Notes on Confidence Intervals - of Elements of Statistics | MA 116, Study notes of Statistics

Download Notes on Confidence Intervals - of Elements of Statistics | MA 116 and more Study notes Statistics in PDF only on Docsity! M116 – NOTES – CH 8 Chapter 8 – Confidence Interval – One Population Section 8.1 - Estimating a Population Mean: σ Known Assumptions:  The sample is a simple random sample  The value of the population standard deviation σ is known.  Either or both of these conditions are satisfied: i) The population is normally distributed, or ii) n ≥ 30 (The sample has 30 or more values) Procedure for Constructing a Confidence Interval for μ (with Known σ) 1. Verify that the required assumptions are satisfied. 2. Find the critical value cz (from table 5). 3. Evaluate the margin of error E. (E = cz n  ) 4. Then using E and the sample mean the confidence interval is: x E x E    or ( , )x E x E  Using the TI-83 to Construct Confidence Intervals for μ: STAT>>TESTS select 7:ZInterval Round-off Rule for Confidence Intervals used to Estimate μ: a) If original data is given: use one more decimal place than original values. b) If you are given summary statistics from a data set, use the same number of decimal places used for the sample mean. 1 CHAPTER 8 Section 8.1 - Estimating a Population Mean: σ Known 1) A simple random sample of size n is drawn from a population whose population standard deviation, σ, is known to be 3.8. The sample mean, x-bar, is determined to be 59.2. a) Compute the 90% confidence interval about µ if the sample size, n, is 45. 90% 1.645cCI z  45 3.8 59.2 n x     x * 3.8 59.2 ± 1.645 * 45 cz n   59.2 ± .9318 → (58.27, 60.13) 58.27< <60.13 For calculator feature use STAT, arrow to TESTS, and select 7:ZInterval, select Stats enter the required information, and CALCULATE b) Compute the 90% confidence interval about µ if the sample size, n, is 55. How does increasing the sample size affect the margin of error E? How does it affect the length of the interval? 90% 1.645cCI z  55 3.8 59.2 n x     x * 3.8 59.2 ± 1.645 * 55 cz n   59.2 ± .8429 → (58.357, 60.043) This is a shorter interval than the one in part (a). It’s more precise. At the same confidence level, a larger sample size produces a more precise (shorter) interval For calculator feature use STAT, arrow to TESTS, and select 7:ZInterval, select Stats enter the required information, and CALCULATE c) Compute the 98% confidence interval about µ if the sample size, n, is 45. Compare the results to those obtained in part (a). How does increasing the level of confidence affect the size of the margin of error, E? How does it affect the length of the interval? 98% 2.33cCI z  (Higher confidence level) 45 3.8 59.2 n x     x * 3.8 59.2 ± 2.33 * 45 cz n   59.2 ± 1.3199 → (57.88, 60.52) This is longer (less precise) than the interval produced in part (a). A higher confidence level produces a longer interval. With a longer interval we are now more confident that the interval contains the population mean. Use calculator feature to check answer: Z-Interval 2 CHAPTER 8 Section 8.2 -Estimating a Population Mean: σ Not Known 2) A simple random sample of size n is drawn from a population. If that sample has a mean of 59.2 and a standard deviation of 3.8, a) Compute the 90% confidence interval about µ if the sample size, n, is 45. 90% 1.684cCI t  (Use the closest degrees of freedom that is smaller than 45) x * 3.8 59.2 ± 1.684 * 45 c s t n  59.2 ± .9539 → (58.25, 60.15) For calculator feature use STAT, arrow to TESTS, and select 8:TInterval, select Stats enter the required information, and CALCULATE b) Compare with your answer to part 1-a. Which interval is longer? Which interval is more precise? Results from part (c): 59.2 ± .9539 → (58.25, 60.15) Results from part (a): 59.2 ± .9318 → (58.27, 60.13) The z interval is smaller, more precise. When we don’t have the standard deviation of the population, σ, we use the standard deviation of the sample, s. In this case we use the t-distribution instead of the z- distribution. Corresponding t-intervals are a little longer than z-intervals. Finding the Point Estimate and the Margin of Error from a Confidence Interval 3) We are 95% confident that the interval from 98.08°F to 98.32°F actually contains the true mean value of the body temperatures of all healthy adults. What is the point estimate in this case? What is the margin of error? 98.32 98.08 98.2 2 98.32 98.08 .12 2 x E       5 CHAPTER 8 Section 8.2 -Estimating a Population Mean: σ Not Known 4) In order to correctly diagnose the disorder of hydrocephalus, a pediatrician investigates head circumferences of two month old babies. Use the sample MHED, augment it with the sample FHED. This will give you a new sample of size 100. Use this sample to estimate the head circumference of all two months old babies. To augment a List we’ll do the following: First: STAT, Edit, go UP and to the RIGHT until you reach a list with no name. Name it BABY. Second: Stay in the editor, press 2nd STAT [LIST], arrow right to OPS, go down and select 9:augment( by pressing enter, select FHED, press “comma”, then select MHED, and press ENTER. The BABY list should fill up with 100 numbers The statement should read augment(FHED, MHED)) a) What is the point estimate? 40.573x  b) Verify that the requirements for constructing a confidence interval about x-bar are satisfied. Since the sample size is large, the distribution of sample means is approximately normally distributed. (Central Limit theorem, section 7.2) c) Construct a 99% confidence interval estimate for the head circumference of all two months old babies. (Are you using z or t? Why?) 100 1.649 40.573 n s x    x * 1.649 40.573 ± 2.639 * 100 c s t n  40.573 ± 0.43517 → (40.14, 41.006) Use calculator feature to check answer: T-Interval (Use DATA option) d) The statement “99% confident” means that, if 100 samples of size __100___ were taken, about __99___ intervals will contain the parameter μ and about __1__ will not. e) Complete the following: We are __99_% confident that the mean head circumference of all two months old babies is between __40.14 cm___ and __41.01 cm____ f) With __99_% confidence we can say that the mean head circumference of all two months old babies is __40.573 cm____ with a margin of error of __0.435 cm_____ g) How can you produce a more precise confidence interval? If we maintain the same degree of confidence, (99%) we’ll get a shorter interval by selecting a larger sample 6 M116 – NOTES – CH 8 Chapter 8 – Confidence Intervals –One Population Section 8.3 - Estimating a Population Proportion Assumptions  The sample is a simple random sample  The conditions for a binomial distribution are satisfied by the sample. That is: there are a fixed number of trials, the trials are independent, there are two categories of outcomes, and the probabilities remain constant for each trial. A “trial” would be the examination of each sample element to see which of the two possibilities it is.  The normal distribution can be used to approximate the distribution of sample proportions because np > 5 and nq > 5 are both satisfied. (q = 1 – p) Procedure for Constructing a Confidence Interval for p 1) Verify that the assumptions are satisfied. 2) Find the critical value cz (from table 5). 3) Evaluate the margin of error E.   c p q E z n   4) Find the interval and write it in one of the following forms p E ;  p E p p E    ;  ( , )p E p E  Round-Off Rule for Confidence Interval Estimates of p 3 significant digits Using the TI-83 to Construct Confidence Intervals for p: STAT>>TESTS choose A:1-propZInt. Notice: x must be a whole number If you have to calculate it, round it to the nearest whole number 7 M116 – NOTES – CH 8 Section 8.5 – Confidence Intervals – Two Populations Means Inferences about Two Means with Unknown Population Standard Deviations Independent Samples Population Standard Deviations not Assumed Equal (Non-Pooled t-Interval) Assumptions  The samples are obtained using simple random sampling  The samples are independent  The populations from which the samples are drawn are normally distributed or the sample sizes are large ( 1 230, 30n n  ) The procedure is robust, so minor departures from normality will not adversely affect the results. If the data have outliers, the procedure should not be used. Use normal probability plots to assess normality and box plots to check for outliers. A normal probability plot plots observed data versus normal scores. If the normal probability plot is roughly linear and all the data lie within the bounds provided by the software (our calculator does not show the bounds), then we have reason to believe the data come from a population that is approximately normal. Using the TI-83/84 to Construct Confidence Intervals for μ1- μ2: STAT>>TESTS select 0:2SampT Interval Use the DATA option if you have been given data 6) – Schizophrenia and Dopamine (with 2 populations we’ll use the calculator only) Previous research has suggested that changes in the activity of dopamine, a neurotransmitter in the brain, may be a causative factor for schizophrenia. In the paper “Schizophrenia: Dopamine b-Hydroxylase Activity and Treatment Response”, Sternberg et al. published the results of their study in which they examined 25 schizophrenic patients who had been classified as either psychotic or not psychotic by hospital staff. The activity of dopamine was measured in each patient by using the enzyme dopamine b-hydroxylase to assess differences in dopamine activity between the two groups. The following are the data, in nanomoles per milliliter-hour per milligram (nmol/ml-h/mg). Psychotic 0.0150 0.0222 0.0204 0.0275 0.0306 0.0270 0.0320 0.0226 0.0208 0.0245 Non-psychotic 0.0104 0.020 0.0210 0.0105 0.0112 0.0230 0.0116 0.0252 0.0130 0.0200 0.0145 0.0180 0.0154 0.0170 0.0156 a) Because the sample sizes are small, for each population we must verify that the variable is normally distributed and the sample does not contain any outliers. Construct a normal probability plot to assess normality and a boxplot in order to check for outliers. For the firs data set, enter the data in L1 (press STAT, select Edit) of the calculator and open two plots, one with a modified box plot (the fourth icon) and another with the normal probability plot, which is the last icon type in the 2nd Y= [STAT PLOT] window. Do the same with the second data set after you enter it into L2. 10 CHAPTER 8 Use the calculator to obtain the following statistics (1-Var-Stats for each list) Dopamine activity (nmol/ml-h/mg) Psychotic Non-Psychotic x1bar = .02426 x2bar = .01643 s1 = .00514 s2 = .0047 n1 = 10 n2 = 15 b) Give the point estimate for the psychotic patients x1bar = .02426 c) Give the point estimate for the non-psychotic patients x2bar = .01643 d) We observe that x1bar is ……………higher than / lower than x2bar. Is it higher by chance; or is it significantly higher? In order to discover this, we’ll do the following: e) Give the point estimate for the difference between dopamine-activity in the two groups: x1bar – x2bar = .02426 - .01643 = .00783 f) Use a calculator feature to obtain a 98% confidence interval for the difference between the two population means. (Are you using z or t? Why?) We are using t because the standard deviations of the population are not known. (We know the standard deviations of the samples) Use a 2-Samp T Interval (non-pooled because we can’t assume 1 2  ) ….00266………. < μ1- μ2 < …… .013………….. g) What does the interval suggest about μ1 and μ2? μ1 < μ2 μ1 > μ2 μ1 = μ2 Explain your choice; be very specific in your explanation The interval gives plausible values for μ1- μ2 . Since the interval is completely ABOVE ZERO, all possible values for μ1- μ2 are positive, which implies than μ1 > μ2 h) We are __98_____% confident that dopamine activity in psychotic patients, on average (choose from the following choices) is higher than, is lower than, may be equal to the dopamine activity in non-psychotic patients. 11 M116 – NOTES – CH 8 Section 8.5 – Confidence Intervals - Two Population Proportions Assumptions  The samples are independently obtained using simple random sampling.  For both samples, the conditions np ≥ 5 and n(1 – p) ≥ 5 are both satisfied. For both samples, the sample size, is no more than 5% of the population size 7) Eating Out Vegetarian A Zogby International poll of 1181 US adults was conducted in March 1999, to gauge the demand for vegetarian meals in restaurants. The study was commissioned by the Vegetarian Resource Group and was published in the September/October 1999 issue of the Vegetarian Journal. In the survey, independent random samples of 747 US men and 434 US women were taken. Of those sampled, 276 men and 195 women said that they sometimes order a dish without meat, fish, or fowl when they eat out. a) Find the point estimate p1hat – p2hat men women n1= 747 n2 = 434 x1 = 276 x2 = 195 p1-hat = .3695 p2-hat = .4493   1 2 .3695 .4493 .0793p p    b) Construct a 90% confidence interval for the difference, p1 – p2, between the proportions of US men and US women who sometimes order a dish without meat, fish, or fowl. Use 2-Prop-Z-Interval to obtain the interval -.1287 < p1 – p2 < -.031 c) Do the data provide sufficient evidence to conclude that, in the United States, the percentage of men who sometimes order a dish without meat, fish, or fowl is smaller than the percentage of women who do the same? Explain why or why not. The interval is completely below zero. This means that any possible value for p1 – p2 is negative, which implies than p1 is smaller than p2. With 90% confidence we can conclude that in the United States, the percentage of men who sometimes order a dish without meat, fish, or fowl is smaller than the percentage of women who do the same. d) We are _90___% confident that, in the United States, the percentage of women who sometimes order a vegetarian meal is___ larger than_________ (larger than, smaller than, may be equal to) the percentage of men who order a vegetarian meal by somewhere between ____3.1_____ and ___12.9______ percentage points 12