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Math 221 **** Example Format **** Week 6 Lab Submitted by: (Insert Name Here **REMOVE THIS NOTE PRIOR TO SUBMITTING**) (Note: Your labs should be well organized, with results clearly identified and in the proper order. When answering questions, be sure to use complete sentences and proper grammar. It is also important for you to fully explain your answers! Please do not answer “yes” (or “no”); you should explain why the answer is “yes”. **REMOVE THIS NOTE PRIOR TO SUBMITTING**) Part 1. Normal Distributions and Birth Weights in America (Insert your answers to the 5 questions on Birth Weights here. Be sure to carefully follow the examples worked in the Normal Ex worksheet from the Week6Lab.xls file. Copy-and-paste the Excel commands and/or Normal Probability Distributions you use to answer Questions 2a, 3a and 4a. You do not need to copy-and-paste anything for the other questions; you must provide your answers, however. **REMOVE THIS NOTE PRIOR TO SUBMITTING**) a) We see from the graph that is between 7 and 7.5 , the only period of gestation having a mean between 7 and 7.5 is: 37 to 39 weeks (is 7.33) Answer: 37 to 39 weeks b) We see from the graph that is between 7.5 and 8 , and the distance between and 7 seems to be 2 times the distance between and 8so is approx 7 2/3= 7.667 so we select the gestation period: 42 weeks and over (having a mean of 7.65) Answer: 42 weeks and over c) We see from the graph that is a little greater than 4 , the only period gestation having that mean is 28 to 31 weeks (it has a mean of 4.07) Answer: 28 to 31 weeks 2) a) X= weight of birth (period: 28 weeks), =1.88 =1.19 P(X<5.5) = NORMDIST(5.5,1.88,1.19,TRUE)= 0.9988 Answer: 99.88 % b) X= weight of birth (period 32 to 35 weeks), =5.73 =1.48 P(X<5.5) =0.4383 Answer: 43.83 % c) X= weight of birth (period 37 to 39 weeks), =7.33 =1.09 P(X<5.5) =0.0466 Answer: 4.66 % d) X= weight of birth (42 weeks or over), =7.65 =1.12 P(X<5.5) =0.0275 Answer: 2.75 % 3) a) X= weight of birth (period 37 to 39 weeks), =7.33 =1.09 We must to find x0.9 where P(X<x0.9)=0.9 x0.9 = NORMINV(0.9,7.33,1.09) = 8.7269 Answer: 8.7269 b) X= weight of birth (period 42 weeks or over), =7.65 =1.12 We must to find x0.9 where P(X<x0.9)=0.9 x0.9= 9.0853 Answer: 9.0853 4) a) X= weight of birth (period 32 to 35 weeks), =5.73 =1.48 P(6<X<9) = NORMDIST(9,5.73,1.48,TRUE) - NORMDIST(6,5.73,1.48,TRUE)= 0.4141 Answer: 41.41 % b) X= weight of birth (period 37 to 39 weeks), =7.33 =1.09 P(6<X<9) = 0.8261 Answer: 82.61 % c) X= weight of birth (period 42 weeks or over ), =7.65 =1.12 P(6<X<9) = 0.8156 Answer: 81.56 % 5) a) X= weight of birth (period 28 weeks), =1.88 =1.19 P(X<3.3) = 0.8836 Answer: 88.36 % b) X= weight of birth (period 32 to 35 weeks), =5.73 =1.48 P(X<3.3) =0.0503 Answer: 5.03 % c) X= weight of birth (period 37 to 39 weeks), =7.33 =1.09 P(X<3.3) =0.0001 Answer: 0.01 % Part 2. Age Distribution in the United States (Insert your answers to the 6 questions on Age Distribution here. See the Age Dist worksheet from the Week6Lab.xls file. That worksheet has the data entered for you, as well as some of the required work. **REMOVE THIS NOTE PRIOR TO SUBMITTING**) 1) Done in excel : Answer: =36.48 2) Answer: Mean of the sample means is 36.209 (See excel file), is near the mean age of the United States found in a) (36.48) and agrees with the Central Limit Theorem that states that the mean of the sample means tends to be the mean of the population () 3) Answer: We can see in the graph that age distribution is not symmetric, right ages (oldest persons) have a small frequency and left ages (youngest persons) have greater values. Then the age of the people distribution is not normal 4) Table of the relative frequency histogram: (See excel file) Relative frequency histogram: Graphs is symmetric since the lowest and the highest ages have lower frequencies, We can see approximately a bell shaped distribution 5) Done in excel: 22.57 6) Answer: Standard deviation is 3.55 (See excel file) and Central limit thorem says that the standard deviation of the sample means tends to /n = 22.57/36=3.76 Standard deviation of the sample means is 3.55 (near 3.76) The result agree with the predict of The Central Limit Theorem Part 3. Finding z- and t-scores for Confidence Intervals 1. Using Excel, find the z-score that corresponds to the following Confidence Levels: a. 80% Answer: 0.842 b. 85% Answer: 1.036 c. 92% Answer: 1.405 d. 97% Answer: 1.881 2. Using Excel, find the t-score that corresponds to the following Confidence Levels and Sample Sizes: a. 95% with n = 25 Answer: 2.064 b. 96% with n = 15 Answer: 2.264 c. 97% with n = 21 Answer: 2.336 d. 91% with n = 10 Answer: 1.899 3. Suppose we wish to estimate the population mean using a confidence interval. When is it appropriate to use a z-score? When is it appropriate to use a t-score? Answer: If sample size is n > 30 results using z-score or t-score are similar, We will use the t score when n<30 or when standard deviation is unknown. If standard deviation is unknown and sample size is large enough the results are so close so on these cases can select z-score or t-score (Answer the above questions. Format your answers so that they are clearly shown. See the worksheet Conf Intervals from the Week6Lab.xls file for examples on finding z- and t-scores. You will find the above questions also posted inside that worksheet. You do not need to copy-and-paste anything from Excel here. **REMOVE THIS NOTE PRIOR TO SUBMITTING**) Part 4. Bob’s Candies 1. Find the sample mean and sample standard deviation of the amount citizens spend per year. Mean = 78.4 Standard deviation = 6.205 2. When finding a confidence interval for the true mean spent of ALL citizens, should we use a z-score or a t-score? Why? If sample size is n > 30 results using z-score or t-score are similar, we will use the t score because: 1) Is more accurate using the t-score when population deviation is unknown 2) Sample size is 40 (not so greater than 30) 3. Find the z/t-values (as appropriate) for a 95% confidence interval and a 92% confidence interval. For 95% and n=40 t-score =2.023 For 92% and n=40 t-score =1.798 4. Find a 95% and a 92% confidence interval for the true mean amount that citizens spend per year. CI = mean ± t-score (s) /n For 95% Mean = 78.4 t-score = 2.023 s = 6.205 n=40 CI at 95% is 78.4±2.023(6.205)/40 = (76.4,80.4) Answer: (76.4,80.4) For 92% Mean = 78.4 t-score = 1.798 s = 6.205 n=40 CI at 92% is 78.4±1.798(6.205)/40 = (76.6,80.2) Answer: (76.6,80.2) 5. What do you think the lowest possible mean amount spent per year is? Why? Looking at the 95% interval (wider) the lowest possible mean amount spent per year is 76.4 6. Do you think Bob has a good customer base for his new business? Explain. Yes, because the wider interval (95%) is (76.4,80.4) and 76.4 is greater than 75 So the confidence interval values are all greater than 75. (Answer the above questions. Format your answers so that they are clearly shown. See the worksheet Candy Business from the Week6Lab.xls file for a complete description of Bob’s situation. Then, use your knowledge of Excel and Confidence Intervals to answer the above questions. You will find the above questions also posted inside the worksheet. You do not need to copy-and-paste anything from Excel here. **REMOVE THIS NOTE PRIOR TO SUBMITTING**)

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