Statistics Homework Solutions 150

# Statistics Homework Solutions 150 - 150 CHAPTER 6 23 n 18 7...

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Unformatted text preview: 150 CHAPTER 6 23. n 18. 7. Â¡ Â¡ Â¢ The sum of the positive signed ranks is S 18 versus H1 : Âµ Â¡ The null and alternate hypotheses are H0 : Âµ Â¡ Since the alternate hypothesis is of the form Âµ Âµ0 , the P-value is twice the area under the signed rank probability mass function corresponding to S 23. Â¡ Â§ 0 1562. Â¡Â¥ Â¢ 2 0 0781 Â¨ Â¨ Â¤ Â¡ From the signed-rank table, P We cannot conclude that the mean concentration differs from 18 g/L. Signed Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Â¡ Â¡ Â¡ Â¡ Â¡ Â¡ Â¡ Â¡ Â¡ Â¡ Â¡ Â¡ Â¡ Â¡ Â¨ Â¨ Â¡ Â¨ Â¨ Â¡ Â¨ Â¡ Â¨ Â¨ Â¨ Â¨ Â¨ Â¡ Â¨ Â¡ Â¨ Â¨ Â¡ Â¨ Â¨ Â¡ Â¨ Â¨ Â¡ Â¨ Â¨ Â¨ Â¨ Â¡ Â¨ Â¨ Â¡ Â¨ Â¨ Â¡ Â¨ Â¨ Â¨ Â¨ Â¡ Â¨ Â¨ Â¨ Â¨ Â¡ Â¨ Â¨ Â¨ Â¨ Â¡ Â¨ Â¨ Â¨ Â¨ Â¨ Â¨ Â¡ Â¨ Â¨ Â¢ 45. 24. 20, compute the z-score of S and use the z table. 0 46. Â¨ Â¡ Â£ Â¥ Â£ Â¥ Â¢ Â¢ Â¤ Â¤Â¥ Â¢ Â¢ Â¤ Â¡ Â¨ Â¡ We cannot conclude that the mean conversion is less than 45. Â¡ Âµ0 , the P-value is the area to the left of z Â¡ Â¡ Â¡ Since the alternate hypothesis is of the form Âµ table, P 0 3228. 0 46. From the z Â¨ Â¡ nn 1 4 1 2n 1 24 Â¡ S nn Â¢ z Â¡ Since n 134. n Â¡ Â¨ Â¡ Â¨ The sum of the positive signed ranks is S 45 versus H1 : Âµ Â§ Â¨ The null and alternate hypotheses are H0 : Âµ Â¡ 45 39 40 50 69 73 77 88 10 4 11 4 11 6 14 4 14 7 14 9 16 2 18 0 18 2 22 8 26 0 27 3 30 7 33 6 35 8 35 9 36 2 Â¡ x Â¡ x 41 1 41 0 40 0 38 1 52 3 37 3 36 2 34 6 33 6 33 4 59 4 30 3 30 1 28 8 63 0 26 8 67 8 19 0 72 3 14 3 78 6 80 8 91 81 2 Â¡ 3. (a) The signed ranks are ...
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