Joong Won Mo Assignment Homework7 due 07/08/2016 at 11:59pm PDT 4. (1 pt) A confidence interval for a population mean has length 22. a) Determine the margin of error. 1. (1 pt) For each statement, express the null hypothesis H0 and alternative hypothesis H1 in symbolic form. 1. The mean annual consumption of beer in the nation's capital is less than the national mean 0 . b) If the sample mean is 61.1, obtain the confidence interval. Confidence interval: ( , ). A. H0 : = 0 , H1 : 6= 0 B. H0 : = 0 , H1 : < 0 . C. H0 : < 0 , H1 : 0 D. H0 : = 0 , H1 : > 0 Answer(s) submitted: Classify the hypothesis test as two-tailed , left-tailed or righttailed . (incorrect) A. Left-tailed B. Right-tailed C. Two-tailed 7. (1 pt) The air in poultry-processing plants often contains fungus spores. Inadequate ventilation can affect the health of the workers. The problem is most serious during the summer. To measure the presence of spores, air samples are pumped to an agar plate and \"colony-forming units (CFUs)\" are counted after an incubation period. Here are data from two locations in a plant that processes 37,000 turkeys per day, taken on four days in the summer. The units are CFUs per cubic meter of air. 2. The mean body temperature for humans differs from 98.6 degrees Farenheit. Math243Summer-Evitts A. H0 : = 98.6, H1 : 6= 98.6 B. H0 : = 98.6, H1 : < 98.6 C. H0 : = 98.6, H1 : > 98.6 D. H0 : > 98.6, H1 : 98.6 Classify the hypothesis test as two-tailed , left-tailed or righttailed . A. Two-tailed B. Left-tailed C. Right-tailed Day 4 1090 224 The spore count is clearly higher in the kill room. Give sample means and a 90% confidence interval to estimate how much higher the spore count is in the kill room. To do this problem correctly you need to think carefully about whether this is really a matched pairs design (one sample with two measurement) or a two sample design. x (kill): x (processing): 90% confidence interval: to Answer(s) submitted: Day 1 Day 2 Day 3 3175 2526 1763 529 141 362 Kill room Processing B A A A (correct) Answer(s) submitted: 3. (1 pt) A confidence interval for a population mean has a margin of error of 3.1. a) Determine the length of the confidence interval. (incorrect) b) If the sample mean is 50.7, obtain the confidence interval. Confidence interval: ( , ). 8. (2 pts) Answer(s) submitted: Weights of 10 red and 36 brown randomly chosen M&M plain candies are listed below. Red: 3.1 5.7-3.1 5.7+3.1 0.909 0.936 (incorrect) 1 0.882 0.924 0.891 0.877 0.897 0.874 0.907 0.933 Brown: 0.905 0.858 0.921 0.902 0.875 0.929 0.912 0.923 0.897 0.955 0.909 0.936 0.877 0.986 0.904 0.856 0.857 0.872 0.918 0.914 0.966 1.001 0.913 0.871 0.92 0.931 0.861 0.931 0.985 0.909 (a) the degree of freedom (b) the critical t value (To do this you will need to use the degree of freedom and the significance level to find the critical value in the table). (c) the t test statistic To make a final conclusion you need to compare the t test statistic to the critical t value. The final conclustion is 0.915 0.92 0.93 0.898 0.867 0.928 A. We can reject the null hypothesis that = 10.1 and accept that < 10.1. B. There is not sufficient evidence to reject the null hypothesis that = 10.1. 1. To construct a 90% confidence interval for the mean weight of red M&M plain candies, you have to use A. The t distribution with 9 degrees of freedom B. The t distribution with 11 degrees of freedom C. The normal distribution D. The t distribution with 10 degrees of freedom E. None of the above Answer(s) submitted: (incorrect) 2. A 90% confidence interval for the mean weight of red M&M plain candies is 13. (1 pt) This problem asks you to do hypothesis testing without calculating precise P-values but using the table of critical values from your textbook instead. When a poultry farmer uses his regular feed, the newborn chickens have normally distributed weights with a mean of 61.8 oz. In an experiment with an enriched feed mixture, ten chickens are born with the following weights (in ounces). << 3. To construct a 90% confidence interval for the mean weight of brown M&M plain candies, you have to use A. The t distribution with 36 degrees of freedom B. The normal distribution C. The t distribution with 37 degrees of freedom D. The t distribution with 35 degrees of freedom E. None of the above 68 65.4 65.2 62.3 67.8 66.1 64.3 63.3 65.3 66.6 Use the = 0.05 significance level to test the claim that the mean weight is higher with the enriched feed. (a) The sample mean is x = (b) The sample standard deviation is s = (c) The test statistic is t = (d) The critical value is t = The conclusion is 4. A 90% confidence interval for the mean weight of brown M&M plain candies is << Answer(s) submitted: C C A. There is sufficient evidence to support the claim that with the enriched feed, the mean weight is greater than 61.8. B. There is not sufficient evidence to support the claim that with the enriched feed, the mean weight is greater than 61.8. (incorrect) 9. (1 pt) The point of this problem is to get you used to using the table of critical values to do hypothesis tests when you don't have software available that will calculate the precise P-value associated to your test statistic. A sample of 5 measurments, randomly selected from a normally distributed population, resulted in a sample mean, x = 9.1 and sample standard deviation s = 1.32. Using = 0.05, test the null hypothesis that the mean of the population is = 10.1 against the alternative hypothesis that the mean of the population satisfies < 10.1. Find: Answer(s) submitted: (incorrect) 2 (a) State the alternative hypothesis in terms of the population mean number of words spoken per day for men (M ) and for women (F ). If necessary, use != to represent 6=. Study 1: M F Study 2: M F (b) Find the two-sample t statistic to test F M . Study 1: Study 2: (c) What degrees of freedom does Option 2 use to get a conservative P-value? Study 1: Study 2: (d) Compare your value of t with the critical values in Table C. What can you say about the P- value of the test? 14. (1 pt) To test the efficacy of a new cholesterol-lowering medication, 10 people are selected at random. Each has their LDL levels measured (shown below as Before), then take the medicine for 10 weeks, and then has their LDL levels measured again (After). Subject Before 1 187 2 122 3 146 4 196 5 189 6 158 7 166 8 133 9 177 10 148 After 176 128 117 205 150 173 168 129 172 163 Study 1: P- value > Study 2: P- value Answer(s) submitted: Give a 97.7% confidence interval for B A , the difference between LDL levels before and after taking the medication. Confidence Interval = at 97.7% confidence. Give your answer as an open interval, in the form (A,B) where A is the lower bound and B is the upper bound. Answer(s) submitted: (incorrect) (incorrect) 15. (2 pts) The following Exercise is based on summary statistics rather than raw data. This information is typically all that is presented in published reports. You can calculate inference procedures by hand from the summaries. Use the conservative Option 2 (degrees of freedom the smaller of n1 1 and n2 1 for two-sample t confidence intervals and P-values. You must trust that the authors understood the conditions for inference and verified that they apply. This isn't always true. ) Equip male and female students with a small device that secretly records sound for a random 30 seconds during each 12.5minute period over two days. Count the words each subject speaks during each recording period, and from this, estimate how many words per day each subject speaks. The published report includes a table summarizing six such studies. Here are two of the six: 16. (2 pts) Coaching companies claim that their courses can raise the SAT scores of high school students. But students who retake the SAT without paying for coaching also usually raise their scores. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were uncoached. Starting with their verbal scores on the first and second tries, we have these summary statistics: Try 1 Try 2 Gain n Coached 427 Uncoached 2733 s x 92 529 101 527 s x 97 29 101 21 s 59 52 Use Table C to estimate a 98% confidence interval for the mean gain of all students who are coached. to at 98% confidence. Sample Size Estimated Average Number (SD) of Words Spoken per Day Study Women Men Women 1 67 67 15652 (7558) 2 30 19 16696 (7840) x 500 506 Now test the hypothesis that the score gain for coached students is greater than the score gain for uncoached students. Let 1 be the score gain for all coached students. Let 2 be the score gain for uncoached students. (a) Give the alternative hypothesis: 1 2 0. (b) Give the t test statistic: (c) Give the appropriate critical value for =5%: . The conclusion is Men 16924 (9224) 12950 (8161) Readers are expected to understand this to mean, for example, the 67 women in the first study had x = 15652 and s = 7558. It is commonly thought that women talk more than men. Does either of the two samples support this idea? For each study: 3 A. There is sufficient evidence to support the claim that with coaching, the mean increase in scores is greater than without coaching. B. There is not sufficient evidence to support the claim that with coaching, the mean increase in scores is greater than without coaching. Answer(s) submitted: (incorrect) c Generated by WeBWorK, http://webwork.maa.org, Mathematical Association of America 4 Table entry for C is the critical value t required for confidence level C. To approximate one- and two-sided P -values, compare the value of the t statistic with the critical values of t that match the P -values given at the bottom of the table. TABLE C Area C t* Tail area 1 C 2 t* t distribution critical values CONFIDENCE LEVEL C DEGREES OF FREEDOM 50% 60% 70% 80% 90% 95% 96% 98% 99% 99.5% 99.8% 99.9% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 50 60 80 100 1000 z 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 0.690 0.689 0.688 0.688 0.687 0.686 0.686 0.685 0.685 0.684 0.684 0.684 0.683 0.683 0.683 0.681 0.679 0.679 0.678 0.677 0.675 0.674 1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.866 0.865 0.863 0.862 0.861 0.860 0.859 0.858 0.858 0.857 0.856 0.856 0.855 0.855 0.854 0.854 0.851 0.849 0.848 0.846 0.845 0.842 0.841 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 1.093 1.088 1.083 1.079 1.076 1.074 1.071 1.069 1.067 1.066 1.064 1.063 1.061 1.060 1.059 1.058 1.058 1.057 1.056 1.055 1.055 1.050 1.047 1.045 1.043 1.042 1.037 1.036 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.341 1.337 1.333 1.330 1.328 1.325 1.323 1.321 1.319 1.318 1.316 1.315 1.314 1.313 1.311 1.310 1.303 1.299 1.296 1.292 1.290 1.282 1.282 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 1.753 1.746 1.740 1.734 1.729 1.725 1.721 1.717 1.714 1.711 1.708 1.706 1.703 1.701 1.699 1.697 1.684 1.676 1.671 1.664 1.660 1.646 1.645 12.71 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 2.120 2.110 2.101 2.093 2.086 2.080 2.074 2.069 2.064 2.060 2.056 2.052 2.048 2.045 2.042 2.021 2.009 2.000 1.990 1.984 1.962 1.960 15.89 4.849 3.482 2.999 2.757 2.612 2.517 2.449 2.398 2.359 2.328 2.303 2.282 2.264 2.249 2.235 2.224 2.214 2.205 2.197 2.189 2.183 2.177 2.172 2.167 2.162 2.158 2.154 2.150 2.147 2.123 2.109 2.099 2.088 2.081 2.056 2.054 31.82 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 2.518 2.508 2.500 2.492 2.485 2.479 2.473 2.467 2.462 2.457 2.423 2.403 2.390 2.374 2.364 2.330 2.326 63.66 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 2.861 2.845 2.831 2.819 2.807 2.797 2.787 2.779 2.771 2.763 2.756 2.750 2.704 2.678 2.660 2.639 2.626 2.581 2.576 127.3 14.09 7.453 5.598 4.773 4.317 4.029 3.833 3.690 3.581 3.497 3.428 3.372 3.326 3.286 3.252 3.222 3.197 3.174 3.153 3.135 3.119 3.104 3.091 3.078 3.067 3.057 3.047 3.038 3.030 2.971 2.937 2.915 2.887 2.871 2.813 2.807 318.3 22.33 10.21 7.173 5.893 5.208 4.785 4.501 4.297 4.144 4.025 3.930 3.852 3.787 3.733 3.686 3.646 3.611 3.579 3.552 3.527 3.505 3.485 3.467 3.450 3.435 3.421 3.408 3.396 3.385 3.307 3.261 3.232 3.195 3.174 3.098 3.091 636.6 31.60 12.92 8.610 6.869 5.959 5.408 5.041 4.781 4.587 4.437 4.318 4.221 4.140 4.073 4.015 3.965 3.922 3.883 3.850 3.819 3.792 3.768 3.745 3.725 3.707 3.690 3.674 3.659 3.646 3.551 3.496 3.460 3.416 3.390 3.300 3.291 One-sided P .25 .20 .15 .10 .05 .025 .02 .01 .005 .0025 .001 .0005 Two-sided P .50 .40 .30 .20 .10 .05 .04 .02 .01 .005 .002 .001 693