Need help with screenshots 1 through 11. All information required can be found below in order for you to be able to solve the problems. Screenshots 1 and 3 include two screenshots of their own, since the graphs would not fit with the prompt(s). Please number your solutions. Do not need any explanations, only solutions. Only answer where needed (the blank spaces). Thanks.

A survey of 2287 adults in a oerlain large oountry aged 18 and older conducted by a reputable polling organization found that 422 have donated blood in the past two years. Complete parts (a) through (c) below. wwwmmmmmgm WWWJJMMMMRaQQZL (3) Obtain a point estimate tor the population proportion of adults in the country aged 18 and older who have donated blood in the past two years. 3: 0.135 (Round to three decimal places as needed.) 0:) Verify that the requirements for constructing a condence interval about p are satised. The sample is stated to be a simple random sample. the value of n6 is , which is greater than or equal to 10. and the sample size is stated to be less than or equal to 5% of the population size. (Round to three decimal places as needed.) - 8(1 - 6) n8 n6(1 6) 8 Help Me Solve This View an Example Get More Help A Clear All A survey of 2287 adults in a certain large country aged 18 and older conducted by a reputable polling organization found that 422 have donated blood in the past two years. Complete parts (a) through (c) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2) (a) Obtain a point estimate for the population proportion of adults in the country peed to p= 0.185 X -X (Round to three decimal places as needed.) Standard Normal Distribution Table (page 1) Standard Normal Distribution Table (page 2) (b) Verify that the requirements for constructing a confidence interval about p ar The sample is stated to be a simple random sample, the value of (Round to three decimal places as needed. 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0.072 0.060 0.0681 0.9772 0.977 0.9783 0.978 0.9793 0.979 0.9808 0.9812 0.9817 -13 0.0968 0.0951 0.0934 0.0918 0.0901 0.088 0.0869 0.0853 0.0838 0.0823 0.9821 0.9826 0.9830 0.983 0.9838 0.984 0.9850 0.9854 0.9857 0.1151 0.1131 0.1112 0.109 0.1075 0.105 0.1038 0.1020 0.1003 0.0985 0.9861 0.986 0.9868 0.9871 0 0901 0.9875 0.9878 0.9884 0.9887 0.9890 0.1357 0.133 0.1314 0.129 0.127 D.125 0.1230 0.1210 0.119 0.1170 0.9893 0.980 0.0904 0.9906 0.990 0 0011 0.9913 0.9916 -1.0 0.1587 0.156 0.1539 0.1515 0.149 0.1469 0.1446 0.1423 ).140 0.9018 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 -09 0.1841 0.181 0.1788 0.176 0.1736 0.1711 0.168 0.1660 0.163 0.1611 0.9938 0.9940 0.9941 .9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1049 0.1922 0.1894 0.1867 0.9953 0.905 0.9956 0.9957 0.9959 0.996 0.9961 0.9962 0.9963 0.9964 0.2420 0.2389 0.2358 0.232 0.2296 0.2266 0.2236 0.2206 0.2177 02148 0.9965 0.9066 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.997 -0.6 0.2743 0.2709 0.2676 0.264 0.2611 0.2578 0.254 0.2514 0.2483 0.2451 0.9074 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.0970 0.9080 0.9081 -0.5 0.3085 0.3050 0 3015 0.2981 0.2946 0.2912 0.2877 0.2843 03810 0.2776 0.9981 0.9082 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986 COOEE CLECO 0.3446 0.3409 0.3372 0.333 0.3300 0.326 0.3228 0.3192 0.3156 0.3121 0.9987 0.9987 0.9987 0.998 0.9988 0.998 0.9989 0.990 -03 0.3821 0.3783 0.3745 0.370 0.3669 0.3632 0.359 0.3557 0.3520 0.3483 0.9990 0.909 0.9091 0.999 0.9992 0.999 0.9992 0.9992 0.9093 0.9993 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 0.9993 0.9993 0.999 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 0.9095 0.9095 0.9095 0.9096 0.9906 0.9996 0.9096 0 0996 0.9096 0.9097 0.500 0.4960 0.4 20 0 4880 0.4840 0.4801 0.4761 0.4721 0 4681 0.4641 0.9097 0.9097 0.9097 0.9997 0.9097 0.9997 9.9097 0.9998 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 Print Done Print Done Help Me Solve This View an Example Get More Help2 The following data represent the pH of rain for a random sample of 12 rain dates. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts a) through d) below. 5.58 5.72 5.24 4.80 5.02 4.57 4.7 4.61 4.76 4.56 69 Click the icon to view the table of critical t-values. (a) Determine a point estimate for the population mean. X A point estimate for the population mean is 5.04 . Table of Critical t-Values (Round to two decimal places as needed.) (b) Construct and interpret a 95% confidence interval for the mean pH of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice. cending order. Round to two decimal places as needed.) A. There is 95% confidence that the population mean pH of rain water is between and. O B. There is a 95% probability that the true mean pH of rain water is between |and right tail C. If repeated samples are taken, 95% of them will have a sample pH of rain water between 4.762 and 5.318 1-Distribution Degrees of Area in Right Tail reedom 0.25 0.20 0.15 0.10 0.05 0.025 0.02 0.01 0.0025 0.001 0.0605 1.000 0.816 -376 1963 5 314 127.32 2.924 3. 408 2.109 3.195 882 82128 2326 3 090 Degrees of 0.25 0.20 0.15 0.10 0.05 0.025 0.02 0.01 0.005 0.0025 0.001 0.0605 Vi Vi More Freedom 1-Distribution Area in Right Tail Help Me Solve This View an Example Get More Help - heck Answer3 The data shown below represent the repair cost for a low-impact collision in a simple random sample of mini- and micro-vehicles (such as the Chevrolet Aveo or Mini Cooper). Complete parts (a) through (c). $3184 $1041 $734 $661 $762 $1748 $3394 $2046 $2699 $1360 Click here to view page 1 of the standard normal distribution table Click here to view page 2 of the standard normal distribution table. Click here to view the table of critical t-values. (a) Draw a boxplot to check for outliers. Choose the correct answer below. A. O B. O C. &D Q MTTTT TTT TTTTTTT o TTT TTT TTT TTT 2000 4 4000 2000 2000 4000 (b) Construct and interpret a 95% confidence interval for the population mean cost of repair. Does the boxplot suggest that it is reasonable to construct a confidence interval for the population mean? O A. No, the boxplot shows there are outliers. B. Yes, the distribution is roughly symmetric and there are no outliers. O C. No, the distribution is roughly symmetric with no outliers. D. Yes, the distribution is highly skewed with outliers. If the boxplot suggests a confidence interval can be constructed, calculate and interpret the lower bound and the upper bound of the confidence interval. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. (Round to the nearest dollar. Use ascending order.) A. There is a 95% probability that the mean cost of repair is between $ and $ O B. We are 95% confident that the mean cost of repair is between $ 1024.35 and $ 2501.45 . O C. The confidence interval should not be constructed. Help Me Solve This View an Example Get More Help - Clear All Check Answer3 (screenshot 2 of 3 because of graphs). The data shown below represent the repair cost for a low-impact collision in a simple random sample of mini- and micro-vehicles (such as the Chevrolet Aveo or Mini Cooper). Complete parts (a) through (c). $3184 $1041 $734 $661 $762 $1748 $3394 $2046 $2699 $1360 Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. Click here to view the table of critical t-values. . X X Table of Critical t-Values Standard Normal Distribution Table (Page 1) Standard Normal Distribution Table (Page 2) right ta Table VII t-Distribution Area in Right Tail Table V (continued) Degrees of Table V Freedom 0.25 0.20 0.15 0.10 0.05 0.025 0.02 0.01 0.005 0.0025 0.001 0.0605 Standard Normal Distribution Standard Normal Distribution 12.706 0.05 0.07 0.08 0.09 1.376 6.314 15,894 3.657 127.321 318.309 636.619 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.00 0.01 0.02 0.03 0.04 0.06 2.353 10.215 12.924 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 4.773 0.0603 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 0.0 0.5596 0.5675 0.5714 0.5753 5.893 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.1 0.5398 0.5438 0.5478 0.5636 0.0004 .0004 0.0003 0.5517 0.5557 0.5793 0.5871 0.5910 0.5948 0.6026 0.6064 0.6103 0.6141 4.317 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 .0005 0.0005 0.6293 0.6331 0.6368 0.6443 0.6517 0.0010 0.0009 0.0009 0.0909 0.0008 0.0008 0.0008 0.0008 ).0007 0.0007 0.3 0.6179 0.6217 0.6252 - 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The data in the first accompanying table, in millions (so that 6.16 represents 6, 160,000 shares traded), represent the volume of a PepsiCo stock traded for a random sample of 38 trading days in 2018. A second random sample of 38 days in 2018 resulted in the data in the second accompanying table. Complete parts (a) through (d). Click here to view the data for sample 1. Click here to view the data for sample 2. Click here to view the table of critical t-values. (a) Use the data from sample 1 to compute a point estimate for the population mean number of shares traded per day in 2018. A point estimate for the population mean number of PepsiCo shares traded per day in 2018 is 5.54 million. (Round to two decimal places as needed.) Table of Critical t-Values - X (b) Using the data from sample 1, construct a 95% confidence interval for the population mean number of shares traded per day in 2018. Interpret the confidence interval. Select the correct choice below and fill in the answer boxes to complete your choice (Round to two decimal places as needed.) O A. The number of shares of PepsiCo stock traded per day is between million and million for % of all days in 2018. O B. There is a % probability that the population mean number of shares of PepsiCo stock traded per day in 2018 is between million and million. O C. One can be % confident that the number of shares of PepsiCo stock traded in all days of 2018 is between million and million. Table VII D. One can be 95 % confident that the population mean number of shares of PepsiCo stock traded per day in 2018 is between 5.07 million and 6.00 million. 1-Distribution (c) Using the data from sample 2, construct another 95% confidence interval for the population mean number of shares traded per day in 2018. Interpret the confidence interval. Degrees of - Area in Right Tail Freedom 0.25 0.20 0.15 0.10 0.05 0.025 0.02 0.01 0.005 0.0025 0.001 0.0605 Select the correct choice below and fill in the answer boxes to complete your choice (Round to two decimal places as needed.) 1000 1376 1963 3.078 6.314 12.706 4.849 31.82% 2925 127.321 318.309 636.619 6:978 10.215 12.924 O A. One can be 95 % confident that the population mean number of shares of PepsiCo stock traded per day in 2018 is between 5.07 million and 6.01 million. O B. There is a % probability that the mean population number of shares of PepsiCo stock traded per day in 2018 is between million and million. 0.718 O C. One can be % confident that the number of shares of PepsiCo stock traded in all days of 2018 is between |million and million. O D. The number of shares of PepsiCo stock traded per day is between |million and million for % of all days in 2018. Sample 1 Volumes (millions) X X Sample 2 Volumes (millions) Sample 1 Sample 2 6.16 4.16 4.89 4.98 754 4.49 10.96 5.55 8.88 8.57 5.27 4.28 7.56 3.65 7.65 7.28 6.10 6.89 4.96 6.39 4.00 5,73 3.79 4.02 6.22 4.54 4.35 4.64 4.85 6.69 6.96 3.65 4.80 8.13 5.32 6.71 5.08 5.75 5.05 4.84 6.85 5.46 4.95 4.13 5.90 4.81 4.42 6.67 4.52 6.69 3.25 6.23 8.37 4.92 7.71 6.04 4.97 7.29 5.07 3.94 5.44 6.58 7.76 6.74 4.80 5.04 4.37 6 92 2:42 3.71 Print Done Print Done Print Done Help Me Solve This View an Example Get More Help -5 The trade magazine QSR routinely checks the drive-through service times of fast-food restaurants. A 95% confidence interval that results from examining 519 customers in Taco Bell's drive-through has a lower bound of 156.2 seconds and an upper bound of 160.0 seconds. Complete parts (a) through (c). (a) What is the mean service time from the 519 customers? The mean service time from the 519 customers is 158.1 seconds (Type an integer or a decimal. Do not round.) (b) What is the margin of error for the confidence interval? The margin of error is 1.9 seconds. (Type an integer or a decimal. Do not round.) (c) Interpret the confidence interval. Select the correct choice below and fill in the answer boxes to complete your choice. (Type integers or decimals. Do not ro O A. One can be % confident that the mean drive-through service time of Taco Bell is seconds. O B. The mean drive-through service time of Taco Bell is seconds % of the time. O C. There is a % probability that the mean drive-through service time of Taco Bell is between seconds and seconds. O D. One can be % confident that the mean drive-through service time of Taco Bell is between seconds and seconds Help Me Solve This View an Example Get More Help - Clear All Check Answer6 In a survey, 31% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so he randomly selected 220 pet owners and discovered that 63 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the a = 0.01 level of significance. Because npo (1 - Po) = 47.1 > 10, the sample size is less than 5% of the population size, and the sample is given to be random, the requirements for testing the hypothesis are satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? Ho: p = 0.31 versus H1: p # 0.31 (Type integers or decimals. Do not round.) Help Me Solve This View an Example Get More Help - Clear All Final Check7 In a previous year, 58% of females aged 15 and older lived alone. A sociologist tests whether this percentage is different today by conducting a random sample of 650 females aged 15 and older and finds that 382 are living alone. Is there sufficient evidence at the a = 0.05 level of significance to conclude the proportion has changed? Because npo (1 - Po ) = 158.3 > 10, the sample size is less than 5% of the population size, and the sample is a random sample, all of the requirements for testing the hypothesis are satisfied. (Round to one decimal place as needed. Identify the null and alternative hypotheses for this test. Ho: p = 0.58 versus H1 : p # 0.58 (Type integers or decimals. Do not round.) Find the test statistic for this hypothesis test. Zo = (Round to two decimal places as needed.) Help Me Solve This View an Example Get More Help - Clear All Check