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- 1. 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.
- 2. Degrees of freedom (fig) <ul><li>The effect of df on the t distribution is illustrated in the four t distributions below. </li></ul><ul><li>Note that the smaller the df, the flatter the shape of the distribution, resulting in greater area in the tails of the distribution. </li></ul>
- 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>

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