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11 SM Ch11

# 11 SM Ch11 - Chapter 11 Comparisons Involving Proportions...

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Chapter 11 Comparisons Involving Proportions and a Test of Independence Learning Objectives 1. Be able to develop interval estimates and conduct hypothesis tests about the difference between the proportions of two populations. 2. Know the properties of the sampling distribution of the difference between two proportions ( ) p p 1 2 - . 3. Be able to conduct a goodness of fit test when the population is hypothesized to have a multinomial probability distribution. 4. For a test of independence, be able to set up a contingency table, determine the observed and expected frequencies, and determine if the two variables are independent. 5. Understand the role of the chi-square distribution in conducting tests of goodness of fit and independence. Solutions: 11 - 1

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Chapter 11 1. a. 1 2 p p - = .48 - .36 = .12 b. 1 1 2 2 1 2 .05 1 2 (1 ) (1 ) p p p p p p z n n - - - ± + .48(1 .48) .36(1 .36) .12 1.645 400 300 - - ± + .12 ± .0614 (.0586 to .1814) c. .48(1 .48) .36(1 .36) .12 1.96 400 300 - - ± + .12 ± .0731 (.0469 to .1931) 2. a. 1 1 2 2 1 2 100(.28) 140(.20) .2333 100 140 n p n p p n n + + = = = + + b. ( 29 ( 29 1 2 1 2 .28 .20 1.44 1 1 1 1 .2333 1 .2333 1 100 140 p p z p p n n - - = = = - + - + p - value = 2(1 - .9251) = .1498 c. p -value > .05; do not reject H 0 . We cannot conclude that the two population proportions differ. 3. a. 1 1 2 2 1 2 200(.22) 300(.16) .1840 200 300 n p n p p n n + + = = = + + ( 29 ( 29 1 2 1 2 .22 .16 1.70 1 1 1 1 .1840 1 .1840 1 200 300 p p z p p n n - - = = = - + - + p - value = 1.0000 - .9554 = .0446 b. p -value .05; reject H 0 . 4. a. Professional Golfers: 1 p = 688/1075 = .64 Amateur Golfers: 2 p = 696/1200 = .58 Professional golfers have the better putting accuracy. b. 1 2 .64 .58 .06 p p - = - = Professional golfers make 6% more 6-foot putts than the very best amateur golfers. 11 - 2
Comparisons Involving Proportions and a Test of Independence c. 1 1 2 2 1 2 .025 1 2 (1 ) (1 ) p p p p p p z n n - - - ± + .64(1 .64) .58(1 .58) .64 .58 1.96 1075 1200 - - - ± + .06 ± .04 (.02 to .10) The confidence interval shows that profession golfers make from 2% to 10% more 6-foot putts than the best amateur golfers. 5. a. 1 p = 115/250 = .46 Republicans 2 p = 98/350 = .28 Democrats b. 1 2 p p - = .46 - .28 = .18 Republicans have a .18, 18%, higher participation rate than Democrats. c. 1 2 2 2 .025 1 2 (1 ) (1 ) p p p p z n n - - + .46(1 .46) .28(1 .28) 1.96 .0777 250 350 - - + = d. Yes, .18 ± .0777 (.1023 to .2577) Republicans have a 10% to 26% higher participation rate in online surveys than Democrats. Biased survey results of online political surveys are very likely. 6. a. 0 : w m H p p : a w m H p p b. w p = 300/811 = .3699 37% of women would ask directions c. m p = 255/750 = .3400 34% of men would ask directions d. 1 2 1 2 300 255 .3555 811 750 w m n p n p p n n + + = = = + + ( 29 ( 29 1 2 1 2 .3699 .3400 1.23 1 1 1 1 .3555 1 .3555 1 811 750 p p z p p n n - - = = = - + - + Upper tail p -value is the area to the right of the test statistic Using normal table with z = 1.23: p -value = 1 - .8907 = .1093 11 - 3

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Chapter 11 p -value > α ; do not reject 0 H We cannot conclude that women are more likely to ask directions. 7. a. p 1 = 256/320 = .80 b. p 2 = 165/250 = .66 c. 1 2 p p - = .80 - .66 = .14 1 1 2 2 .025 1 2 (1 ) (1 ) .14 p p p p z n n - - ± + .80(1 .80) .66(1 .66) .14 1.96 320 250 - - ± + .14 ± .0733 (.0667 to .2133) 8. Let p 1 = the population proportion of delayed departures at Chicago O’Hare p 2 = the population proportion of delayed departures at Atlanta Hartsfield-Jackson a. H 0 : p 1 - p 2 = 0 H a : p 1 - p 2 0 b.
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