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Chapter 7 – Hypothesis Testing There is a claim about a population parameter: mean (mu) or proportion (p). Collect sample data and compare the observed statistic with the claimed value. (is the difference due to chance, or to something else?) 1. Is the difference observed due to chance, sample variability? 2. Or, is the difference observed statistically significant (too large to happen by chance alone) to provide evidence against the null hypothesis and in favor of the alternative hypothesis? We assume the null hypothesis is true We use sample data to evaluate the credibility of the null hypothesis The data will either provide support for Ho or tend to reject Ho We analyze the distribution of sample means, for all samples of size n to determine what sample means are consistent with Ho and what sample means are at odds with Ho. The distribution is divided into two sections: 1. Sample means that are likely to be obtained if Ho is true, that are “close” to Ho, that are “consistent with” Ho, that are “usual”, that have “high probability”, that “can happen by chance”, that can be “due to sample fluctuation” 2. Sample means that are very unlikely to be obtained if Ho is true, that are “very different” from Ho, that are “unusual” that have a “low probability”, that can not be explained by chance alone. The extremely unlikely values, in the direction of the alternative hypothesis, make up the critical region. The significance level, α, is the probability of rejecting Ho when in fact it is true, it is the area of the critical region. It is an indication of the strength of the evidence against the null hypothesis. Critical values are the boundary z or t scores that separate sample statistics that are likely to occur (usual scores, with large probabilities) from those that are unlikely (unusual, with very small probability). Test Statistic: z- or t- score of the sample mean x bar. Under the assumption that Ho is true, the p-value is the probability of obtaining a test statistic at least as extreme as the one observed. If the p-value = 0.02, this means that there is a 2% chance that the observed value of the sample mean x-bar is due to random variation (to chance). The observed sample mean x-bar is expected to occur by chance 2 out of 100 times. To reach conclusions we use two methods: 1. In the Critical value method we compare the test statistic and the critical value 2. In the p-value method we compare the p-value to the significance level α Test results are statistically significant (in favor of the alternative hypothesis) o Reject Ho and support H1 Test results are not statistically significant. The difference between the test statistic and the hypothesized mean is probably due to chance, sample variability. o Do not reject Ho, we don’t have sufficient evidence to support H1 Reporting results In a scientific journal you will not be told explicitly that the null hypothesis has been rejected. You will see the statement: The treatment with medication had a significant effect on people’s depression scores. Z = 2.45, p < 0.05 (Significant: the result is different from what would be expected due to chance. The hypothesis test has ruled out chance as a plausible explanation for the results.) There was no evidence that the medication had an effect on depression scores, z = 1.30, p > 0.05 The treatment was not statistically significant. (This tells you that the obtained result is relatively likely to occur by chance. Some examples: 1) In an advertisement, a pizza shop claims that its mean delivery time is less than 30 minutes. A random selection of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. Is there enough evidence to support the claim at α = 0.01? (z = -2.57, p = 0.0051) 2) What decision should ou make for the following Minitab printout, using a level of significance of a) α = 0.05, b) α = 0.01 Test of mu = 6.200 vx mu not = 6.200 The assumed sigma = 0.470 Variable Sample N 53 Mean 6.0666 Practice Interpreting results: Page 398: 17, 18 Page 406: 15, 16 Page 417: 23, 24 StDev 0.4146 SE Mean 0.0646 Z -2.07 P 0.039