The T Test and the Chi-square Test - Article Example

The T Test and the Chi-square Test T test: The t-distribution can be used to compare the mean of a sampled population to some fixed, known value. This statistical test of hypothesis is referred to as a "one-sample t-test." This one sample t-test statistic is calculated as follows: Test statistic: t = (ẋ - µ) / (σ /) where x represents the mean of the sampled data, µ represents the hypothesized value of this mean, σ represents the hypothesized population standard deviation and n represents the total number of sampled measures. The t-statistic has n−1 degrees of freedom. When two independents populations are involved, they are compared using the two-sample t-test. That is the significant differences in the means of two independently sampled populations can be evaluated. The two assumptions are independence of population and within each population the variable of interest is normally distributed with equal variances. The mathematical derivation of this test statistic is as follows: Test statistic: t = (ẋ1 - ẋ2) / (sp2 /) where n1 and n2 are the number of observations in each of the two groups; x1 and x2 are the means of the two groups. is the estimate of the standard deviation of (ẋ1 - ẋ2) and is calculated using the formula: Where sp is the pooled sample deviation where s12 and s22 represent the sample variances in the two groups. The total number of degrees of freedom associated with this t-test is n1+n2-2. Let us consider the application of this test statistic in health care: Consider the situation when we would like to determine whether there was significant difference in the infant birth weight between mothers who smoked during pregnancy and those who did not. The mean birth weight measured among ten infants whose mothers smoked was 5 lbs., while the mean birth weight measured among the same number of infants whose mothers did not smoke was 8 lbs. Based on the weight measurements of the 20 infants, the pooled sample variance was obtained as 3 lbs. Using the above formula, the calculated t-test statistic was approximately 3.9 with 18 degrees of freedom. The two-sided p-value associated with this test is approximately 0.0006. In other words, even if there was no connection between birth weight and maternal smoking, it could be said that there is a 6 out of 10,000 chance of observing a difference at least as large as 3 lbs. by chance alone. Hence we would conclude that the observed mean difference of 3 lbs. was statistically significant as it could not be explained as to be by chance. Chi-square test: The chi-square test is used to compare the observed data with data expected to obtain according to a specific hypothesis. The data involves categorical variables only. It requires the observed data in a tabular format containing the categories in rows and columns. Thus chi-square test statistic is designed to test the null hypothesis that there is no association between the rows and columns of a contingency table. It can also be used to interpret what proportion of one variable differs from the other variable. The expected count is calculated from the table as follows: The formula for the test statistic is as follows: X2= (O-E)2/E That is, chi-square is the sum of the squared difference between observed (O) and the expected (E) data (or the deviation, d), divided by the expected data in all possible categories. The degrees of freedom = (r-1)*(c-1) where r and c are the number of rows and columns respectively. Based on size of test statistic the conclusion is made. If the observed count deviates from the expected cell count, then we can conclude that there is significant association between the two categorical variables. Let us consider the following application of the test in health care: A study designed to determine whether there is an association between cigarette smoking and asthma might accumulate the collected data into a 2x2 table (Table 1). In this case, the rows could represent whether or not the subject experienced symptoms of asthma. The two columns could be defined by whether the subject smoked or not. The cells of the table would contain the number of observations or patients as defined by these two variables. Based on the above formula, the expected cell count is calculated for the table. The chi-square statistic is obtained. Table 1 Symptoms of asthma Ever smoke cigarettes Total Yes No Yes 20 30 50 No 22 30 100 Total 42 108 150 The table above shows, among asthmatics the proportion of smokers was 40 percent (20/50), while the corresponding proportion among asymptomatic individuals was 22 percent (22/100). The observed cell counts are 20, 30, 22, and 78 and the corresponding expected counts are 14, 36, 28, and 72. The observed and expected counts can then be used to calculate the chi-square test statistic. The degree of freedom is one. The chi-square test statistic is found to be approximately 5.36. The corresponding p-value for this chi-square distribution is obtained as 0.02. In other words, if there was truly no association between smoking and asthma, there is a 2 out of 100 probability of observing a difference in proportions that is at least as large as 18 percent (40%–22%) by chance alone. We would therefore conclude that the observed difference in the proportions is unlikely to be explained by chance alone, and consider this result statistically significant. Reference: http://www.enotes.com/public-health-encyclopedia/t-test http://www.enotes.com/public-health-encyclopedia/chi-square-test Comparison between t test and chi-square test: No. T test Chi-square 1 The t test is used to compare the mean of a sampled population to some hypothesized value of the population mean. The chi-square test is used to compare the observed data with data expected to obtain according to a specific hypothesis 2 It requires the sample data and some fixed value for comparison. The sample mean, sample standard deviation and sample size are then calculated from the data. It requires the observed data in a tabular format containing the categories in rows and columns. The expected data is calculated from the table. 3 Test statistic: t = (ẋ - µ ) /(s / ) Where ẋ is the sample mean, µ is some hypothesized value, s is the sample deviation and n is the sample size Test statistic: X2= (o-e)2/e That is, chi-square is the sum of the squared difference between observed (o) and the expected (e) data (or the deviation, d), divided by the expected data in all possible categories. 4 The degrees of freedom = n-1 The degrees of freedom = (r-1)*(c-1) where r and c are the number of rows and columns respectively 5 The test statistic is then compared with tabulated t distribution critical values and the conclusion is made. Based on size of test statistic the conclusion is made. Read More