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barcode font reporting services Tests of Hypotheses and Significance in Software
CHAPTER 7 Tests of Hypotheses and Significance Decode QR Code In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. QR Code ISO/IEC18004 Creator In None Using Barcode printer for Software Control to generate, create Denso QR Bar Code image in Software applications. Table 76 Cumulative Relative Frequency (%) 5.0 23.0 65.0 92.0 100.0 QR Code Recognizer In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. Drawing QR In Visual C# Using Barcode drawer for .NET framework Control to generate, create QR Code image in VS .NET applications. Height (inches) Less than 61.5 Less than 64.5 Less than 67.5 Less than 70.5 Less than 73.5
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Create DataMatrix In None Using Barcode generation for Software Control to generate, create Data Matrix image in Software applications. Bar Code Generation In None Using Barcode maker for Software Control to generate, create barcode image in Software applications. The work may be organized as in Table 77. In calculating z for the class boundaries, we use z (x x)>s # where the mean x and standard deviations s have been obtained respectively in Problems 5.35 and 5.40. # Table 77 Heights (inches) 60 62 63 65 66 68 69 71 72 74 Class Boundaries (x) 59.5 62.5 65.5 68.5 71.5 74.5 z for Class Boundaries 2.72 1.70 0.67 0.36 1.39 2.41 Area under Normal Curve from 0 to z 0.4967 0.4554 0.2486 0.1406 0.4177 0.4920 Encode UCC  14 In None Using Barcode creator for Software Control to generate, create UCC  14 image in Software applications. DataMatrix Generation In Visual Studio .NET Using Barcode printer for Reporting Service Control to generate, create Data Matrix image in Reporting Service applications. Area for Each Class 0.0413 0.2086
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In the fourth column, the areas under the normal curve from 0 to z have been obtained by using the table in Appendix C. From this we find the areas under the normal curve between successive values of z as in the fifth column. These are obtained by subtracting the successive areas in the fourth column when the corresponding CHAPTER 7 Tests of Hypotheses and Significance
z s have the same sign, and adding them when the z s have opposite signs (which occurs only once in the table). The reason for this is at once clear from a diagram. Multiplying the entries in the fifth column (which represent relative frequencies) by the total frequency n (in this case n 100) yields the theoretical or expected frequencies as in the sixth column. It is seen that they agree well with the actual or observed frequencies of the last column. The goodness of fit of the distribution is considered in Problem 7.44. 7.33. Table 78 shows the number of days f in a 50day period during which x automobile accidents occurred in a city. Fit a Poisson distribution to the data. Table 78 Number of Accidents (x) 0 1 2 3 4 TOTAL Number of Days ( f ) 21 18 7 3 1 50 The mean number of accidents is l a fx af (21)(0) 45 50 Then, according to the Poisson distribution, P(x accidents) (0.90)xe x! (18)(1) (7)(2) 50 (3)(3) (1)(4) In Table 79 are listed the probabilities for 0, 1, 2, 3, and 4 accidents as obtained from this Poisson distribution, as well as the theoretical number of days during which z accidents take place (obtained by multiplying the respective probabilities by 50). For convenience of comparison, the fourth column giving the actual number of days has been repeated. Table 79 Number of Accidents (x) 0 1 2 3 4 P (x accidents) 0.4066 0.3659 0.1647 0.0494 0.0111 Expected Number of Days 20.33 or 20 18.30 or 18 8.24 or 8 2.47 or 2 0.56 or 1 Actual Number of Days 21 18 7 3 1 Note that the fit of the Poisson distribution to the data is good. For a true Poisson distribution, s2 l. Computation of the variance of the given distribution gives 0.97. This compares favorably with the value 0.90 for l, which can be taken as further evidence for the suitability of the Poisson distribution in approximating the sample data. CHAPTER 7 Tests of Hypotheses and Significance
The chisquare test 7.34. In 200 tosses of a coin, 115 heads and 85 tails were observed. Test the hypothesis that the coin is fair using a level of significance of (a) 0.05, (b) 0.01. (c) Find the P value of the test. Observed frequencies of heads and tails are, respectively, x1 115, x2 85. Expected frequencies of heads and tails if the coin is fair are np1 100, np2 x2 (x1 np1)2 np1 (x2 np2)2 np2 (115 100)2 100 2, n k (85 1 100, respectively. Then 100)2 100 2 1 4.50 1. Since the number of categories or classes (heads, tails) is k x2 0.95
(a) The critical value for 1 degree of freedom is 3.84. Then since 4.50 that the coin is fair at a 0.05 level of significance.

