Transcript 9.1 Day 3
Section 9.1 CI for a Mean Day 2 Do Computations A confidence interval for the population mean, , is given by: s x t* n Can use calculator: STAT, TESTS 8: TInterval Using Original Data Using Summary Statistics Give Interpretation in Context Good interpretation is of this form: “I am 95% confident that the population mean, , is in this interval.” Be specific about the confidence interval and describe the population you are talking about. Give Interpretation in Context Suppose we want to construct a 95% CI for the mean number of hours students study each night. The interval is (1.7, 3.5). Give Interpretation in Context Suppose we want to interpret a 95% CI for the mean number of hours students study each night. The interval is (1.7, 3.5). Consider three different surveys: 1) random sample of 40 students from EHS 2) random sample of 120 students from IL 3) random sample of 500 students from US Give Interpretation in Context Suppose we want to interpret a 95% CI for the mean number of hours students study each night. The interval is (1.7, 3.5). 1) random sample of 40 students from EHS I’m 95% confident that the mean number of hours students at EHS study each night is in the interval (1.7, 3.5). Give Interpretation in Context Suppose we want to interpret a 95% CI for the mean number of hours students study each night. The interval is (1.7, 3.5). 2) random sample of 120 students from IL I’m 95% confident that the mean number of hours students in IL study each night is between 1.7 hours to 3.5 hours. Give Interpretation in Context Suppose we want to interpret a 95% CI for the mean number of hours students study a night. The interval is (1.7, 3.5). 3) random sample of 500 students from US I’m 95% confident that the mean number of hours students in the US study each night is in the interval (1.7, 3.5). Page 574, P3 (a) Page 574, P3 (a) TInterval Inpt: Data Stats Page 574, P3 (a) TInterval Inpt: Data Stats Page 574, P3 (a) TInterval Inpt: Data Stats x: 27 sx: 12 n: 4 C-Level: .95 Calculate Page 574, P3 (a) TInterval Inpt: Data Stats x: 27 sx: 12 n: 4 C-Level: .95 Calculate (7.9053, 46.095) Page 576, E3 Should Jack and Jill readjust the machine? Use a statistical argument to support your advice. Page 576, E3 Check conditions: Page 576, E3 Check conditions: 1) told bottles are random sample from the day’s production Page 576, E3 Check conditions: 1) told bottles are random sample form the day’s production 2) told the distribution of number of ounces of water in the bottle is approx. normal (Which distribution: sample or population?) Page 576, E3 Check conditions: 1) told bottles are random sample form the day’s production 2) told the distribution of number of ounces of water in the bottle is approx. normal Population Page 576, E3 Check conditions: 1) told bottles are random sample form the day’s production 2) told the distribution of number of ounces of water in the bottle is approx. normal 3) day’s production is most likely more than 100 bottles, which is 10 times the sample size Page 576, E3 TInterval Inpt: Data Stats Page 576, E3 TInterval Inpt: Data Stats List: L1 Freq: 1 C-Level: .95 Calculate Page 576, E3 TInterval Inpt: Data Stats List: L1 Freq: 1 C-Level: .95 Calculate 95% CI is (15.916, 16.031) Page 576, E3 I’m 95% confident that the mean weight of the water bottles produced that day is between 15.917 oz and 16.031 oz. Page 576, E3 I’m 95% confident that the mean weight of the water bottles produced that day is between 15.917 oz and 16.031 oz. Because 16 oz is one of the plausible values for the population mean, there is no need to adjust the machine. Page 575, P8 Page 575, P8 a) False. A CI is a statement about plausible values for a population mean, not about the individual values within a population. Page 575, P8 b) False. A CI is a statement about plausible values for a population mean, not about the values in the sample. Page 575, P8 c) False. The method has a 95% chance of success, but after this particular interval is calculated, either it contains the population mean or it doesn’t. Page 575, P8 d) False. The sample mean is the center of the CI. Page 575, P8 e) True. 95% of 200 is 190. So about 190 of the 200 CIs would include the population mean and approximately 10 would not. Page 579, E13 x ± t*● s n x ± t*● s n (a) standard deviation of sample increased? margin of error _________ x ± t*● s n (a) standard deviation of sample increased? margin of error increases x ± t*● s n (b) sample size is increased? margin of error _______ x ± t*● s n (b) sample size is increased? margin of error decreases x ± t*● s n (c) confidence level is increased? margin of error _______ x ± t*● s n (c) confidence level is increased? margin of error increases Page 577, E6 Page 577, E6 a) This is not a random sample of bags of fries. Also, because all the bags came from one restaurant, even with a random sample, you could only infer about bags of fries from this restaurant. Page 577, E6 a) Distribution of number of fries is fairly symmetric, so there is no indication that the underlying population is not normal (See Display 9.13, page 578) Page 577, E6 a) This McDonald’s restaurant probably sells more than 320 bags of fries, which would be more than 10 times the sample size. So, random sample condition not met but the other conditions are. Page 577, E6 b) 95% CI is (44.888, 51.362) Page 577, E6 b) 95% CI is (44.888, 51.362) If this were a random sample, Page 577, E6 b) 95% CI is (44.888, 51.362) If this were a random sample, then I would be 95% confident that the mean number of fries in a small bag at this McDonald’s restaurant is between about 45 and 51 fries. Page 574, P7 For aldrin data in P4, 95% CI is (4.229, 5.809) Page 574, P7 D. You are 95% confident that the mean aldrin level of the Wolf River falls in the confidence interval (4.229, 5.809) Questions?