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APSTAT - Unit 4b Inference Continued Chapter 23 Inference about means One Sample Z-Test for Mean One Sample T-Test for Mean One Sample T-Interval for Mean One Sample Z - Test for Mean Example. The 2005 SAT: Mean 1080, Standard Deviation 180 Priory Students n = 32, mean 1200 Two Possibilities (Just like for Proportions…) Higher WPS scores just happened by chance (natural variation of a sample) The likelihood of 50 people averaging 1150 is so remote we must conclude that Priory Students are better at SAT than national average. Let’s Do It! WPS SAT Example Step 1 Define Parameter: m, the true mean score of WPS SAT test-takers Step 2 Hypotheses H0: m = 1080, Priory students perform at the same level as the National Average Ha: m > 1080, Priory students perform better than the National Average WPS SAT Example Continued Step 3 Assumptions: SRS Normal Distribution s is known Assume/not stated n>30 Yep! Step 4 Name It, Show it, DO IT One Sample Z-Test for a Mean x - m 1200 -1080 z= = = 3.77 s 180 n 32 WPS SAT Example Continued Step 5 P-value and sketch of normal curve: P(z>3.77)= 0.0000816 Z=3.77 P=0.000082 1080 1250 Step 6 Interpret P-value and Conclusion A P-Value of .0000816 indicates that there is less than a 1 in 10000 chance that a result this distant from the m happened merely by chance. Therefore, reject H0 in favor of Ha. It is very likely that WPS students performed far better on average than the National Average on the 2005 SAT Two-Sided Example Mr R is convinced that his blood-sugar count is not normal. He tests himself 35 times and finds his count to be 90. If the healthy mean for blood sugar count is 84 with a standard deviation of 23, is Mr. R’s blood sugar count abnormal? Let’s Do It! WPS SAT Example Step 1 Define Parameter: m, Riebhoff’s true mean blood sugar level Step 2 Hypotheses H0: m = 84, Boff’s blood sugar level is normal Ha: m ≠ 84, Boff’s blood sugar level is abnormal WPS SAT Example Continued Step 3 Assumptions: SRS Normal Distribution s is known Assume/not stated n>30 Yep! Step 4 Name Test and DO IT One Sample Z-Test for a Mean x m 90 84 z 1.54 s 23 n 35 WPS SAT Example Continued Step 5 P-value and sketch of normal curve: 2.P(z≠1.54) = 2(.06178) = 0.1236 1.54 m +1.54 Step 6 Interpret P-value and Conclusion A P-Value of .1236 indicates that a sample mean this different from the true mean would occur about once in every eight samples of this size simply by chance. Therefore we fail to reject H0. There is not enough evidence that Mr R’s blood sugar level is different than the population mean. Confidence and Significance… Hmm, think about a significance level of a = .05 Now think about a confidence interval of C = .95 P = 0.025 P = 0.95 P = 0.025 Z = -1.96 Z = 1.96 Or -2 (empirical) Or 2 (empirical) Apples Problem, CI Style Step 1 Define Parameter m, Step 2 Hypotheses H0: Ha: Step 3 Assumptions Recall: x 122.5 n 49 m 120 s 12 Apples Problem, CI Style Step 4 Name Test and DO IT Confidence Interval for Sample Mean x z* s n 12 122.5 1.96 49 122.5 3.36 x 122.5 n 49 m 120 s 12 Apples Problem, CI Style Step 5 Sketch of Interval: m = 120 122.5 125.86 Step 6 Interpret119.14 Interval and Conclusion We are 95% confident that the true population mean falls between 119.14 and 125.86. Because the population mean of 120 IS included in this interval, we must fail to reject H0. The mean weight of this shipment cannot be considered abnormal. Inference as a decision Example: I am the Major League Baseball equipment checker dude. In each shipment of baseballs, I select 36 balls at random to make sure they comply with MLB Standards. The rules state that the balls are to be 5.2 oz each Over the course of a great number of years, the standard deviation of weights (of accepted balls) has been 0.15 (we will accept this as s) Here is the start of my test m: the true mean weight of the balls in the shipment Hypothesis H0: m = 5.2, Shipment mean is 5.2 Ha: m ≠ 5.2, Shipment mean is not 5.2 My decision will be to accept or reject the shipment. If I reject, the balls will be sent back to the manufacturer. I will use an a of 0.05 Let’s Continue My ball sample had a mean of 5.15, Do I accept or reject the shipment? Assumptions: SRS Normal Distribution s is known Stated 36 sample size should give a normal-ish distribution Yep! Mas Step 4 Name Test and DO IT One Sample Z-Test for a Mean x m 5.2 5.15 z 2.00 s .15 n 36 Mas Mas Step 5 P-value and sketch of normal curve: 2.P(z>2.00) = 2(.02275) = 0.0455 Z=-1.96 Z=1.96 Accept Z-Score = 2.0 Step 6 Interpret P-value and Conclusion We will reject this shipment of balls since .0455<.05. There is enough evidence that this shipment of balls is not acceptable by MLB standards T-Procedures – No s given In most situations, knowing m and s is not likely. We must find a way to estimate s What we do have: x-bar: Our unbiased estimate of m s: the standard deviation from our sample. An unbiased estimate of s So…… Before: z x m s n Now: x m0 t s n Like z, with a twist. You’ll see Standard Error (SE) of the sample mean What is this t thang, baby? Like z, Shows how far from m in standard deviation units. Bell Shaped BUT, unlike z Curve changes as sample size changes If n is small, more probability rests in tail of tdistribution curve As n increases, becomes more like z-distribution (standard normal) T-distribution curve Standard Normal T (n=3) T (n=9) m UH OH!!! If curve changes as n changes, we will need a crazy chart To use it we need something new: DEGREES OF FREEDOM (k) k = n – 1 We describe the t-distribution curve with the following: t(k) – means a t-distribution with k degrees of freedom Lets do it Last page in book or Table B on AP handout Critical Value t* EX: n = 10, P = .025 to right of t* Sketch it! P=.025 df= m t*= Lets do it again! Find Critical Value t* EX: n = 18, P = .98 to left of t* P=.98 Sketch it! df= m t*= Assumptions SRS Normal population or large n Our Book’s Rule Of Thumb n<15, ok if pretty close to normal n>15, ok if no outliers or super skew n>40, all good in my ‘hood If outliers, you may eliminate them Other texts n>30 if non-normal population (but always check for outliers and skew if given raw data) s is unknown Significance test. Does body fat increase after 1 wk of McDonalds-only eating? Sample changes in body fat % from a Put in random sample of 8 people. List on TI, 1.3 0.8 You’ll see Find: n= why 2.1 df= 1.6 xbar= -0.2 s= 1.5 -1.0 2.0 Test is on! PARAMETERm: __________________________ HYPOTHESESH0: m = 0, _____________________ Ha: m > 0, _____________________ More… ASSUMPTIONS Sketch graph! SRS Normalness s is not given TEST – One sample t-test for mean x m0 1.0125 0 t 2.616 s 1.0947 n 8 More…. P(VALUE) P(t>2.616)= Look on t-dist table df=7, t=2.616 Find what probability this is between ***NOTE - T-distribution table reverse of normal distribution table Probability on the OUTSIDE, t-score on the INSIDE TI version of t-distribution table Z test we would do normalcdf(blah) T test we do: Tcdf(low, high, df) That is it. More……. Interpretation A p-value between .01 and .02 would indicate that the sample mean would occur roughly 1 in 50 to 1 in 100 samples of this size simply by chance if H0 were true. We will therefore reject H0 in favor of Ha. It is likely that eating only fast food would be a factor in increasing a person’s body fat count. Now to make you crazy and angry Remember, to do a test correctly you must throw down all the PHAT PI action. There is no shortcut in what you need to show. Now, hit STAT>TESTS>T-TEST Input:Data, m0=0, List:L1, >, calculate OH MY!!!!!! Mas TI LOVE! Go back to STAT>TESTS>T-TEST You can also enter stats, instead of drawing on data from a list Boo ya! Confidence Intervals s x t* n Based on t(k) Let’s hop right in We do a random sample of the length (in inches) of 8 senior male feet and find the following: N=8 x-bar=12 s=2.4 Construct a 95% CI Do Assumptions (no need to do a Hypothesis test, why?) Senior Feet Test – One sample confidence interval for mean of a population s x t* n 2.4 12 (2.365) 8 12 2.01 (9.99,14.01) t* is from table, C=.95, df=7 Senior Feet… Interpret CI We are 95% confident that the true mean of WPS senior foot size is between 9.99 and 14.01 inches. Chapter 24 Comparing Means Two Sample Z-Test for Mean Two Sample T-Test for Mean Two Sample T-Interval for Mean Comparing two means Two sample problems Compare characteristics of two populations Separate samples Random Sample from Population 1 C O M P A R E Random Sample from Population 2 Comparing two means Another example: Group 1 Treatment A C O M P A R E Group 2 Treatment B ***Matched pairs is different, not a 2 sample problem… Assumptions for two sample t-test 2 SRS from different and independent populations Normal Population or large enough sample size Typically n1 + n2 >40 s and m are unknown Two Sample Z Statistic ONE SAMPLE z x m0 s n Looking at difference of means TWO SAMPLE z ( x1 x2 ) ( m1 m2 ) Remember, we cant add standard deviations… s 2 1 n1 + s 2 2 n2 Two Sample T Statistic ONE SAMPLE TWO SAMPLE x m0 t s n t ( x1 x2 ) ( m1 m2 ) 2 1 2 2 s s + n1 n2 Important stuff A Two-Sample t statistic does NOT have a t distribution We are replacing 2 standard deviations with 2 standard errors It’s ok though we can use it if: We change our degrees of freedom a bit 1 way – Big Ugly Formula (P.468) Your TI does this Easier Way – Just use smaller of n1-1 or n2-1 Two Sample t* Confidence Interval 2 1 2 2 s s ( x1 x2 ) + t * + n1 n2 •If you use CI as part of a test of significance: •If m0 is included in the interval, fail to reject H0 •If m0 is included in the interval, reject H0 in favor of Ha Testing Hypotheses In a Two Sample Test: Null Hypothesis H0 : m1=m2 OR m1- m2 = 0 Basically, there is NO difference between the means Alternative Hypothesis Ha : m1> (or < or ≠) m2 OR m1- m2 > (or < or ≠) 0 Basically, there is a difference between the means or one is greater than the other Pooled Use Pooled Formula if data have exactly the same variance: s s 2 1 2 2 Honestly, since a pooled t-test is so sensitive to slight differences in variances, JUST USE REGULAR TWO-SAMPLE T-TEST. Do girls take more AP classes? An SRS is taken, here’s the data: Boys Girls P H A n 29 25 x 2.9 3.2 s 1.1 .9 Girls in AP classes T P I Do it with A 95% Confidence Interval P – Same H – Same A – Same T – 2-sample 95% t* Confidence Interval P – Don’t get one I– Chapter 25 Paired Samples Paired T-Test Paired T-Interval Matched Pairs Looking at change/difference Before training/after training Left hand/right hand So…we will find the difference/change and it will become our data! Then….We are basically performing a One Sample T-Test on these differences. Easy. Raw data 10 Students Before/After SAT Tutoring, is there a positive effect? 1 2 3 4 5 6 7 8 9 10 before 500 535 600 605 575 560 525 400 415 550 after 525 550 590 635 550 575 525 450 410 575 Diff 25 15 -10 30 -25 15 0 50 -5 25 LET’S DO IT! P H A LET’S DO IT! – Use the Calc T T-Test for difference of matched pairs P I Paired T-Interval Construct and interpret a 90% Confidence Interval for the true mean difference in the previous problem. Paired Wrap-up Not too hard, huh? Tough part is determining when to use Paired Procedures Simple Signs: 2 groups of data, same exact size Before/after data One person doing 2 things Chapter 26 Chi-Square Procedures Comparing Counts (Categorical Data) Three Tests Goodness of Fit Independence Homogeneity Chi Square – Goodness of Fit Remember M&M’s, we did 1 Prop t-test for all 6 colors. 2 Problems Took a loooooong time Doesn’t give us an overall picture of how WAC the package was overall Chi-Square Goodness of Fit “How well does our “observations” fit what we “expect” My M&M data: Observed Expected 17 13 .24(58) .20(58) 10 8 3 .14(58) .14(58) .13(58) 7 .10(58) Chi Squared is the SUM of… Observed Expected 17 .24(58) 13 .20(58) 10 .14(58) 8 .14(58) 3 .13(58) 7 .10(58) (O E ) 2 E X 2 What to do with X2 value Remember… T-score? Z-Score? X2 is same (but different) Need X2 value Degrees of Freedom (df) (Number of Categories – 1) M&M’s Example X2 = _________ df = 6-1=5 Look at X2 distribution table Note, curve is not normal Skewed right Gets “normaler’ as df raises P(X2 >___)= Between ____ and ____ Goodness of fit – on TI-83 Unfortunately no “TEST” on TI-83 Can use LISTS to make it easier Can also use X2cdf(low,high, df) For M&M’s Example P-Value=_____ Significance Test Mostly the same Parameters: Define proportions Let p1,p2,…p6 = Proportion of each color of M&Ms Hypotheses: H0 – Proportions are as stated p1=.24, p1=.20… Ha – H0 is NOT true Significance test Assumptions Observed counts – SRS Large enough sample (all expected counts are above 5 Test – We just did it P-Value - Same as before Interpretation – Same as before Let’s Do One! Are students more likely to miss school on certain days? Data from a random sample of 5 Mondays, Tuesdays…is taken. Observed MON 18 TUE 15 WED 12 THU 16 FRI 19 Absence days Parameters Let p1,p2,…p5 = Proportion of absences on each day Monday through Friday Hypotheses: H0 – p1,p2,p3,p4,p5=____ Ha – H0 is NOT true Absence Days Assumptions Observed counts – SRS Large enough sample (all expected counts are above 5) – Will Show Below Test – X2 Goodness of Fit Test (O E ) X E 2 2 Absence Days Day Observed Expected MON 18 16 TUE 15 16 WED 12 16 THU 16 16 FRI 19 16 (O-E)2/E Absence Days P VALUE X2 = ______ df = ______ P( )= Interpretation: X2 for Homogeneity and Independence We just looked at 1 category ie. Color of M&M, Day of Week Now 2 Categories. Yeah! Two Tests (done same way) Homogeneity – No difference in proportions within a category Independence – Is one variable independent of the other? Drinking Habits Does there appear to be a gender difference with respect to drinking behavior of college students? 2017 male and female students were asked to monitor their drinking over the course of a week. Levels were classified as None, Low (1-7), Medium (8-24), High (25+). Drinking Habits – Observed GENDER Drinking Level None Male Female 140 186 Low 478 661 Medium 300 173 High 63 16 Significance Test – X2 Homogeneity Parameters: NOT NEEDED HYPOTHESES H0: True Category Proportions are the same for all populations Ha: True Category Proportions are NOT the same for all populations Assumptions: Same as other X2: SRS and Expected Cell Counts > 5 Significance Test – X2 Homogeneity Test: Pretty close to Goodness-Of-Fit, but sum ALL cells. Expected cell counts: Row Marginal x Column Marginal Grand Total P-Value: Same as before with X2 distribution chart or X2cdf(). BUT df is different: df=(#Rows-1)(#Columns-1) Interpretation: Same ol’ Same ol’ DO IT! – Drinking Example First Find Expected Counts – Fill in chart (just do 2) GENDER DRINKING LEVEL MALE FEMALE NONE 140 186 LOW 478 661 MEDIUM 300 173 HIGH 63 16 Column Marginal Row Marginal ENTER THE MATRIX Matrix – Choose Edit [A] Choose (r x c) – Plug in observed #s Should Look Like Your Table Stat>Test>X2 Test Calculate (You can DRAW later for your sketch) Ignore output (for a bit) and go look at Matrix [B] (press enter) – Those are Expected Counts WRITE THEM IN! Drinking Example Hypotheses H0: True Category Proportions are the same for all populations Ha: True Category Proportions are NOT the same for all populations Assumptions SRS Large enough sample size Drinking Example TEST: Don’t write all the way out, Do This: 2 ( O E ) (__ __) (__ __) 2 X + ... + E __ __ P-Value: df = Interpret: NOW X2 - Independence Looking to see if certain category is independent of another. Example: Do blondes have more fun? Do exactly like homogeneity, but thinking (and Hypothesis and Interpretation) is a bit different. Recall: If A and B are Independent: P(A&B) = P(A) x P(B) Observed Data – from an SRS of 70 people Hair Color Fun Level Blonde Non-Blonde Always 12 12 24/70 = .34 Sometimes 9 12 21/70 = .3 Never 4 21 25/70 = .36 25/70 = .36 45/70 = .64 70 Expected Value = Row Proportion x Column Proportion x Grand Total Sig Test Differences Hypotheses: H0: Two variables are independent Ha: Two variables are NOT independent Interpretation: answer question, is there evidence against the hypotheses that the variables are independent? Everything else is same Do it! Do blondes have more fun? Is fun level independent of hair color? Hypotheses: H0: Ha: Assumptions Blonde = Fun TEST:1st - Show Completed Chart w/ Expected counts too! Hair Color Fun Level Blonde Non-Blonde Always 12 12 24/70 = .34 Sometimes 9 12 21/70 = .3 Never 4 21 25/70 = .36 25/70 = .36 45/70 = .64 70 Blonde = Fun Now conduct X2 test 2 ( O E ) (__ __) (__ __) 2 X + ... + E __ __ P-Value df = Interpret THAT’S IT! JUST ONE MORE CHAPTER TO GO AFTER THIS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Chapter 27 Inference For Regression Final Chapter!!!!! The Main Idea Inference – We take a sample Use procedures to find Could the results happen by chance variation of the sample Is there evidence that this result might not have happened by chance Now we apply this to linear relationships between 2 variables Let’s Review Bivariate Data 1st Mother’s Age and Birth Weight Age Weight (lbs) 15 5.0 17 6.3 18 7.3 15 5.8 16 6.8 19 6.6 17 6.9 16 5.6 18 7.5 19 8.1 Chuck these into L1 and L2 and let’s roll! Stuff to do with Bivariate Data Graph it (don’t forget labeling) Find LRSL (graph it and label it too!) Stuff to do with Bivariate Data Coefficient of Correlation (interpret) Coefficient of Determination (interpret) Stuff to do with Bivariate Data Graph residuals (don’t forget labels) Stuff to do with Bivariate Data Overall comment on Bivariate data. Direction, Strength, Linear-ness, Weirdness: Stuff to do with Bivariate Data Normalness of residuals (use boxwhisker, stem-leaf, or histogram) Stuff to do with Bivariate Data Predict BW for the birth weight of a child born to a 17 year-old mother. Stuff to do with Bivariate Data Interpret slope of regression line in the context of the problem. Interpret intercept of regression line in the context of the problem. Inference for an LSRL - Data Two major FR problem types Do inference based on given set of bivariate data (not common) Do inference based on the output from a statistics software program (very common) We will do both, but raw data first… Basics Our LSRL is simply an estimator of the true LSRL (which would be based on a census of the entire population). Estimated y-int Predicted y value Estimated slope ŷ a + bx Basics Real LSRL (from a census) Mean y-value for that given x-value True y-intercept True Slope (Pretty drawing stolen with love from Gabriel Tang who stole it from someone else) Basics For a given x-value, the y’s will vary normally about a my. (standard error of the residuals) Basics So….to do inference, we need the following statistics: a – an estimate of a, the y-int of my (LinReg) b – an estimate of b, the slope of my (Lin Reg) sb – the standard error of the slope of the regression line (on computer output, not TI, I will show you a trick to find it) s – the standard error of the residuals (on computer and TI output) Basics yˆ b b x 0 1 Check out the formulas on formula sheet: Make sure you can use them/know what they are Great multiple choice fodder… i.e. given r, sy, and b1 find sx b1 blob b0 blob r blob b1 blob Sb1 blob Basics - SEResiduals 1 2 s Residual n2 ( y yˆ ) Why n-2? Just deal. (TI will calculate if you have the raw data!) Basics – SE Slope SE of Residuals sb s x x 2 (TI will NOTcalculate – Boff has a trick) Standard Error of Slope Trick When we do LinRegTTest on T1, it gives us: T-value s (SE of Resids) Now the t-value formula is: b b t therefore sb sb t Yee haw, plug and chug! Significance Test for Slope of a Regression Line – From Data Idea: we are checking to see if this LSRL is a good fit for our data Usual test is whether the slope of the regression line is ZERO (no relationship) or NOT ZERO (a relationship) – Two Sided Sometimes we look to see if relationship is positive or negative (>0 or <0, respectively) – One Sided Let’s do it – Mom Age/Baby wt Parameters: Let b represent the true slope of the regression line. Hypotheses: H0: b = 0, No relationship between birth weight and mother’s age Ha: b ≠ 0, There is a relationship between birth weight and mother’s age Let’s do it – Mom Age/Baby wt Assumptions: Linear Relationship (show resids, talk about r) Variance about the line is both constant and normal (show resids and comment) WHAT TO LOOK FOR: YUCK – Curvy! YUCK – Not constant variance (Better check for outlier here) Let’s do it – Mom Age/Baby wt Assumptions: Linear Relationship Variance about the line is both constant and normal How do ours look? Let’s do it – Mom Age/Baby wt TEST – T Test for a Linear Regression On TI – STAT>TEST>LINREGTTEST Fill in Blanks b t sb Let’s do it – Mom Age/Baby wt Total P P-Value df = 10-2 = 8 2 x P(t > )= b1 b=0 b2 t t=0 t Interpretation: P-Value is very small so we will reject H0 in favor of the alternative. There is almost assuredly a relationship between birth weight and a mother’s age. Remember, we can never say one causes the other unless we have a well designed and controlled experiment Confidence Intervals (from data) Similar to other CI, but remember we are looking at slope… b t * Sb1 Our estimated slope from LinReg Get from table, need df (n-2) and confidence level Remember Riebhoff’s sneaky trick? Part 2, Computer Output Data collected by counting cricket chirps in 15 seconds and noting current temperature. Output (from some statistical software): TEMP = 44.0 + 0.993 NUMBER Predictor Coef STDev T-Ratio P CONSTANT 44.01 1.827 24.09 .000 NUMBER 0.993 0.065 15.23 .000 S = 1.538 R-sq = 95.9% R-sq(adj) = 95.5% Interpreting Computer Output TEMP = 44.0 + 0.993 NUMBER Predictor Coef STDev T-Ratio P CONSTANT 44.01 1.827 24.09 .000 NUMBER 0.993 0.065 15.23 .000 S = 1.538 R-sq = 95.9% R-sq(adj) = 95.5% This is the LSRL for predicting the temperature based on the number of chirps. Interpreting Computer Output TEMP = 44.0 + 0.993 NUMBER Predictor Coef STDev T-Ratio P CONSTANT 44.01 1.827 24.09 .000 NUMBER 0.993 0.065 15.23 .000 S = 1.538 R-sq = 95.9% R-sq(adj) = 95.5% The Coef for the CONSTANT is the y-intercept (a) of the LSRL The Coef for the NUMBER is the Slope (b) of the LSRL Interpreting Computer Output TEMP = 44.0 + 0.993 NUMBER Predictor Coef STDev T-Ratio P CONSTANT 44.01 1.827 24.09 .000 NUMBER 0.993 0.065 15.23 .000 S = 1.538 R-sq = 95.9% R-sq(adj) = 95.5% STDev of NUMBER is the standard error of the slope (Sb). Interpreting Computer Output TEMP = 44.0 + 0.993 NUMBER Predictor Coef STDev T-Ratio P CONSTANT 44.01 1.827 24.09 .000 NUMBER 0.993 0.065 15.23 .000 b S = 1.538 t R-sq = 95.9% sb R-sq(adj) = 95.5% T-ratio of NUMBER is the value of the t test statistic. Check it: b t sb P is the P-value associated with the t-ratio (assuming a 2-sided test – if you are using a 1sided test, divide by 2): Interpreting Computer Output TEMP = 44.0 + 0.993 NUMBER Predictor Coef STDev T-Ratio P CONSTANT 44.01 1.827 24.09 .000 NUMBER 0.993 0.065 15.23 .000 S = 1.538 R-sq = 95.9% R-sq(adj) = 95.5% S is the standard error of the residuals (s) Interpreting Computer Output TEMP = 44.0 + 0.993 NUMBER Predictor Coef STDev T-Ratio P CONSTANT 44.01 1.827 24.09 .000 NUMBER 0.993 0.065 15.23 .000 S = 1.538 R-sq = 95.9% R-sq(adj) = 95.5% R-sq is r-squared (coefficient of determiniation – r2 ) Interpreting Computer Output TEMP = 44.0 + 0.993 NUMBER Predictor Coef STDev T-Ratio P CONSTANT 44.01 1.827 24.09 .000 NUMBER 0.993 0.065 15.23 .000 S = 1.538 R-sq = 95.9% R-sq(adj) = 95.5% Ignore these, we do not need! Significance Test – From Output Same as before, but maybe easier… Here we go… Let’s do it – Chirps/Temp Parameters: Let b represent the true slope of the regression line. Hypotheses: H0: b = 0, No relationship between # of chirps and temperature Ha: b ≠ 0, There is a relationship between # of chirps and temperature Let’s do it – Chirps/Temp Assumptions: Linear Relationship (no resids, talk about r) Variance about the line is both constant and normal (we must assume….) Let’s do it – Mom Age/Baby wt TEMP = 44.0 + 0.993 NUMBER Predictor Coef STDev T-Ratio P CONSTANT 44.01 1.827 24.09 .000 NUMBER 0.993 0.065 15.23 .000 S = 1.538 R-sq = 95.9% R-sq(adj) = 95.5% TEST – T Test for a Linear Regression Fill in Blanks From Output b t sb Let’s do it – Chirps/Temp Total P P-Value df = not given 2 x P(t > )= Interpretation: b1 b=0 b2 t t=0 t DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE DONE YEAH!