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19-25 (not including 21, 22).pdf

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Susan Kamp

Where are we? • Ch. 2–9: Data set and its distribution, statistics, Normal model. • Ch. 11–13: Collecting data, random samples, randomized experiments. • Ch. 14–16: Probability, random variables, probability distribution models. • Ch. 18–28: Statistical inference: what does sample data tell us about the underlying population. Inferences about parameters (proportions, means, etc.) in a model for the population distribution. Quick Review of Ch 18: The Sampling Distribution of a Sample Proportion Suppose we are just interested in one characteristic occurred in the population of interest. For convenience, we will call the outcome we are looking for “Success”. The population proportion, p, is obtained by taking the ratio of the number of successes in a population to the total number of elements in the population. 1 of 79 1)  p is the mean of the sampling distribution of pˆequals p:  p p   p  Notation: pˆ is also denoted as 2) The standard deviation of the sampling distribution of p ,  p, is: p(1 p)  p n Notation:  p is also denoted as SD p  . 3) When n is large and p is not too close to 0 or 1, the sampling distribution of pˆ is approximately normal. Rule of thumb: The sample size is considered to be sufficiently large if np ≥ 10 and n(1 – p) ≥ 10 The Sampling Distribution of a Sample Mean If y is the sample average of an SRS of size n drawn from a population with mean μ and standard deviation σ, then: 1. The population mean of y , y , is equal to μ.  y  2 of 79   y NOTATION: textbook denoted y as 2. The population standard deviation of y, denoted  y, is    y n , NOTATION: textbook denoted y as SD y  3. If random samples of n observations are drawn from any normal distributed population with mean μ and standard deviation σ, then the sampling distribution of the mean y is  normal distributed, with mean μ and standard deviation . n 4. CLT: If random samples of n observations are drawn from any population with mean μ and standard deviation σ, then for large n (ie. n ≥ 30), the sampling distribution of the mean y is normal distributed, with mean μ and standard deviation  . n Ch 19 Confidence Intervals for Proportions We will be starting now to cover inferential statistics! Its objective is to use sample data to obtain results about the whole population. 3 of 79 In a first step, the goal is to describe an underlying population. Since the populations are described in form of models that are characterized by parameters (mean μ and standard deviation σ or probability p for the event of interest). At this time we will estimate those characteristics. There are two different approaches for estimating: Point Estimation and Interval Estimation. For Point Estimation, you give one value for a characteristic, which is hopefully close to the true unknown value. For Interval Estimation, you give an interval of likely values, where the width of the interval will depend on the confidence you require to have in this interval. Since we base our statement just on a sample, we see later how to give a measure of accuracy or confidence for the estimate. Point Estimation A point estimate of a possible characteristic is a single number that is based on sample data and represents the population parameter. 4 of 79 Example: - The sample mean y is a point estimate for the population mean μ. - The sample proportion p is a point estimate of p the population probability for Success. A point estimate gives a single value that is supposed to be close to the true value of the characteristic but it does NOT tell how close the estimate is. Considering we know that we would observe different values of a point estimate from sample to sample, point estimates are not enough to describe a parameter. Thus, we introduce the second type of estimate – interval estimate. Confidence Interval As an alternative to point estimation we can report not just a single value for the population characteristic, but an entire interval of reasonable values based on sample data. These intervals take into account of error and uncertainty. We often 5 of 79 associate interval estimate with some level of confidence and the result is called a confidence interval.  Recall: Both of the sampling distributions for proportions and means are Normal. p(1 p)  For proportions :  p n  For means:    y n  When we don’t know p or σ, we’re stuck, right?  No, we will use sample statistics to estimate these population parameters.  Whenever we estimate the standard deviation of a sampling distribution, we call it a standard error.  The standard error for a sample proportion: SE(p)  p(1 p) n s  The standard error for the sample mean: SE y  n 6 of 79 By the 68-95-99.7% Rule, we know - about 68% of all samples will have p ’s within 1 SE of p - about 95% of all samples will have p ’s within 2 SEs of p pˆ - about 99.7% of all samples will have ’s within 3 SEs of p p  We can look at this from ’s point of view… Consider the C = 95% level: - There’s a 95% chance that p is no more than 2 SEs away pˆ from . - So, if we reach out 2 SEs, we are 95% sure that p will be in that interval. In other words, if we reach out 2 SEs in either direction of pˆ, we can be 95% confident that this interval contains the true proportion.  This is called a 95% confidence interval. What does “95% Confidence” really mean?  Being "95% confident" means, if you were to construct 100 95% confidence intervals from 100 different samples. Of the 100 intervals, you expect 95 to capture the true mean, and 5 not to capture the mean. 7 of 79  How can this happen?  Each confidence interval uses a sample statistic to estimate a population parameter, but since samples vary, the statistics we use, and thus the confidence intervals we construct, vary as well.  In conclusion, you cannot be sure that a specific confidence interval captures the true proportion p.  Our confidence is in the process of constructing the interval, not in any one interval itself.  The following figure shows that some of our confidence intervals (from 20 random samples) capture the true proportion (the horizontal line), while others do not: 8 of 79 Margin of Error - Most confidence intervals are of the form: Point estimate ± margin of error = point estimate ± critical value × SE(estimate) - The more confident we want to be, the larger our margin of error needs to be (makes the interval wider). - We need more values in our confidence interval to be more certain. - Because of this, every confidence interval is a balance between certainty and precision. - The tension between certainty and precision is always there. - Fortunately, in most cases we can be both sufficiently certain and sufficiently precise to make useful statements. - The most commonly chosen confidence levels (C) are 90%, 95%, and 99% (but any percentage can be used). Critical Values - The critical value is how far we need to deviate from the estimate to capture the central 100C% of the values on the sample distribution. 9 of 79 - The ‘2’ in p  2SE(p) (our 95% confidence interval) came from the 68-95-99.7% Rule. - Using a table or technology, we find that a more exact value for our 95% confidence interval is 1.96 instead of 2. - We call 1.96 the critical value and denote it z*. Example: To find the central 95% region on a standard normal curve, you need to cut off 2.5% at each end. The z* value for C = 0.95 has 97.5% of the area to the left. Using z-table, we find z* = 1.96. - For any confidence level, we can find the corresponding critical value. - Commonly used critical values Confidence Coefficient C 1 – C (1 – C)/2 z* 0.90 0.1 0.05 1.645 0.95 0.05 0.025 1.96 0.99 0.01 0.005 2.58 Example: Show that for a 90% confidence interval, the critical value is 1.645. 10 of 79 Example: Consider flipping an unbiased coin 1000 times. The results showed that you flipped 400 heads. Based on this result, what interval captures the most likely 95% of the values of the actual proportion of heads? Then check if the coin is fair. A 100C% Large Sample Confidence Interval for a Population Proportion p. Assumption:  Here are the assumptions and the corresponding conditions you must check before creating a confidence interval for a proportion: 1) Independence Assumption: You cannot check this by looking at the data. Instead, we check two conditions to decide whether independence is reasonable.  Randomization Condition: Were the data sampled at random or generated from a properly randomized experiment? Proper randomization can help ensure independence.  10% Condition: Is the sample size no more than 10% of the population? 11 of 79 2) Sample Size Assumption: The sample needs to be large enough for us to be able to use the CLT. - Success/Failure Condition: We must expect at least 10 “successes” and at least 10 “failures” - When the conditions are met, we are ready to find the confidence interval for the population proportion, p. - The confidence interval is p(1 p) p  z*SE(p)  p  z* n - The critical value, z*, depends on the particular confidence level, C, that you specify. Example (cont): To answer this question, we will calculate a 95% confidence interval from this data and check if 0.5 (the probability for HEAD, when tossing an unbiased coin) is in the confidence interval. 12 of 79 NOTE: Do NOT say: - There is a probability of 0.95 that p is between 0.37 and 0.43. - Between 37% and 43% of all tosses appear as HEADs. - 95% of all random samples of coin tosses will show between 37% and 43% of HEADs. - We can be 95% confident that all random samples will show 40% of HEADs. - There is a 95% chance that the true proportion of HEADs is between 37% and 43%. 13 of 79 Example: For a project, a student randomly sampled 182 other students at a large university to determine if the majority of students were in favor of a proposal to build a field house. He found that 75 were in favor of the proposal. Find the 95% confidence interval for p. Find the 99% confidence interval for p. Remark: In order to have a higher confidence, we need to accept a larger margin of error, ie. a wider interval. 14 of 79 A Confidence Interval for Small Samples  When the Success/Failure Condition fails, all is not lost.  A simple adjustment to the calculation lets us make a confidence interval anyway.  All we do is add four observations, two successes and two failures. y  So instead of p  , we use the adjusted proportion n y 2 p  n4 ~ p(1 p)  Now the adjusted interval is p  z* n  The adjusted form gives better performance overall and works much better for proportions of 0 or 1. Choosing the sample size p(1 p) Recall: the margin of error in the CI for p is: ME  z* n We may like to choose the sample size n to achieve a certain margin of error, so we solve for n: 2 n   z*  p(1 p) ME  15 of 79 - p is based on a pilot study or on past experience, but we may not have prior sampling done! o use p = 0.50. This is conservative as it gives a margin of error bigger than the true margin of error. Example: If a TV executive would like to find a 95% confidence interval estimate within 0.03 for the proportion of all households that watch NYPD Blue regularly. How large a sample is needed if a prior estimate for p was 0.15? Example (Revisited): Suppose a TV executive would like to find a 95% confidence interval estimate within 0.03 for the proportion of all households that watch NYPD Blue regularly. How large a sample is needed p if we have no reasonable prior estimate for ? 16 of 79 Example (Please try it on your own): To conduct a political poll that is 99% sure of finding the level of support for the Conservative party to within 0.01 of margin of error, how large a sample would we need?  z* 2 2.576 2 n    p*(1 p*)    0.5(10.5) 16589.44 ME   0.01  Thus, to be 99% confidence of finding the level of support for the Conservative party to within 0.01, we need 16590 samples. 17 of 79 Ch 23 Inference About Mean Now that we know how to create confidence intervals and test hypotheses about proportions, it’d be nice to be able to do the same for means. Just as we did before, we will base both our confidence interval and our hypothesis test on the sampling distribution model. Recall: If we use the statistic for estimating the population mean μ, we can use the following information from the CLT in order to obtain a confidence interval for μ.   y  ,   y standard deviation of y, n s The standard error of y isSE(y)  AND n If the population distribution is originally normal, then the sampling distribution is also normal OR If the population distribution is non normal, but it has n ≥ 30, then we can assume that the sampling distribution of is approximately normal. 18 of 79 Gosset’s t Until now, all statistical tools that were introduced were based on the assumption that population standard deviation  is known. In practice, this assumption is very artificial and is never fulfilled in any real live situation. All procedures introduced until now are based on the normal distribution, which requires the population standard deviation . In most situations,  is unknown and has to be replaced by the sample standard deviation s, it causes variability in the result. In order to calculate a confidence interval, we need to fix the problem of variability by introducing another distribution called the Student’s t-distribution. The t-distribution only depends on one parameter, which is called the degrees of freedom (df). Properties of the t-distribution: - its density curves look quite similar to the standard normal curve. They are symmetric about 0, single-peaked, and bell-shaped. - The spread of the t-distributions is a bit larger than that of the standard normal curve. (As we are now using an 19 of 79 estimate for the population standard deviation, we must accept slightly more error in our estimation.) - As degrees of freedom (d.f.) gets bigger, the t-density curve gets closer to the standard normal density curve. (NOTE: Table t) In another words, as degrees of freedom increases, the spread of the corresponding t density curve decreases. - In fact, the t-model with infinite df is exactly normal. Remark: The structure of the table is different than the table for the standard normal distribution. - It is giving you the upper tail probabilities! - the probabilities are the label of the columns instead of being inside the table. 20 of 79 Example: Find t* (the critical value). a) The t-distribution with 5 df has probability 0.05 to the right of t*. b) The t-distribution with 5 df and confidence level of 90%. c) The one sample t statistic from an SRS with 20 observations has a probability of 0.9 to the left of t*. A Confidence Interval for a Population Mean  (when σ is unknown) Assumptions for using the t-statistics: Independence Assumption: - Independence Assumption. The data values should be independent. - Randomization Condition: The data arise from a random sample or suitably randomized experiment. Randomly sampled data (particularly from an SRS) are ideal. - 10% Condition: When a sample is drawn without replacement, the sample should be no more than 10% of the population. 21 of 79 Normal Population Assumption: - We can never be certain that the data are from a population that follows a Normal model, but we can check the Nearly Normal Condition: The data come from a distribution that is unimodal and symmetric. - Check by making a histogram or Normal probability plot. Nearly Normal Condition: - The smaller the sample size (n < 15 or so), the more closely the data should follow a Normal model. - For moderate sample sizes (15 ≤ n ≤ 40 or so), the t works well as long as the data are unimodal and reasonably symmetric. - For large sample sizes (n > 40 or 50), the t methods are safe to use unless the data are extremely skewed. A confidence interval for the population mean  (when σ is unknown) is given by s y t* n 22 of 79 where t* is the critical value for the t distribution with df = n – 1 1C confidence level C. In other words, t* is the upper 2 critical value for the t(n – 1) distribution. NOTE:  When Gosset corrected the model for the extra uncertainty, the margin of error got bigger.  Your confidence intervals will be just a bit wider and your P-values just a bit larger than they were with the Normal model.  By using the t-model, you’ve compensated for the extra variability in precisely the right way. Example: (Using the battery lifetime example from Ch3) We have a random sample of n = 4 observations on y = battery lifetime (hrs): 5.9, 7.3, 6.6, 5.7 NOTE: y = 6.375, s = 0.7274 (calculated in Ch3) Find the 95% confidence interval for the mean battery lifetime. 23 of 79 Example: A scientist interested in monitoring chemical contaminants in food, and thereby the accumulation of contaminants in human diets, selected a random sample of n = 50 male adults. It was found that the average daily intake of dairy products was y = 756grams with a standard deviation of s = 35grams. Find a 95% confidence interval for the mean daily intake of dairy products for men. 24 of 79 Example: IQ test scores The SRS IQ test scores of 31 girls in Region A as follows: 113 102 105 … 95 This has a sample mean y 105.84 and a sample standard deviation of s = 15. The shape of the population distribution is unimodel and relatively symmetric. a) Give a 99% confidence interval for the true mean IQ  of all girls in the district.  b) Give a 90% confidence interval for the true mean IQ of all girls in the district. 25 of 79 c) If the sample mean of IQ test scores of 20 girls in Region A is 105.84, give a 90% confidence interval for the true  mean IQ of all girls in the district. Remark: Margin of error m  t* s gets smaller when n - t* gets smaller, which is the same as smaller (1 – α). To obtain a smaller margin of error, you must accept lower confidence. - n gets larger. Increasing the sample size gives more accuracy. -  gets smaller. The less inherent variation in the population you are studying, the more accurate your estimate will be. NOTE: we can control t* and n, but we cannot control  . 26 of 79 Example: (Please try it on your own) Bank Mean Number of Phone Calls/hour A 15.6 B 11.9 C 11.7 Suppose that each sample mean was based on an SRS of n = 50 working hours and that s = 5 is known. a) Compute a 95% CI for  A the true mean number of phone calls per hour to Bank A. With C = 95% and df = 49 (round down to 45), t* = 2.014. s 5 y t* 15.6 2.014 15.61.424 n 50 We are 95% confident that A is between 14.176 and 17.024 phone calls per hour. b) Compute a 95% for the other 2 banks. 95% CI for  B: s 5 y t* 11.9 2.014 11.91.424  (10.476,13.324) n 50 95% CI for  C: y t* s 11.7  2.0145 11.7 1.424  (10.276,13.124) n 50 27 of 79 c) A survey claims that Bank A receives more phone calls than the other Banks. Based on the confidence intervals from parts (a) and (b), do you agree? The yvalue for Bank A is so large that its confidence interval lies entirely to the right of all other CIs. Even taking random variation into account, the number of phone calls received is clearly larger than other Banks. Example: A researcher found that a 98% confidence interval for the mean hours per week spent studying by college students was (13, 17). Which is true? a) There is a 98% chance that the mean hours per week spent studying by college students is between 13 and 17 hours b)We are 98% confident that the mean hours per week spent studying by college students is between 13 and 17 hours c) Students average between 13 and 17 hours per week studying on 98% of the weeks d)98% of all students spend between 13 and 17 hours studying per week. 28 of 79 Ch 23 Inference about the Mean Previously, population parameters were described, now we will be checking if claims about the population parameters are true, or plausible to a given degree. Example: A company is advertising that the mean lifetime of their light bulbs is 1000 hours with standard deviation of 5 hours. A person suspects the mean lifetime of the light bulbs is less than 1000 hours (company is lying in their advertisement), so he picks a sample of 100 light bulbs and find the average lifetime of these 100 light bulbs is y  998 . Based on this result, can we state that: i) the mean lifetime of this company’s light bulb, on average, is less than 1000 hours (so this company is lying in their advertisement)? OR ii) the difference between 1000 hours (the average lifetime for the population) and 998 hours (the average lifetime for the sample) may have occurred because of sampling variability? 29 of 79 A hypothesis test is a method for using sample statistics to decide between two competing claims on hypotheses about a population parameter. It follows the following procedure: 1) Define the variable, the parameter(s) of interest, and any relevant assumptions. 2) State the null hypothesis H and alternative hypothesis H . 0 a 3) Gather the evidence (sample). Based on the data in the sample, we will calculate a test statistic. 4) Assess the strength of the evidence against the null hypothesis in favor of the alternative. This will be done by finding p-value. 5) Make a decision based on Step 4. 6) State the conclusion. State the hypotheses: The null hypothesis H is 0 claim about a population parameter that is assumed to be true until it is declared false. It is generally the hypothesis of “no effect.”  We usually write down the null hypothesis in the form H : 0 parameter = hypothesized value. 30 of 79 The alternative hypothesis H is aaclaim about a population parameter that will be true ONLY when we reject the null hypothesis. In another words, this is the hypothesis that we are trying to find evidence for. Remark: Common choices of hypotheses are: - Two-tailed Test: o H :0population characteristic = specific value versus o H :apopulation characteristic ≠ specific value - Upper-tailed Test: o H :0population characteristic = specific value versus o H :apopulation characteristic > specific value - Lower-tailed Test: o H : population characteristic = specific value versus 0 o H : population characteristic < specific value a Examples: - H : μ = 100 versus H : μ < 100 0 a - H : p = 0.25 versus H : p ≠ 0.25 0 a - We cannot test H : μ0= 100 versus H : μ >a150 - We cannot test H : 0 y 100 versus H :a y 100 31 of 79 Example: You are considering moving to Richmond Hill, and are concerned about the average one-way commute time to downtown Toronto. Does the average one-way commute time exceed 25 minutes? You take a random sample of 50 Richmond Hill residents and find an average commute time of 29 minutes with a standard deviation of 7 minutes. Which set of hypotheses should you test? A) H 0 μ = 25 vs H A μ > 25 B) H 0 μ = 25 vs H A μ < 25 C) H : μ = 29 vs H : μ > 29 0 A D) H 0 μ = 25 vs H A μ ≠ 25 Example: You want to see if the number of minutes cell phone users use each month has changed from its mean of 120 minutes 2 years ago. You take a random sample of 100 cell phone users and find an average of 135 minutes used. Which set of hypotheses should you test? A) H : μ = 120 vs H : μ > 120 0 A B) H 0 μ = 120 vs H A μ ≠ 120 C) H 0 μ = 120 vs H A μ < 120 D) H 0 μ = 135 vs H A μ ≠ 135 32 of 79 Example: According to a June 2004 Gallup poll, 28% of Americans “said there have been times in the last year when they haven’t been able to afford medical care.” Is this proportion higher for black Americans than for all Americans? In a random sample of 801 black Americans, 38% reported that there had been times in the last year when they had not been able to afford medical care. Which type of hypothesis test would you use? A. One-tail upper tail B. One-tail lower tail C. Two-tail D. Both A and B Example: A statistics professor wants to see if more than 80% of her students enjoyed taking her class. At the end of the term, she takes a random sample of students from her large class and asks, in an anonymous survey, if the students enjoyed taking her class. Which set of hypotheses should she test? A. H 0 p < 0.80 H A p > 0.80 B. H 0 p = 0.80 H :Ap > 0.80 33 of 79 C. H 0 p > 0.80 H : A = 0.80 D. H 0 p = 0.80 H : A < 0.80 Example: An online catalog company wants on-time delivery for 90% of the orders they ship. They have been shipping orders via UPS and FedEx but will switch to a new, cheaper delivery service (ShipFast) unless there is evidence that this service cannot meet the 90% on-time goal. As a test the company sends a random sample of orders via ShipFast, and then makes follow-up phone calls to see if these orders arrived on time. Which hypotheses should they test? A. H : p < 0.90 H : p > 0.90 0 A B. H 0 p = 0.90 H : A > 0.90 C. H 0 p > 0.90 H : A = 0.90 D. H 0 p = 0.90 H : A < 0.90 Testing H v0. H : a - H 0ill be rejected only if the sample evidence strongly suggests that H i0 false. - Otherwise H will not be rejected. 0 34 of 79 So there are two possible conclusions: - reject H (0ccept H ) a - do not reject H (W0en H is not0being rejected, it doesn't mean strong support for H , b0t lack of strong evidence for H a) Note: these decisions are not symmetric, there is NO way you can say you accept H . 0 Idea: Compare the process to a criminal trial. The fact is that a person accused of a crime is either guilty or not guilty.  To prove someone is guilty, we start by assuming they are innocent.  H :0innocent  We retain that hypothesis until the facts make it unlikely beyond a reasonable doubt.  H :aguilty Rejection of H : 0 Nonrejection of H : 0 35 of 79 Example (con’t): A company is advertising that the average lifetime of their light bulbs is 1000 hours with standard deviation of 5 hours. You might question this, and want to show that in fact the lifetime is shorter. State the hypothesis. Interpret rejection and nonrejection of H for this example. 0 You would test: Rejection of H : 0 Nonrejection of H : 0 How to make the decision (reject H or do 0ot reject H ) 0 The decision to reject, or not to reject H i0 based on information contained in a sample drawn from the population of interest. Use the sample to: 36 of 79 - Calculate a test statistic (a number that measures how many standard deviations away the estimate in the sample is from the hypothesized value of the parameter in H ),0 - p-value: o Use the value of the test statistic and its distribution to calculate the p-value (the probability of observing the value of the test statistic as extreme or more extreme than the one observed, if H 0s true). In other words, we try to find out how likely the observed results could have happened if the null hypothesis were true. In general: - When the data are consistent with the model from the null hypothesis, the P-value is high and we fail to reject the null hypothesis. - If the P-value is low enough, we’ll “reject the null hypothesis,” since what we observed would be very unlikely were the null model true.  H A parameter ≠ value (a two-sided alternative)  we are equally interested in deviations on either side of the null hypothesis value. 37 of 79  For two-sided alternatives, the P-value is the probability of deviating in either direction from the null hypothesis value.  The other two alternative hypotheses are called one-sided alternatives.  A one-sided alternative focuses on deviations from the null hypothesis value in only one direction.  Thus, the P-value for one-sided alternatives is the probability of deviating only in the direction of the alternative away from the null hypothesis value. P-Values and Decisions:  How small should the P-value be in order for you to reject the null hypothesis?  It turns out that our decision criterion is context-dependent.  When we’re screening for a disease and want to be sure we treat all those who are sick, we may be willing to reject the null hypothesis of no disease with a fairly large P-value. 38 of 79  A longstanding hypothesis, believed by many to be true, needs stronger evidence (and a correspondingly small P-value) to reject it.  Your conclusion about any null hypothesis should be accompanied by the P-value of the test.  If possible, it should also include a confidence interval for the parameter of interest.  Don’t just declare the null hypothesis rejected or not rejected.  Report the P-value to show the strength of the evidence against the hypothesis.  This will let each reader decide whether or not to reje
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