hw5.sol

hw5.sol - the conduction velocities are significantly...

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AMS007, Biostatistics Week 6 Homework Solutions Chapter 6 1. The hypotheses are Ho : 2 = 400g 2 versus Ha: 2 < 400g 2 . Under Ho we expect the test statistics ± 2 to be n-1 (29 here). Calculate ± 2 = 21.75 and the corresponding P -value is 0.17 (The value is not available on the back cover of the test book, but you can find it online). In words we would say that the probability of getting a chi-square statistics as large as 21.75 or larger by chance when the population variance is 400g 2 is 0.17, which is not extreme enough to reject Ho and accept Ha . So we can’t reach the conclusion that the breeder has been successfully reduce the variance. 3. (a). The hypotheses are Ho : ² = 65 versus Ha: ² < 65. For a t test, the c.v. = -1.753 Calculate t = -5.7 and the corresponding P -value is <<0.0005. Since –5.7<-1.753 and the P -value is also extremely small, reject Ho . That is, the data indicates that
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Unformatted text preview: the conduction velocities are significantly slower in the poisoned individuals. (b). The hypotheses are Ho : &amp; 2 = 25 versus Ha: &amp; 2 &gt; 25. Here s = 7 and n=16. Testing at = 0.05, the c.v. = 25.0. Calculate 2 = 29.4 which is bigger than the c.v. And 0.01&lt; P-value &lt; 0.025, so reject Ho . That is, the data indicates that the poisoned individuals are more variable in their sciatic nerve conduction velocities than normal individuals. 7. s 2 = 1.33 (calculated from the given data) The hypotheses are Ho : &amp; 2 = 0.92 versus Ha: &amp; 2 &gt; 0.92. Here n=16. Testing at = 0.05, the c.v. = 25.0. Calculate 2 = 21.68 which is less than the c.v. And P-value &gt; 0.1, so there is no enough evidence to reject Ho . That is, the data doesnt provide enough evidence that the population variance is greater than 0.92....
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