E Test: POST UNIT 7 - TEST - Ch. 10 Question 10 of 13 This test: 13 point(s) possible This question: 1 point(s) possible Submit test A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 702.2. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 31 high-income individuals and found the sample mean credit score to be 718.4 with a standard deviation of 84.6. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the a = 0.05 level of significance. State the null and alternative hypotheses. Ho: H H1 : H (Type integers or decimals. Do not round.) Identify the t-statistic. to = (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) Make a conclusion regarding the hypothesis. the null hypothesis. There sufficient evidence to claim that the mean credit score of high-income individuals isTest: POST UNIT 7 - TEST - Ch. 10 Question 11 of 13 This test: 13 point(s) possible This question: 1 point(s) possible Submit test The mean waiting time at the drive-through of a fast-food restaurant from the time an order is placed to the time the order is received is 86.4 seconds. A manager devises a new drive-through system that she believes will decrease wait time. As a test, she initiates the new system at her restaurant and measures the wait 107.6 80.3 time for 10 randomly selected orders. The wait times are provided in the table to the right. Complete parts (a) and (b) below. 66.1 93.5 59.2 B4.9 75.3 70.5 66.4 86.2 Click the icon to view the table of correlation coefficient critical values. (a) Because the sample size is small, the manager must verify that the wait time is normally distributed and the sample does not contain any outliers. The normal probability plot is shown below and the sample correlation coefficient is known to be r= 0.981. Are the conditions for testing the hypothesis satisfied? the conditions satisfied. The normal probability plot | linear enough, since the correlation coefficient is |than the critical value. In addition, a boxplot does not show any outliers Expected 2-Score Q 1- 105 Time (sec) - X (b) Is the new system effective? Conduct a hypothesis test using the P-value approach and a level of significance of a = 0.01. Critical values First determine the appropriate hypotheses. Ho : V V 86.4 Sample Size, n Critical Value Sample Size, n Critical Value 0.880 16 0.941 Hy : V V 86.4 0.888 17 0.944 0.898 0.946 Find the test statistic 0.906 0.949 0.912 0.951 to = 10 0.918 0.952 (Round to two decimal places as needed.) 0.923 0.954 0.928 0.956 Find the P-value. 0.932 0.957 0.935 0.959 The P-value is 15 0.939 0.960 (Round to three decimal places as needed.) Use the a = 0.01 level of significance. What can be concluded from the hypothesis test? O A. The P-value is less than the level of significance so there is sufficient evidence to conclude the new system is effective. Print Done O B. The P-value is greater than the level of significance so there is not sufficient evidence to conclude the new system is effective. O C. The P-value is greater than the level of significance so there is sufficient evidence to conclude the new system is effective. O D. The P-value is less than the level of significance so there is not sufficient evidence to conclude the new system is effective.E Test: POST UNIT 7 - TEST - Ch. 10 Question 12 of 13 This test: 13 point(s) possible This question: 1 point(s) possible Submit test It has long been stated that the mean temperature of humans is 98.6"F. However, two researchers currently involved in the subject thought that the mean temperature of humans is less than 98.6 F. They measured the temperatures of 44 healthy adults 1 to 4 times daily for 3 days, obtaining 200 measurements. The sample data resulted in a sample mean of 98.3"F and a sample standard deviation of 1.1"F. Use the P-value approach to conduct a hypothesis test to judge whether the mean temperature of humans is less than 98.6 F at the or = 0.01 level of significance. State the hypotheses. Ho 98.6F 98.6F Find the test statistic. to = 0 (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.) What can be concluded? O A. Reject Ho since the P-value is less than the significance level. O B. Do not reject Ho since the P-value is less than the significance level. O C. Reject Ho since the P-value is not less than the significance level. O D. Do not reject Ho since the P-value is not less than the significance level.E Test: POST UNIT 7 - TEST - Ch. 10 Question 13 of 13 This test: 13 point(s) possible KO This question: 1 point(s) possible Submit test One year, the mean age of an inmate on death row was 39.3 years. A sociologist wondered whether the mean age of a death-row inmate has changed since Next question selects 32 death-row inmates and finds that their mean age is 37.8, with a standard deviation of 9.2. Construct a 95% confidence interval about the mean age. What does the interval imply? Click the icon to view the table of critical t-values. Table of Critical t-Values Choose the correct hypotheses. Ho Area in right tail (Type integers or decimals. Do not round.) t-Distribution Construct a 95% confidence interval about the mean age. Area in Right Tail Degrees of Freedom 0.25 0.20 0.15 0.10 0.05 0.025 0.02 0.01 0.005 0.0025 0.001 0.9005 With 95% confidence, the mean age of a death row inmate is between | years and | years. (Round to two decimal places as needed.) 1.000 1.376 1.963 3.078 6.314 12.706 15.894 31.821 63.657 127.321 318.309 636.619 0.816 1.061 1.386 1.886 2.920 4.303 4.849 6.965 9.925 14.089 22.327 31.599 What does the interval imply? 0.765 0.978 1.250 1.638 2.353 3.182 3.482 4.541 5.841 7453 10.215 12.924 0.741 0.941 1.190 1.533 2.132 2.776 2.999 3.747 4.604 5.598 7173 8.610 0.727 0.920 1.156 1.476 2.015 2 571 2.757 3.365 4.032 4,773 5.893 6.869 O A. Since the mean age from the earlier year is contained in the interval, there is sufficient evidence to conclude that the mean age had changed. 0.718 0.906 1.134 1.440 1.943 2.447 2.612 3.143 3.707 4.317 5.208 5.959 0.711 0.896 1.119 1.415 1.895 2.365 2 517 2.998 3.499 4.029 4.785 5.408 O B. Since the mean age from the earlier year is not contained in the interval, there is not sufficient evidence to conclude that the mean age had changed. 0.706 0.889 1.108 1.397 1.860 2.306 2.449 2.896 3.355 3.833 4.501 5.041 0.703 0.883 1.100 1.383 1.833 2.262 2.398 2.821 3.250 3.690 4.207 4.781 O C. Since the mean age from the earlier year is not contained in the interval, there is sufficient evidence to conclude that the mean age had changed. 10 0.700 0.879 1093 1.372 1.812 2.228 2.359 2.764 3.169 3.581 4.144 4.587 11 0.697 0.876 1.088 1.363 1.796 2.201 2.328 2.718 3.106 3.497 4.025 4.437 O D. Since the mean age from the earlier year is contained in the interval, there is not sufficient evidence to conclude that the mean age had changed 0.695 0.873 1.083 1.356 1.782 2.179 2.303 2.681 3.055 13 3.428 3.930 318 0.694 0.870 1.079 1.350 1.771 2.160 2.282 2.650 3.012 3.372 3.852 4.221 14 0.692 0.868 1.076 1.345 1.761 2.145 2.264 2.624 2.977 3.326 3.787 4.140 15 0.691 0.866 1.074 1.341 1.753 2.131 2.249 2.602 2.947 3.286 3.733 4.073 16 0.690 0.865 1.071 1.337 1.746 2.120 2.23 2.583 2.921 3.252 3.686 4.01 17 0.689 0.863 1.069 1.333 1.740 2.110 2.224 2.567 2.898 3.222 3.646 3.965 18 0.688 0.862 1.067 1.330 1.734 2.101 2.214 2.552 2.878 3.197 3.610 3.92 19 0.688 0.861 1.066 1.328 1.720 2.093 2.205 2.539 2.861 3.174 3.579 3.883 0.687 0.860 1.064 1.325 1.725 2.086 2.197 2.528 2845 3.153 3.552 3.850 0.686 0.859 1.063 1.323 1.721 2.080 2.189 2.518 2.831 3.135 3.527 3.819 ).686 0.858 1061 1.321 1.717 2.074 2.183 2.506 2.819 3.119 3.505 3.792 A aFA Statcrunch Print Done