UB MGQ 301 - mgqhomework3 lecturenotes (5 pages)

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mgqhomework3 lecturenotes



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mgqhomework3 lecturenotes

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Pages:
5
School:
University at Buffalo, The State University of New York
Course:
Mgq 301 - Stat Decisions in Mgt
Stat Decisions in Mgt Documents
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first claim then collect data the null hypothesis always what we test and assume it s true gives us a value for parameter statement of equality miu 10 alpha increase boundary lines move toward the center three different natures of the test 1 two sided test assume change could have happened in either direction assume evidence fall away from the center of the distribution expected data value to fall 95 area greater than or less than current claimed value in the center of the distribution two rejection zones alpha risks determines boundary lines critical values alpha will be divided by two if we reject alternative hypothesis says evidence is not going to be within the region the null hypothesis says the value falls within the LOC somewhere if the evidence fall within we say do not reject null hypothesis keep null hypothesis is true alternative hypothesis evidence falls outside of the LOC either greater than or less than the hypothesis value there is enough evidence to support 2 one sided test right 3 left just a increase or decrease when we are interested in deviation in which side you first give a claim then you collect data test assume to be true within LOC the null hypothesis test is what we always test assumed to be true defendants in trail even if your claim is in alternative you always test null null hypothesis give value for parameter statement of equality equal to less than and equal to greater than and equal to the equal value will work everytime z xbar miu sigma sqrt n null hypothesis always assume to be true nature of the test one tail to two tail 3 a always be miu claim is always about parameter his claim is an alternative hypothesis Ha find evidence far enough to the left of 2 to be statistically significant null would be opposite from the claim test the null hypothesis assume the opposite is true assume null hypothesis is true given Ha miu 2 test Ho miu 2 we ll have a negative CV value alpha regoin determine boundary line show sample mean xbar definitely fall less than the critical value enough evidence show the average is significantly less than 2 reject null hypothesis if he find x bar inside loc within the margin of error do not reject the null hypo not enough evidence move beyond the boundary of the cv reduced one tail any affect two tails reduced less than nothing mentioned about one tail people can do two tail test two tail not equal to one tail stronger greater alternative or null critical value for two tail test boundary line based on alpha risk toward or away middle of the curve based on value of alpha alpha increase toward decrease away toward tail less area of the tail area of the tail alpha region evidence fall beyond the fences reject the null isn t enough evidence to say significant test statistic p value don t come up with CV jump in evidence test statistics Ho any numerical value but when turn in to z t score middle becomes zero what is the z score or t score location of the evidence the test of the statistics z deviation standard error how far away is the evidence from the middle the claimed value then divided by the units every curve changes bases on standard error standard error decreases the curve taller and narrower standard error wider shorter with more variation this particular curve the unit depend on standard error the probability of being more extreme than the evidence assuming the null is true why because based on this curve based on this distribution centered at the null hypothesis is the p value the area under the curve t dis t score t dist t df true normdist z score left tail test normsdist z dist area under the curve how do you get probability distribution function how do you get critical value inverse function if you have sigma or not sigma normsdist test statistics correspond to the evidence positive negative greater than 1 normsdist z t dist rt t df two tail test Ha not equal to two tail test alpha 2 double p value then compare to all the alpha positive 1 normsdist z 2 t dist 2t t df negative normsdist z 2 t dist 2t t df inverse function give you CV test statistics you need a formula p value innocent small p value stronger evidence to reject jury enough evidence to reject the null not enough evidence to reject the null not accpet p alpha risk p alpha risk p small reject null something increased decreased or changed there is enough evidence statistically significantly change has occur different from what it assumed to be smaller strong evidence to reject the null therefore accept alternative more power to reject reduces the risk changes how small smaller than the alpha risk small p value stronger evidence to reject null 5 assume alpha risk if they don t offer if you reject the null you accept alternatives or then you keep the null dist function for z right T alternative null p value evidence z s right 1 normsdist z t dist rt t df left tail test Ha Ho alpha risk on the left tail negative test statistics normsdist z t dist t df true two tail test Ha not equal to Ho equal to alpha 2 in each tail like a level of confidence to compare to the p value because you are compare the entire alpha risk you have to double alpha 2 2 z 2 normsdist z t t dist 2t absolute value of t excel doesn t accept negative t value df evidence fall greater than the middle of the curve positive z t positive or negative t 2 1 normsdist z t dist 2t absolute t df all z s what function are you going to use p value not reject null p alpha risk reject not reject p smaller than alpha reject p larger than alpha do not reject NEVER 1 normsinv Right tail What s really happening where 44 evidence falls Add me positive cv based on alpha risk Does 44 fall in rejection zone or not rejection zone Until come out with CV we then know Critical z score inv function How to come up with critical value normsinv 1 alpha never 1 normsinv 1 put in the probability Positive critical value normsinv 1 alpha Negative cv left normsinv alpha Two tail test normsinv alpha 2 then add sign Test statistics where 44 falls Z 44 42 13 sqrt31 42 claimed value


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