1 / 12

p-Values for Hypothesis Testing About  With  Known

p-Values for Hypothesis Testing About  With  Known. Hypothesis Testing (Revisited). Five Step Procedure 1. Define Opposing Hypotheses. 2. Choose a level of risk ( ) for making the mistake of concluding something is true when its not. 3. Set up test (Define Rejection Region).

thomas-jordan
Download Presentation

p-Values for Hypothesis Testing About  With  Known

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. p-Values for Hypothesis Testing About  With  Known

  2. Hypothesis Testing(Revisited) • Five Step Procedure 1. Define Opposing Hypotheses. 2. Choose a level of risk () for making the mistake of concluding something is true when its not. 3. Set up test (Define Rejection Region). 4. Take a random sample. 5. Calculate statistics and draw a conclusion.

  3. p-Value The probability of getting an value at least as far away as the observed value, if H0 were true. Concept of a p-value • Ignore step 3 • Calculate

  4. Alternate Hypothesisp-value “>” Area to the right of “<” Area to the left of “” 2* Area in the “tail” (to the right or left of ) Calculating p-Values • A p-value is a probability whose definition varies depending on the type of test we are doing (i.e. the form of the alternate hypothesis.)

  5. p-value = p = 1-.9798 = .0202 25 p-value For “> Tests” = P(Getting a value greater than When H0 is true) H0:  = 25 HA:  > 25  = 4.2, n = 49 .9798 0Z 2.05 z = (26.23-25)/.6

  6. p-value = p = .1003 27 p-value For “< Tests” = P(Getting a value less than When H0 is true) H0:  = 27 HA:  < 27  = 4.2, n = 49 0Z -1.28 z = (26.23-27)/.6

  7. p-value = Area above 26.23 + Area below 25.77 = 2*Area above 26.23 p = 2(.3520) = .7040 .3520 .6480 .3520 26 .23 below 26 .23 above 26 p-value For “ Tests” = P(Getting a value at least as far away as When H0 is true) H0:  = 26 HA:   26  = 4.2, n = 49 0Z .38 z = (26.23-26)/.6

  8. P-VALUES AND α • Consider HA:  > 25 • Here we got z = 2.05 • Since  = .05, z.05 = 1.645 Can Accept HA • Suppose  = .01; z.01 = 2.326 Cannot Accept HA • What about  = .02? z.02 = 2.054 Cannot Accept HA • What about  = .03? z.03 = 1.88 Can Accept HA • There is some value of  that is the “break-point” between accepting and not accepting HA-- this is the p-value. If p  α, Accept HA If p > α,Do Not Accept HA LOW p-values are SIGNIFICANT!!

  9. =AVERAGE(A2:A50) =(C6-C3)/(C2/SQRT(49)) =1 – NORMSDIST(C7)

  10. =AVERAGE(A2:A50) =(C6-C3)/(C2/SQRT(49)) =NORMSDIST(C7)

  11. =AVERAGE(A2:A50) =(C6-C3)/(C2/SQRT(49)) =2*(1-NORMSDIST(C7)) Note: If z were negative, the p-value would have been: =2*NORMSDIST(C7)

  12. REVIEW • p-values measure the strength of the test • lower p-values indicate more strongly that HA is true • p-values • “>” tests -- Area in upper tail (to the right of ) • “<” tests -- Area in lower tail (to the left of ) • “” tests -- twice the area in a “tail” • If z >0 -- twice the area in the upper tail • If z< 0 -- twice the area in the lower tail

More Related