Stat 217 – Day 19RemindersLast Time: Test of Significance for a Population Mean (p. 395, 402)Activity 20-1 (p. 397)Other explanationsLast Time – One sample t-interval (p. 375, 384)Last Time – One Sample t IntervalPreviously- Statistical InferencePreviously – Tests of SignificancePreviously - Confidence IntervalsData CollectionHandoutLab 3: JFK vs. JFKCSolution approach 1Solution Approach 2: Central Limit TheoremButSolution Approach 2Chocolate chip vs butterscotchBUTSlide 20Stat 217 – Day 19Two-sample proceduresRemindersAdditional examples in Blackboard and Self-check ActivitiesLast Time: Test of Significance for a Population Mean (p. 395, 402)1. Let  represent a population mean2. State competing hypotheses about  Ho:  = 0Ha:  <,>, ≠ 03. Check technical conditions- Random sample from population of interest- Large sample or normal population (check by looking at the sample)4. Test statistic5. p-value (computer)6. Conclusion about Ho, summarize in Englishnsxt/0Activity 20-1 (p. 397)With a p-value of .0023, we have strong evidence to reject the null hypothesis(n) Because p-value is small, we would reject at the .10 level, the .05 level, the .01 level, the .005 level!Pretty surprising for a random sample of 25 games from a population with  =183.2 to give a sample mean of 195.88 or larger.About .23% of random samples from a population with  = 183.2 will have a sample mean of 198.55 or largerHave eliminated “random chance” as a plausible explanation for the difference we observedOther explanationsSample size was not especially large, our p-value calculation could be off (CLT)But sample was reasonably normalNot a random sample from entire seasonGames early in the season may be higher scoring than later in seasonNot an experimentEven if decide it’s convincing that scoring has gone up on average, can’t say it was because of the rule changesLast Time – One sample t-interval (p. 375, 384)We are 95% confident that (after the rule change) the population mean points scored per game () is between 187.5 and 204.25Meaning 95% of all intervals constructed this way will succeed in capturing Last Time – One Sample t IntervalNot a prediction interval (p. 381)Effects of confidence level, sample size, and now sample variability (s) as wellLess variability between samples gives “better” (more precise) estimate of population meanLess variability between observational units gives “better” (more precise) estimate of population meanPreviously- Statistical InferenceConfidence Intervals: Goal is to estimate population parameter based on sample statisticInterval of believable values for parameterTests of Significance: Goal is to assess how much evidence have against a particular claim about the parameter based on sample statisticp-value => strength of evidence against H0Previously – Tests of SignificancePopulation proportion H0: Technical conditionsSRS Test statisticp-value: computerPopulation mean H0: Technical conditionsSRSn>30 or normal populationTest statisticp-value: computer10)1(,1000nnnpz/)1(ˆ000nsxt/0 % of samples with statistic at least this extreme when Ho truePreviously - Confidence IntervalsPopulation proportion Common z* values95%: z*  2 Technical conditionsSRS Population mean t* > z*Technical conditionsSRSn>30 or normal population (look at sample))/(*nstx nppzp /)ˆ1(ˆ*ˆ10)ˆ1(,10ˆ pnpnData CollectionFlip a coinHeads = chocolate chipTails = butterscotchPlace the chip on the top of your tongue and hold it to the roof of your mouth and record how long it takes to melt (in seconds)StopwatchHandout(a) How would we decide whether this is convincing evidence that the average melting time among all Cal Poly students differs from one minute?one-sample t-test(b) How would we decide whether one type of chip takes longer to melt, on average?Lab 3: JFK vs. JFKC1. Define parameter of interest2. State Ho and Ha about parameterSolution approach 1“Randomization test” (Lab 3)Difference in means ( )21xx Should center around zero (Null hypothesis)Rather symmetric shape!7.29 pretty surprisingSolution Approach 2: Central Limit TheoremThe randomization/sampling distribution of the difference in sample means will be 1. approximately normal2. mean equal to 1-23. standard deviation equal toas long asrandom assignment or random samplesNormal populations or large sample sizes222121nnButWill instead useAnd compare to the t distributionTest of Significance Calculator: Two means222121nsnsSolution Approach 2Chocolate chip vs butterscotch1. Define parameter of interest2. State Ho and Ha about parameter3. Check the technical conditions4. Calculate the test statistic5. Calculate the p-value6. Draw conclusions (Reject Ho? Answer research question!)If given a “level of significance ,” can use that as the cut-off to determine whether your p-value is “small”BUTIs there a better way to compare chocolate chip vs. butterscotch chip melting times?(is there a better way to decide whether strength shoes improve jumping ability)Why better?Key: Getting rid of a (uninteresting) source of variability…To turn in with partner:(h) Interpret the p-value (what is it the probability of?). Then state your final decision in context.(i) Interpret the confidence interval (Hint: Remember what our parameter is)For ThursdaySelf-check Activity 22-4 (not part d)No pre-labBy MondayTopic 23 (23-1, 23-5), Review Lab