z Decision rule using the p value Reject H if p value significance level 339

Z decision rule using the p value reject h if p value

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z Decision rule using the p-value: Reject H 0 if p -value < significance level 339
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Calculating the p-value 1. Calculate the z-score 2. Using the Normal chart, find the probability of getting ( more than / less than ) this score, based on the direction of the equality expression in the ALTERNATE hypothesis 3. If it is a two-tailed test, multiple the result by 2. z NOTE: The final calculated p-value can be between 0.00 and 1 (0% and 100%) Calculating the p-value - Details z If H A : ȝ > # Find probability Z > Z-value z If H A : ȝ < # Find probability Z < Z-value z If H A : ȝ ¡ # a. If Z-value is positive , Find probability Z > Z-value b. If Z-value is negative , Find probability Z < Z-value Multiply that answer by 2
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p -Value in Hypothesis Testing - Example Recall the last problem where the hypothesis and decision rules were set up as: H 0 : P ¢ 200 H 1 : P > 200 Reject H 0 if Z > Z D where Z = 1.55 and Z D =2.33 Reject H 0 if p-value < D 0.0606 is not < 0.01 Conclude: Fail to reject H 0 339 What does it mean when p-value < D ? (a) .10 , we have some evidence that H 0 is not true. (b) .05 , we have strong evidence that H 0 is not true. (c) .01 , we have very strong evidence that H 0 is not true. (d) .001 , we have extremely strong evidence that H 0 is not true. 339
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LYING with Statistics z Create Hypothesis at 0.05 z Run the test and find INSIGNIFICANT z Find p-value = 0.07 z Rerun at 0.10 to state SIGNIFICANT THIS is BAD statistics z NOTE: Not bad to run a test and state p-value (even if you do not reject Null) ? Testing for the Population Mean: Population Standard Deviation Unknown z When the population standard deviation ( ı ) is unknown, the sample standard deviation ( s ) is used in its place z The t -distribution is used as test statistic, which is computed using the formula: 341
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Testing for the Population Mean: Population Standard Deviation Unknown - Example The McFarland Insurance Company Claims Department reports the mean cost to process a claim is $60 . An industry comparison showed this amount to be larger than most other insurance companies, so the company instituted cost-cutting measures. To evaluate the effect of the cost-cutting measures, the Supervisor of the Claims Department selected a random sample of 26 claims processed last month. The sample information is reported below. At the 0.01 significance level is it reasonable a claim is now less than $60 ? 342 Testing for a Population Mean with a Known Population Standard Deviation- Example Step 1: State the null hypothesis and the alternate hypothesis. H 0 : P ¡ $60 H 1 : P < $60 (note: keyword in the problem “now less than”) Step 2: Select the level of significance. Į = 0.01 as stated in the problem Step 3: Select the test statistic. Use t -distribution since ı is unknown 342
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t -Distribution Table (portion) 343 Testing for a Population Mean with a Known Population Standard Deviation- Example Step 5: Make a decision and interpret the result.
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