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Null and Alternative Hypothesis | Real Statistics Using …

The Central Limit Theorem having come to our rescue, we can now set aside the caveat that the populations shown in are non-normal and proceed with our analysis. From we can see that the center of the theoretical distribution (black line) is 11.29, which is the actual difference we observed in our experiment. Furthermore, we can see that on either side of this center point, there is a decreasing likelihood that substantially higher or lower values will be observed. The vertical blue lines show the positions of one and two SDs from the apex of the curve, which in this case could also be referred to as SEDMs. As with other SDs, roughly 95% of the area under the curve is contained within two SDs. This means that in 95 out of 100 experiments, we would expect to obtain differences of means that were between “8.5” and “14.0” fluorescence units. In fact, this statement amounts to a 95% CI for the difference between the means, which is a useful measure and amenable to straightforward interpretation. Moreover, because the 95% CI of the difference in means does not include zero, this implies that the -value for the difference must be less than 0.05 (i.e., that the null hypothesis of no difference in means is not true). Conversely, had the 95% CI included zero, then we would already know that the -value will not support conclusions of a difference based on the conventional cutoff (assuming application of the two-tailed -test; see below).

If the null hypothesis is true, P-values are random values, uniformly distributed between 0 and 1.

It is also worth pointing out that there is another way in which the -test could be used for this analysis. Namely, we could take the ratios from the first three blots (3.33, 3.41, and 2.48), which average to 3.07, and carry out a one-sample two-tailed -test. Because the null hypothesis is that there is no difference in the expression of protein X between wild-type and backgrounds, we would use an expected ratio of 1 for the test. Thus, the -value will tell us the probability of obtaining a ratio of 3.07 if the expected ratio is really one. Using the above data points, we do in fact obtain = 0.02, which would pass our significance cutoff. In fact, this is a perfectly reasonable use of the -test, even though the test is now being carried out on ratios rather than the unprocessed data. Note, however, that changing the numbers only slightly to 3.33, 4.51, and 2.48, we would get a mean of 3.44 but with a corresponding -value of 0.054. This again points out the problem with -tests when one has very small sample sizes and moderate variation within samples.

Fail to reject the null hypothesis and ..

proportions or distributions refer to data sets where outcomes are divided into three or more discrete categories. A common textbook example involves the analysis of genetic crosses where either genotypic or phenotypic results are compared to what would be expected based on Mendel's laws. The standard prescribed statistical procedure in these situations is the test, an approximation method that is analogous to the normal approximation test for binomials. The basic requirements for multinomial tests are similar to those described for binomial tests. Namely, the data must be acquired through random sampling and the outcome of any given trial must be independent of the outcome of other trials. In addition, a minimum of five outcomes is required for each category for the Chi-square goodness-of-fit test to be valid. To run the Chi-square goodness-of-fit test, one can use standard software programs or websites. These will require that you enter the number of expected or control outcomes for each category along with the number of experimental outcomes in each category. This procedure tests the null hypothesis that the experimental data were derived from the same population as the control or theoretical population and that any differences in the proportion of data within individual categories are due to chance sampling.

Regardless of the method used, the -value derived from a test for differences between proportions will answer the following question: What is the probability that the two experimental samples were derived from the same population? Put another way, the null hypothesis would state that both samples are derived from a single population and that any differences between the sample proportions are due to chance sampling. Much like statistical tests for differences between means, proportions tests can be one- or two-tailed, depending on the nature of the question. For the purpose of most experiments in basic research, however, two-tailed tests are more conservative and tend to be the norm. In addition, analogous to tests with means, one can compare an experimentally derived proportion against a historically accepted standard, although this is rarely done in our field and comes with the possible caveats discussed in . Finally, some software programs will report a 95% CI for the difference between two proportions. In cases where no statistically significant difference is present, the 95% CI for the difference will always include zero.

The “fail to reject the null hypothesis” carries with it ..

Entire books are devoted to the statistical method known as . This section will contain only three paragraphs. This is in part because of the view of some statisticians that ANOVA techniques are somewhat dated or at least redundant with other methods such as (see ). In addition, a casual perusal of the worm literature will uncover relatively scant use of this method. Traditionally, an ANOVA answers the following question: are any of the mean values within a dataset likely to be derived from populations that are truly different? Correspondingly, the null hypothesis for an ANOVA is that all of the samples are derived from populations, whose means are identical and that any difference in their means are due to chance sampling. Thus, an ANOVA will implicitly compare all possible pairwise combinations of samples to each other in its search for differences. Notably, in the case of a positive finding, an ANOVA will not directly indicate which of the populations are different from each other. An ANOVA tells us only that at least one sample is likely to be derived from a population that is different from at least one other population.

The basis for many nonparametric tests involves discarding the actual numbers in the dataset and replacing them with numerical rankings from lowest to highest. Thus, the dataset 7, 12, 54, 103 would be replaced with 1, 2, 3, and 4, respectively. This may sound odd, but the general method, referred to as a , is well grounded. In the case of the Mann-Whitney test, which is used to compare two unpaired groups, data from both groups are combined and ranked numerically (1, 2, 3, … ). Then the rank numbers are sorted back into their respective starting groups, and a is tallied for each group. If both groups were sampled from populations with identical means (the null hypothesis), then there should be relatively little difference in their mean ranks, although chance sampling will lead to some differences. Put another way, high- and low-ranking values should be more or less evenly distributed between the two groups. Thus for the Mann-Whitney test, the -value will answer the following question: Based on the mean ranks of the two groups, what is the probability that they are derived from populations with identical means? As for parametric tests, a -value ≤ 0.05 is traditionally accepted as statistically significant.

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Define Null and Alternative Hypotheses

i) State the null alternative hypotheses, explaining your choice.
ii)Calculate the P-value
iii)Represent your results on a graph that marks clearly the rejection and non rejection regions. What do you concludes.
iv) Construct a 90% confidence interval around the March sample mean. Comment on the, relative to your conclusion in part (iii)

Hypothesis Testing - Six Sigma Material

The rationale behind using the paired -test is that it takes meaningfully linked data into account when calculating the -value. The paired -test works by first calculating the difference between each individual pair. Then a mean and variance are calculated for all the differences among the pairs. Finally, a one-sample -test is carried out where the null hypothesis is that the mean of the differences is equal to zero. Furthermore, the paired -test can be one- or two-tailed, and arguments for either are similar to those for two independent means. Of course, standard programs will do all of this for you, so the inner workings are effectively invisible. Given the enhanced power of the paired -test to detect differences, it is often worth considering how the statistical analysis will be carried out at the stage when you are developing your experimental design. Then, if it's feasible, you can design the experiment to take advantage of the paired -test method.

Statistical hypothesis testing - Wikipedia

A-squared (A2) refers to a numerical value produced by the Anderson-Darling test for normality. The test ultimately generates an approximate P-value where the null hypothesis is that the data are derived from a population that is normal. In the case of the data in , the conclusion is that there is

Fisher's null hypothesis testing Neyman–Pearson decision theory; 1

Interestingly, there is considerable debate, even among statisticians, regarding the appropriate use of one- versus two-tailed -tests. Some argue that because in reality no two population means are ever identical, that all tests should be one tailed, as one mean must in fact be larger (or smaller) than the other (). Put another way, the null hypothesis of a two-tailed test is always a false premise. Others encourage standard use of the two-tailed test largely on the basis of its being more conservative. Namely, the -value will always be higher, and therefore fewer false-positive results will be reported. In addition, two-tailed tests impose no preconceived bias as to the direction of the change, which in some cases could be arbitrary or based on a misconception. A universally held rule is that one should never make the choice of a one-tailed -test after determining which direction is suggested by your data In other words, if you are hoping to see a difference and your two-tailed -value is 0.06, don't then decide that you really intended to do a one-tailed test to reduce the -value to 0.03. Alternatively, if you were hoping for no significant difference, choosing the one-tailed test that happens to give you the highest -value is an equally unacceptable practice.

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