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A B C D E F G H K L M N O P Q R S T U 108 Questions 5-7 are related to the following information 109 In the population of employees in your company, 32 percent contribute to the annual United Way campaign. 110 You plan to take a sample of size n = 250 and calculate the sample proportion who contribute to United Way. 111 5 In random samples of size n = 250, the expected value of the sample proportion is 112 a 0.35 113 0.32 114 0.30 115 d 0.021 116 117 6 In the previous question, the standard error of the sample proportion is: 118 a 0.0295 119 b 0.0354 120 C 0.0425 121 d 0.0510 122 123 7 95% percent of sample proportions from samples of size n = 250 deviate from the population proportion of 124 It = 0.32 by no more that _(or_percentage points). 125 0.058 (5.8 percentage points.) 126 0.046 (4.6 percentage points.) 127 0.037 (3.7 percentage points.) 128 0.030 (3.0 percentage points.) 129 130 8 There is a population of 15 families in a small neighborhood. You plan to take a random sample of 4 131 families (without replacement). How many samples of size n = 4 are possible? 132 a 32760 133 b 1365 134 C 240 135 60A B C D E IF G H 136 K L M N 0 P Q R S T 137 9 138 The expression pu - 1.96 -S s u + 1.96-) =0.95 means: 139 140 a In repeated sampling the probability that x is within #1.960/Vn from the population mean is 0.95. 141 b 95% of sample means deviate from the population mean by no more than 1.960/Vn in 142 either direction. 143 144 c In repeated sampling the probability that population mean is within #1.960/Vn from x is 0.95. 145 d Both a and b are correct. 146 148 147 10 The population proportion of Americans with diabetes is 12 percent (it = 0.12). In repeated random samples of 149 n = 950 Americans, 95% of sample proportions of people with diabetes would fall within +_ 0.032 (3.2 percentage point) from It. 150 0.026 (2.6 percentage point) 151 0.021 (2.1 percentage point) 152 0.017 (1.7 percentage point) 153A B C D E F G H 153 K L M N 0 P R S T 154 Questions 11 and 12 are related 155 11 As part of a statistics assignment in October to estimate the percentage of voters who would vote for a mayoral 156 candidate, each of 500 students collects his or her own random sample of likely voters. There are 400 voters in 157 each student's random sample. 158 159 Each student then constructs a 95 percent confidence interval for the population proportion who will vote for the candidate using his or her own random sample. Considering the 500 intervals constructed by the students, 160 the expected number of intervals that will contain the population proportion who will vote for that candidate 161 will be approximately: 162 500 163 475 164 400 165 380 166 167 12 Suppose the sample proportion of one of the students in the previous question, Beth's sample, is p = 0.54. 168 169 Beth's 95% interval estimate of proportion of the population of voters voting for the candidate is 0.42 0.66 170 0.45 0.63 171 0.47 0.61 172 0.49 0.59 173 174 13 It is estimated that 85% of Americans go out to eat at least once per week, with a margin of error of 0.035 (for 175 95% confidence). A 95% confidence interval for the population proportion of Americans who go out to eat 176 once per week or more is 177 a 0.788 0.912 178 0.803 0.897 179 0.815 0.885 180 d 0.824 0.876A B D E G H M N 0 Q R 88 2 In the previous question the variance of the sample mean is. 89 539.506 90 134.877 91 33.719 92 26.975 93 94 3 Using the population mean and standard deviation in question 1, what is the probability that the mean of a 95 random sample of n = 16 days exceeds 486 gizmos? 96 0.4766 97 0.4289 98 0.3860 99 0.3474 100 101 4 Using the population data above, in repeated samples of size n = 16, 95% of sample means fall within that fall 102 within + gizmos from the population mean. 103 a 9.48 104 b 11.38 105 C 13.66 106 d 16.3936 37 Note: 38 If your answer does not exactly match the correct choice, it is due to rounding of intermediate calculations. To 39 avoid the discrepancy, do your calculations in Excel without rounding. 40 41 You run a gizmo factory. The following population data shows the daily output of your factory for the past year. 42 43 486 463 470 460 181 456 474 198 481 498 44 484 471 472 526 470 520 455 495 487 438 45 466 465 459 470 514 456 532 488 481 485 46 494 148 503 439 469 155 446 491 495 458 47 494 470 462 477 194 476 459 443 496 487 48 437 498 474 514 496 500 483 189 466 459 49 501 494 512 485 494 512 510 480 532 465 50 517 503 480 464 463 523 490 512 485 465 51 484 497 489 480 478 488 458 468 458 456 52 508 451 495 511 489 493 524 493 457 450 53 478 515 463 484 509 439 485 479 504 473 54 515 199 499 467 422 435 499 497 447 505 55 482 467 488 506 483 471 484 472 487 529 56 490 199 484 531 480 475 513 487 493 507 57 453 180 479 512 510 490 500 554 526 499 58 509 503 449 470 497 516 471 487 528 487 59 474 473 486 471 154 504 513 510 505 479 60 478 501 476 489 492 176 481 447 460 479 61 487 492 453 460 502 502 468 509 463 497 62 470 513 513 495 448 492 501 516 468 506C D E F G H K L M N 0 P Q R S T A B 470 513 513 495 148 492 501 516 468 506 484 503 486 438 483 482 479 470 491 437 460 181 505 444 469 494 490 513 508 473 493 184 476 480 508 486 496 441 492 484 458 199 452 456 478 426 474 467 559 505 521 499 450 485 453 486 484 466 481 478 526 514 514 491 488 487 519 474 443 456 513 489 485 501 514 464 491 534 502 458 70 498 502 498 461 187 172 488 448 486 494 492 562 462 494 463 445 505 454 464 518 480 439 504 454 476 486 441 435 440 472 73 467 476 502 483 504 451 468 470 456 472 74 461 189 481 489 445 478 479 470 463 491 75 488 537 516 504 189 486 474 465 498 477 76 512 501 512 471 490 507 478 454 503 470 77 502 500 490 480 187 477 456 503 496 461 78 501 475 499 488 445 504 432 528 487 460 79 80 Use the data above to answer questions 1-4 below. 81 You take repeated random samples of size n = 16 days from the above population data. 82 1 The expected value of the sample means is 83 493.544 84 488.608 85 483.722 86 478.885A B C D E F G H K L M N P 181 R S T 182 14 You are the manager of a political campaign. You think that the population proportion of voters who will vote 183 for your candidate is 0.50 (use this for a planning value). Your candidate wants to know what proportion of 184 the population will vote for her. Your candidate wants to know this with a margin-of-error of 1 0.02 (at 95% 185 confidence). How big of a sample of voters should you take? 186 a 2401 187 2038 188 1734 189 1474 190 191 15 Your candidate changes her mind and now wants a margin-of-error of 1 0.03 (but still 95% confidence). Which 192 of the following options would you follow? 193 a The margin-of-error does not have anything to do with the sample size 194 b Use the same sample, but adjust the standard error. 195 c Select a larger sample 196 d Select a smaller sample