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11. Do one of the following, as appropriate: (a) Find the critical value z a, (b) find the critical value (ipoint tap, (c) state that neither the normal nor the t distribution applies. 99%; n = 17, σ is unknown; population appears to be normally distributed. O zan 2.583 O tan a/2 -2.898 O ta/2-2921 O za/2#2567 12. Do one of the following, as appropriate: (a) Find the critical value za/2, (b) find the critical value (point) tan,(c) state that neither the normal nor the t distribution applies. 9896, n-7; ơ-27; population appears to be normally distributed. Oe-2.05 ta2-2.575 ta2-1.96 ze 2-233 13. Use the given degree of confidence and sample data to construct a confidence interval for the population mean u. Assume that the population has a normal distribution. 4 point) A sociologist develops a test to measure attitudes towards public transportation, and 27 randomly selected subjects are given the test. Their mesn score is 76.2 and thair standard deviation is 21.4. Construct the 95% confidence interval for the mean score of all such subjects. 67.7 K แ K 847 69.2 83.2 74.6 uje 77,8 64.2 K < 88.2
(1 poi 14. Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution. A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean checking account balance for local customers. O $455.65 < μ < S872.63 〇$492.52 < μ < $ 835.76 O $493.71 u $834.57 $453.59 < μ < S874.69 S. Assume that a sample is used to estimate a population mean μ. Use the given confidence level and (1 point) 1 sample data to find the margin of error. Assume that the sample is a simple random sample and the population has a normal distribution. Round your answer to one more decimal place than the sample standard deviation. 95% confidence; n=12;X= 76.4: s = 5.9 4.50 О 3.38 3.75 審2.81 16. Use the given degree of confidence and sample data to find a confidence interval for the population 4 point standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. To find the standard deviation of the diameter of wooden dowels, the manufacturer messures 19 randomly selected dowels and finds the standard deviation of the sample to bes 0.16. Find the 9395 confidence interval for the population standard deviation 0.11*0* 0.25 0.13-0 0.22 0.12 0.24 0.15 к 0.21
You want to be 99% confident that the sample standard deviation s is within 596 of the population standard deviation. 0 2638 O 2434 0 1336 O 923 18. Solve the problem. (1 point) Find the critical value X R corresponding to a sample size of 3 and a confidence level of 95% 0 7.378 O 0.0506 5.991 0.103 19. Use the given degree of confidence and sample data to find a confidence interval for the population (ipoint standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data. The daily intakes of milk (in ounces for ten randomly selected people were: 26.013013.5 14.8 19.1 13.6 216 30.8 15.8 14.4 Find s 99% confidence interval for the population standard devistion c. 3.65 oz«o < 12.5102 3.78 oz< σ< 12.51 oz 0,88 oz < ơ < 3.38 oz 3.78 oz<a-1392 02 20. Solve the problem. Find the chi-square value x corresconding to a sample size of 23 and a corfidence level of sos 12.338 1 69 16.013
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Answer Date: 24/01/2019 The t-critical value is, The sample size is small and two-tailed test. Look in the column headed α-001 and the row headed in the t distribution table by using degree of freedom is, - 16 The t-critical value is 2.921 12) The z-critical value is, From the standard normal table, the z-critical value for two-tailed at 98% confidence is 2.33The z-critical value is+ 2.33 13) The 95% confidence interval for the mean score of all such subjects is, First, compute t-critical value then find confidence interval. The t critical value for the 95% confidence interval is. The sample size is small and two-tailed test. Look in the column headed α-0.05 and the row headed in the t distribution table by using degree of freedom is, - 27-1 - 26 The t critical value for the 95% confidence interval is 2.056.76.2±2.056 × 21.4/M - 76.2+8.5 67.7 to 84.7 The 95% confidence interval for the mean score ofall such subjects is 67.7 to 847 . 14) A 98% confidence interval for the true mean checking account balance for local customers is, First, compute t-critical value then findconfidence interval The t critical value for the 98% confidence interval is. The sample size is small and two-tailed test. Look in the column headed α-0.02 and the row headed in the t distribution table byusing degree of freedom is, d.f.-n-1 = 14-1 - 13The t critical value for the 98% confidence interval is 2.650. -664.14 2.650 x 297.29/14 664.14210.55 = 453.59 to 874.69 A 98% confidence interval for the true mean checking account balance for local customers is453.59 to 874.69 15 The margin oferror at 95% confidence level is. First, compute t-critical value then find margin of error The t critical value for the 95 % confidence interval is The two-tailed test. Look in the column headed α=0.05 and the row headed in the t distribution table byusing degree of freedom is,-12-1 The t critical value for the 95% confidence interval is 2.201 . Vn 5.9 - 2.201x V12) - 3.75 The margin oferror at 95% confidence level is 13.75 16) The 95% confidence interval for the population standard deviation is. First, compute chi-square variance critical value then find confidence interval. The chi-square variance critical value at 5% significance level isFrom the chi-square distribution table, degrees of freedom is 18 (n-1) and the significance level 0.025 left-tailed chi-square value is 31.526 and the significance level 0.9750 right- tailed value is 8.231 (7-1) × s2 (n-1)xs 95% CI. ん0.0250 ん0.9750 (19-1)×0.16 < σ < (19-1)×0.162 8.231 31.526 V0.0 146 < σ < No.0560 =0.12 < σ < 0.24 The 95% confidence interval for the population standard deviation is 0.12 to 0.24 17) The sample sizeis From the table of sample size to population standard deviation, show the confidence level 99% and estimate within value of population standard deviation equal 5% and the sample size is 1336The sample size is 1336 18) The chi-square critical value for right-tailed is, From the chi-square distribution table, degrees of freedom are 2 (n-1) the significance level 0.9750 right-tailed value is 0.0506 The chi-square critical value for right-tailedis l0.0506 19) A 99% confidence interval for the population standard deviation ofdaily milk intake is, aware, find confidence interval with the help of following steps By using MegaStat so 1) Import the data. 2) Select Chi-Square Vaiiance Test from Hypothesis Tests in MegaStat option from Add-Ins men 3) Select Sinnary input and enter Hypothesized variance 4) Select Alternative and confidence interval. 5) Click Ok.

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