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 unfluctuating evidence 9.22 When scrutinying a hypothesis most populations mean to find whether to use a z test or a test simply comes down to whether o is cognize 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 pop out a chi-square test using these selective information because it is gener everyy agree that n should be considered large if any 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 reason that I cannot stand firm the pi values found in our problem, since it is much larger. Depreciation method and country are dependant 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 relieve oneself a full essay, order it on our website:
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