Confidence Intervals and Significance Testing in the World of T Unless you live in my animated world, Z-Testing with population σ isn’t reality… So, let’s get to the real world…. Welcome to My World Sucka!! T

Z-test for population mean Conditions we need for my Z-procedure: Data from a SRS and Population is approximately NORMAL (Large n) Yeah, Fooh… But you have to assume the POPULATION standard deviation, σ. That’s UNREALISTIC!!!! Look’s like fancy boy needs a taste of my world!! Don’t end up like Caped Boy… Do a T- Test Fooh!!

Why a T-Test? What’s T-like?  T is a density curve  Symmetric about Zero, single peaked, “bell” shaped  T’s variation depends on sample size  Remember, samples become less variable as they get larger  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!! Let me break this down really simply for you… For T- Testing, we’re still testing for Population Means, but we only need SAMPLE data!! Let’s See What Makes T, T… Check Me Out!! Wow! A tailor made T for each sample!!

T’s Statistic Standard Error Sample Standard Dev. With n-1 Degrees of Freedom Since you’re not using a Z distribution anymore, you’ll need a different statistic!! NO more Z!! One Sample T Statistic

Reading the T-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 Let’s Practice Using the Table…

Table Practice  Find the t-statistic for the following:  1) 5 dof; p =.05 (right)  2) n = 22; p =.99 (left)  3) 80% CI; n = 18  Find the p-value for the following:  1) 5 dof; t =  2) n = 12; t =  3) n = 67; t = t = t = t = p = < p < < p <.025 Notice the t-statistic is limited to certain values on your table!!! What happens if you get a T that’s not on your table? Then What? You will simply say you’re p-value is BETWEEN 2 values!!

Confidence Intervals in the T - Distribution Confidence Interval = This Confidence Interval will be approximately correct for large n. The t * comes from p area on the right half of the Confidence %... Ex: 90% CI for n = 10 P =.05 T = With n-1 Degrees of Freedom

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. Find a 95% Confidence Interval for the average Golden Circle bling weight. P =.025 for 95% CI df = 31 (round down to 30 w/ chart) I am 95% confident the mean Golden Circle bling weight is between & lbs.

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 ≤  – Reject the null; accept the H a  p >  – Fail to reject the null H a : µ > µ 0 H a : µ < µ 0 H a : µ ≠ µ 0

Significance Testing for Population Mean  For T – Tests…  Assumptions  SRS  Approx normal; large sample size (*show normality with graphs)  Conditions  Testing for Population Mean w/ unknown σ We use the Same Basic steps as in all Hypothesis Testing State the Ho and Ha in symbols and context Find the T-Statistic Find the p-value from the t-statistic w/ n-1 degrees of freedom Compare your p-value to the specified , and make your decision in context

Investigating T’s Ice Cream Mr. T has asked his factories to be sure the average ounceage of Mr. T icing on a Mr. T Ice Cream cone is 4.5 ounces. Because Mr. T is so hard, he has decided to take a simple random sample of 50 cones from his Lexington factory to see if they’re falling in line. His sample produces an average of 4.42 ounces with a standard deviation of.235 ounces. Does Mr. T have enough evidence to prove Lexington’s been skimpin’ on their T and shut them down? ( < 4.5oz) Ho: µ = 4.5oz Ha: µ < 4.5oz t = ( n – 1) df = 49 (round down to 40 for table) P is between.01 &.02… Since p is between.01 and.02, which is less than.05, I have statistically significant evidence to reject that the Lexington Factory is putting an average of 4.5 oz of T on their cones, and SQUAK them for putting on less.

Matched Pairs T-Testing  Matched Pairs Test  Match data values of different distributions based on similar characteristics  “Difference” between values is THE data  H o = µ diff = 0 [µ diff = (µ 1 - µ 2 )]  H a = µ diff or ≠ Male Rat Weight (g)Female Rat Weight (g)Difference (M – F) g

When to Use T (or not to)  Most important assumption is that the data is from an SRS  Except for small samples sizes (then you need normality)  Sample size < 15  Data close to normal - USE T  Data has outliers, skewness - DON’T USE T or state questionability of results  Sample Size ≥ 15  USE T – except for strong outliers or strong skewness  Large Samples (USE T)  Even if it is skewed, you can use T as long as n ≥ 40 This means you’ll have to add 1 more step to your Significance Tests!!! Graph your distribution for samples less than 40 to determine level of normality!!! If your HISTOGRAM doesn’t look is skewed, either scrap the t-test or talk about the questionability of the results!!

Practice  Today’s Work  Height Revisited  11.1 Worksheet  Ch 11 #’s 27 – 33 Get To Work