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Hypothesis testing Is it statistically significant? Most of the slides are adopted from:

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Example In a trial a jury must decide between two hypotheses. – Null hypothesis H 0 : The defendant is innocent – Alternative hypothesis H 1 : The defendant is guilty The jury must make a decision on the basis of evidence presented. In the language of statistics convicting the defendant is called rejecting the null hypothesis in favor of the alternative hypothesis. That is, the jury is saying that there is enough evidence to conclude that the defendant is guilty. If the jury acquits, then it is stating that there is not enough evidence to support the alternative hypothesis.

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Goal Assess the evidence provided by data in favor of some claim (null hypothesis) about the population. The procedure begins with the assumption that the null hypothesis is true. The goal is to determine whether there is enough evidence to infer that the alternative hypothesis is true, or the null is not likely to be true. Results of the test are expressed in terms of a probability – This probability measures how well the data and the hypothesis agree.

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Example usage in our work Usually we need to find if the performance of our proposed algorithm is statistically superior than that of a base line algorithm. General procedure – For i=1:n Generate a random train/test dataset. Run both algorithms and get performance numbers. Record the difference between the performance of the proposed and base line algorithms. Let it be x. – Null hypothesis H 0 : x <= 0 i.e. base line better than proposed algorithm. – Alternate hypothesis H1 : x > 0 Our algorithm is better. We want to reject H 0 hypothesis in favor of H 1 with certain confidence value.

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Statistical Example Consider mean demand for computers during assembly lead time. The operations manager wants to know whether the mean is different from 350 units. Our test statistics is over mean demand. Null hypothesis: H 0 = 350 Our alternative hypothesis becomes: H 1 ≠ 350 Assume σ = 75 and the sample size n = 25, and the sample mean is calculated to be A large value of (say, 600) provides enough evidence. If is close to 350 (say, 355), then this does not provide a great deal of evidence to infer that the population mean is different than 350.

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Statistical Example Assume H 0 : μ = 350 is true. Determine the sampling distribution of sample mean. Assume has mean 350. Variance of is 75/sqrt(25) = 15 Is the Sample Mean in the Guts of the Sampling Distribution?

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Hypothesis Test 1: Unstandardized test For the guts to be the center 95% of the distribution [α = 0.05]. The critical values that define the guts will be 1.96 standard deviations of X-Bar UCV = *15 = LCV = 350 – 1.96*15 = 320.6

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Hypothesis Test 2: Z-score Calculate z score of the sample mean. Z = ( - )/ = ( – 350)/15 = Is this Z-Score in the guts of the sampling distribution?

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Hypothesis Test 3: p-value Increase “Rejection Region” until it “captures” sample mean. – P( > ) = P(Z > 1.344) = – p-value = double of this area for two tailed test = Since rejection region is defined to be 5% and our sample mean is in the 18.02% region, it in not in rejection region.

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Statistical Conclusions Unstandardized Test Statistic – Since LCV (320.6) < (370.16) < UCV (379.4), we reject the null hypothesis at a 5% level of significance. Standardized Test Statistic – Since -Z /2 (-1.96) < Z(1.344) < Z /2 (1.96), we fail to reject the null hypothesis at a 5% level of significance. P-value – Since p-value (0.1802) > 0.05 [ ], we fail to reject the null hypothesis at a 5% level of significance

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Interpreting the p-value Overwhelming Evidence (Highly Significant) Strong Evidence (Significant) Weak Evidence (Not Significant) No Evidence (Not Significant)

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Assumptions Distribution of test statistics follows standard deviation. – Generally one appeals to the central limit theorem to justify assuming that a test statistic varies normally. – Central limit theorem (CLT): mean (x bar) of a sufficiently large number of independent random variables (x), each with finite mean and variance, will be approximately normally distributed. – If the variation of the test statistic is strongly non-normal, a Z-test should not be used. should be known or can be estimated reliably. – when is same for each x and all x are independent from each other. [Follows from var(c.x) = c 2 var(x) and var (x1 + x2) = var (x1)+ var (x2) if x1 and x2 are independent]. A Z-test is appropriate when you are handling moderate to large samples (n > 30).

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Hypothesis Test 4: Student’s t-test The test statistic (x bar) follows a Student’s T-distribution if the null hypothesis is true. T-test is appropriate for small sample size (n < 30). Great if the populations’ standard deviation is unknown. s= sample standard deviation. Once a t value is determined, a p-value can be found using a table of values from Student's t-distribution The Student t-distribution is symmetric and bell-shaped, like the normal distribution, but has heavier tails

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Student’s t-test Two sample t-test can be used to compare means of two independent and identically distributed samples.

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Hypothesis Test 5: Wilcoxon Signed Test A non-parametric statistical hypothesis test. Alternative to the paired Student's t-test when the population cannot be assumed to be normally distributed. H 0 : μ = 0 Rank x i s based on their magnitude. Let their rank be R i. Wilcoxon signed rank statistics φ i = I(x i > 0) Score Find the critical value for the given sample size n and the wanted confidence level. Compare S to the critical value, and reject H 0 if S is less than or is equal to the critical value.

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Thank you

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Two types of Errors Type I error – When we reject a true null hypothesis. That is, jury convicts an innocent person. – P(Type 1 error) = α [usually 0.05 or 0.01] Type II error – When we don’t reject a false null hypothesis. That is, when a guilty defendant is acquitted. – P(Type 2 error) = β H0H0 TF Reject I II The two probabilities are inversely related. Decreasing one increases the other, for a fixed sample size.

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