The Mann Whitney U statistic is defined as:
– where samples of sizeandare pooled and Ri are the ranks of sample 1.
U can be resolved as the number of times observations in one sample precede observations in the other sample in the ranking.
In most circumstances a two sided test is required; here the alternative hypothesis is that 1 values tend to be distributed differently to 2 values. For a lower side test the alternative hypothesis is that 1 values tend to be smaller than 2 values. For an upper side test the alternative hypothesis is that 1 values tend to be larger than 2 values.
Assumptions of the Mann-Whitney test:
A confidence interval for the difference between two measures of location is provided with the sample medians. The assumptions of this method are slightly different from the assumptions of the Mann-Whitney test:
Example
The following data represent fitness scores from two groups of boys of the same age, those from homes in the town and those from farm homes.
Farm Boys |
Town Boys |
14.8 11.1 |
12.7 18.9 |
12.2 |
17.4 |
14.2 |
Pooling and sorting gives 11.1, 12.2, 12.7, 14.2, 14.8, 17.4, 18.9.
The Farm Boys have ranks1, 2 and 5.andso
Comparison with values in statistical tables causes us not to reject the null hypothesis of no difference at the 10% level.
]]>Example: Two samples are taken from two populations.
Sample 1: 11, 23, 43, 12, 14, 32, 22, 43, 35, 45
Sample 2: 15, 44, 21, 26, 30, 39, 48, 50
Comparing this with the 5% level from thedistribution, 3.677, we see that 1.052<3.677 so we do not reject the hypothesis that
]]>There exists a similar formula for the variance. The pooled sample varianceofsurveys, each of sizeand sample varianceis more complicated:
.
The standard deviation is the square root of this.
The pooled standard deviation is most useful when used in the two sample t test, when conducting hypothesis tests for the means of two samples.
In this case
The sample variance is the same as if the two samples had been pooled and the pooled data used to find the variance, with the difference that the denominator in the root is notbutThis is because one degree of freedom is lost for each sample.
Proof:
Example: Find the pooled standard deviation of two samples of size 40, sample standard deviation 7 and sample size 55, sample size 12 respectively.
]]>Write down the null hypothesisfor exampleor
Write down the alternative hypothesisfor exampleor
Specify the significance level
Write down the number of degrees of freedom
Write down the critical value (one tailed test), values (two tailed test) or the critical region (one tailed test) or critical regions (two tailed test).
Calculate the sample meansand variances,and
Calculated the pooled variance
Calculate the value of the test statistic
Interpret the test statistic. If the value of the test statistic is in the critical region,can be rejected at the significance level of the test.
Example: Conduct a hypothesis test at the 10 % level for the equality of the means of the populations from which the following samples are drawn.
Population 1: 4, 2, 4, 8, 2, 6
Population 2: 7, 3, 6, 1, 9, 9, 10
For a significance level of 10% with 6+7-2 =11 degrees of freedom, (we are conducting a two tailed test since we are testing for equality) the critical values are
This is not in the critical region since
]]>A simple example is shown in calculating the mean of a sample. Supposeis a sample from a population.
Let
Theare estimates of the residualswhereis the true mean of the population.
The sum of the residuals (unlike the sum of the errors) is necessarily 0. If we know the values of anyof the residuals, we can find the last residual, so that even though there are n residuals, onlyof these are free to vary. For this reason we say that the residuals of a sample of size n hasdegrees of freedom.
For the same reason, the sample standard deviationincludes an expression n-1 in the denominator.
We can write
is the sum of the squares of the residuals and the residuals only have n-1 degrees of freedom, so that
]]>An estimatorfor a population parameteris an unbiased estimator forif
An unbiased estimator is always better than a biased estimator, and an estimator is better in general if the bias is small and/or the variance is small.
For exampleis a biased estimator forbutis an unbiased estimator forThe relationship betweenandisso that The bias ofis then
If we have two estimators for the mean of a population,andwith thenand so bothandare unbiased estimators forbut
and
The variance ofis less than the variance ofsois a better estimator for
]]>A number of situations can be modelled by the exponential distribution. In these examples it must be assumed however that each event is independent.
The lifetimes of electric bulbs. Some bulbs will fail almost immediately and some will last a very long time.
The time intervals between cars passing a certain point.
The the time intervals between earthquakes or volcanoes.
The mean of the exponential distribution is
and the variance is
The effect size is the difference between the true value and the value specified in the null hypothesis.
Effect size = True value – Hypothesized value
For example, suppose the null hypothesis states that a population mean is equal to 100. A researcher might ask: What is the probability of rejecting the null hypothesis if the true population mean is equal to 90? In this example, the effect size would be 90 – 100, which equals -10. Obviously if the true value is far from the hypothesised value then the null hypothesis is more likely to be rejected so the probability of committing a Type II error is reduced. With this made clear we can make the following summary.
The power of a hypothesis test is affected by three factors.
In addition, the probability of committing a Type II error increases with decreasing probability of committing a Type I test. It is impossible to simultaneously decrease the probability of a Type I test and Type II test.
]]>We can do this using the fact thatthedistribution with degrees of freedom.
Denoting byandthe upper and lowerpoints of thedistribution withdegrees of freedom we have thatwith a certainty of
We can separate this into two inequalities:
We can combine these two into a single inequalitywith a certainty ofThe confidence interval is
Example: The standard deviation of a sample of 15 tomato plants is 5.8 cm. Find a 95% confidence interval for the variance of the tomato plant population.
The upper and lower 2.5% points of the %chi^2 distribution with (15-1)=14 degrees of freedom are 5.63 and 26.12 respectively. The confidence interval is
]]>It is quite easy to calculate the probability of committing a Type I error – rejecting the null hypothesis when the null hypothesis is true. If the test is conducted at the% level then the probability of rejecting the null hypothesis issince in conducting the hypothesis test we always assume the null hypothesis is true, and so the probability of committing a Type I error is also
If the test is conducted so that the null hypothesis is rejected if values less than a certain valueare observed then the power of the test isand the power function isFor example, the number of tornadoes to hit a particular town historically follows a Poisson distribution with mean,Suppose we now want to asses whether climate change has decreased the frequency of hurricanes. In the last year there were 3 hurricanes.
The null hypothesis is
The alternative hypothesis is
The power function is
The power of the test is here
Ifthen the power is
The power of the test increases in this case if %lambda increase. This means the null hypothesis is more likely to be rejected ifis fixed at 3 and the probability of a Type II error is reduced. This is not necessarily a benefit as it means that the probability of a Type I error is increased. It is in fact impossible to decrease the probability of a Type I and Type II error simultaneously. The probability of each error must be traded so that an optimum is reached.
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