Reporting Test Results as p Values How to Decide Whether to Reject H 1 Choose

# Reporting test results as p values how to decide

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Reporting Test Results asp-Values: How to Decide Whether to Reject H1.Choose the maximum value of that you are willing to tolerate.2.If the observed significance level (p-value) of the test is less than the chosen value of reject the null hypothesis. Otherwise, do not reject the null hypothesis. 0 ,
Two-Tailed zTest p-Value Example Does an average box of cereal contain 368grams of cereal? random sample of 25boxes showed x= 372.5. The company has specified to be 15grams. Find the p-value. How does it compare to = .05? A
Two-Tailed z Test p -Value Solution z 0 1.50 z value of sample statistic (observed) z x n 372 .5 368 15 25 1 .50
Two-Tailed Z Test p -Value Solution 1/2 p -Value 1/2 p -Value z value of sample statistic (observed) p -Value is P ( z –1.50 or z 1.50) z 0 1.50 –1.50 From z table: lookup 1.50 .4332 .5000 – .4332 .0668
Two-Tailed z Test p -Value Solution 1/2 p -Value .0668 1/2 p -Value .0668 p -Value is P ( z –1.50 or z 1.50) = .1336 z 0 1.50 –1.50
Two-Tailed z Test p -Value Solution 0 1.50 –1.50 z Reject H 0 Reject H 0 1/2 p -Value = .0668 1/2 p -Value = .0668 1/2 = .025 1/2 = .025 p -Value = .1336  = .05 Do not reject H 0 . Test statistic is in ‘Do not reject’ region
One-Tailed z Test p -Value Example Does an average box of cereal contain more than368grams of cereal? A random sample of 25boxes showed x= 372.5The company has specified to be 15grams. Find the value. How does it compare to = .05? . p -
One-Tailed z Test p -Value Solution z 0 1.50 z value of sample statistic z x n 372 .5 368 15 25 1 .50
One-Tailed z Test p -Value Solution Use alternative hypothesis to find direction p -Value is P ( z 1.50) z value of sample statistic p -Value z 0 1.50 From z table: lookup 1.50 .4332 .5000 – .4332 .0668
One-Tailed z Test p -Value Solution p -Value .0668 z value of sample statistic From z table: lookup 1.50 Use alternative hypothesis to find direction .5000 – .4332 .0668 p -Value is P ( z 1.50) = .0668 z 0 1.50 .4332
= .05 One-Tailed z Test p -Value Solution 0 1.50 z Reject H 0 p -Value = .0668 ( p -Value = .0668) ( = .05). Do not reject H 0 . Test statistic is in ‘Do not reject’ region
p -Value Thinking Challenge You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is less than 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. What is the p - value? How does it compare to = .01?
Use alternative hypothesis to find direction p -Value Solution* z 0 –2.65 z value of sample statistic From z table: lookup 2.65 .4960 p -Value .004 .5000 – .4960 .0040 p -Value is P ( z -2.65) = .004. p -Value < ( = .01). Reject H 0 .
Converting a Two-Tailed p -Value from a Printout to a One- Tailed p -Value if H a is of the form > and z is positive or H a is of the form < and z is negative if H a is of the form > and z is negative H a is of the form < and z is positive p Reported  p -value 2 p 1 Reported  p -value 2
7.4 Test of Hypotheses about a Population Mean: Normal ( z ) Statistic
Large-Sample Test of Hypothesis about µ One-Tailed Test Two-Tailed Test H 0 : µ = µ 0 H 0 : µ = µ 0 H a : µ < µ 0 H a : µ µ 0 (or H a : µ > µ 0 ) Test Statistic : Test Statistic : z x µ 0 x x µ 0 s n z x µ 0 x x µ 0 s n
Large-Sample Test of Hypothesis about µ One-Tailed Test Rejection region : z < – z (or z > z  when H a : µ > µ 0 ) where z is chosen so that P ( z > z ) =
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