{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# ch 16 - Chapter 16 Solutions 16.1(a Yes(1.96(\$1125 958 =...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 16 Solutions 16.1. (a) Yes: ( 1 . 96 )( \$1125 / 958 ) . = \$71, and \$8740 ± \$71 = \$8669 to \$8811. (b) No: Because the numbers are based on voluntary response rather than an SRS, the confidence interval methods of this chapter cannot be used; the interval does not apply to the whole population. 16.2. (a) x ± 1 . 96 σ/ 880 = 1 . 92 ± 0 . 1209 = 1 . 80 to 2.04 motorists. (b) The large sample size means that, because of the central limit theorem, the sampling distribution of x is roughly Normal even if the distribution of responses is not. (c) Only people with listed phone numbers were represented in the sample, and the low response rate (10 . 9% . = 5029 45 , 956 ) means that even that group may not be well represented by this sample. 16.3. (a) Not included: This is a ﬂaw in the sampling design. (b) Not included: This error arises from the sampling process. (c) Included: This random error is the only error addressed by confidence interval methods. 16.4. (a) The central limit theorem tells us that the sampling distribution of x is approximately Normal because we have a reasonably large sample. (b) The 99% confidence interval is x ± 2 . 576 σ/ 250 = \$237 ± 10 . 59 = \$226 . 4 to \$247.6. (c) The sample may be badly biased: At best, this sample represents only the opinions of those who shop in malls similar to the one where the sample was chosen. 16.5. (a) z = 544 518 114 / 50 . = 1 . 61—not significant at the 5% level ( P = 0 . 0537). (b) z = 545 518 114 / 50 . = 1 . 67—significant at 5% level ( P = 0 . 0475), 182

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Solutions 183 16.6. The full applet output for n = 5 is below on the left; on the right are the Normal curves drawn for n = 15 and n = 40. The reported P -values (0.1867, 0.0606, and 0.0057) agree with the “hand-computed” values given in the solution to the next exercise. 16.7. The z statistics and the P -values are given in the table on the right. In each case, the computation is z = 4 . 8 5 0 . 5 / n and P = P ( Z < z ). n z P (a) 5 0.89 0.1867 (b) 15 1.55 0.0606 (c) 40 2.53 0.0057 16.8. The confidence intervals are given in the table on the right. In each case, the interval is 4 . 8 ± 1 . 960 0 . 5 n . n = 5 4.362 to 5.238 n = 15 4.547 to 5.053 n = 40 4.645 to 4.955 16.9. (a) “Statistically insignificant” means that the differences observed were no more than might have been expected to occur by chance even if SES had no effect on LSAT results. (b) If the results are based on a small sample, then even if the null hypothesis were not true, the test might not be sensitive enough to detect the effect. Knowing the effects were small tells us that the test was not insignificant merely because of a small sample size. 16.10. (a) No: In a sample of size 500, we expect to see about 5 people who have a P -value of 0.01 or less [5 = ( 500 )( 0 . 01 ) ]. These four might have ESP, or they may simply be among the “lucky” ones we expect to see. (b) The researcher should repeat the procedure on these four to see whether they again perform well. 16.11. Recall that effect size = true mean response hypothesized response standard deviation of response . The hypothesized response is 115, and the standard deviation is 30. (a) The effect size is 130 115 30 = 0 . 5, so from the table in the text, we need a sample of size n = 35 to have 90% power against the alternative µ = 130. (b) The effect size is 139 115 30 = 0 . 8, so we need a sample of size n =
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

ch 16 - Chapter 16 Solutions 16.1(a Yes(1.96(\$1125 958 =...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online