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barcode font reporting services Tests of Hypotheses and Significance in Software
CHAPTER 7 Tests of Hypotheses and Significance QR Code Scanner In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. QR Generator In None Using Barcode creation for Software Control to generate, create QR Code ISO/IEC18004 image in Software applications. Let p probability of heads in a single toss of the coin. Under the hypothesis (H0: p 0.5) (i.e., the coin is fair), f (x) P(x heads in 6 tosses) Denso QR Bar Code Scanner In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. QR Code Drawer In Visual C#.NET Using Barcode creation for .NET Control to generate, create QR Code ISO/IEC18004 image in Visual Studio .NET applications. 1 6 6 1 Then the probabilities of 0, 1, 2, 3, 4, 5, and 6 heads are given, respectively, by 64, 64, 15, 20, 15, 64, and 64. 64 64 64 Creating QR Code 2d Barcode In VS .NET Using Barcode generation for ASP.NET Control to generate, create Quick Response Code image in ASP.NET applications. Encoding QR Code In .NET Framework Using Barcode drawer for .NET Control to generate, create Denso QR Bar Code image in .NET applications. 6Cx 2
Print QR Code ISO/IEC18004 In VB.NET Using Barcode printer for .NET Control to generate, create Quick Response Code image in .NET applications. Code 39 Full ASCII Drawer In None Using Barcode generation for Software Control to generate, create Code 3 of 9 image in Software applications. 1 2 Encoding GS1  13 In None Using Barcode creator for Software Control to generate, create GTIN  13 image in Software applications. Create Barcode In None Using Barcode encoder for Software Control to generate, create barcode image in Software applications. Onetailed test Here we wish to decide between the hypotheses (H0: p 0.5) and (H1: p 0.5). Since P(6 heads) 1 6 1 0.01562 and P(5 or 6 heads) 64 0.1094, we can reject H0 at a 0.05 but not a 0.01 level 64 64 (i.e., the result observed is significant at a 0.05 but not a 0.01 level). Twotailed test Here we wish to decide between the hypotheses (H0: p 0.5) and (H1: p 2 0.5). Since P(0 or 6 heads) 1 1 0.03125, we can reject H0 at a 0.05 but not a 0.01 level. 64 64 UPCA Generation In None Using Barcode creator for Software Control to generate, create GS1  12 image in Software applications. ECC200 Maker In None Using Barcode generator for Software Control to generate, create ECC200 image in Software applications. 7.59. Work Problem 7.58 if the coin comes up heads 5 times.
MSI Plessey Encoder In None Using Barcode printer for Software Control to generate, create MSI Plessey image in Software applications. Paint UPCA Supplement 5 In None Using Barcode generation for Word Control to generate, create UPCA image in Word applications. Onetailed test Since P(5 or 6 heads) Barcode Printer In .NET Using Barcode creator for Reporting Service Control to generate, create bar code image in Reporting Service applications. Drawing ANSI/AIM Code 39 In Java Using Barcode encoder for Android Control to generate, create ANSI/AIM Code 39 image in Android applications. 6 64 1 64 7 64 Paint Code 128A In Visual C#.NET Using Barcode drawer for VS .NET Control to generate, create Code 128 Code Set C image in Visual Studio .NET applications. Reading Bar Code In C#.NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in VS .NET applications. 0.1094, we cannot reject H0 at a level of 0.05 or 0.01. 0.2188, we cannot reject H0 at a level of 0.05 or 0.01. GS1  12 Decoder In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. Reading Bar Code In VS .NET Using Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications. Twotailed test Since P(0 or 1 or 5 or 6 heads) 2 A 64 B
7.60. Show that a chisquare test involving only two categories is equivalent to the significance test for proportions (page 216). If P is the sample proportion for category I, p is the population proportion, and n is the total frequency, we can describe the situation by means of Table 723. Then by definition, x2 (nP np n2(P np p)2 n2(P nq p)2 np)2 [n(1 P) nq n(P n(1 p)]2 1 p)2 p 1 q n(P pq p)2 (P p)2 pq>n which is the square of the Z statistic (5) on page 216. Table 723 I Observed Frequency Expected Frequency nP np II n(l n(1 p) P) nq
TOTAL n n
7.61. Suppose X1, X2, c, Xk have a multinomial distribution, with expected frequencies np1, np2, c, npk, respectively. Let Y1, Y2, c, Yk be mutually independent, Poissondistributed variables, with parameters l1 np1, l2 np2, c, lk npk, respectively. Prove that the conditional distribution of the Y s given that Y1 Y2 c Yk n
is precisely the multinomial distribution of the X s.
For the joint probability function of the Y s, we have (1) P(Y1 y1, Y2 y2, c, Yk yk) B RB R c B R
(np1)y1e y1! (np2)y2e y2! (npk)yke yk! y2 c ykpy1py2 1
c pyk k e cy! y1!y2! k
CHAPTER 7 Tests of Hypotheses and Significance
where we have used the fact that p1 given by (2) P(Y1 y1, Y2 p2 c pk
1. The conditional distribution we are looking for is c
y2, c, Yk P(Y1 y1, Y2
yk uY1
n) Y2 n) c Yk n) y2, c, Yk P(Y1 Y2
yk and Y1 c Y
Now, the numerator in (2) has, from (1), the value nnpy1py2 c pyk 1 2 k e y !y ! c y ! 1 2 k n
As for the denominator, we know from Problem 4.94, page 146, that Y1 Y2 c Yk is itself a Poisson variable with parameter np1 np2 c npk n. Hence, the denominator has the value nne n! Therefore, (2) becomes P(Y1 y1, Y2 y2, c,Yk yk uY1 Y2 c Yk n) n! py1py2 c pyk k y1!y2! c yk! 1 2 which is just the multinomial distribution of the X s [compare (16), page 112]. 7.62. Use the result of Problem 7.61 to show that x2, as defined by (21), page 220, is approximately chisquare distributed. As it stands, (21) is difficult to deal with because the multinomially distributed X s are dependent, in view of the restriction (22). However, Problem 7.61 shows that we can replace the X s by the independent, Poissondistributed Y s if it is given that Y1 Y2 c Yk n..Therefore, we rewrite (21) as (1) x2 Y1 l1 !l1 As n S ` , all the l s approach ` , and the central limit theorem for the Poisson distribution [(14), page 112] gives (2) x2 < Z 2 1 Z2 2 c Z2 k where the Z s are independent normal variables having mean 0 and variance 1 whose distribution is conditional upon the event (3) !l1Z1 !l2Z2 c !lkZk 0 or !p1Z1 !p2Z2 c !pkZk 0

