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2/5/13 • The One-Sample z Interval for a Population Mean To calculate a 95% confidence interval for µ , we use the familiar formula: estimate ± (critical value) • (standard deviation of statistic) 10.2 Estimating a Population Mean x ± z *⋅ σ 20 = 240.79 ± 1.96⋅ n 16 = 240.79 ± 9.8 = (230.99,250.59) Estimating a Population Mean In Section 10.1, we estimated by constructing a confidence interval using the sample mean. Example mean = 240.79,n=16, SD=20.! One-‐Sample z Interval for a Popula3on Mean Choose an SRS of size n from a population having unknown mean µ and known standard deviation σ. As long as the Normal and Independent conditions are met, a level C confidence interval for µ is x ± z* σ n The critical value z* is found from the standard Normal distribution. • When σ is Unknown: The t Distributions Estimating a Population Mean When the sampling distribution of x is close to Normal, we can find probabilities involving x by standardizing : x −µ z= σ n We must estimate the standard deviation using the sample standard deviations S. This is called the standard error. When we don’t know σ, we can estimate it using the sample standard deviation sx. What happens when we standardize? ?? = x −µ sx n This new sta3s3c does not have a Normal distribu3on! 1 2/5/13 When we standardize based on the sample standard deviation sx, our statistic has a new distribution called a t distribution. It has a different shape than the standard Normal curve: S1: Clear Y1, Y2, Y3 and turn off all stat plots. ü It is symmetric with a single peak at 0, ü However, it has much more area in the tails. S2: Define Y1 = normalpdf(x). Change graph style To thick line. S3: Set window to X[-‐3,3], Y[-‐0.1, 0.4] S4: Define Y2 = tpdf(X,2) Found in DISTR – the 2 refers to degree of freedom (df) S5: Turn off Y1. Define Y3 = tpdf(X,9). Change graph style to dashed Like any standardized statistic, t tells us how far x is from its mean µ in standard deviation units. S6: Define Y4 = tpdf(X,30) Describe the change in the graph as n increases. Activity 10C p644 • Compare Normal Distribu1on and t distribu1on. Estimating a Population Mean • When σ is Unknown: The t Distributions However, there is a different t distribution for each sample size, specified by its degrees of freedom (df). • Distribu3on Parameters The t-‐distribu1on has only one parameter: the number of degrees of freedom (n-‐1) The t-‐distribu1on is a distribu1on for the devia1on of a sample mean from its popula1on value. The sample mean does NOT have a t-‐ distribu1on. However the sample mean has a Normal distribu1on and the standardized sta1s1c “t” has a t-‐distribu1on. • The t Distributions; Degrees of Freedom When we perform inference about a population mean µ using a t distribution, the appropriate degrees of freedom are found by subtracting 1 from the sample size n, making df = n - 1. We will write the t distribution with n - 1 degrees of freedom as tn-1. The t Distributions; Degrees of Freedom Draw an SRS of size n from a large population that has a Normal distribution with mean µ and standard deviation σ. The statistic t= x −µ sx n Estimating a Population Mean • Recall that the Normal Distribu1on had two parameters that characterized its shape: mean (mu) and standard devia1on (sigma) has the t distribution with degrees of freedom df = n – 1. The statistic will have approximately a tn – 1 distribution as long as the sampling distribution is close to Normal. 2 2/5/13 Suppose you want to construct a 95% confidence interval for the mean µ of a Normal population based on an SRS of size n = 12. What critical t* should you use? ü The density curves of the t distributions are similar in shape to the standard Normal curve. ü The spread of the t distributions is a bit greater than that of the standard Normal distribution. ü The t distributions have more probability in the tails and less in the center than does the standard Normal. ü As the degrees of freedom increase, the t density curve approaches the standard Normal curve ever more closely. Upper-tail probability p df .05 .025 .02 .01 10 1.812 2.228 2.359 2.764 11 1.796 2.201 2.328 2.718 12 1.782 2.179 2.303 2.681 z* 1.645 1.960 2.054 2.326 90% 95% 96% 98% Confidence level C • DISTR (2nd Vars) – 4. invT area: .025 df: 11 t* = 2.200985143 • Constructing a Confidence Interval for µ When the conditions for inference are satisfied, the sampling distribution for x has roughly a Normal distribution. Because we don’t know σ , we estimate it by the sample standard deviation sx . Definition: s The standard error of the sample mean x is x , where sx is the n sample standard deviation. It describes how far x will be from µ, on average, in repeated SRSs of size n. To construct a confidence interval for µ, Estimating a Population Mean T-‐value from calculator • df = 12 – 1 = 11 • Area = (1 -‐ .95)/2 = .025 We move across that row to the entry that is directly above 95% confidence level. The desired cri3cal value is t * = 2.201. We can use Table B in the back of the book to determine critical values t* for t distributions with different degrees of freedom. • Find cri3cal value t* for sample size n=12 and 95% confidence interval In Table B, we consult the row corresponding to df = n – 1 = 11. Estimating a Population Mean • Using Table B to Find Critical t* Values When comparing the density curves of the standard Normal distribution and t distributions, several facts are apparent: Estimating a Population Mean • The t Distributions; Degrees of Freedom ü Replace the standard deviation of x by its standard error in the formula for the one - sample z interval for a population mean. ü Use critical values from the t distribution with n - 1 degrees of freedom in place of the z critical values. That is, statistic ± (critical value)⋅ (standard deviation of statistic) s = x ± t* x n 3 2/5/13 Except in the case of small samples, the condition that the data come from a random sample or randomized experiment is more important than the condition that the population distribution is Normal. Here are practical guidelines for the Normal condition when performing inference about a population mean. Condi3ons The One-‐Sample for Inference t Interval about for aa PPopula3on opula3on M Mean ean •Random: Choose an The SRSdata of size come n from fromaapopulation random sample havingofunknown size n from mean theµ. population A level C confidence of interest orinterval a randomized for µ is experiment.s x ± t* x • Normal: The population has a Normal distribution or the sample size is large n (n ≥ 30). where t* is the critical value for the tn – 1 distribution. • Independent: The method for calculating a confidence interval assumes that Use this interval only when: individual observations are independent. To keep the calculations accurate wheniswe sample replacement from(na ≥finite (1) reasonably the population distribution Normal orwithout the sample size is large 30), population, we should check the 10% condition: verify that the sample size (2) is thenopopulation is at least 10 times as large as the sample. more than 1/10 of the population size. William S. GosseR (1876-‐1937) worked for the Guinness brewery and his goal was to make beder beer. He needed sta1s1cal methods to develop cheap quality control using small samples of beer. He used his new t-‐procedures to find the best varie1es of barley and hops. When he was tes1ng the beer, he no1ced that when he standardized the values of xbar from his samples using the sample standard devia1on, he got values beyond +/= 3 standard devia1ons much more ohen than was expected. This led him to explore the distribu1on. He published under the name “Student”. • Sample size at least 15: The t procedures can be used except in the presence of outliers or strong skewness. • Large samples: The t procedures can be used even for clearly skewed distributions when the sample is large, roughly n ≥ 30. • As part of their final project for AP Stats, Chris1na and Rachel randomly selected 18 rolls of a generic brand of toilet paper to measure how well this brand could absorb water. To do this, they poured ¼ cup of water onto a hard surface and counted the number of squares it took to completely absorb the water. Here are the results from their 18 rolls: • -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ • 29 20 25 29 21 24 27 25 24 • 29 24 27 28 21 25 26 22 23 -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ Construct and interpret a 99% confidence interval for the mean of this problem. Example 1 • • • • • • Using One-‐Sample t Procedures: The Normal Condi3on • Sample size less than 15: Use t procedures if the data appear close to Normal (roughly symmetric, single peak, no outliers). If the data are clearly skewed or if outliers are present, do not use t. Estimating a Population Mean • Using t Procedures Wisely The one-sample t interval for a population mean is similar in both reasoning and computational detail to the one-sample z interval for a population proportion. As before, we have to verify three important conditions before we estimate a population mean. Rearrange to use - SIN Estimating a Population Mean • One-Sample t Interval for a Population Mean 4 2/5/13 • Step 1: Parameters • Find the mean number of squares of a generic brand of toilet paper needed to absorb ¼ cup of water using a 99% confidence interval. • Calculate 1-‐Var Stats for data • xbar = 24.94 sample standard devia1on = 2.86 • Step 2: Condi1ons • Standard devia+on unknown : t-‐distribu+on S – The problem states that 18 rolls were randomly selected I – Since we are sampling without replacement, we must check the 10% rule. It is reasonable to assume that there are moe than 10(18)= 180 rolls of generic toilet paper. N – Since the sample size is small ( greater than 15 but less than 30) and we are not told that the popula1on is Normally distributed, we must check the data. • Since there are no outliers or strong skewness it is reasonable to assume that the popula1on distribu1on is approximately normal. Step 3: Calculations df = 18 – 1 = 17 t* = 2.898 à invT(.005,17) Confidence interval = 24.94 +/- (2.898)(2.86/(18^.5)) = 24.94 +/- 1.95 (22.99, 26.89) Step 4: Interpretation We are 99% confident that the interval from 22.99 squares to 26.89 squares captures the true mean of squares of generic toilet paper needed to absorb ¼ cup of water. 5 2/5/13 Example 2 • The principal at a large high school claims that students spend at least 10 hours per week doing homework, on average. To inves1gate this claim, an AP Stats class selected a random sample of 250 students from their school and asked them how long they spent doing homework the last week. The sample mean was 10.2 hours and the sample standard devia1on was 4.2 hours. • Construct and interpret a 95% confidence interval for the mean 1me that students in this school spent doing homework in the last week. Based on your interval, what can you conclude about the principal’s claim? • Step 2: Condi1ons • Standard devia+on unknown : t-‐distribu+on S – The problem states that 250 students were randomly selected I – Since we are sampling without replacement, we must check the 10% rule. It is reasonable to assume that there are more than 10(250)= 2500 students in this large school. N – Since the sample size is large ( greater than 30), it is safe to use t-‐procedures. *Assume there are not outliers. • Environmentalists, government officials, and vehicle manufacturers are all interested in studying auto exhaust emissions produced by motor vehicles. The major pollutants in auto exhaust from gasoline engines are hydrocarbons, monoxide, and nitrogen oxides (NOX). Table 10.2 gives the NOX levels (g/m) for a random sample of light duty engines of the same type. See p 647 • Construct and interpret a 95% confidence interval for the amount of NOX emided by light-‐ duty engines of this type. Example 3 (10.9 p646) Step 3: Calculations df = 250 – 1 = 249 t* = 1.970 à invT(.025,249) Confidence interval = 10.2 +/- (1.970)(4.2/(250^.5)) = 10.2 +/- 0.5233 (9.6767, 10.7233) Step 4: Interpretation We are 95% confident that the interval from 9.6767 hours to 10.7233 hours captures the true mean of hours spent doing homework at this school last week. Since the plausible values for mu contain values less than 10, the interval does not provide convincing evidence to support the principal’s claim. • Step 1: Parameters • Find the mean number hours spent doing homework at this school last week using a 95% confidence interval. 6 2/5/13 • Step 1: Parameters • Find the mean amount of NOX emided by light-‐duty engines of this type using a 95% confidence interval. • n = 46 • Step 2: Condi1ons • Standard devia+on unknown : t-‐distribu+on S – The problem states that data was from a random sample (n=46). I – Since we are sampling without replacement, we must check the 10% rule. It is reasonable to assume that there are more than 10(46)= 460 engines of this type. N – Since the sample size is large ( greater than 30), it is safe to use t-‐procedures – but check for outliers. • Normality Plot – NOX Emmissions • This plot is somewhat linear but there I there is an apparent outlier. • Proceed with cau1on – examine • impact of outlier later. • Step 3 – Calcula1ons • Calculate 1-‐var stats • xbar = 1.329 g/m sample standard devia1on = 0.484 g/m df = 46 – 1 = 45 t* = 2.014 Confidence Interval = 1.329 +/-‐ (2.014)(0.484/(46^.5)) = 1.329 +/-‐ 0.1437 (1.1853, 1.4727) • Step 4 – Interpreta1on • We are 95% confident that the interval between 1.1853 g/m and 1.4727 g/m captures the true mean level of NOX emiRed by this type of light-‐duty engine. • **When using a table to determine the correct cri1cal value of t and the actual df does not appear in the table, use the greatest df available that is less than your desired df. 7 2/5/13 • Using t Procedures Wisely Definition: An inference procedure is called robust if the probability calculations involved in the procedure remain fairly accurate when a condition for using the procedures is violated.! Estimating a Population Mean The stated confidence level of a one-sample t interval for µ is exactly correct when the population distribution is exactly Normal. No population of real data is exactly Normal. The usefulness of the t procedures in practice therefore depends on how strongly they are affected by lack of Normality. Fortunately, the t procedures are quite robust against non-Normality of the population except when outliers or strong skewness are present. Larger samples improve the accuracy of critical values from the t distributions when the population is not Normal. Comparative studies are more convincing than single sample investigations. For this reason, one-sample inference is less common than comparative inference. Reminder: When we have the results of two samples, the first task is to understand whether the samples are indepenndent or dependent. A sampling method is independent when the individuals selected for one sample do not in any way dictate which individuals are to be selected for the second. Otherwise, the samples are dependent. Dependent samples are often referred to as matched pairs. e.g. A researcher might wonder if people who marry tend to have similar IQs. 36 married couples are selected and the IQs are measured. The researcher then tests whether the IQ difference between the male sample and the female sample is significantly different from 0. In this example, the samples of men and women are dependent, since the choice of one sample dictates the choice of the second sample. 8 2/5/13 e.g. A business owner wonders whether female customers tend to make larger purchases than males. She randomly selects 40 receipts taken from female customers during the last week and 40 from male customers and compares the sample means. In this example, the two samples are independent, since membership in one sample has no influence on membership in the other Matched Pairs can also be one person receiving 2 different treatments. Independent Samples vs. Paired Samples.doc The parameter mu in a paired t procedure is • The mean difference in the responses to the 2 treatments within the matched pairs of subjects in the entire population, or • The mean difference in response to the 2 treatents for individuals in the population (when the same subjet receives both treatments) or • The mean difference between before-and-after measurements for all individuals in the population Example 5 • Data is from Beck Depression Inventory – higher scores show more symptoms of depression. “Beats” is the number of beats per minute the subject achieved when asked to press a budon 200 1mes as quickly as possible. We are interested in whether being deprived of caffeine affects these outcomes. Construct and interpret at 90% confidence interval for the mean change in depressions score. Is caffeine dependence real? • Our subjects are 11 people diagnosed as being dependent on caffeine. Each subject was barred from coffee, colas, and other substances containing caffeine. Instead, they took capsules containing their normal caffeine intake. During a different 1me period, they took placebo capsules. The order in which subjects took caffeine and the placebo was randomized. • Data can be found in Table 10.3 p652 • Step 1 – Parameter The popula1on of interest is all people who are dependent on caffeine. We want to es1mate the mean difference = mean(placebo – mean) in depression score that would be reported if all individuals in the popula1on took both the caffeine capsule and the placebo. Step 2 -‐ Condi1ons Popula1on standard devia1on is unknown: t-‐distribu1on S – The problem does not say that this is an SRS, so they are probably volunteers. They may not be representa1ve of the popula1on, which might make it difficult to generalize for the popula1on. However, since placebo or caffeine capsules are randomly assigned, any consistent differences should be due to the treatments. 9 2/5/13 I – It seems reasonable that the difference (placebo – caffeine) values for the 11 subjects are independent based on this experiment. N – The problem does not state that the popula1on distribu1on is normal and the sample size is small. n = 11. We need to check the data. There are no obvious Outliers or strong Skewness. Step 3 – Calculations Create lists for both caffeine and placebo. Create an Additional list which is Placebo – Caffeine Calculate 1-var stats for (Placebo – Caffeine) list xbar(placebo – caffeine) = 7.36 s(placebo – caffeine) = 6.918 df = 11 – 1 = 10 t* = 1.812 [invT(.05,10)] Conf Int = 7.36 +/- (1.812)(6.918/11^.5) = 7.36 +/- 3.78 (3.58, 11.14) Step 4 – Interpretation We are 90% confident that the interval between 3.58 and 11.14 points captures the true mean difference in depression score for the population. That is, we estimate that caffeine-dependent individuals would score, on average, between 3.58 and 11.14 points higher on the Beck Depression Inventory when they are given a placebo instead of caffeine. This study provides evidence that withholding caffeine from caffeine-dependent individuals may lead to depression. However, the fact that the subjects in this study were not an SRS prevents us from generalizing further. 10 2/5/13 t-procedures are NOT robust against outliers. Percent of residents aged 65 and over in the US Can we use t-procedure to estimate mean age over 65?? Can you use inference to estimate mean time of day for 1st lightening strike? Can you use inference to find the mean number of word length in Shakespeare’s plays. 11 2/5/13 T-interval on Calculator: STAT – TESTS – 8.Tinterval – DATA – choose List for data – choose C-Level – Calculate 12
