed is the probability of obtaining an absolute value greater than or equal to the observed t statistic, if the difference between the sample means is purely random. The 95% Confidence Interval of the Difference provides an estimate of the boundaries between which the true mean difference lies in 95% of all possible random samples of 84 students. Table 1: Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper length of stay in hospital Equal variances assumed .448 .505 .339 82 .736 .881 2.599 -4.290 6.052 Equal variances not assumed .365 62.067 .716 .881 2.411 -3.939 5.701 Conclusion and Discussion Since the significance value of the t- test is greater than 0.05 (P = 0.736 > 0.05, Table 2), we fail to reject the null hypothesis. That is there is no difference between male and female in regard to their length of stay in hospital. The independent-samples t test is appropriate whenever two means drawn from independent samples are to be compared. As with all t tests, the independent-samples t test assumes that each sample mean comes from a population that is reasonably normally distributed, especially with respect to Skewness. Test variables with extreme or outlying values should be carefully checked; or non-parametric test can be used to be sure. 2) One-Way ANOVA Assumptions 1. Observations are independent (the value of one observation is not related to any other observation. In other words, one person’s score should not provide any clue as to how any of the other people should score). 2. Variances on the dependent variable are equal across groups. 3. The dependent variable is normally distributed for each group Evaluation of the Assumptions Because ANOVA is robust, it can be used when variances are only approximately equal if the number of subjects in each is equal. ANOVA also is robust if the dependent variable data are even approximately normally distributed. Thus, if assumption # 2, or, even more so, #3 is not fully met, you may still be able to use ANOVA. There are also several choices of post hoc tests to use depending on whether the assumption of equal variances has been violated. Hypothesis Statement Using ANOVA analysis, we aim to examine whether there are differences among the three age groups on the length of stay in hospital. H0: There are no differences among the three age groups on the length of stay in hospital H1: There are differences among the three age groups on the length of stay in hospital Analysis Table 2: Descriptives length of stay in hospital N Mean Std. Deviation Std. Error 95% Confidence Interval for Mean Minimum Maximum Lower Bound Upper Bound 0 - 59 17 19.53 13.210 3.204 12.74 26.32 8 50 60 - 69 39 18.97 9.441 1.512 15.91 22.03 3 49 70 and Older 28 24.14 11.411 2.156 19.72 28.57 10 49 Total 84 20.81 11.067 1.207 18.41 23.21 3 50 Table 4 present the Levene Test of Homogeneity of Variances. The Levene statistic fail to reject the null hypothesis that the group variances are equal (P = 0.118 > 0.05). Table 3: Test of Homogeneity of Variances length of stay in hospital Levene Statistic df1 df2 Sig. 2.199 2 81 .118 The significance value of the F test in the ANOVA table is 0.147. Thus, we fail to reject the hypothesis that there is no difference among the three age groups on the length of stay in hospital. In this case, Bonferroni is unnecessary since the null is not rejected. Table 4: ANOVA length of stay in hospital Sum of Squares df Mean Square F Sig. Between Groups 470.314 2 235.157 1.965 .147 Within Groups 9694.638 81 119.687 Total 10164.952 83 Conclusion and Discussion One way analysis of variance (ANOVA) revealed no difference among the three age groups on length of stay in hospital. Post hoc Tests was not conducted because the null is not rejected. 3) Mann-Whitney U test Assumptions Unlike the parametric t-test, this non-parametric makes no assumptions about the distribution of the data (e.g., normality). This test is an alternative to the independent group t-test, when the assumption of normality or equality of variance is not met. This, like many non-parametric tests, uses the ranks of the data rather than their raw values to calculate the statistic. Since this test does not make a distribution assumption, it is not as powerful as the t-test. Hypothesis Statement Using, the Mann-Whitney U test analysis we aim to examine whether male and female differ significantly in regard to their length of stay in hospital. H0: There is no difference between male and female in regard to their length of stay in hospital. H1: There is a difference between male and female in regard to their length of stay in hospital. Analysis Because the test variables are assumed to be nominal, the Mann-Whitney and Wilcoxon tests are based on ranks of the original values and not on the values themselves. The Ranks table shows the mean or average ranks for males and females on the length of stay in hospital. SPSS ranks the 84 students from 84 (highest) to 1 (lowest) so that, in contrast to the typical ranking procedure, a high mean rank indicates the group scored higher. Table 5: Ranks gender N Mean Rank Sum of Ranks length of stay in hospital male 27 47.15 1273.00 female 57 40.30 2297.00 Total 84 Table 8 provides the Mann-Whitney U, z-scores, and the Sig. (significance) level or p. Note that the mean ranks of the genders did not differ significantly on the length of stay in hospital, as was the case for the similar t tests. Table 6: Test Statisticsa length of stay in hospital Mann-Whitney U 644.000 Wilcoxon W 2297.000 Z -1.204 Asymp. Sig. (2-tailed) .229 a. Grouping Variable: gender Conclusion and Discussion Using the Mann-Whitney test, we fail to reject the hypothesis that there is no difference among the three age groups on the length of stay in hospital. The Mann-Whitney test is only slightly less powerful than the t test, so it is a good alternative if the assumptions of the t test are violated, as was the case with length of stay in hospital. 4) Kruskal-Wallis H test Assumptions The Kruskal–Wallis test does not make assumptions about normality. Like most non-parametric tests, it is performed on ranked data, so the measurement observations are converted to their ranks in the overall data set: the smallest value gets a rank of 1, the next smallest gets a rank of 2, and so on. The loss of information involved in substituting ranks for the original values can make this a less powerful test than an anova, so the anova should be used if the data meet the assumptions. If the test is , we can make multiple comparisons between the samples. We may choose the level of significance for these comparisons (default is α = 0.05). All pairwise comparisons are made and the probability of each presumed "non-difference" is indicated (Conover, 1999; Critchlow and Fligner, 1991; Hollander and Wolfe, 1999). Hypothesis Statement The purpose of Kruskal-Wallis H analysis is to examine whether there are differences among the three age groups on the length of stay in hospital. H0: There is no difference among the three age groups on the length of stay in hospital H1: There is a difference among the three age groups on the length of stay in hospital Analysis The Kruskal-Wallis test uses ranks of the original values and not the values themselves (Table 9). That's appropriate in this case, because the scale used by the age group is ordinal. First, each case is ranked without regard to group membership. Cases tied on a particular value receive the average rank for that value. After ranking the cases, the ranks are summed within groups. Table 7: Ranks AGE (Grouped) N Mean Rank length of stay in hospital 0 - 59 17 34.94 60 - 69 39 39.82 70 and Older 28 50.82 Total 84 The Kruskal-Wallis statistic measures how much the group ranks differ from the average rank of all groups. The chi-square value is obtained by squaring each group's distance from the average of all ranks, weighting by its sample size, summing across groups, and multiplying by a constant. The degrees of freedom for the chi-square statistic are equal to the number of groups minus one. The asymptotic significance estimates the probability of obtaining a chi-square statistic greater than or equal to the one displayed, if there truly are no differences between the group ranks. A chi-square of 5.377 with 2 degrees of freedom should occur only about 68 times per 1,000. Like the F test in standard ANOVA, Kruskal-Wallis does not tell us how the groups differed, only that they are different in some way. The Mann-Whitney test could be used for pairwise comparisons. Table 8: Test Statisticsa,b length of stay in hospital Chi-Square 5.377 df 2 Asymp. Sig. .068 a. Kruskal Wallis Test b. Grouping Variable: AGE (Grouped) Conclusion and Discussion Using Kruskal-Wallis H test analysis, we fail to reject the hypothesis that there is no difference among the three age groups on the length of stay in hospital. Like the F test in standard ANOVA, Kruskal-Wallis does not tell us how the groups differed, only that they are different in some way. The Mann-Whitney test could be used for pairwise comparisons. 5) Chi Square test Assumptions The formula for chi square yields a statistic that is only approximately a chi square distribution. In order for the approximation to be adequate, the total number of subjects should be at least 20. Some authors claim that the correction for continuity should be used whenever an expected cell frequency is below 5. Research in statistics has shown that this practice is not advisable. For example, see: Bradley, Bradley, McGrath, & Cutcomb (1979). Hypothesis Statement The aim of the chi-square analysis is to examine the relationship between two categorical variables (‘alive discharge’ and ‘preexisting hypertension’). That is, we hypnotize that: H0: The two variables ‘alive discharge’ and ‘preexisting hypertension’ are independent. H1: The two variables ‘alive discharge’ and ‘preexisting hypertension’ are not independent. Analysis Table 11 shows the cross tabulation analysis between ‘alive discharge’ and ‘preexisting hypertension’. The cross tabulation shows the frequency of each level of ‘preexisting hypertension’ at levels of ‘alive discharge’. If each levels of ‘preexisting hypertension’ have a similar level of frequency, the pattern of responses should be similar across level of ‘alive discharge’. At each levels of ‘preexisting hypertension’, the majority of frequencies occur in the yes category of ‘alive discharge’. From the cross tabulation alone, it is impossible to examine whether these differences are real or due to chance variation. We check the chi-square test to be sure. Table 9: preexisting hypertension * alive discharge Crosstabulation Count alive discharge Total no yes preexisting hypertension no 4 26 30 yes 8 46 54 Total 12 72 84 Chi-Square Test The chi-square test measures the discrepancy between the observed cell counts and what you would expect if the rows and columns were unrelated. The two-sided asymptotic significance of the chi-square statistic is greater than 0.05 (P = 0.853), so it is safe to conclude that the differences are due to chance variation, which implies that there is no relationship between ‘alive discharge’ and ‘preexisting hypertension’. That is, it is not necessary that a patient who has preexisting hypertension is more likely to be alived discharge. While the chi-square test is useful for determining whether there is a relationship, it does not tell the strength of the relationship. Systematic and Directional measures (Lambda and Gamma) attempt to quantify this. Table 10: Chi-Square Tests Value df Asymp. Sig. (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided) Pearson Chi-Square .035a 1 .853 Continuity Correctionb .000 1 1.000 Likelihood Ratio .035 1 .852 Fisher's Exact Test 1.000 .564 Linear-by-Linear Association .034 1 .853 N of Valid Casesb 84 a. 1 cells (25.0%) have expected count less than 5. The minimum expected count is 4.29. b. Computed only for a 2x2 table Lambda Statistics Lambda defines error as the misclassification of cases, and cases are classified according to the modal (most frequent) category. The Lamda value of 0.00 means that there is a 0% reduction in misclassification (Table 13). Table 11: Directional Measures Value Asymp. Std. Errora Approx. T Approx. Sig. Nominal by Nominal Lambda Symmetric .000 .000 .b .b preexisting hypertension Dependent .000 .000 .b .b alive discharge Dependent .000 .000 .b .b a. Not assuming the null hypothesis. b. Cannot be computed because the asymptotic standard error equals zero. c. Based on chi-square approximation A pairwise comparison is considered concordant if the case with the larger value in the row variable also has the larger value in the column variable. Concordance implies a positive association between the row and column variables. A pairwise comparison is considered discordant if the case with the larger value in the row variable has the smaller value in the column variable. Discordance implies a negative association between the row and column variables. Gamma is P(concordant) - P(discordant), ignoring comparisons that are tied. Unfortunately, this often excludes a number of pairwise comparisons. The approximate significance value of Gamma is equal to 0.851. Since this is greater than 0.05, we conclude there is no relationship between’ alive discharge’ and ‘preexisting hypertension’. Table 12: Symmetric Measures Value Asymp. Std. Errora Approx. Tb Approx. Sig. Ordinal by Ordinal Gamma -.061 .329 -.188 .851 N of Valid Cases 84 a. Not assuming the null hypothesis. b. Using the asymptotic standard error assuming the null hypothesis. Conclusion and Discussion Using cross tabulation, chi-square, lambda and gamma analysis, we fail to reject the hypothesis that there is no relationship between ‘alive discharge’ and ‘preexisting hypertension’. We initially found that at each levels of ‘preexisting hypertension’, the majority of frequencies occur in the yes category of ‘alive discharge’. However, the chi-square analysis revealed no relationship between ‘alive discharge’ and ‘preexisting hypertension’. Further, the Lamda value analysis revealed a 0% reduction in misclassification. Finally, The approximate significance value of Gamma was greater than 0.05, which further confirmed that there is no relationship between’ alive discharge’ and ‘preexisting hypertension’. 6) Regression analysis. Assumptions The linear regression analysis assumes the following: 1) The error term has a normal distribution with a mean of 0. 2) The variance of the error term is constant across cases and independent of the variables in the model. 3) The value of the error term for a given case is independent of the values of the variables in the model and of the values of the error term for other cases. Evaluation of the Assumptions Figure 1 presents the scatter plot between length of stay in hospital and age. The resulting scatterplot appears to be suitable for linear regression. Figure 1: Scatter Plot Figure 2 present the histogram of regression standardized residuals. A residual is the difference between the observed and model-predicted values of the dependent variable. The residual for a given variable is the observed value of the error term for that variable. A histogram or P-P plot of the residuals helps to check the assumption of normality of the error term. The shape of the histogram should approximately follow the shape of the normal curve. This histogram is acceptably close to the normal curve. Figure 2: Residual Histogram Figure 17 presents the P-P plot of the regression standardized residuals. The P-P plotted residuals should follow the 45-degree line. Neither the histogram nor the P-P plot indicates that the normality assumption is violated. Figure 3: Normal P-P Plot of Regression Standardized Residual Figure 18 shows the plot of residuals by the predicted values of regression equation. The plot of residuals by the predicted values shows that the variance of the errors does not increases with increasing predicted length of stay in hospital. There is a good scatter. Figure 4 Hypothesis Statement The aim of regression analysis is to analyze the relationship between age and length of stay in hospital. That is, we hypnotize that: H0: There is no relationship between age and length of stay in hospital. H1: There is a relationship between age and length of stay in hospital. Analysis Table 15 shows the coefficients of the regression line. It states that the expected length of stay in hospital is equal to (0.123 * age) - 12.720. Table 13: Coefficientsa Model Unstandardized Coefficients Standardized Coefficients t Sig. B Std. Error Beta 1 (Constant) 12.720 7.466 1.704 .092 age .123 .112 .120 1.098 .275 a. Dependent Variable: length of stay in hospital The ANOVA table (Table 16) tests the acceptability of the model from a statistical perspective. The Regression row displays information about the variation accounted for by your model. The Residual row displays information about the variation that is not accounted for by your model. The regression sums of squares is less than residual sums of squares, which indicates that not much of the variation in length of stay in hospital is explained by age. The significance value of the F statistic is greater than 0.05, which means that the variation explained by the model is due to chance. While the ANOVA table is a useful test of the model's ability to explain any variation in the dependent variable, it does not directly address the strength of that relationship. Table 14: ANOVAb Model Sum of Squares df Mean Square F Sig. 1 Regression 147.281 1 147.281 1.206 .275a Residual 10017.672 82 122.167 Total 10164.952 83 a. Predictors: (Constant), age b. Dependent Variable: length of stay in hospital Table 17 reports the strength of the relationship between the model and the dependent variable. R, the multiple correlation coefficient, is the linear correlation between the observed and model-predicted values of the dependent variable. Its large value indicates a strong relationship. R Square, the coefficient of determination, is the squared value of the multiple correlation coefficient. It shows that about one percent of the variation in length of stay in hospital is explained by age. Table 15: Model Summaryb Model R R Square Adjusted R Square Std. Error of the Estimate 1 .120a .014 .002 11.053 a. Predictors: (Constant), age b. Dependent Variable: length of stay in hospital Conclusion and Discussion Using regression analysis, we fail to reject the hypothesis that there is no relationship between age and length of stay in hospital (P = 0.275). Knowing the age for each patient might help the hospital to predict the patients’ length of stay in hospital. However, the regression analysis did not reveal any relationship between age and length of stay in hospital. Using linear regression, the hospital cannot make use of the relationship between the age and length of stay to predict the patients’ length of stay in hospital. Overall Summary Analysis of t-test revealed no difference between male and female in regard to their length of stay in hospital. The independent-samples t test is appropriate whenever two means drawn from independent samples are to be compared. One way analysis of variance (ANOVA) revealed no difference among the three age groups on length of stay in hospital. Using the Mann-Whitney test, no difference among the three age groups on the length of stay in hospital was found significant. The Mann-Whitney test is only slightly less powerful than the t test, so it is a good alternative if the assumptions of the t test are violated, as was the case with length of stay in hospital. Similar to ANOVA result, Kruskal-Wallis H test proved no difference among the three age groups in the mean length of stay in hospital. The chi-square, lambda and gamma analysis showed no relationship between ‘alive discharge’ and ‘preexisting hypertension’. Finally, regression analysis revealed no relationship between age and length of stay in hospital. Bibliography Bradley, D. R., Bradley, T. D., McGrath, S. G., & Cutcomb, S. D. (1979) Type I error rate of the chi square test of independence in r x c tables that have small expected frequencies. Psychological Bulletin, 86, 1290-1297. Brown, M. B., and A. B. Forsythe. 1974b. Robust tests for the equality of variances. Journal of the American Statistical Association, 69:, 364-367. Conover WJ, Practical Nonparametric Statistics (3rd edition). Wiley 1999. Conover, W. J. 1980. Practical Nonparametric Statistics, 2nd ed. New York: John Wiley and Sons. Critchlow DE, Fligner MA. On distribution-free multiple comparisons in the one-way analysis of variance. Communications in Statistics - Theory and Methods 1991;20:127-139. Hays, W. L. 1981. Statistics, 3rd ed. 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