$ solve these
The following data are annual disposable income and total annual consumption for 22 families selected at random from a large metropolitan area. Income data - SAS outputs. Useful information about the data he following data are annual disposable income and total annual consumption for 22 families selected at random from a large metropolitan area. Both income and consumption are expressed in U$ 1,000, i.e. multiply the values by 1,000 to get the real values. 2 Variables: income cons Simple Statistics Fig. 1 Variable N Mean Std Dev Sum Minimum Maximum income 22 30.72727 19.10667 676.00000 12.00000 70.00000 cons 22 29.03364 14.76920 638.74000 12.00000 63.25400 Pearson Correlation Coefficients, N = 22 Prob > |r] under HO: Rho=0 Income cons income 1.00000 0.81618 .0001 cons 0.81618 1.00000 4.0001Fig. 2 Number of Observations Read 22 Number of Observations Used 22 Analysis of Variance Sum of Mean Source OF Squares Square F Value Pr > F Model 3051.45057 30001 Error 76.46305 Corrected Total The black spaces must be computed, according to the question. Fig. 3 Parameter Estimates Parameter Standard Variable OF Estimate Error t Value Pr > (t| 95%% Confidence Limits Intercept 3.59062 income 1 0.63090 0.09987 The black spaces must be computed, according to the question.Fig. 4: Output Statistics Std Error Dependent Predicted Mean Obs Variable Value Predict 95% CL Mean 95% CL Predict Residual 12.0 2.6407 -1.8353 36.2720 2 34.8 2.6407 -1.8353 36.2726 3 16.0 18.4805 2.5033 13.2508 23.7021 -0.4926 37.4535 -2.4485 A 20.8 18.4805 2.5033 13.2588 23.7021 -0.4926 37.4535 2.2875 14.0 19.7422 2.3746 14.7889 24.6956 0.8413 38.6432 -5.7422 6 17.0 19.7422 2.3746 14.7889 24.6956 0.8413 38.6432 -2.7422 7 34.8 19.7422 2.3746 14.7889 24.6956 0.8413 38.6432 15.0378 8 17.0 19.7422 2.3746 14.7889 24.6956 0.8413 38.6432 -2.7422 9 20.8 19.7422 2.3746 14.7889 24.6956 0.8413 38.6432 1.0258 10 20.8 19.7422 2.3746 14.7889 24.6956 0.8413 38.6432 1.0258 11 20.8 1.9817 6.0866 43.4923 12 22.4 26.0512 1.9231 22.0396 30.0628 7.3750 44.7275 -3.6652 13 24.5 28.5748 1.8657 24.6830 32.4666 9.9239 47 2257 -4.0298 14 34.8 28.5748 1.8657 24.6830 32 4666 9.9239 47 2257 6.2052 15 24.5 28.5748 1.8657 24.6830 32 4666 9.9239 47 2257 -4,0298 16 36. 8 36.7765 2.2311 32.1224 41 4305 17.9518 55.6012 -0.000463 17 16.0 36.7765 2.2311 32.1224 41 4305 17 9518 55 6012 -20.7445 18 49.5 41.1927 2.6796 35.6032 46.7823 22.1152 60.2703 8.3553 19 10.2 3.1378 25.5990 64.3573 20 34.8 3.1378 25.5990 64.3573 21 13.3 4.3427 33.4448 74.1765 22 63. 4.3427 33.4448 74.1765 Calculate the error when Income is equal to 31.66 and consumption is equal to 37.27. Use 2 decimal places for all calculations, include zero before the decimal and the negative sign when needed, e.g. -0.12. No credit will be given for rounding mistakes.For the year 2010, 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized O deductions on their federal Income tax return. The mean amount of deductions for this population of O taxpayers was $16,642. Assume that the standard deviation is a = $2,600. If required, round your answer to two decimal places. (a) What are the sampling distributions of x for itemized deductions for this population of taxpayers for each of the following sample sizes: 30, 50, 100, and 400? E(x) = n 30 50 100 400 (b) What is the advantage of a larger sample size when attempting to estimate the population mean? A larger sample - Select your answer - : the standard error and results in a(n) - Select your answer - * precise estimate of the population mean.round your answer to four decimal places. (a) Suppose a sample of 300 primary care doctors was taken. Show the distribution of the sample proportion of doctors who think their patients receive unnecessary medical care. np = n(1-p) = E(p) = (b) Suppose a sample of 500 primary care doctors was taken. Show the distribution of the sample proportion of doctors who think their patients receive unnecessary medical care. np = n(1-p) = E(p) = (c) Suppose a sample of 1,000 primary care doctors was taken. Show the distribution of the sample proportion of doctors who think their patients receive unnecessary medical care. np = n(1-p) = E(p) = (d) In which of the preceding three cases, part (a) or part (b) or part (c), Is the standard error of p smallest? Why? The standard error is the smallest in | - Select your answer - : because , isOne of the questions on a survey of 1,000 adults asked if today's children will be better off than their parents. Representative data are shown In the file named ChildOutlook. A response of Yes indicates that the adult surveyed did think today's children will be better off than their parents. A response of No indicates that the adult surveyed did not think today's children will be better off than their parents. A response of Not Sure was given by 23% of the adults surveyed. Click on the datafile logo to reference the data. DATA file (a) What is the point estimate of the proportion of the population of adults who do think that today's children will be better off than their parents? If required, round your answer to two decimal places. (b) At 95% confidence, what is the margin of error? If required, round your answer to four decimal places. (c) What is the 95% confidence interval for the proportion of adults who do think that today's children will be better off than their parents? If required, round your answers to four decimal places. Do not round your intermediate calculations. to (d) What is the 95% confidence interval for the proportion of adults who do not think that today's children will be better off than their parents? If required, round your answers to four decimal places. Do not round your intermediate calculations. to (e) Which of the confidence intervals in parts (c) and (d) has the smaller margin of error? Why? The confidence interval in part - Select your answer - : | has the smaller margin of error. This is because p is- Select your answer - + 0.5.The Pew Research Center Internet Project conducted a survey of 1,057 Internet users. This survey provided a variety of statistics on them. If required, round your answers to four decimal places. (a) The sample survey showed that 90% of respondents said the Internet has been a good thing for them personally. Develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally. to (b) The sample survey showed that 67% of Internet users said the Internet has generally strengthened their relationship with family and friends. Develop a 95% confidence interval for the proportion of respondents who say the Internet has strengthened their relationship with family and friends. to (c) Fifty-six percent of Internet users have seen an online group come together to help a person or community solve a problem, whereas only 25% have left an online group because of unpleasant interaction. Develop a 95% confidence interval for the proportion of Internet users who say online groups have helped solve a problem. to (d) Compare the margin of error for the interval estimates in parts (a), (b), and (c). How is the margin of error related to the sample proportion? The margin of error | - Select your answer - : as p gets closer to .50.\fLast year, 46% of business owners gave a holiday gift to their employees. A survey of business owners Indicated that 45% plan to provide a holiday gift to their employees. Suppose the survey results are based on a sample of 60 business owners. (a) How many business owners in the survey plan to provide a holiday gift to their employees? (b) Suppose the business owners in the sample do as they plan. Compute the p value for a hypothesis test that can be used to determine if the proportion of business owners providing holiday gifts has decreased from last year. If required, round your answer to four decimal places. If your answer is zero, enter "0". Do not round your intermediate calculations. (c) Using a 0.05 level of significance, would you conclude that the proportion of business owners providing gifts has decreased? We - Select your answer - * the null hypothesis. We - Select your answer - * conclude that the proportion of business owners providing gifts has decreased. What is the smallest level of significance for which you could draw such a conclusion? If required, round your answer to four decimal places. If your answer is zero, enter "0". Do not round your intermediate calculations. The smallest level of significance for which we could draw this conclusion Is ; because p-value is - Select your answer - : |the corresponding o, we - Select your answer - : the null hypothesis.am # 2 - Online ports X + https:/www.webassign.net/web/Student/Assignment-Responses/lastidep=19411305 E -45 points ASWSBE 13 7.E 026. My Notes You may need to use the appropriate appendix table or technology to answer this question. For the year 2010, 33% of taxpayers with adjusted gross Incomes between $30,000 and $60,000 Itemized deductions on their federal Income tax return. The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is o = $2,400. (a) What is the probability that a sample of taxpayers from this Income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 30, 60; 150, and 3007 (Round your answers to four decimal places.) sample size n = 30 sample size n = 60 sample size n = 150 sample size n = 300 (b) What is the advantage of a larger sample size when attempting to estimate the population mean? O A larger sample has a standard error that is closer to the population standard deviation. O A larger sample Increases the probability that the sample mean will be a specified distance away from the population mean. O A larger sample lowers the population standard deviation. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.
