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IPS6e.ISM.Ch07

# IPS6e.ISM.Ch07 - Chapter 7 Solutions 7.1(a The standard...

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Chapter 7 Solutions 7.1. (a) The standard error of the mean is s n = \$570 15 . = \$27 . 1109. (b) The degrees of freedom are df = n 1 = 14. 7.2. In each case, use df = n 1; if that number is not in Table D, drop to the lower degrees of freedom. (a) For 95% con±dence and df = 11, use t = 2 . 201. (b) For 99% con±dence and df = 23, use t = 2 . 807. (c) For 90% con±dence and df = 199, we drop to df = 100 and use t = 1 . 660. (Software gives t = 1 . 6525 for df = 199.) 7.3. For the mean monthly rent, the 95% con±dence interval is \$570 ± ( 2 . 145 )( \$105 / 15 ) = \$570 ± \$58 . 15 = \$511.85 to \$628.15. 7.4. The margin of error for 90% con±dence would be smaller—and the interval would be narrower—because we are taking a greater risk that the interval does not include the true mean µ . 7.5. (a) Yes, t = 2 . 35 is signi±cant when n = 15. This can be determined either by comparing to the df = 14 line in Table D (where we see that t > 2 . 264, the 2% critical value) or by computing the two-sided P -value (which is P = 0 . 0340). (b) No, t = 2 . 35 is not signi±cant when n = 6, as can be seen by comparing to the df = 5 line in Table D (where we see that t < 2 . 571, the 2.5% critical value) or by computing the two-sided P -value (which is P = 0 . 0656). (c) Student sketches will likely be indistinguishable from Normal distributions; careful students may try to show that the t ( 5 ) distribution is shorter in the center and heavier to the left and right than the t ( 14 ) distribution (as is the case here), but in reality, the difference is nearly imperceptible. 123 –1 –2 –3 0 2.35 –2.35 –1 –2 –3 0 2.35 –2.35 t (14) t (5) 7.6. For the hypotheses H 0 : µ = \$550 vs. H a : µ> \$550, we ±nd t = 570 550 105 / 15 . = 0 . 7377 with df = 14, for which P . = 0 . 2364. We do not reject H 0 ;wedo not have suf±cient evidence to conclude that the mean rent is greater than \$550. Note: Computing a P-value is not really necessary; a t ( 14 ) distribution is similar enough to a Normal distribution that we can quickly determine that t . = 0 . 74 gives no reason to doubt the null hypothesis. 7.7. Software will typically give a more accurate value for t than that given in Table D, but otherwise the details of this computation are the same as what is shown in the textbook: df = 7, t = 2 . 3646, 5 ± t ( 3 . 63 / 8 ) = 5 ± 3 . 0348 = 1.9652 to 8.0348, or about 2.0 to 8.0 hours per month. 198

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Solutions 199 7.8. We wish to test H 0 : µ = 0 vs. H a : µ 6= 0, where µ is the mean of (drink A rating) minus (drink B rating). Compute the differences for each subject ( 2, 1, 5, 11, and 8), then Fnd the mean and standard deviation of these differences: x . = 4 . 6 and s . = 5 . 2249. Therefore, t = 4 . 6 0 5 . 2249 / 5 . = 1 . 9686 with df = 4, for which P = 0 . 1204. We do not have enough evidence to conclude that there is a difference in preference. 7.9. Using the mean and standard deviation from the previous exercise, the 95% conFdence interval is 4 . 6 ± ( 2 . 7764 )( 5 . 2249 / 5 ) = 4 . 6 ± 6 . 4876 = -1.8876 to 11.0876. 7.10. See also the solutions to Exercises 1.30, 1.64 and 1.144. The CO 2 data are sharply right-skewed (clearly non-Normal). However, the robustness of the t procedures should make them safe for this situation because the sample size is large ( n = 48). The bigger question is whether we can treat the data as an SRS; we have recorded CO 2 emissions for every country with a population over 20 million, rather than a random sample.
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IPS6e.ISM.Ch07 - Chapter 7 Solutions 7.1(a The standard...

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