This is why many statisticians always recommend a two tailed alternative

This is why many statisticians always recommend a two

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type of hypothesis test being evaluated. This is why many statisticians always recommend a two- tailed alternative, because you have made a stronger case for rejecting H 0. Probability Values (P-Values) and Hypothesis Testing The level of significance chosen is rather arbitrary. For that reason, most statisticians use the concept of probability values ( p-values) to report the outcome of a hypothesis test. The p-value for a hypothesis test is the smallest value of α for which the data indicates rejection of H 0 . Thus, to reject the null hypothesis the p-value must be <= α. If it is > α, we fail to reject the H 0 . H a : μ < μ 0 H 0: μ = μ 0 H a : μ μ 0 |𝑥 ̅ μ 0 | >= 1.96σ/ √𝑛
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5 | P a g e The p-value may also be interpreted as the probability of observing a value of the test statistic at least as extreme as the observed value of the test statistic if H 0 is true . In other words, it’s the probability of incorrectly rejecting the null hypothesis. If we let 𝑿 ̅ represent the random variable for the sample mean under H 0 and x be the observed value of 𝑥̅ , then the p-value for the one sample Z-test is computed as follows: P-Values for One Sample Z-Test All probabilities are computed under the assumption that H 0 is true. In our Cooley High School example, the p-value for the two-tailed test is: 2*Prob( |𝑿 ̅ -75|>=4) = 2*Prob( 𝑿 ̅ >=79), which can be computed as the following in Excel: 2*(1-NORM.DIST(79,75,15/ √49 ), True) = 2*(0.030974) = 0.061948. Because our p-value of 0.06 > our alpha value of .05, we fail to reject H 0 . The p-value for a one-tailed test is: Prob( 𝑿 ̅ >=79) = 0.030974. Because our p-value of 0.03 < than our alpha value of .05, we reject H 0 . One Sample Hypothesis for Mean: Small Sample, Normal Population, Variance Unknown Suppose we are interested in testing a hypothesis about the mean of a normal population where the population variance is unknown, and the sample size n is <30. Then, it can be shown that ( 𝑥̅ - µ )/(s/√n) follows a t -distribution with n-1 degrees of freedom. Here s = sample standard deviation. Like the standard normal distribution, the t-distribution has a density symmetric around 0. As shown below, the t-distribution has fatter tails than the standard normal density, and as n increases the t-distribution approaches the standard normal density. Hypotheses P-Value H 0: μ <= μ 0 H a : μ > μ 0 Prob( 𝑿 ̅ >= x) H 0: μ >= μ 0 H a : μ < μ 0 Prob( 𝑿 ̅ <= x) H 0: μ = μ 0 H a : μ μ 0 Prob( |𝑿 ̅ − µ 0 | ≥ 𝑥 )
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6 | P a g e As shown in the Hypothesis Testing.xlsx spreadsheet, T random variable worksheet shows the percentiles of the t-distribution and how they can be computed using the T.INV function. We let t (α,n -1) represent the α percentile of a t -distribution with n-1 degrees of freedom. Below we find, for example, t (.025,28) = -2.04841. Basically, one sample t-tests look just like one sample z- tests with s replacing σ and the t percentiles replacing the z percentiles. Critical Region for One Sample t-tests If we let T n-1 stand for a t-distribution with n-1 df and t represent the observed value of ( 𝑥̅ - µ )/(s/√n) , then the p-values for a one sample t-test may be computed as follows: P-Values for One Sample t-Test 0 0.1 0.2 0.3 0.4 0.5 -4 -3 -2 -1 0 1 2 3 4 T and Normal Densities Normal T 5 df T 15 df T 30 df 6 7 8 9 D E F G H 2.5 %ile 28 df -2.04841 =T.INV(0.025,28)
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