Mean bone density amongst two groups of 50 people each is compared with each other. The test of significance used would be:
Correct Answer: Student t-test
Description: Ans. is 'b' i.e. Student t-test Let's first revise the various statistical tests:Chi-Square test:Chi-square test is used for qualitative data; data in which there is no notion of magnitude or size of the attribute.The data are classified by counting the individuals having the same characteristic or attribute (i.e. persons with the same characteristic are counted to form two groups such as vaccinated or not vaccinated, cured or not cured, on drug or not on drug, taken iron supplements or not taken iron supplements)The data is in the form of proportions.Student t-testIs used for Quantitative data i.e. the data has a magnitude and is normally distributed in the population (i.e. continuous data) e.g. each individual has one measurement from a continuous spectrum or range such as body temperature, height, weight, blood pressure, mean bone density etc.Student t-test is of three types:Student t-test for single small sampleStudent t-test for independent samples (also k/a unpaired test)Student t-test for paired sample (paired t - test)Paired t - testIt compares the means of two paired samples (when two readings, are taken from a same group, before and after an intervention)Analysis of variance (ANOVA)It is used for same datas as the t-testWhereas t-test is appropriate for making just one comparison (between two sample means, or between a sample mean and hypothesized population mean), ANOVA is used when more than one comparison is to be made (when means of more than two groups are being compared).In the question since the attribute (mean bone density) is a quantitative data and we have only two samples the test of significance used would be student t-test
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