When you set up a hypothesis test to determine the validity of a statistical claim, you need to define both a null hypothesis and an alternative hypothesis.
Typically in a hypothesis test, the claim being made is about a population (one number that characterizes the entire population). Because parameters tend to be unknown quantities, everyone wants to make claims about what their values may be. For example, the claim that 25% (or 0.25) of all women have varicose veins is a claim about the proportion (that’s the ) of all women (that’s the ) who have varicose veins (that’s the — having or not having varicose veins).
Therefore, there is not sufficient evidence to reject the null hypothesis that the two correlation coefficients are equalClearly, this test can be modified and applied for test of hypothesis regarding population correlation based on observed r obtained from a random sample of size n:provided | r | 1, and | | 1, and n is greater than 3.
Under the null hypothesis and normality condition, the test statistic is: where: and n= sample size associated with r, and n =sample size associated with r.
Given that two populations have normal distributions, we wish to test for the following null hypothesis regarding the equality of correlation coefficients:H: = , based on two observed correlation coefficients r, and r, obtained from two random sample of size n and n, respectively, provided | r | 1, and | r | 1, and n, n both are greater than 3.
Some statistical tests, such as testing equality of the means by the t-test and ANOVA, assume that the data come from populations that have the same variance, even if the test rejects the null hypothesis of equality of population means.
However, applying the goodness-of-fit, at 0.05, under the null hypothesis that there are no differences in the number of accidents in three shifts, one expects 5, 5, and 5 accidents in each shift.
For this goodness-of-fit test, we formulate the null and alternative hypothesis as H0: fY(y) = fo(y)
Ha: fY(y) fo(y) At the level of significance, H0 will be rejected in favor of Ha if C is greater than .
Among three possible scenarios, the interesting case is in testing the following null hypothesis based on a set of n random sample observations: H: Variation is about the claimed value.
H: The variation is more than what is claimed, indicating the quality is much lower than expected.
For the goodness-of-fit sample test, we formulate the null and alternative hypothesis as H : fY(y) = fo(y)
H : fY(y) fo(y) At the level of significance, H will be rejected in favor of H if is greater than However, it is possible that in a goodness-of-fit test, one or more of the parameters of fo(y) are unknown.
If the value you get from calculating the Chi-square statistic is sufficiently high (as compared to the values in the Chi-square table), it tells you that your null hypothesis is probably wrong.
The test for homogeneity, on the other hand, is designed to test the null hypothesis that two or more , according to some criterion of classification applied to the samples.
In other words, the simplest correction is to move the cut-off point for the continuous distribution from the observed value of the discrete distribution to midway between that and the next value in the direction of the null hypothesis expectation.
Following the above process for this test, the K-S statistic is 0.421 with the p-value of 0.0009, indicating a strong evidence against the null hypothesis.
The p-value is equal to 0.062, indicating that there is moderate evidence against the null hypothesis that the three populations are statistically identical.