Homework

9.13 A. Ho : u<42 Ho : u>42 B. z= 42.95-42/2.64/ square root of 65 = 0.95/0.33=2.87. .10 =Ho>Ha .05=Ho>Ha .01=Ho>Ha .001=Ho<Ha C. 1-0.9979=0.0021 .10 =Ho>Ha .05=Ho>Ha .01=Ho>Ha .001=Ho<Ha D. in that location is extremely ironlike evidence 9.22 When scrutinying a hypothesis most populations mean to shape whether to use a z test or a test simply comes down to whether o is cognise or not. If it is unknown you use a t test is it is known you use a z test. 12.10 A. It is withdraw to carry bulge out a chi-square test using these selective information because it is gener everyy hold that n should be considered large if all(a) of the expected booth frequencies are at least 5. In this problem all of the Ei values are greater than or equal to 5. B. I reason out that I cannot winnow out the pi values found in our problem, since it is much larger. 12.18 (a) A. I conclude that I cannot avert the pi values found in our probl em, since it is much larger. Depreciation method and country are cut back and the test is valid. You can see this because since the p-value is much greater than .05 we cannot reject the hypothesis of normality at the .05 level of significance.If you want to stool a full essay, order it on our website: OrderCustomPaper.com

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