8 questions please

On January 20, 2020, the Diamond Princess cruise ship departed Yokohama, Japan, carrying 3711 passengers and crew. On January 25, a symptomatic passenger disembarked the ship in Hong Kong, where he tested positive for the SARS-CoV-2 virus, which causes a disease commonly known as COVID-19. On February 3, the ship returned to Japan, and on February 5, passengers were quarantined in their cabins. Initially, only those passengers with fever or respiratory symptoms were tested for COVID-19, but later, testing was expanded to include older passengers, those with underlying medical conditions, and those in internal cabins with no access to the outdoors. People who had negative test results or no respiratory symptoms, and no close contact with a person who had a confirmed case of COVID-19, were allowed to disembark after being quarantined for 14 days. Among the 3,711 Diamond Princess passengers and crew, 712 had positive test results for COVID-19 and nine tragically died as a result. Of those who tested positive, 331 were asymptomatic at the time of testing. Assuming that the passengers on the ship are a random sample from the population of people on the planet (they are not, because people who go on cruises tend to be older and more affluent than the general population), construct a 95% confidence interval for the proportion of people who test positive for COVID-19 but have no symptoms. Because the number symptomatic and asymptomatic cases are both greater than 10, use large-sample confidence interval. Give your answer in the form (lower, upper). Give the limits as a proportion, not a percentage, to at least four decimal places. Confidence Interval:Suppose the math department chair at a large state university wants to estimate the average overall rating, out of five points, that students taking Introductory Statistics gave their lecturers on their end-of-term evaluations. She selects a random sample of 38 evaluations and records the summary statistics shown. Sample Sample Sample standard Population standard Standard Confidence size mean deviation deviation deviation of x level X ON C 38 4.0414 1.0501 0.9903 0.1606 90% The department chair determines the standard deviation of the sampling distribution of x using o, the standard deviation calculated from all student evaluations submitted to her department. Assume the Statistics lecturers' ratings have the same standard deviation. Use this information to find the margin of error and the lower and upper limits of a 90% confidence interval for #, the mean overall rating students gave their Introductory Statistics lecturers. Round your answers to the nearest hundredth. Margin of error = 0.26 Lower limit = 3.78 Upper limit = 4.31 Complete the following sentence to state the interpretation of the depuruncan cun a confidence interval. The that the of the overall ratings given on at the end of the term Answer Bank is 4.0414 probability is 90% mean standard deviation standard deviation of x department chair is 90% confident confidence interval the student evaluations in the sample is between the lower and upper limits all student evaluations of Statistics lecturersSuppose the math department chair at a large state university wants to estimate the average overall rating, out of five points, that students taking Introductory Statistics gave their lecturers on their end-of-term evaluations. She selects a random sample of 34 evaluations and records the summary statistics shown. Sample Sample Sample standard Population standard Standard Confidence size mean deviation deviation error level X Ox C 34 4.0930 1.0512 1.1780 0.2020 90% The department chair determines the standard error using o, the standard deviation calculated from all student evaluations submitted to her department. Assume the Statistics lecturers' ratings have the same standard deviation. Use this information to find the margin of error and the lower and upper bounds of a 90% confidence interval for #, the mean overall rating students gave their Introductory Statistics lecturers. Round your answers to the nearest hundredth. Margin of error = Lower bound = Upper bound = Complete the following sentence to state the interpretation of the departnom cnan's confidence interval. The that the of the overall ratings given on at the end of the term Answer Bank is 4.0930 the student evaluations in the sample department chair is 90% confident probability is 90% mean standard error standard deviation is between the lower and upper bounds confidence interval all student evaluations of Statistics lecturersThe Weschler Intelligence Scale for Children (WISC) is an intelligence test designed for children between the ages of 6 and 16. The test is standardized so that the mean score for all children is 100 and the standard deviation is 15. Suppose that the administrators of a very large and competitive school district wish to estimate the mean WISC score for all students enrolled in their programs for gifted and talented children. They obtained a random sample of 40 students currently enrolled in at least one program for gifted and talented children. The test scores for this sample are as follows: 117 142 112 99 107 109 109 121 121 125 116 123 110 105 110 111 95 128 103 106 155 98 100 123 107 137 115 127 102 106 121 113 109 128 103 105 130 134 112 102 Click to download the data in your preferred format. CrunchIt! CSV Excel JMP Mac Text Minitab PC Text R SPSS TI Calc Use this data to calculate the mean WISC score, x, for these 40 students. Next, compute the standard deviation, SD, of the sampling distribution of the sample mean, assuming that the standard deviation of WISC scores for students in the district is the same as for the population as a whole. Finally, determine both the lower and upper limits of a 99% z-confidence interval for #, the mean score for all students in the school district who are enrolled in gifted and talented programs. Give x and the limits of the confidence interval precise to one decimal place, but give the standard deviation to at least three decimal places in order avoid rounding errors when computing the limits. X = 114.9 13.111 SD = Incorrect 109.3 120.5 Lower limit = Upper limit =Suppose 55% of people in Georgia support a special transportation tax. Alejandro is not confident that this claim is correct. To investigate the claim, he surveys 150 people in his community and discovers that 54 people support a special transportation tax. Calculate Alejandro's sample proportion, p. Give your answer precise to two decimal places. p = 0.36 Calculate the standard error of the sample proportion, SE. Give your answer precise to three decimal places. SE = 0.041 Calculate the standard error of the sample proportion estimate, SEest. Give your answer precise to three decimal places. 0.040 SEest = IncorrectSuppose a veterinarian wants to estimate the difference between the proportion of cat owners who are single and the proportion of dog owners who are single. Of the pet owners that visit any of the veterinarian clinics in her city regularly, she identifies 3872 pet owners that exclusively have cats and 4108 pet owners that exclusively have dogs. From this list of pet owners, she surveys 158 randomly selected cat owners and 139 randomly selected dog owners and asks each of them if they are single or married. Her findings are summarized in the table. Population Population Sample Number Sample description size of successes proportion cat owners n1 = 158 X1 = 63 P1 = 0.39873 N dog owners n2 = 139 X2 = 52 P2 = 0.37410 Calculate the upper and lower limits (bounds) for a large sample 90% z-confidence interval for the difference in two population proportions, P1 - p2. Give each of your answers with three decimal places of precision. lower limit = upper limit = Then, complete the following sentence to state the interpretation of the confidence interval. The that the who are single is between # and Answer Bank probability is 90% 0.866 -0.039 0.415 difference in the proportions of all cat and dog owners -0.048 difference in the proportions of cat and dog owners in the samples veterinarian can be 90% confident mean number of dog owners proportion of cat owners 0.092 0.089 0.335 0.680 0.333 -0.068 difference in the mean numbers of cat and dog owners 0.097 -0.043 mean number of cat owners 0.118 -0.118 0.463 0.068 proportion of dog owners 0.156 -0.1078. Suppose a researcher wants to test whether the proportion of people that reflect a certain characteristic in a population, population A, is lower than the proportion of people with that characteristic in another population, population B. The researcher plans to perform a two-sample z-test for two proportions, testing the null hypothesis, Ho : PA - PB 2 0, against the alternative hypothesis, H1 : PA - PB