Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Transcript

Sampling Distributions •Numerical descriptive measures calculated from the sample are called statistics. •Statistics vary from sample to sample and hence are random variables. •The probability distributions for statistics are called sampling distributions. •In repeated sampling, they tell us what values of the statistics can occur and how often each value occurs. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Sampling Distributions Definition: The sampling distribution of a statistic is the probability distribution for the possible values of the statistic that results when random samples of size n are repeatedly drawn from the population. Population: 3, 5, 2, 1 Draw samples of size n = 3 without replacement p(x) x Possible samples 10 / 3 3.33 3, 5, 2 3, 5, 1 9/3 3 3, 2, 1 6/3 2 5, 2, 1 8 / 3 2.67 Each value of x-bar is equally likely, with probability 1/4 1/4 2 3 x Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Sampling Distributions Sampling distributions for statistics can be Approximated with simulation techniques Derived using mathematical theorems The Central Limit Theorem is one such theorem. Central Limit Theorem: If random samples of n observations are drawn from a nonnormal population with finite m and standard deviation s , then, when n is large, the sampling distribution of the sample mean x is approximately normally distributed, with mean m and standard deviation s / n. The approximation becomes more accurate as n becomes large. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Applet Toss a fair coin n = 1 time. The distribution of x the number on the upper face is flat or uniform. m xp( x) 1 1 1 1( ) 2( ) ... 6( ) 3.5 6 6 6 s ( x m ) 2 p( x) 1.71 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Applet Toss a fair coin n = 2 time. The distribution of x the average number on the two upper faces is moundshaped. Mean : m 3.5 Std Dev : s/ 2 1.71 / 2 1.21 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Applet Toss a fair coin n = 3 time. The distribution of x the average number on the two upper faces is approximately normal. Mean : m 3.5 Std Dev : s/ 3 1.71 / 3 .987 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Why is this Important? The Central Limit Theorem also implies that the sum of n measurements is approximately normal with mean nm and standard deviation s n . Many statistics that are used for statistical inference are sums or averages of sample measurements. When n is large, these statistics will have approximately normal distributions. This will allow us to describe their behavior and evaluate the reliability of our inferences. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. How Large is Large? If the sample is normal, then the sampling distribution of x will also be normal, no matter what the sample size. When the sample population is approximately symmetric, the distribution becomes approximately normal for relatively small values of n. When the sample population is skewed, the sample size must be at least 30 before the sampling distribution of x becomes approximately normal. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Sampling Distribution of the Sample Mean A random sample of size n is selected from a population with mean m and standard deviation s. The sampling distribution of the sample mean have mean m and standard deviation s / n . x will If the original population is normal, the sampling distribution will be normal for any sample size. If the original population is nonnormal, the sampling distribution will be normal when n is large. The standard deviation of x-bar is sometimes called the STANDARD ERROR (SE). Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Finding Probabilities for the Sample Mean If the sampling distribution of x is normal or approximately normal, standardize or rescale the interval of interest in terms of xm z s/ n Find the appropriate area using Table 3. Example: A random sample of size n = 16 from a normal distribution with m = 10 and s = 8. 12 10 P( x 12) P( z ) 8 / 16 P( z 1) 1 .8413 .1587 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Applet Example A soda filling machine is supposed to fill cans of soda with 12 fluid ounces. Suppose that the fills are actually normally distributed with a mean of 12.1 oz and a standard deviation of .2 oz. What is the probability that the average fill for a 6-pack of soda is less than 12 oz? P (x 12) x m 12 12.1 P( ) s / n .2 / 6 P( z 1.22) .1112 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Sampling Distribution of the Sample Proportion The Central Limit Theorem can be used to conclude that the binomial random variable x is approximately normal when n is large, with mean np and standard deviation . x The sample proportion, pˆ n is simply a rescaling of the binomial random variable x, dividing it by n. From the Central Limit Theorem, the sampling distribution of p̂ will also be approximately normal, with a rescaled mean and standard deviation. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Sampling Distribution of the Sample Proportion A random sample of size n is selected from a binomial population with parameter p. The sampling distribution of the sample proportion, x pˆ n will have mean p and standard deviation pq n If n is large, and p is not too close to zero or one, the sampling distribution of p̂ will be approximately normal. The standard deviation of p-hat is sometimes called the STANDARD ERROR (SE)Copyright of p-hat. ©2003 Brooks/Cole A division of Thomson Learning, Inc. Finding Probabilities for the Sample Proportion If the sampling distribution of p̂ is normal or approximately normal, standardize or rescale the interval of interest in terms of z pˆ p pq n Find the appropriate area using Table 3. .5 .4 Example: A random P( pˆ .5) P( z ) .4(.6) sample of size n = 100 100 from a binomial population with p = .4. P( z 2.04) 1 .9793 .0207 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example The soda bottler in the previous example claims that only 5% of the soda cans are underfilled. A quality control technician randomly samples 200 cans of soda. What is the probability that more than 10% of the cans are underfilled? n = 200 S: underfilled can p = P(S) = .05 q = .95 np = 10 nq = 190 OK to use the normal approximation P( pˆ .10) .10 .05 P( z ) P( z 3.24) .05(.95) 200 1 .9994 .0006 This would be very unusual, if indeed p =©2003 .05! Brooks/Cole Copyright A division of Thomson Learning, Inc. Types of Inference • Estimation: – Estimating or predicting the value of the parameter – “What is (are) the most likely values of m or p?” • Hypothesis Testing: – Deciding about the value of a parameter based on some preconceived idea. – “Did the sample come from a population with m 5 or p = .2?” Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Types of Inference • Examples: – A consumer wants to estimate the average price of similar homes in her city before putting her home on the market. Estimation: Estimate m, the average home price. –A manufacturer wants to know if a new type of steel is more resistant to high temperatures than an old type was. Hypothesis test: Is the new average resistance, mN equal to the old average resistance, mO? Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Definitions • An estimator is a rule, usually a formula, that tells you how to calculate the estimate based on the sample. – Point estimation: A single number is calculated to estimate the parameter. – Interval estimation: Two numbers are calculated to create an interval within which the parameter is expected to lie. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Measuring the Goodness of an Estimator • The distance between an estimate and the true value of the parameter is the error of The distance between the bullet and estimation. the bull’s-eye. • In this chapter, the sample sizes are large, so that our unbiased estimators will have normal distributions. Because of the Central Limit Theorem. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Margin of Error • For unbiased estimators with normal sampling distributions, 95% of all point estimates will lie within 1.96 standard deviations of the parameter of interest. •Margin of error: The maximum error of estimation, calculated as 1.96 std error of the estimator Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Estimating Means and Proportions •For a quantitative population, Point estimatorof populationmean μ : x M argin of error (n 30) : 1.96 s n •For a binomial population, Point estimatorof populationproportionp : pˆ x/n pˆ qˆ Margin of error (n 30) : 1.96 n Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example • A homeowner randomly samples 64 homes similar to her own and finds that the average selling price is $252,000 with a standard deviation of $15,000. Estimate the average selling price for all similar homes in the city. Point estimatorof μ : x 250,000 s 15,000 Margin of error : 1.96 1.96 3675 n 64 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example A quality control technician wants to estimate the proportion of soda cans that are underfilled. He randomly samples 200 cans of soda and finds 10 underfilled cans. n 200 p proportionof underfilled cans Point estimatorof p : pˆ x/n 10 / 200 .05 pˆ qˆ (.05)(.95) Margin of error : 1.96 1.96 .03 n 200 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Interval Estimation • Create an interval (a, b) so that you are fairly sure that the parameter lies between these two values. • “Fairly sure” is means “with high probability”, measured using the confidence coefficient, 1a. Usually, 1-a .90, .95, .98, .99 • Suppose 1-a = .95 and that the estimator has a normal distribution. Parameter 1.96SE Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Confidence Intervals for Means and Proportions •For a quantitative population, Confidence interval for a population mean μ : x za / 2 s n •For a binomial population, Confidence interval for a population proportion p : pˆ za / 2 pˆ qˆ n Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example • A random sample of n = 50 males showed a mean average daily intake of dairy products equal to 756 grams with a standard deviation of 35 grams. Find a 95% confidence interval for the population average m. s 35 x 1.96 756 1.96 7 56 9 . 70 n 50 or 746.30 m 765.70 grams. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example • Find a 99% confidence interval for m, the population average daily intake of dairy products for men. s 35 x 2.58 756 2.58 n 50 756 12.77 or 743.23 m 768.77 grams. The interval must be wider to provide for the increased confidence that is does indeed Copyright ©2003 Brooks/Cole enclose the true value of m. A division of Thomson Learning, Inc. Example • Of a random sample of n = 150 college students, 104 of the students said that they had played on a soccer team during their K-12 years. Estimate the porportion of college students who played soccer in their youth with a 98% confidence interval. 104 .69(.31) pˆ qˆ 2.33 pˆ 2.33 150 150 n or .60 p .78. .69 .09 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Estimating the Difference between Two Means •Sometimes we are interested in comparing the means of two populations. •The average growth of plants fed using two different nutrients. •The average scores for students taught with two different teaching methods. •To make this comparison, A random sample of size n1 drawn from A random of size s n2 2drawn from population 1 with mean μsample and variance . 1 1 population 2 with mean μ2 and variance s 22 . Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Estimating the Difference between Two Means •We compare the two averages by making inferences about m1-m2, the difference in the two population averages. •If the two population averages are the same, then m1-m2 = 0. •The best estimate of m1-m2 is the difference in the two sample means, x1 x2 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Sampling Distribution of x1 x2 1. The mean of x1 x2 is m1 m 2 , the difference in the population means. 2. The standard deviation of x1 x2 is SE s 12 n1 s 22 n2 . 3. If the sample sizes are large, the sampling distributi on of x1 x2 is approximat ely normal, and SE can be estimated as SE s12 s22 . n1 n2 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Estimating m1-m2 •For large samples, point estimates and their margin of error as well as confidence intervals are based on the standard normal (z) distribution. Point estimate for m1 - m 2 : x1 x2 2 1 2 2 s s Margin of Error : 1.96 n1 n2 Confidence interval for m1 - m 2 : ( x1 x2 ) za / 2 s12 s22 n1 n2 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Avg Daily Intakes Men Women Sample size 50 50 Sample mean 756 762 Sample Std Dev 35 30 • Compare the average daily intake of dairy products of men and women using a 95% confidence interval. s12 s22 ( x1 x2 ) 1.96 n1 n2 35 30 (756 762) 1.96 50 50 6 12.78 or -18.78 m1 m2 6.78. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example, continued -18.78 m1 m2 6.78 • Could you conclude, based on this confidence interval, that there is a difference in the average daily intake of dairy products for men and women? • The confidence interval contains the value m1-m2= 0. Therefore, it is possible that m1 = m2. You would not want to conclude that there is a difference in average daily intake of dairy products for men and women. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Estimating the Difference between Two •We compare the two proportions by making inferences about p1-p2, the difference in the two population proportions. •If the two population proportions are the same, then p1-p2 = 0. •The best estimate of p1-p2 is the difference in the two sample proportions, x1 x2 pˆ1 pˆ 2 n1 n2 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Sampling Distribution of pˆ1 pˆ 2 1. The mean of pˆ 1 pˆ 2 is p1 p2 , the difference in the population proportion s. 2. The standard deviation of pˆ 1 pˆ 2 is SE p1q1 p2 q2 . n1 n2 3. If the sample sizes are large, the sampling distributi on of pˆ 1 pˆ 2 is approximat ely normal, and SE can be estimated as SE pˆ 1qˆ1 pˆ 2 qˆ 2 . n1 n2 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Estimating p1-p2 •For large samples, point estimates and their margin of error as well as confidence intervals are based on the standard normal (z) distribution. Point estimate for p -p : pˆ pˆ 1 2 1 2 pˆ1qˆ1 pˆ 2 qˆ 2 Margin of Error : 1.96 n1 n2 Confidence interval for p1 p2 : ( pˆ1 pˆ 2 ) za / 2 pˆ1qˆ1 pˆ 2 qˆ 2 n1 n2 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Youth Soccer Male Female Sample size 80 70 Played soccer 65 39 • Compare the proportion of male and female college students who said that they had played on a soccer team during their K-12 years using a 99% confidence interval. pˆ1qˆ1 pˆ 2 qˆ2 ( pˆ1 pˆ 2 ) 2.58 n1 n2 65 39 .81(.19) .56(.44) ( ) 2.58 .25 .19 80 70 80 70 or .06 p1 p2 .44. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example, continued .06 p1 p2 .44 • Could you conclude, based on this confidence interval, that there is a difference in the proportion of male and female college students who said that they had played on a soccer team during their K-12 years? • The confidence interval does not contains the value p1-p2 = 0. Therefore, it is not likely that p1= p2. You would conclude that there is a difference in the proportions for males and females. A higher proportion of males than females played Copyright soccer©2003 in their youth. Brooks/Cole A division of Thomson Learning, Inc. One Sided Confidence Bounds • Confidence intervals are by their nature two-sided since they produce upper and lower bounds for the parameter. • One-sided bounds can be constructed simply by using a value of z that puts a rather than a/2 in the tail of the z LCB : Estimator za (Std Error of Estimator) distribution. UCB : Estimator za (Std Error of Estimator) Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Sampling Distribution of the Sample Mean • When we take a sample from a normal x population, the sample mean has a normal distribution for any sample size n, and xm xm is not normal! z s/ n s/ n • has a standard normal distribution. • But if s is unknown, and we must use s to estimate it, the resulting statistic is not Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Student’s t Distribution • Fortunately, this statistic does have a sampling distribution that is well known to statisticians, called the Student’s t distribution, with n-1 degrees of freedom. xm t s/ n •We can use this distribution to create estimation testing procedures for the population Copyright ©2003 Brooks/Cole mean m. A division of Thomson Learning, Inc. Properties of Student’s t •Mound-shaped and symmetric about 0. •More variable than z, with “heavier tails” • • Shape depends on the sample size n or the degrees of freedom, n-1. Applet As n increases the shapes of the t and z distributions become almost identical. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Using the t-Table • • Table 4 gives the values of t that cut off certain critical values in the tail of the t distribution. Index df and the appropriate tail area a to a random size n = find ta,the value of For t with areasample a toofits 10, find a value of t that cuts off right. .025 in the right tail. Row = df = n –1 = 9 Column subscript = a = .025 t.025 = 2.262 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Small Sample Inference for a Population Mean m • The basic procedures are the same as those used for large samples. For a test of hypothesis: Test H 0 : m m 0 versus H a : one or two tailed using the test statistic x m0 t s/ n using p - values or a rejection region based on a t - distributi on with df n 1. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Small Sample Inference for a Population Mean m • For a 100(1a)% confidence interval for the population mean m: s x ta / 2 n where ta / 2 is the value of t that cuts off area a/2 in the tail of a t - distributi on with df n 1. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example A sprinkler system is designed so that the average time for the sprinklers to activate after being turned on is no more than 15 seconds. A test of 5 systems gave the following times: 17, 31, 12, 17, 13, 25 Is the system working as specified? Test using 0 : m 15 (working as specified) a =H .05. H a : m 15 (not worki ng as specified) Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Data: 17, 31, 12, 17, 13, 25 First, calculate the sample mean and standard deviation, using your calculator or the formulas in Chapter 2. xi 115 x 19.167 n 6 ( x) 115 x 2477 n 6 7.387 s n 1 5 2 2 2 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Data: 17, 31, 12, 17, 13, 25 Calculate the test statistic and find the rejection region for a =.05. Test statistic : Degrees of freedom : x m 0 19.167 15 t 1.38 df n 1 6 1 5 s / n 7.387 / 6 Rejection Region: Reject H0 if t > 2.015. If the test statistic falls in the rejection region, its p-value will be less than a = .05. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Conclusion Data: 17, 31, 12, 17, 13, 25 Compare the observed test statistic to the rejection region, and draw Test statistic : t 1.38 conclusions. H 0 : m 15 Rejection Region : H a : m 15 Reject H if t 2.015. 0 Conclusion: For our example, t = 1.38 does not fall in the rejection region and H0 is not rejected. There is insufficient evidence to indicate that the average activation time is greater than 15. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Testing the Difference between Two Means Asin Chapter9, independent random samplesof sizen1 and n2 are drawn from populations1 and 2 with means μ1 and μ2 and variancess 12and s 22 . Since the samplesizesare small,the twopopulations must be normal. •To test: •H0: m1m2 D0 versus Ha: one of three where D0 is some hypothesized difference, usually 0. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Testing the Difference between Two Means •The test statistic used in Chapter 9 x1 x2 z s12 s22 n1 n2 •does not have either a z or a t distribution, and cannot be used for small-sample inference. •We need to make one more assumption, that the population variances, although unknown, are equal. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Testing the Difference between Two Means •Instead of estimating each population variance separately, we estimate the common variance with 2 2 •And the resulting ( n 1 ) s ( n 1 ) s 1 2 2 s2 1 test statistic, n1 n2 2 t x1 x2 D0 1 1 s n1 n2 2 has a t distribution with n1+n2-2 degrees of freedom. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Estimating the Difference between Two Means •You can also create a 100(1-a)% confidence interval for m1-m2. Remember the three ( x1 x2 ) ta / 2 1 1 s n1 n2 2 2 2 ( n 1 ) s ( n 1 ) s 1 2 2 with s 2 1 n1 n2 2 assumptions: 1. Original populations normal 2. Samples random and independent 3. Equal population variances. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example • Two training procedures are compared by measuring the time that it takes trainees to assemble a device. A different group of trainees are taught using each method. Is there a difference in the two methods? Use a = .01. Time to Assemble Method 1 Method 2 Sample size 10 12 Sample mean 35 31 Sample Std Dev 4.9 4.5 H 0 : m1 m 2 0 H a : m1 m 2 0 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Applet Example • Solve this problem by approximating the pvalue using Time to Method 1 Method 2 Assemble Table 4. Calculate : Sample size 10 12 Sample mean 35 31 Sample Std Dev 4.9 4.5 2 2 ( n 1 ) s ( n 1 ) s 1 2 2 s2 1 n1 n2 2 9(4.9 2 ) 11(4.52 ) 21.942 20 Test statistic : t 35 31 1 1 21.942 10 12 1.99 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example p - value : P(t 1.99) P(t 1.99) 1 P(t 1.99) ( p - value) 2 df = n1 + n2 – 2 = 10 + 12 – 2 = 20 .025 < ½( p-value) < .05 .05 < p-value < .10 Since the p-value is greater than a = .01, H0 is not rejected. There is insufficient evidence to indicate a difference in the population means. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Testing the Difference between Two Means •How can you tell if the equal variance assumption is reasonable? Rule of Thumb : larger s 2 If the ratio, 3, 2 smaller s the equal variance assumption is reasonable . larger s 2 If the ratio, 3, 2 smaller s use an alternativ e test statistic.Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Testing the Difference between Two Means •If the population variances cannot be assumed equal, the test statistic t x1 x2 s12 s22 n1 n2 2 s s n1 n2 df 2 ( s1 / n1 ) 2 ( s22 / n2 ) 2 n1 1 n2 1 2 1 2 2 •has an approximate t distribution with degrees of freedom given above. This is most easily done by computer. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Paired-Difference Test •Sometimes the assumption of independent samples is intentionally violated, resulting in a matched-pairs or paired-difference test. •By designing the experiment in this way, we can eliminate unwanted variability in the experiment by analyzing only the differences, di = x1i – x2i •to see if there is a difference in the two population means, m1m2. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Car 1 2 3 4 5 Type A 10.6 9.8 12.3 9.7 8.8 Type B 10.2 9.4 11.8 9.1 8.3 • One Type A and one Type B tire are randomly assigned to each of the rear wheels of five cars. Compare the average tire wear for types A and B using a test of hypothesis. • But the samples are not independent. The pairs of H 0 : m1 m 2 0 responses are linked because H a : m1 m 2 0 measurements are taken on the same car. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Paired-Difference Test To test H 0 : m1 m 2 0 we test H 0 : m d 0 using the test statistic d 0 t sd / n where n number of pairs, d and sd are the mean and standard deviation of the difference s, d i . Use the p - value or a rejection region based on a t - distributi on with df n 1. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Car 1 2 3 4 5 Type A 10.6 9.8 12.3 9.7 8.8 Type B 10.2 9.4 11.8 9.1 8.3 Difference .4 .4 .5 .6 .5 H 0 : m1 m 2 0 H a : m1 m 2 0 Test statistic : di Calculated .48 n di d 0 .48 0 t 12.8 sd / n .0837 / 5 2 sd d 2 i n 1 n .0837 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Car 1 2 3 4 5 Type A 10.6 9.8 12.3 9.7 8.8 Type B 10.2 9.4 11.8 9.1 8.3 Difference .4 .4 .5 .6 .5 Rejection region: Reject H0 if t > 2.776 or t < -2.776. Conclusion: Since t = 12.8, H0 is rejected. There is a difference in the average tire wear for the two types of tires. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Inference Concerning a Population Variance •Sometimes the primary parameter of interest is not the population mean m but rather the population variance s2. We choose a random sample of size n from a normal distribution. •The sample variance s2 can be used in its standardized form: 2 (n 1) s 2 s2 • which has a Chi-Square distribution with n - 1 degrees of freedom. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Inference Concerning a Population Variance •Table 5 gives both upper and lower critical values of the chi-square statistic for a given df. For example, the value of chi-square that cuts off .05 in the upper tail of the distribution with df = 5 is 2 =11.07. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Inference Concerning a Population Variance To test H 0 : s 2 s 02 versus H a : one or two tailed we use the test statistic 2 (n 1) s s 2 2 0 with a rejection region based on a chi - square distributi on with df n 1. Confidence interval : (n 1) s 2 a / 2 2 s 2 (n 1) s 2 2 (1a / 2 ) Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example •A cement manufacturer claims that his cement has a compressive strength with a standard deviation of 10 kg/cm2 or less. A sample of n = 10 measurements produced a mean and standard deviation of 312 and 13.96, respectively. A test of hypothesis: H0: s2 = 10 (claim is correct) Ha: s2 > 10 (claim is wrong) uses the test statistic: 2 2 ( n 1 ) s 9 ( 13 . 96 ) 2 17.5 2 10 100 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example •Do these data produce sufficient evidence to reject the manufacturer’s claim? Use a = .05. Rejection region: Reject H0 if 2 16.919 a .05. Conclusion: Since 2= 17.5, H0 is rejected. The standard deviation of the cement strengths is more than 10. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Inference Concerning Two Population Variances •We can make inferences about the ratio of two population variances in the form a ratio. We choose two independent random samples of size n1 and n2 from normal distributions. •If the two population variances are equal, the statistic s12 F 2 s2 •has an F distribution with df1 = n1 - 1 and df2 = n2 - 1 degrees of freedom. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Inference Concerning Two Population Variances •Table 6 gives only upper critical values of the F statistic for a given pair of df1 and df2. For example, the value of F that cuts off .05 in the upper tail of the distribution with df1 = 5 and df2 = 8 is F =3.69. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Inference Concerning Two Population Variances To test H 0 : s s versusH a : one or twotailed 2 1 2 2 weuse the test statistic s12 F 2 wheres12 is the larger of the twosamplevariances. s2 with a rejectionregion basedon an F distribution with df1 n1 1 and df 2 n2 1. Confidence interval : 2 1 2 2 s s F s s 2 2 1 1 df 2 , df1 2 2 df1 , dfCopyright 2 2 2 ©2003 Brooks/Cole s 1 s F A division of Thomson Learning, Inc. Example •An experimenter has performed a lab experiment using two groups of rats. He wants to test H0: m1 m2, but first he wants to make sure that the population variances are equal. Standard (2) Experimental (1) Sample size 10 11 Sample mean 13.64 12.42 Sample Std Dev 2.3 5.8 Preliminar y test : H 0 : s 12 s 22 versus H a : s 12 s 22 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Standard (2) Experimental (1) Sample size 10 11 Sample Std Dev 2.3 5.8 H0 : s s 2 1 Ha :s s 2 1 2 2 2 2 Test statistic : 2 1 2 2 2 s 5.8 F 6.36 2 s 2.3 We designate the sample with the larger standard deviation as sample 1, to force the test statistic into the upper tail of the F distribution. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example H0 : s s 2 1 Ha :s s 2 1 2 2 2 2 Test statistic : s12 5.82 F 2 6.36 2 s2 2.3 The rejection region is two-tailed, with a = .05, but we only need to find the upper critical value, which has a/2 = .025 to its right. From Table 6, with df1=10 and df2 = 9, we reject H0 if F > 3.96. CONCLUSION: Reject H0. There is sufficient evidence to indicate that the variances are unequal. Do not rely on the assumption of equal variances for your t test! Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Parts of a Statistical Test 1. The null hypothesis, H0: – Assumed to be true until we can prove otherwise. 2. The alternative hypothesis, Ha: – Will be accepted as true if we can disprove H0 Court trial: Pharmaceuticals: H0: innocent H0: m does not exceeds allowed amount Ha: guilty Ha: m exceeds allowed amount Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Parts of a Statistical Test 3. The test statistic and its p-value: • A single statistic calculated from the sample which will allow us to reject or not reject H0, and • A probability, calculated from the test statistic that measures whether the test statistic is likely or unlikely, assuming H0 is true. 4. The rejection region: – A rule that tells us for which values of the Copyright ©2003 Brooks/Cole division of Thomsonthe Learning, Inc. test statistic, or for which Ap-values, null Parts of a Statistical Test 5. Conclusion: – Either “Reject H0” or “Do not reject H0”, along with a statement about the reliability of your conclusion. How do you decide when to reject H0? – – Depends on the significance level, a, the maximum tolerable risk you want to have of making a mistake, if you decide to reject H0. Usually, the significance level is a .01 or a .05. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Large Sample Test of a Population Mean, m • Take a random sample of size n 30 from a population with mean m and standard deviation s. • We assume that either 1. s is known or 2. s s since n is large • The hypothesis to be tested is – H0:m = m0 versus Ha: mCopyright m0©2003 Brooks/Cole A division of Thomson Learning, Inc. Test Statistic • Assume to begin with that H0 is true. x The sample mean is our best estimate of m, and we use it in a standardized form as the test statistic: x m0 x m0 z s / n s/ n since x has an approximate normal distribution with mean m0 and standard deviation s / n . Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Applet Example • The daily yield for a chemical plant has averaged 880 tons for several years. The quality control manager wants to know if this average has changed. She randomly selects 50 days and records an average yield of 871 tons with a standard deviation ofH21: mtons. Test statistic : 880 0 x m 871 880 0 H a : m 880 z 3.03 s/ n 21 / 50 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Applet Example What is the probability that this test statistic or something even more extreme (far from what is expected if H0 is true) could have happened just by chance? p - value : P( z 3.03) P( z 3.03) 2 P( z 3.03) 2(.0012) .0024 This is an unlikely occurrence, which happens about 2 times in 1000, assuming m = 880! Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example • To make our decision clear, we choose a significance level, say a = .01. If the p-value is less than a, H0 is rejected as false. You report that the results are statistically significant at level a. If the p-value is greater than a, H0 is not rejected. You report that the results are not significant at level a. Since our p-value =.0024 is less than, we reject H0 and conclude that the average yield has changed. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Using a Rejection Region If a = .01, what would be the critical value that marks the “dividing line” between “not rejecting” and “rejecting” H0? If p-value < a, H0 is rejected. If p-value > a, H0 is not rejected. The dividing line occurs when p-value = a. This is called the critical value of the test statistic. Test statistic > critical value implies p-value < a, H0 is rejected. Test statistic < critical value implies p-value > a, H0 is not rejected. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Applet Example What is the critical value of z that cuts off exactly a/2 = .01/2 = .005 in the tail of the z distribution? For our example, z = -3.03 falls in the rejection region and H0 is rejected at the 1% significance level. Rejection Region: Reject H0 if z > 2.58 or z < -2.58. If the test statistic falls in the rejection region, its p-value will be less than a = .01. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. One Tailed Tests • Sometimes we are interested in a detecting a specific directional difference in the value of m. • The alternative hypothesis to be tested is one tailed: – Ha:m m0 or Ha: m < m0 • Rejection regions and p-values are calculated using only one tail of the sampling distribution. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Applet Example • A homeowner randomly samples 64 homes similar to her own and finds that the average selling price is $252,000 with a standard deviation of $15,000. Is this sufficient evidence to conclude that the average selling price is greater than $250,000? Use a = .01. Test statistic : H 0 : m 250,000 x m 0 252,000 250,000 1.07 H a : m 250,000 z s/ n 15,000 / 64 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Critical Value Approach What is the critical value of z that cuts off exactly a= .01 in the right-tail of the z distribution? For our example, z = 1.07 does not fall in Applet the rejection region and H0 is not rejected. There is not enough evidence to indicate that m is greater than $250,000. Rejection Region: Reject H0 if z > 2.33. If the test statistic falls in the rejection region, its p-value will be less than a = .01. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. p-Value Approach • The probability that our sample results or something even more unlikely would have occurred just by chance, when m = 250,000. p - value : P( z 1.07) 1 .8577 .1423 Applet Since the p-value is greater than a = .01, H0 is not rejected. There is insufficient evidence to indicate that m is greater than $250,000. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Statistical Significance • If the p-value is less than .01, reject H0. The results are highly significant. • If the p-value is between .01 and .05, reject H0. The results are statistically significant. • If the p-value is between .05 and .10, do not reject H0. But, the results are tending towards significance. • If the p-value is greater than .10, do not reject H0. The results are not statistically significant. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Two Types of Errors There are two types of errors which can occur in a statistical test. Actual Fact Guilty Jury’s Decision Innocent Actual Fact H0 true Your (Accept H0) Decision H0 false (Reject H0) Guilty Correct Error Correct Type II Error Innocent Error Correct H0 true (Accept H0) H0 false (Reject H0) Type I Error Correct Define: a = P(Type I error) = P(reject H0 when H0 is true) b P(Type II error) = P(accept H0 when H0 is false) Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Testing the Difference between Two Means A random sample of size n1 drawn from population 1 with mean μ1 and variance s 12 . A random sample of size n2 drawn from population 2 with mean μ2 and variance s 22 . •The hypothesis of interest involves the difference, m1m2, in the form: •H0: m1m2 D0 versus Ha: one of three where D0 is some hypothesized difference, usually 0. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Sampling Distribution of x1 x2 1. The mean of x1 x2 is m1 m 2 , the difference in the population means. 2. The standard deviation of x1 x2 is SE s 12 n1 s 22 n2 . 3. If the sample sizes are large, the sampling distributi on of x1 x2 is approximat ely normal, and SE can be estimated as SE s12 s22 . n1 n2 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Testing the Difference between Two Means H 0 : m1 m 2 D0 versus H a : one of three alternativ es Test statistic : z x1 x2 s12 s22 n1 n2 with rejection regions and/or p - values based on the standard normal z distributi on. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Avg Daily Intakes Men Women Sample size 50 50 Sample mean 756 762 Sample Std Dev 35 30 • Is there a difference in the average daily intakes of dairy products for men versus women? Use a = .05. H0 : m1 m 2 0 (same) Ha : m1 m 2 0 (different ) Test statistic : 756 762 0 x1 x2 0 .92 z 2 2 2 2 35 30 s1 s2 50 50 n1 n2 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. p-Value Approach • The probability of observing values of z that as far away from z = 0 as we have, just by chance, if indeed m1m2 = 0. p - value : P( z .92) P( z .92) Since the p-value is 2(.1788) .3576 greater than a = .05, H0 is not rejected. There is insufficient evidence to indicate that men and women have different average daily intakes. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Critical Value Approach What is the critical value of z that cuts off exactly a= .05 in the left-tail of the z distribution? For our example, z = -1.25 does not fall in the rejection region and H0 is not rejected. There is not enough evidence to indicate that p is less than .2 for people over 40. Rejection Region: Reject H0 if z < -1.645. If the test statistic falls in the rejection region, its p-value will be less than a = .05. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Testing the Difference between Two Proportions •To compare two binomial proportions, A random sample of size n1 drawn from binomial population 1 with parameter p1. A random sample of size n2 drawn from binomial population 2 with parameter p2 . •The hypothesis of interest involves the difference, p1p2, in the form: H0: p1p2 D0 versus Ha: one of three •where D0 is some hypothesized difference, usually 0. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. The Sampling Distribution of pˆ1 pˆ 2 1. The mean of pˆ 1 pˆ 2 is p1 p2 , the difference in the population proportion s. 2. The standard deviation of pˆ 1 pˆ 2 is SE p1q1 p2 q2 . n1 n2 3. If the sample sizes are large, the sampling distributi on of pˆ 1 pˆ 2 is approximat ely normal. 4. The standard error is estimated differentl y, depending on the hypothesiz ed difference , D 0 . Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Testing the Difference between Two Proportions H 0 : p1 p2 0 versus H a : one of three alternativ es pˆ 1 pˆ 2 Test statistic : z 1 1 pˆ qˆ n1 n2 with pˆ x1 x2 to estimate the common val ue of p n1 n2 and rejection regions or p - values based on the standard normal z distributi on. Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Youth Soccer Male Female Sample size 80 70 Played soccer 65 39 • Compare the proportion of male and female college students who said that they had played on a soccer team during their K-12 years using a test of hypothesis. H0 : p1 p2 0 (same) Ha : p1 p2 0 (different ) Calculate pˆ1 65 / 80 .81 pˆ 2 39 / 70 .56 x1 x2 104 pˆ .69 n1 n2 150 Copyright ©2003 Brooks/Cole A division of Thomson Learning, Inc. Example Youth Soccer Male Female Sample size 80 70 Played soccer 65 39 Test statistic : .81 .56 pˆ 1 pˆ 2 0 3.30 z 1 1 1 1 .69(.31) pˆ qˆ 80 70 n1 n2 p - value : P( z 3.30) P( z 3.30) 2(.0005) .001 Since the p-value is less than a = .01, H0 is rejected. The results are highly significant. There is evidence to indicate that the rates of participation are differentCopyright for boys and girls. ©2003 Brooks/Cole A division of Thomson Learning, Inc.