So, let’s get Unless you real live in to the myworld…. animated world, Z-Testing Welcome to with population σ T Myisn’t World Sucka!! reality… Confidence Intervals and Significance Testing in the World of T Z-test for population mean Don’t Conditions we need for my end up like Caped Z-procedure: Data from a SRS and Boy… Do a TPopulation is approximately Test Fooh!! NORMAL (Large n) Yeah, Fooh… But you boy have to Look’s like fancy That’s assume the POPULATION needs a taste of my world!! UNREALISTIC!!!! standard deviation, σ. Why a T-Test? What’s T-like? Wow! A tailor made T for each Check Me Out!! sample!! me break ForLet T- Testing, we’re T is a density curve Let’s See still testing this downfor Symmetric about Zero, single peaked, “bell” shaped What Makes Population Means, really simply T’s variation depends on sample size but T, T…SAMPLE webecome only need Remember, samples less variable as they get larger for you… data!! Degrees of Freedom T makes an adjustment for each sample size with by changing the degrees of freedom Basically gives us a new T to work with for each sample size!! T’s Statistic Since you’re not using a Z distribution anymore, you’ll need a different statistic!! NO more Z!! One Sample T Statistic __ Sample Standard Dev. x s n Standard Error With n-1 Degrees of Freedom Reading the T-Table Let’s Practice Using the Table… Degrees of Freedom (df) Left hand column of chart Different T-Distribution for each sample size Larger the sample, the closer to Normal the T distribution T-Statistic Leads to the p-value or vice versa P-Value Area to the right of t Area to the left of –t 2(P) for two-sided Table Practice What happens if you get a T Find the t-statistic for the following: that’s not on your table? 1) 5 dof;Then p = .05What? (right) t = 2.015 2) n = 22; p = .99 (left) t = 2.518 t = 1.333 3) 80% CI; n = 18 Find the p-value for the following: 1) 5 dof; t = 3.365 2) n = 12; t = 1.856 3) n = 67; t = 2.056 p = .01 .025 < p <.05 .02 < p <.025 the t-statistic YouNotice will simply say you’re is p-value limited to certain2values on is BETWEEN values!! your table!!! Confidence Intervals in the T - Distribution s x t n Confidence Interval = __ With n-1 Degrees of Freedom * This Confidence Ex: 90% CIInterval for will be approximately correct for nlarge = 10n. The t* Pcomes = .05from p area on the right half of = 1.833%... the T Confidence Confidence in T’s Bling?!! Mr. T’s looking to get into the Golden Circle bling business. He’s doing some research to find the overall average weight of Golden Circle bling so he can plan his gold needs. He bought a random sample of 32 Golden Circles from different stores and found they had an average weight of 4.6 lbs with a standard deviation of .45 lbs. IFind am 95% confident the mean Goldenfor Circle a 95% Confidence Interval thebling weight is between 4.376 & 4.7625 lbs. average Golden Circle bling weight . P = .025 for 95% CI df = 31 (round down to 30 w/ chart) .45 4.6 2.042 32 4.376 - 4.7624 Significance Testing for Population Mean (unknown σ) With these tests you are given an alpha level against which you test your p-value or you use .05 if nothing’s given: p ≤ a – Reject the null; accept the Ha p > a – Fail to reject the null Ha: µ > µ0 Ha: µ < µ0 Ha: µ ≠ µ0 Significance Testing for Population Mean For T – Tests… Assumptions Compare Find State the the your p-value Hop-value and from Hato SRS the in t-statistic symbols and w/size n-1 the specified a, and make Find the T-Statistic Approx normal; large sample (*show normality with graphs) your degrees decision context of freedom in context Conditions Testing for Population Mean w/ unknown σ We use the Same Basic steps as in all Hypothesis Testing Investigating T’s Ice Cream Mr. T has asked his factories to be sure the average ounceage of Mr. icing on .01 a Mr. Ice Cream cone is 4.5 Since p isTbetween andT.02, ounces. which Because Mr. than T is so hard, he has decided to take a is less .05, I have simple statistically random sample of 50 cones from his Lexington significant evidence factory to see if they’re falling in line. His sample reject that the Lexington produces an average of 4.42 ounces with Factory is putting an average of a standard deviation ounces. Does Mr.and T have enough evidence 4.5of.235 oz of T on their cones, to prove Lexington’s been skimpin’ their T and shut SQUAK them for putting on on less. them down? ( < 4.5oz) 4.42 4.5 (n – 1) df = 49 Ho: µ = 4.5oz t Ha: µ < 4.5oz .235 (round down to 40 for table) 50 t = -2.4072 P is between .01 & .02… Matched Pairs T-Testing Matched Pairs Test Match data values of different distributions based on similar characteristics “Difference” between values is THE data Ho = µdiff = 0 [µdiff = (µ1 - µ2)] Ha = µdiff < or > or ≠ Male Rat Weight (g) Female Rat Weight (g) Difference (M – F) g 12 9.5 2.5 11 9 8 5 3 4 When to Use T (or not to) Most important assumption that the If your HISTOGRAM This means you’ll ishave to data is from an doesn’t lookstep is skewed, add 1 more to your Except for small samples sizes (then you need normality) either scrap the t-test Significance Tests!!! Sample size or< 15 talk about the Data close toyour normaldistribution - USE T Graph for questionability of the Data has outliers, skewness - DON’T USE T or state samples results less than 40 to questionability of results!! determine level of Sample Size ≥ 15 normality!!! USE T – except for strong outliers or strong skewness SRS Large Samples (USE T) Even if it is skewed, you can use T as long as n ≥ 40 Practice Today’s Work Height Revisited 11.1 Worksheet Ch 11 #’s 27 – 33 Get To Work