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Ch7 Statistics Dr. Deshi Ye [email protected] Outline State what is estimated Point estimation Interval estimation Compute sample size Tests of Hypotheses Null Hypothesis and Tests of Hypotheses Hypotheses concern one mean 2/139 Why Statistics? The purpose of most statistical investigations is to generalize from information contained in random samples about the populations from which the samples were obtained. How: estimation and tests of Hypothesis 3/139 Thinking Suppose you’re interested in the average amount of money that students in this class (the population) have on them. How would you find out? 4/139 Introduction to Estimation 5/139 Statistical Methods Statistical Statistical Methods Methods Descriptive Descriptive Statistics Statistics Inferential Inferential Statistics Statistics Estimation Estimation Hypothesis Hypothesis Testing Testing 6/139 Estimation Process Population Mean, , is unknown Random Sample Mean X = 50 I am 95% confident that is between 40 & 60. Sample 7/139 Estimation Methods Estimation Point Estimation Interval Estimation 8/139 Point estimation We want to know the mean of a population. However, it is unavailable. Hence, we choose a sample data, and calculate from the choosing sample data. Then estimate that the population is also the same mean. Point estimation: concerns the choosing of a statistic, that is, a single number calculated from sample data for which we have some expectation, or assurance, that is reasonably close the parameter it is supposed to estimate. 9/139 Point Estimation 1. Provides Single Value Based on Observations from 1 Sample 2. Gives No Information about How Close Value Is to the Unknown Population Parameter 3. Example: Sample MeanX = 3 Is Point Estimate of Unknown Population Mean 10/139 Point estimation of a mean Parameter: Population mean Data: A random sample X1 , X 2 , , X n Estimator: X Estimate of standard error: S n 11/139 EX Scientists need to be able to detect small amounts of contaminants in the environment. Sample data is listed as follows: 2.4 2.9 2.7 2.6 2.9 2.0 2.8 2.2 2.4 2.4 2.0 2.5 12/139 EX of point estimation Compute the point estimator X and estimate its standard deviation n (also called the estimated standard error of X ). Solution: 29.8 x 2.483 12 n s2 ( xi x ) i 1 n 1 n 2 2 2 x nx i i 1 n 1 75.08 (29.8) 2 /12 0.09788 12 1 s 0.09788 /12 0.090 Hence the estimated standard deviation is 13/139 n Bias or Unbias Question: is the estimation good enough? EX. In previous ex, as a check on the current capabilities, the measurements were made on test specimens spiked with a known concentration 1.25 ug/l of lead. That is the readings should average 1.25 if there is no background lead in the samples. There appears to be either a bias due to laboratory procedure or some lead already in the samples before they were spiked. 14/139 Unbiased estimator Let be the parameter of interest and ˆ be a statistic. A statistic ˆ is said to be an unbiased estimator, or its value an unbiased estimate, if and only if the mean of the sampling distribution of the estimator equals , whatever the value of . 15/139 More efficient unbiased estimator Estimator is not unique: for example: it can be shown that for a random sample of size n=2, the mean X 2 X as well as the weighted mean aXa bX where a, b are b positive constants, are unbiased estimates of the mean of the population. 1 1 2 2 A statistics ˆ1 is said to be a more efficient unbiased estimator of the parameter ˆ2 than the statistics if 1. ˆ1 and ˆ2 are both unbiased estimators of 2. the variance of the sampling distribution of the first estimator is no larger than that of the second and is smaller for at least one value of . 16/139 Finding Sample Sizes I don’t want to sample too much or too little! The error The error between the estimator and the quantity it is supposed to estimate is: X X / n is a random variable having approximately the standard normal distribution We could assert with probability 1 inequality z / 2 Remind that that the X z / 2 / n P(Z z / 2 ) 1 / 2 18/139 Maximum error of estimate The error | X | will be less than E z / 2 with probability Specially, 1 n . z / 2 1.96, when 0.05 z / 2 2.575, when 0.01 19/139 EX An industrial engineer intends to use the mean of a random sample of size n=150 to estimate the average mechanical aptitude of assembly line workers in a large industry. If, on the basis of experience, the engineer can assume that 6.2 for such data, what can he assert with probability 0.99 about the maximum size of his error. 6.2 E 2.575 1.3 150 Thus, the engineer can assert with probability 0.99 that his error is at most 1.3 20/139 Determine of sample size Suppose that we want to use the mean of a large random sample to estimate the mean of a population, and want to be able to assert with probability 1 that the error will be at most some prescribed quantity E. As before, we get z / 2 2 n ( ) E 21/139 EX A research worker want to determine the average time it takes a mechanic to rotate the tires of a car, and she wants to be able to assert with 95% confidence that the mean of her sample is off by at most 0.5 minute. If she can presume from past experience that 1.6 minutes, how large a sample will she have to take? 1.96 1.6 2 n( ) 39.3 0.5 22/139 Continuous The method discussed so far requires be known or it can be approximated with the sample standard deviation s, thus requiring that n be large. Another approach: if it is reasonable to assume that we are sampling from a normal population, we get X t S/ n is a random variable having the t distribution with n-1 degree of freedom. 23/139 EX The first example, n=12, t0.01 2.718 s 0.09788 /12 0.090 n For n=11 degrees of freedom s t0.01 2.718 0.09 0.2 n Thus, one can assert with 98% confidence that the maximum error is within 0.2 X 2.483 1.25 1.233 0.2 24/139 Estimation Methods Estimation Point Estimation Interval Estimation 25/139 Interval Estimation 1. Provides Range of Values Based on Observations from 1 Sample 2. Gives Information about Closeness to Unknown Population Parameter Stated in terms of Probability 3. Example: Unknown Population Mean Lies Between 50 & 70 with 95% Confidence 26/139 Interval Estimation Sample statistic (point estimate) 27/139 Interval Estimation Confidence interval Confidence limit (lower) Sample statistic (point estimate) Confidence limit (upper) 28/139 Interval Estimation A probability that the population parameter falls somewhere within the interval. Confidence interval Confidence limit (lower) Sample statistic (point estimate) Confidence limit (upper) 29/139 Confidence Interval Estimates Confidence Intervals Mean Known Proportion Variance Unknown 30/139 Confidence Interval Mean ( Known) Assumptions Population standard deviation is known Population is normally distributed If not normal, can be approximated by normal distribution (n 30) Confidence Interval Estimate X Z / 2 n X Z / 2 n 31/139 Interval Estimation Interval estimation: with intervals for which we can assert with a reasonable degree of certainty that they will contain the parameter under consideration. For a large random sample (n > 30) from a population with the unknown mean and the known variance. When the observed value x becomes available, we obtain x z / 2 n x z / 2 n 32/139 Confidence interval We can claim with that the interval [ x z / 2 (1 )100% n , x z / 2 confidence n ] Contains It is customary to refer to an interval of this kind as a confidence interval for having the degree of (1 )100% confidence 33/139 Confidence Interval Estimates Confidence Intervals Mean Known Proportion Variance Unknown 34/139 Solution for Small Samples 1. Assumptions Population of X Is Normally Distributed Use Student’s t Distribution 1. Define variable X T s/ n 2. T has the Student distribution with n -1 degrees of freedom (When X is normally distributed) • There’s a different Student distribution for different degrees of freedom • As n gets large, Student distribution approximates a normal distribution with mean = 0 and sigma = 1 35/139 Small sample (n<30) Small sample and we assume to get sampling from a normal distribution population. We get the (1 )100% confidence interval formula s s x t / 2 x t / 2 n n 36/139 EX. The mean weight loss of n=16 grinding balls after a certain length of time in mill slurry is 3.42 grams with a standard deviation of 0.68 gram. Construct a 99% confidence interval for the true mean weight loss of such grinding balls under the stated condition. 37/139 Confidence Depends on Interval (z) X = ± Zx x_ -2.58x -1.65x -1.96x +1.65x +2.58x +1.96x X 90% Samples 95% Samples 99% Samples 38/139 Confidence Level 1. Probability that the Unknown Population Parameter Falls Within Interval 2. Denoted (1 - Is Probability That Parameter Is Not Within Interval 3. Typical Values Are 99%, 95%, 90% 39/139 Intervals & Confidence Level Sampling Distribution /2 of Mean x_ 1 - /2 x = _ X (1 - ) % of intervals contain . Intervals extend from X - ZX to X + ZX % do not. Intervals derived from many samples 40/139 Confidence interval & level It is useful to think confidence intervals as a range of "plausible" values for the parameter. confidence interval is different from interpreting the confidence level suppose we've taken a random sample of 10 ice-cream cones, and determined that a 95% confidence interval for the mean caloric contents of a single scoop of ice-cream is (260,310). Interpret the confidence level: If we repeatedly took samples of size 10 and then formed confidence intervals, we would expect 95% of them to contain the true (but unknown) mean. Interpret this particular confidence interval: we are 95% confident that the true mean caloric content lies between 260 and 310. 41/139 Confidence interval & level The wider the confidence interval you are willing to accept, the more certain you can be that the whole population answers would be within that range. For example, if you asked a sample of 1000 people in a city which brand of cola they preferred, and 60% said Brand A, you can be very certain that between 40 and 80% of all the people in the city actually do prefer that brand, but you cannot be so sure that between 59 and 61% of the people in the city prefer the brand. 42/139 Factors Affecting Interval Width 1. Data Dispersion Measured by 2. Sample Size Intervals Extend from X - ZX toX + ZX x / n 3. Level of Confidence (1 - ) Affects Z © 1984-1994 T/Maker Co. 43/139 Statistical Methods Statistical Methods Descriptive Statistics Inferential Statistics Estimation Hypothesis Testing 44/139 Hypothesis Testing Population I believe the population mean age is 50 (hypothesis). Reject hypothesis! Not close. Random sample Mean X = 20 45/139 What’s a Hypothesis? A belief about a population parameter I believe the mean GPA of this class is 3.8! Parameter is Population mean, proportion, variance Must be stated before analysis © 1984-1994 T/Maker Co. 46/139 Tests of Hypotheses Suppose that a consumer protection agency wants to test a paint manufacturer’s claim that the average drying time of his new “fast-drying” paint is 20 minutes. It instructs a member of its research staff to paint each of 36 boards using a different 1-gallon can of the paint, with the intention of rejecting the claim if the mean of the drying time exceeds 20.75 minutes. Otherwise, it will accept the claim. Question: Is it a infallible criterion for accepting or rejecting the claim? 47/139 EX cont. Assuming that it is known from past experience that the standard deviation 2.4 The probability of erroneously rejecting the hypothesis that 20 Let us investigate the possibility that the sample may exceed 20.75 minutes even though the true mean is 20 minutes z 20.75 20 1.875 2.4 / 36 Accept the claim that 20 Reject the claim that 20 0.0304 x Minutes 20 20.75 48/139 Another possibility The procedure fails to detect that 20 Suppose that the true mean of drying time is 21 Reject the claim that 20 Accept the claim that 20 z 20.75 21 0.625 2.4 / 36 0.2660 20.75 21 49/139 Type of errors H: the hypothesis. Ex. Accept H 20 Reject H H is true Correct decision Type I error H is false Type II error Correct decision 50/139 Type I error: If hypotheses is true but rejected. Denoted by the letter EX. 0.0304 Type II error: If hypotheses is false but not rejected. Denoted by the letter EX. 0.2660 51/139 7.4 Null Hypotheses Question: Can we formulate 20 minutes, where can take on more than one possible value? Null Hypotheses H 0 : (Pronounced H-nought) Usually require that we hypothesize the opposite of what we hope to prove. EX. If we want to show that one method of teaching computer programming is more efficient than another, we hypothesize that the two methods are equally effective. The null hypothesis proposes something initially presumed true. It is rejected only when it becomes evidently false. 52/139 Null Hypothesis 1. What is tested 2. has serious outcome if incorrect decision made 3. Designated H0 4. Specified as H0: Some Numeric Value Specified with = Sign Even if , or Example, H0: 3 53/139 Alternative Hypothesis 1. Opposite of Null Hypothesis 2. Always Has Inequality Sign: ,, or 3. Designated Ha 4. Specified Ha: < Some Value Example, Ha: < 3 will lead to two-sided tests <, > will lead to one-sided tests 54/139 Alternative One-sided alternative: In the drying time example, the null hypothesis is 20 minutes and the alternative hypothesis is 20 Two-sided alternative: 0 where 0 is the value assumed under the null hypothesis. 55/139 Selecting the null hypothesis Guideline for selecting the null hypothesis: When the goal of an experiment is to establish an assertion, the negation of the assertion should be taken as the null hypothesis. The assertion becomes the alternative hypothesis. 56/139 Identifying Hypotheses Steps 1. Example Problem: Test That the Population Mean Is Not 3 2. Steps State the Question Statistically ( 3) State the Opposite Statistically ( = 3) Must Be Mutually Exclusive & Exhaustive Select the Alternative Hypothesis ( 3) Has the , <, or > Sign State the Null Hypothesis ( = 3) 57/139 What Are the Hypotheses? Is the population average amount of TV viewing 12 hours? State the question statistically: = 12 State the opposite statistically: 12 Select the alternative hypothesis: Ha: 12 State the null hypothesis: H0: = 12 58/139 What Are the Hypotheses? Is the population average amount of TV viewing different from 12 hours? State the question statistically: 12 State the opposite statistically: = 12 Select the alternative hypothesis: Ha: 12 State the null hypothesis: H0: = 12 59/139 What Are the Hypotheses? Is the average cost per hat less than or equal to $20? State the question statistically: 20 State the opposite statistically: 20 Select the alternative hypothesis: Ha: 20 State the null hypothesis: H0: 20 60/139 What Are the Hypotheses? Is the average amount spent in the bookstore greater than $25? State the question statistically: 25 State the opposite statistically: 25 Select the alternative hypothesis: Ha: 25 State the null hypothesis: H0: 25 61/139 Decision Making Risks Errors in making decision 1. Type I Error Reject True Null Hypothesis Has Serious Consequences Probability of Type I Error Is (Alpha) Called Level of Significance 2. Type II Error Do Not Reject False Null Hypothesis Probability of Type II Error Is (Beta) 63/139 Decision Results H0: Innocent Jury Trial H0 Test Actual Situation Actual Situation Verdict Innocent Guilty Decision H0 True Innocent Correct Guilty Error Error Do Not Reject H0 Correct Reject H0 1- H0 False Type II Error () Type I Power Error () (1 - ) 64/139 Decision Results H0: Innocent Jury Trial H0 Test Actual Situation Verdict Innocent Guilty Decision H0 True Innocent Correct Guilty Actual Situation Error Error Accept H0 Correct Reject H0 1- H0 False Type II Error () Type I Power Error () (1 - ) 65/139 & have an inverse relationship You can’t reduce both errors simultaneously! 66/139 Factors Affecting 1. True Value of Population Parameter Increases When Difference With Hypothesized Parameter Decreases 2. Significance Level, Increases When Decreases 3. Population Standard Deviation, Increases When Increases 4. Sample Size, n Increases When n Decreases 67/139 Hypothesis Testing Population I believe the population mean age is 50 (hypothesis). Reject hypothesis! Not close. Random sample Mean X = 20 68/139 Basic Idea 69/139 Basic Idea Sampling Distribution = 50 H0 Sample Mean 70/139 Basic Idea Sampling Distribution It is unlikely that we would get a sample mean of this value ... 20 = 50 H0 Sample Mean 71/139 Basic Idea Sampling Distribution It is unlikely that we would get a sample mean of this value ... ... if in fact this were the population mean 20 = 50 H0 Sample Mean 72/139 Basic Idea Sampling Distribution It is unlikely that we would get a sample mean of this value ... ... therefore, we reject the hypothesis that = 50. ... if in fact this were the population mean 20 = 50 H0 Sample Mean 73/139 Hypothesis testing 1. We formulate a null hypothesis and an appropriate alternative hypothesis which we accept when the null hypothesis must be rejected. 2. We specify the probability of a Type I error. If possible, desired, or necessary, we may also specify the probabilities of Type II errors for particular alternatives. 3. Based on the sampling distribution of an appropriate statistic, we construct a criterion for testing the null hypothesis against the given alternative. 4. We calculate from the data the value of the statistic on which the decision is to be based. 5. We decide whether to reject the null hypothesis or whether to fail to reject it. 74/139 Level of Significance 1. Probability 2. Defines Unlikely Values of Sample Statistic if Null Hypothesis Is True Called Rejection Region of Sampling Distribution 3. Designated (alpha) 4. Selected by Researcher at Start 75/139 Level of significance The probability of a Type I error is also called the level of significance. Usually, we set . 0.05 or 0.01 Step 2 can often be performed even when the null hypothesis specifies a range of values for the parameter. Ex. The null hypothesis: 20 Then we can claim that 0.0304 In general, we can only specify the maximum probability of a Type I error, and by again the worst possibility. 76/139 Criterion One-sided criterion (one-sided test or onetailed test): ex. One-sided alternative 20 Two-sided criterion (two-sided test or twotailed test): ex. two-sided alternative 4 77/139 Rejection Region (One-Tail Test) 78/139 Rejection Region (One-Tail Test) Sampling Distribution Level of Confidence Rejection Region 1- Nonrejection Region Critical value H0 value Sample Statistic 79/139 Rejection Region (One-Tail Test) Sampling Distribution Level of Confidence Rejection Region 1- Nonrejection Region Critical Value Ho Value Sample Statistic Observed sample statistic 80/139 Rejection Region (One-Tail Test) Sampling Distribution Level of Confidence Rejection Region 1- Nonrejection Region Ho Value Critical Value Observed sample statistic Sample Statistic 81/139 Rejection Region (Two-Tailed Test) 82/139 Rejection Region (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region Rejection Region 1- 1/2 1/2 Nonrejection Region Critical Value Ho Sample Statistic Value Critical Value 83/139 Rejection Region (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region Rejection Region 1- 1/2 1/2 Nonrejection Region Critical Value Ho Sample Statistic Value Critical Value 84/139 Observed sample statistic Rejection Region (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region 1/2 Rejection Region 1- 1/2 Nonrejection Region Ho Sample Statistic Critical Value Critical Value Value 85/139 Observed sample statistic Rejection Region (Two-Tailed Test) Sampling Distribution Level of Confidence Rejection Region 1/2 Rejection Region 1- 1/2 Nonrejection Region Ho Sample Statistic Critical Value Critical Value Value 86/139 Observed sample statistic H0 Testing Steps State H0 Set up critical values State Ha Collect data Choose Compute test statistic Choose n Make statistical decision Choose test Express decision 87/139 One Population Tests One Population Mean Proportion Variance Z Test t Test Z Test c2 Test (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) 88/139 7.5 Hypothesis concerning one mean Suppose we want to test on the basis of n=35 determinations and at the 0.05 level of significance whether the thermal conductivity of a certain kind of cement brick is 0.340, as has been claimed. We can expect that the variability of such determinations is given by 0.010 89/139 Solution 1) Null hypothesis: 0.340 alternative hypothesis: 0.340 2) Level of significance: 0.05 3) Criterion: X Z / n z z / 2 z / 2 90/139 Criterion Region for testing (Normal population and known) Alternative hypothesis Reject null hypothesis if 0 Z z 0 Z z 0 0 Z z / 2 or Z z / 2 91/139 Critical values 0.05 One-sided alternatives -1.645 1.645 Two-sided alternatives -1.96 1.96 0.01 -2.33 2.33 -2.575 2.575 92/139 EX. Cont. 3) Criterion: Reject the null hypothesis if Z 1.96 or Z 1.96 where 4) Calculations: X Z / n 0.343 0.34 z 1.77 0.01/ 35 5) Decision: The null hypothesis cannot be rejected. 93/139 P-value P value (or tail probability): the probability of getting difference between x and 0 greater than or equal to that actually observed. P(Z z ) EX. In above example 2(1 0.9616) 0.078 ½ P-value 1.77 ½ P-value 1.77 94/139 P-Value 1. Probability of Obtaining a Test Statistic More Extreme (or than Actual Sample Value Given H0 Is True 2. Called Observed Level of Significance Smallest Value of H0 Can Be Rejected 3. Used to Make Rejection Decision If p-Value , Do Not Reject H0 If p-Value < , Reject H0 95/139 P value P value for a given test statistic and null hypothesis: The P value is the probability of obtaining a value for the test statistic that is as extreme or more extreme than the value actually observed. Probability is calculated under the null hypothesis. 96/139 EX A process for producing vinyl floor covering has been stable for a long period of time, and the surface hardness measurement of the flooring produced has a normal distribution with mean 4.5 and standard deviation 1.5. A second shift has been hired and trained and their production needs to be monitored. Consider testing the hypothesis H 0 : 4.5 versus H1 : 4.5 97/139 A random sample of hardness measurements is made of n=25 vinyl specimens produced by the second shift. Calculate the P value when using the test statistic X Z If / n X 3.9 98/139 Solution The observed value of the test statistic is z 3.9 4.5 2.0 1.5 / 25 Since the alternative hypothesis is two-sided, we must consider large negative value for Z as well as large positive values P( Z 2.0) P( Z 2.0) 0.0228 Consequently, the P value is 0.0456 0.05 The small P value suggests the mean of the second 99/139 shift is not at the target value of 4.5 P value To understand P values, you have to understand fixed level testing. With fixed level testing, a null hypothesis is proposed (usually, specifying no treatment effect) along with a level for the test, usually 0.05. All possible outcomes of the experiment are listed in order to identify extreme outcomes that would occur less than 5% of the time in aggregate if the null hypothesis were true. This set of values is known as the critical region. They are critical because if any of them are observed, something extreme has occurred. Data are now collected and if any one of those extreme outcomes occur the results are said to be significant at the 0.05 level. The null hypothesis is rejected at the 0.05 level of significance and one star (*) is printed somewhere in a table. Some investigators note extreme outcomes that would occur less than 1% of the time and print two stars (**) if any of those are observed. Many researchers quickly realized the limitations of reporting only whether a result achieved the 0.05 level of significance. Was a result just barely significant or wildly so? Would data that were significant at the 0.05 level be significant at the 0.01 level? At the 0.001 level? Even if the result are wildly statistically significant, is the effect large enough to be of any practical importance? 100/139 P value Observed significance level (or P value)--the smallest fixed level at which the null hypothesis can be rejected. If your personal fixed level is greater than or equal to the P value, you would reject the null hypothesis. If your personal fixed level is less than to the P value, you would fail to reject the null hypothesis. For example, if a P value is 0.027, the results are significant for all fixed levels greater than 0.027 (such as 0.05) and not significant for all fixed levels less than 0.027 (such as 0.01). A person who uses the 0.05 level would reject the null hypothesis while a person who uses the 0.01 level would fail to reject it. 101/139 P-Value Thinking Challenge You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. What is the value of the observed level of significance (p-Value)? 102/139 p-Value Solution* p-Value is P(Z -2.65) = .004. p-Value < ( = .01). Reject H0. Use alternative hypothesis to find direction p-Value .004 .5000 - .4960 .0040 .4960 -2.65 0 Z value of sample statistic Z From Z table: lookup 2.65 103/139 One Population Tests One Population Mean Proportion Variance Z Test t Test Z Test c2 Test (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) (1 & 2 tail) 104/139 t Test for Mean ( Unknown) 1. Assumptions Population Is Normally Distributed If Not Normal, Only Slightly Skewed & Large Sample (n 30) Taken 2. Parametric Test Procedure 105/139 Statistic for small sample The test of null hypothesis 0 statistic on the X T S/ n 106/139 0 Criterion Region for testing (Statistic for small sample ) Alternative hypothesis Reject null hypothesis if 0 T t 0 T t 0 T t / 2 or T t / 2 107/139 Two-Tailed t Test Finding Critical t Values Given: n = 3; = .10 0 t 108/139 Two-Tailed t Test Finding Critical t Values Given: n = 3; = .10 /2 = .05 0 t /2 = .05 109/139 Two-Tailed t Test Finding Critical t Values Given: n = 3; = .10 df = n - 1 = 2 /2 = .05 0 t /2 = .05 110/139 Two-Tailed t Test Finding Critical t Values Given: n = 3; = .10 df = n - 1 = 2 /2 = .05 Critical Values of t Table (Portion) v t.10 t.05 t.025 1 3.078 6.314 12.706 2 1.886 2.920 4.303 0 t 3 1.638 2.353 3.182 /2 = .05 111/139 Two-Tailed t Test Finding Critical t Values Given: n = 3; = .10 df = n - 1 = 2 /2 = .05 Critical Values of t Table (Portion) v t.10 t.05 t.025 1 3.078 6.314 12.706 2 1.886 2.920 4.303 -2.920 0 2.920 t /2 = .05 3 1.638 2.353 3.182 112/139 One-Tailed t Test You’re a marketing analyst for Wal-Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was: 8 11 0 4 7 8 10 5 8 3. At the .05 level, is there evidence that the average bear sales per store is more than 5 ($ 00)? 113/139 One-Tailed t Test Solution* H0: Ha: = df = Critical Value(s): Test Statistic: Decision: Conclusion: 114/139 One-Tailed t Test Solution* H0: = 5 Ha: > 5 = df = Critical Value(s): Test Statistic: Decision: Conclusion: 115/139 One-Tailed t Test Solution* H0: = 5 Ha: > 5 = .05 df = 10 - 1 = 9 Critical Value(s): Test Statistic: Decision: Conclusion: 116/139 One-Tailed t Test Solution* H0: = 5 Ha: > 5 = .05 df = 10 - 1 = 9 Critical Value(s): Test Statistic: Decision: Reject .05 0 1.8331 Conclusion: t 117/139 One-Tailed t Test Solution* H0: = 5 Ha: > 5 = .05 df = 10 - 1 = 9 Critical Value(s): Test Statistic: X 6.4 5 t 1.31 S 3.373 n 10 Decision: Reject .05 0 1.8331 Conclusion: t 118/139 One-Tailed t Test Solution* H0: = 5 Ha: > 5 = .05 df = 10 - 1 = 9 Critical Value(s): Reject .05 0 1.8331 Test Statistic: X 6.4 5 t 1.31 S 3.373 n 10 Decision: Do not reject at = .05 Conclusion: t 119/139 One-Tailed t Test Solution* H0: = 5 Ha: > 5 = .05 df = 10 - 1 = 9 Critical Value(s): Reject .05 0 1.8331 t Test Statistic: X 6.4 5 t 1.31 S 3.373 n 10 Decision: Do not reject at = .05 Conclusion: There is no evidence average is more than 5 120/139