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Hyp B
Hyp B
Hyp B
Hyp B
Hyp B
Hyp B
Hyp B
Hyp B
Hyp B
Hyp B
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Hyp B

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  1.  
  2. THE t DISTRIBUTION DEFINITION The t distribution is a theoretical probability distribution. It is symmetrical, bell-shaped, and similar to the standard normal curve. It differs from the standard normal curve, however, in that it has an additional parameter, called degrees of freedom, which changes its shape.
  3. Student t- distribution We knows the significant of the difference between the mean of the sample and the mean of the population t = x-   /  n
  4. If the S.D is not known we have to estimate it from the sample . <ul><li>If s is the S.D of the sample is known then the Standard Error is s/  n. </li></ul><ul><li>t = x-  </li></ul><ul><li>s/  n -1 if n is small(i.e. n <=25) </li></ul><ul><li>Here we will take n-1 degrees of freedom for finding the significance level(if n is small) </li></ul>
  5. Testing of mean population on the basis of Small and Large Samples <ul><li>If  is known we will take the value of  in the standard normal distribution table </li></ul><ul><li>If  is not known and n is small we will take the value of n-1 (degrees of freedom) in the standard normal distribution table. </li></ul><ul><li>If n value is high in both the situation we will take value from  only </li></ul>
  6. If the sample is large the test is Z test <ul><li>z = x-  </li></ul><ul><li> /  n </li></ul><ul><li>Here we will take  degrees of freedom for finding the significance level </li></ul>
  7. Problems <ul><li>In a production process ,the target value of  is 50 and  is not known. The sample measurement on a day are 45,54,51,47,52,50,41,51,43 and 53.Test H0:  =50 against H1:  <50 with ∞=.05) Ans: t=-.924,s=4.45 </li></ul><ul><li>D.O.F=n-1 = 9 is 1.833 So rejection region is R:t<=-1.833 The value of t comes under acceptance region.i,e H0 is accepted at 5% level of significance. </li></ul><ul><li>Student’s t-distribution- one tail test </li></ul><ul><li>(Q.T by Srivastava </li></ul><ul><li> </li></ul><ul><li>-1.833 </li></ul>
  8. Problem 2 <ul><li>The kairali restaurant has been arranging sales of 300 lunch packets per day at Brigade road. Because of the construction of the new building and other complexes , it expects to increase the sale. During the first 16 days after the occupation of these building ,the daily sales were 304,367,385,386,262,329,302,292,350,320,298,258,364,294,276 and 333. On the basis of this information will you conclude that Kairali’s sales have increased. </li></ul><ul><li>Ans.Let x be the daily sales </li></ul><ul><li>H0 is  <=300(We consider the sales have not increased unless proved) </li></ul><ul><li>H1 is  >300 </li></ul><ul><li>t=1.94 </li></ul><ul><li>R:t>=1.75 t(.05,15) n-1=15 Two tail Test </li></ul><ul><li>We reject the hypothesis </li></ul><ul><li>.05 </li></ul><ul><li>1.75 </li></ul>
  9. Home work <ul><li>A certain stimulus administered to each of 10 patients resulted in the following increases of blood pressure.8,8,7,5,4,1,0,0,-1,-1. Can it be concluded that the stimulus was responsible for the increase in blood pressure. </li></ul><ul><li>Ans: H0:  =  0 H 1:  0 </li></ul><ul><li>s=3.53 </li></ul><ul><li>t =2.63 </li></ul><ul><li>D.o.f=9 , </li></ul><ul><li>. </li></ul><ul><ul><li>. 025 .025 </li></ul></ul><ul><li>-2.26 2.26 </li></ul>
  10. Homework <ul><li>Illustration 1: A manufacturer of vitamin B-complex tables wants to check the quality of his product. on the basis of his experiences, he has estimated mean content (  ) of a batch as 100 units and the standard deviation as 10 units. For a new batch, if he tests 25 tables, what will be the distribution for the sample mean  ? </li></ul><ul><li>  </li></ul>

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