A researcher investigated a set of SNPs at a gene-locus by typing a series of random patients with a disease and a series of random controls without the disease. Then statistical tests were performed to assess whether the allele Skip to content. Author : N. Video Audio icon An illustration of an audio speaker. Audio Software icon An illustration of a 3. Software Images icon An illustration of two photographs. Images Donate icon An illustration of a heart shape Donate Ellipses icon An illustration of text ellipses.
EMBED for wordpress. Want more? The sample standard deviation s and the range w are first determined. The test statistic is q and critical values are available for q from Table If the observed value of q lies outside the two critical values, the sample distribution cannot be considered as a normal distribution.
Example For this test of normality we produce the ratio of sample range divided by sample standard deviation and compare with critical values from Table We have two samples for consideration. They are taken from two fluid injection processes.
The two test statistics, q1 and q2 , are both within their critical values. Hence we accept the null hypothesis that both samples could have been taken from normal distributions. Such tests are particularly relevant to quality control situations. Hence the null hypothesis cannot be rejected. It is assumed that the K samples are taken from normally distributed populations.
Each sample is of equal size. Critical values of C are available from Table The null hypothesis that the large variance does not differ significantly from the others is rejected if the observed value of C exceeds the critical value. Example In a test for the equality of k means analysis of variance it is assumed that the k populations have equal variances.
In this situation a quality control inspector suspects that errors in data recording have led to one variance being larger than expected. She performs this test to see if her suspicions are well founded and, therefore, if she needs to repeat sampling for this population a machine process line.
Since the test statistic is less than the critical value she has no need to suspect data collection error since the largest variance is not statistically different from the others. Limitations This test is applicable when the population distribution function is continuous. Method From the sample, the cumulative distribution Sn x is determined and plotted as a step function.
The cumulative distribution F x of the assumed population is also plotted on the same diagram. Example As part of the calibration of a traffic flow model, a traffic engineer has collected a large amount of data. In one part of his model he wishes to test an assumption that traffic arrival at a particular road intersection follows a Poisson model, with mean 7. Can he reasonably assume that this assumption is true?
His test statistic, maximum D is 0. So he cannot assume such an arrival distribution and must seek another to use in his traffic model. Without a good distributional fit his traffic model would not produce robust predictions of flow. Numerical calculation To test the hypothesis that the data constitute a random sample from a Poisson population with mean 7. Limitations The best results are obtained when the samples are sufficiently large — say, 15 or over. Method Given samples of size n1 and n2 from the two populations, the cumulative distribution functions Sn1 x and Sn2 y can be determined and plotted.
Hence the maximum value of the difference between the plots can be found and compared with a critical value obtained from Table If the observed value exceeds the critical value the null hypothesis that the two population distributions are identical is rejected. Example A quality control engineer uses an empirical distribution as a calibration for a particular machining process.
He finds that he has better results using this than any other theoretical model. Since this is less than the 5 per cent critical value of Table 16 he can assume that both samples came from the same population. Do not reject the hypothesis. Both samples come from the same population. The observed and theoretical distributions should contain the same number of elements.
The division into classes must be the same for both distributions. The expected frequency in each class should be at least 5. The observed frequencies are assumed to be obtained by random sampling. For example, to fit data to a normal distribution may require the estimation of the mean and variance from the observed data.
Example The experiment of throwing a die can be regarded as a general application and numerous specific situations can be postulated. In this case we have six classes or outcomes, each of which is equally likely to occur.
The outcomes could be time 10 minute intervals of the hour for an event to occur, e. Since the calculated value is less than the critical value from the table it can be reasonably assumed that the die is not biased towards any side or number. That is, it is a fair die. The outcomes are equally likely.
Numerical calculation A die is thrown times. Can we consider the die to be fair at the 5 per cent level of significance? Hence there are no indications that the die is not fair. Limitations The counts must be obtained under comparable conditions. Method Method a Times for counts are all equal Let the ith count be denoted by Ni. Method b Times for counts are not all equal Let the time to obtain the ith count be ti.
This is compared with Table 5 as above. Example A lime producing rotary kiln is operated on a shift-based regime with four shift workers. A training system is to be adopted and it is desired to have some idea about how the workers operate in terms of out-of-control warning alerts. Do all the four shift workers operate the kiln in a similar way?
The number of alerts over a full shift is recorded and the test statistic is calculated. Do the four workers operate the kiln in a similar way? The answer to this question has bearing on the sort of training that could be implemented for the workers. The chi-squared test statistic is Since the calculated chi-squared value exceeds the 5 per cent critical value we reject the null hypothesis that the counts are effectively the same and conclude that the four counts are not consistent with each other.
The four workers operate the kiln differently. Hence different training schemes may be appropriate. The four counts are not consistent with each other. Limitations This test is applicable if the classification is dichotomous and the elements originate from two sources. It is usually applied when the number of elements is small or the expected frequencies are less than 5.
If p is less than the significance level chosen, we may reject the null hypothesis of independence between samples and classes, i. Example A medical officer has data from two groups of potential airline pilot recruits. Two different reaction tests have been used in selection of the potential recruits.
He uses a more sophisticated test and finds the results given for the two samples. Class 1 represents quick reactions and class 2 represents less speedy reactions. Is there a difference between the two selection tests? This is greater than 0. We add the three probabilities of the three schemes according to hyper-geometric distributions. Limitations It is necessary that the two sample sizes are large enough.
In other words, the two samples were not drawn from one common population. Example In this case the medical officer in Test 39 has a larger sample and can use the chisquared test. He obtains a chi-squared value of 4. Hence he rejects the null hypothesis and now concludes that the two selection methods do produce when compared with the sophisticated reaction test.
Limitations It is necessary that the K sample sizes are large enough. This is usually assumed to be satisfied if the cell frequencies are equal to 5. Sample i.. Example Our medical officer as in Test 39 now has four different colour-blindness selection tests and wishes to see if they produce differences when compared with the recruitment standards. Her data produce a chi-squared value of 5. She does not rejects the null hypothesis and concludes that the colour blindness tests do not differ in their outcome.
It is assumed that there are K series of observations on the same n elements. The observations are dichotomous and the observations in the two classes are represented by 0 or 1. The number of elements must be sufficiently large — say, greater than Let S denote the total score, i. The null hypothesis that the K samples come from one common dichotomous distribution is rejected if Q is larger than the tabulated value.
Example A panel of expert judges assess whether each of four book cover formats is acceptable or not. Each book cover format, therefore, receives an acceptability score. The Cochran Q statistic is calculated as It seems that the book covers are not equally acceptable to the judges. The two samples are sufficiently large. The K classes when put together form a complete series.
Sample 1 Sample 2 n11 n21 n12 n Class j K Total n1j n2j Example Our medical officer from Tests 39 and 40 now wishes to have a third or middle class which represents a reserve list of potential recruits who do not quite satisfy the stringent class1 requirements.
She can still use the chi-squared test to compare the reaction selection tests. Her data produce a chi-squared value of 4. She concludes that the reaction tests do not differ when a third classification of an intermediate reaction is introduced.
Limitations The sample should be sufficiently large. This condition will be satisfied if each cell frequency is greater than 5. Method The sample, of size N, can be categorized into p classes by the first attribute and into q classes by the second.
The frequencies of individuals in each classification can be shown symbolically by the table: Second attribute First attribute She calculates a chi-squared value of Limitations It is assumed that the observations in the sample are independent of each other.
Any sample values equal to M0 should be discarded from the sample. Method A count is made of the number n1 of sample values exceeding M0 , and also of the number n2 below M0.
The null hypothesis is that the population median equals M0. If the alternative hypothesis is that the population median does not equal M0 then the test statistic, T , is the smaller of n1 and n2 with n taken as the sum of n1 and n2.
The null hypothesis is rejected if T is greater than the critical value obtained from Table Example It is assumed that the median value of a financial ratio is 0. A random sample of ten new builds is taken and the ratios computed. Can it be assumed that the sample has been taken from a population of ratios with median 0. Since the calculated T value of 4 is less than the critical value of 7 from Table 17 this assumption is accepted.
Limitations The observations in the two samples should be taken in pairs, one from each distribution. Each one of a pair of observations should be taken under the same conditions, but it is not necessary that different pairs should be taken under similar conditions.
It is not necessary to take readings provided the sign of the difference between two observations of a pair can be determined. Method The signs of the differences between each pair of observations are recorded. The test statistic, r, is the number of times that the least frequent sign occurs. If this is less than the critical value obtained from Table 18 the null hypothesis that the two population medians are equal is rejected.
Example A quality engineer takes two samples from a production line, one before a maintenance modification and one after. Has the modification altered the median value of a critical measurement standard units from the production items? For each pair of values the production machine settings are the same. The maintenance has altered the median value since the critical value is less than the calculated value.
Numerical calculation xi yi Sign 0. Limitations This is a distribution-free test and requires a symmetrical population. The observations must be obtained randomly and independently from a continuous distribution. Rank numbers are now assigned to the differences. Where ties occur among differences, the ranks are averaged among them.
The sum of the ranks with a positive sign and the sum of the ranks with a negative sign are calculated. The test statistic T is the smaller of these two sums. Critical values of this statistic can be found from Table When the value of T falls in the critical region, i.
Example The mean deposit rate GBP per savings level for a sample of ten investors is examined to see if mail advertising has altered this from a value of 0. The signed rank test is used and produces a T value of Since this calculated value is greater than the tabulated value we do not reject the null hypotheses. It would appear that the advertising has not altered the mean deposit level. Any pair of observations giving equal values will be ignored in the analysis.
Method The differences between pairs of observations are formed and these are ranked, irrespective of sign. Where ties occur, the average of the corresponding ranks is used.
Then each rank is allocated the sign from the corresponding difference. When the value of T is less than the critical value, the null hypothesis of equal population means is rejected. Example A manually operated component punch produces two springs at each operation. It is desired to test if the mean component specification differs between the two springs. The sample of pairs of springs produces a signed rank test statistic, T , of 11, which is less than the tabulated value of Hence the null hypothesis of no difference is rejected.
The punch needs re-setting. Numerical calculation xi yi 1. Limitations It is assumed that the two frequency distributions are continuous and that the two samples are random and independent. Method Samples of size n1 and n2 are taken from the two populations. When the two samples are merged and arranged in ascending order, there will be a number of jumps or inversions from one series to the other.
The smaller of the number of inversions and the number of non-inversions forms the test statistic, U. The null hypothesis of the same frequency distribution is rejected if U exceeds the critical value obtained from Table Example An educational researcher has two sets of adjusted reading scores for two sets of five pupils who have been taught by different methods.
It is possible that the two samples could have come from the same population frequency distribution. The collected data produce a calculated U value of 4. Since the sample U value equals the tabulated critical value the educational researcher rejects the null hypothesis of no difference. The data suggest that the two reading teaching methods produce different results.
Below these underlines, the corresponding number of inversions, i. The sample value of U is equal to the critical value. The null hypothesis may be rejected; alternatively, the experiment could be repeated by collecting a second set of data. Limitations The two samples are assumed to be reasonably large. Half of the houses in the estate are maintained by one maintenance company and the other half by another company.
Do the repair regimes of the two companies produce similar results from the residents? Samples of 15 residents are taken from each half. The calculated chi-squared value is 0. The housing officer does not reject the null hypothesis and concludes that the two maintenance companies produce similar results from their repair regimes. Limitations The K samples are assumed to be reasonably large — say, greater than 5.
Method The K samples are first amalgamated and treated as a single grand sample, of which the median is found. Then, for each of the K samples, the number of elements above and below this median can be found. K Total Above median Below median a11 a21 a12 a Example The housing officer in test 50 has a larger estate which is maintained by five maintenance companies. He has sampled the residents receiving maintenance from each company in proportion to the number of houses each company maintains.
The officer now produces a chi-squared value of 0. The tabulated chi-squared value is 9. Limitations It is assumed that the two populations have continuous frequency distributions with the same shape and spread.
Method The results of the two samples x and y are combined and arranged in order of increasing size and given a rank number. In cases where equal results occur the mean of the available rank numbers is assigned. The rank sum R of the smaller sample is now found. Let N denote the size of the combined samples and n denote the size of the smaller sample. The values R and R1 are compared with critical values obtained from Table If either R or R1 is less than the critical value the null hypothesis of equal means would be rejected.
Note If the samples are of equal size, then the rank sum R is taken as the smaller of the two rank sums which occur. Example A tax inspector wishes to compare the means of two samples of expenses claims taken from the same company but separated by a period of time the values have been adjusted to account for inflation.
Are the mean expenses for the two periods the same? He calculates a test statistic, R1 of and compares this with the tabulated value of Since the calculated value is greater than the tabulated critical value he concludes that the mean expenses have not changed.
Numerical calculation Total x Rank Hence there is no difference between the two means. Limitations It is assumed that the two populations have continuous frequency distributions and that the sample sizes are not too small, e. Method The results of the two samples are combined and arranged in order of increasing size. This procedure continues, working from the ends towards the centre, until no more than one unranked value remains.
That is to say, if the number of values is odd, the middle value has no rank assigned to it. Let R1 be the rank sum of the series of size n1. The null hypothesis of equal variance is rejected if Z falls in the critical region. Example A catering manager wants to know if two types of pre-prepared sauce give the same spread or variability of values. This is because he has to set his dispensers to a fixed value and an unusually large value will cause problems.
He takes a sample of ten sauces of each type and compares them using the Siegel—Tukey rank sum dispersion test. He rejects the null hypothesis of no difference and concludes, in this case, that sauce type y has greater dispersion than type x. The K frequency distributions should be continuous. Method The K samples are combined and arranged in order of increasing size and given a rank number.
Where ties occur the mean of the available rank numbers is used. The rank sum for each of the K samples is calculated. Let Rj be the rank sum of the jth sample, nj be the size of the jth sample, and N be the size of the combined sample. The null hypothesis of equal means is rejected when H exceeds the critical value. Example A cake preference score is a combination of four components, viz.
The minimum score is 0 and the maximum Three cake formulations are compared using these scores by three panels of accredited tasters. The results produce an H test statistic of 2. This is less than the tabulated value of 4.
The catering manager concludes the three cake formulations are equally preferred. Numerical calculation Combined rank assignment of three sample data x1 , x2 , x3 : Sample Value Rank x1 1.
Limitations The K samples must have the same size, and the frequency distributions of the population are assumed continuous.
Method The K samples are combined and arranged in order of increasing size and then given a rank number. The highest raw value is assigned rank 1. For each sample the rank sum is determined. Critical values of this test statistic can be obtained from Table Example A perfume manufacturer has four floral fragrances and wishes to compare each one against the others in a preference test.
Selected perfume testers can give a perfume a score between 1 and For each of these four fragrances four testers are used and the results are shown. The critical value from Table 23 is Fragrances 1 and 2 and 1 and 3 are viewed as different, with fragrance 1 generally preferred. Hence samples 1 and 2 and samples 1 and 3 are significantly different. Limitations It is assumed that the populations have continuous frequency distributions and that the K samples are of equal size n.
Method The K samples are merged together and rank numbers allocated to the Kn observations. The sum of the rank numbers of the observations belonging to a particular sample is formed. This is repeated for each sample and the test statistic is the largest of these rank sums.
When the test statistic exceeds the critical value obtained from Table 24 the mean of the population generating the maximum rank sum is said to be significantly large. Example As an alternative to Test 55 the perfume manufacturer uses the rank sum maximum test for the largest 4 population means.
The largest R value is R1 at 58 which is greater than the tabulated value of Hence fragrance 1 is significantly greater in preference than the other fragrances. This is a similar result to that found with Test Numerical calculation Combined rank assignment of four samples, i.
The calculated value of R1 is greater than the critical value. Hence the sample 1 is statistically significantly greater than the others. Limitations The K samples, one from each treatment and one from the control, should all be of the same size. Method Each of the treatment samples is compared with the control sample in turn.
To test the jth sample, it is merged with the control sample and rank numbers are allocated to the 2n observations. This provides two rank sums and the smallest of these is used as the test statistic if a two-tailed test is desired. To test the alternative hypothesis that treatment j has a smaller effect than the control treatment, the rank sum for the jth control sample forms the test statistic.
In both cases, the null hypothesis that there is no difference between the jth treatment and the control is rejected if the test statistic is less than the critical value obtained from Table Example Four different sprain relief creams are compared with controls. Treatments are allocated at random and each is compared with its control. The results show that rank sums for controls 1 and 4 are less than the critical tabulated value of 76 Table A survey of the statistical power of research in We estimated the statistical power of the first and last statistical test presented in Bradley DR , Russell RL, Reeve The importance of statistical power.
Reporting statistics in clinical trials published Jaykaran, 1 ND Kantharia, 1 Y as well as the number for which statistical test used to see Reporting of statistics of clinical trials published in western.
Iverson to conclude that her testing Shawna's IQ score was calculated using the formula developed by Lewis Which of the following statistics is Observations on the use of statistical methods in many software packages are available that facilitate statistical analysis Lewis; Outliers in statistical Statistical tests and graphics using R.
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