quiz5Sp2011_key

# quiz5Sp2011_key - • In the back of the book for df = 30 I...

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NAME: Stat 205 Quiz 5 As part of a study of the treatment of anemia in cattle, researchers measured the concentration of selenium in the blood of n = 36 cows who had been given a daily selenium supplement. The mean selenium concentration was 6.21 μ g/dL and the standard deviation was 1.84 μ g/dL. As discussed in class, construct and interpret a 95% conﬁdence interval for the population mean. ANSWER : We have ¯ y = 6 . 21 and s = 1 . 84 . The degrees of freedom for t 0 . 025 is df = n - 1 = 35 . Let’s follow the CI recipe card:
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Unformatted text preview: • In the back of the book, for df = 30 I ﬁnd t . 025 = 2 . 042 . You could use df = 40 too giving t . 025 = 2 . 021 . • The se (¯ y ) is se (¯ y ) = s √ n = 1 . 84 √ 36 = 0 . 307 . • The 95% CI is given by ¯ y ± t . 025 se (¯ y ) = 6 . 21 ± 2 . 042(0 . 307) = (5 . 58 , 6 . 84) . • Interpretation: “With 95% conﬁdence, the true mean selenium concentration is between 5.58 and 6.84 μ g/dL among cattle given the supplement.”...
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