Chapter 6 paradox 6: Suppose that the population threadbare divagation (?) for a normally distri furthered standardized stress of work is know to be 7.20. What would the standard error of the sample ridiculous ([pic]) be if we were to draw a random sample of 16 test scores? firmness: For this problem, since the population standard outcome is given, we utilise the canon on varlet 193. [pic] =[pic] Therefore: [pic] = [pic] =[pic] = 1.8 enigma 7: Suppose that the random sample from riddle 6 yielded these ascertained scores: 656125101113 121092023202818 a. Find the 95% boldness detachment for the specify. b. Find the 99% arrogance detachment for the mean. Answer: The commencement exercise thing we need to do is set out the mean for these entropy (we will employ the population standard deviation but if we did not know it we would have to calculate that and use the t distribution). Using Excel the mean is 13 (you should ins ist this rank for practice). a. now the 95% confidence breakup is raise using the formula on page 196: 95% confidence interval = [pic] ± 1.96 [pic] so we then have: 13 ±1.96 (1.8) = 13 ± 3.528 = 9.47 to 16.53 b. Now the 99% confidence interval is show using the formula on page 197: 99% confidence interval = [pic] ± 2.58 [pic] so we then have: 13 ±2.58 (1.8) = 13 ± 4.644 = 8.36 to 17.
64 riddle 8 Now suppose that we do not know wanton assuming that ? = 7.20. Use the scores in Problem 7 to: a. estimate the standard error of the sample mean ([pic]). b. mystify the 95% confidence interval for the mean. c. lift the 99% confidence interval for the! mean. Answer: First we need to find the standard deviation from our sample. I used Excel and found it to be (S = 6.899) but please verify! Now we would get the precedent Error of the mean using the formula on page 202: [pic] =[pic] a. Therefore: [pic] = [pic] =[pic] = 1.78 The general formula for the confidence interval is shown on page 205: Confidence...If you want to get a large essay, direct it on our website: BestEssayCheap.com
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