How do I solve this problem: Computer Software generated 500 random numbers that

Question

How do I solve this problem: Computer Software generated 500 random numbers that shouldlook like they are from the Standard Normal distribution. They arecategorized in five groups: (1)less than or equal to -0.6 (2)greater than -0.6 and less than or equal to -.1 (3)greater than-0.1 and greater than or less than 0.1 (4) greater than .1 and lessthan or equal to 0.6 (5) greater than 0.6. The counts in the fivegroups are 139, 102,41,78,140 respectively. Find the probabilitiesfor these five intervals using Table A from the book ofIntroduction to the Practice of Statistics (5th). Then compute theexpected number for each interval for a sample of 500.Perform thegoodness of fit test. How do I solve this problem: Computer Software generated 500 random numbers that shouldlook like they are from the Standard Normal distribution. They arecategorized in five groups: (1)less than or equal to -0.6 (2)greater than -0.6 and less than or equal to -.1 (3)greater than-0.1 and greater than or less than 0.1 (4) greater than .1 and lessthan or equal to 0.6 (5) greater than 0.6. The counts in the fivegroups are 139, 102,41,78,140 respectively. Find the probabilitiesfor these five intervals using Table A from the book ofIntroduction to the Practice of Statistics (5th). Then compute theexpected number for each interval for a sample of 500.Perform thegoodness of fit test.

Explanation / Answer

Null Hypothesis: There is no significancedifference between the observed values and the expectedvalues Alternative Hypothesis: There is a significancedifference between the observed values and the expectedvalues Consider the following table Groups Counts(Oi) p(x) E=N*p(x) (Oi-Ei)^2 (Oi-Ei)^2/Ei 1 139 0.2 100 1521 15.21 2 102 0.2 100 4 0.04 3 41 0.2 100 3481 34.81 4 78 0.2 100 484 4.84 5 140 0.2 100 1600 16 Total(N) 500 70.9 Groups Counts(Oi) p(x) E=N*p(x) (Oi-Ei)^2 (Oi-Ei)^2/Ei 1 139 0.2 100 1521 15.21 2 102 0.2 100 4 0.04 3 41 0.2 100 3481 34.81 4 78 0.2 100 484 4.84 5 140 0.2 100 1600 16 Total(N) 500 70.9
THe degrees of freeedom = 5-1 =4df and thecritical value of of the chi-square at 0.05, the level ofsignificance is 9.49 Since the calculated value of chi-squaers isgreater than the critical value of chi-squaer so we conclude thatthere is a significance difference between the observed values andthe expected values Groups Counts(Oi) p(x) E=N*p(x) (Oi-Ei)^2 (Oi-Ei)^2/Ei 1 139 0.2 100 1521 15.21 2 102 0.2 100 4 0.04 3 41 0.2 100 3481 34.81 4 78 0.2 100 484 4.84 5 140 0.2 100 1600 16 Total(N) 500 70.9

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