2Formula: Margin of error Standard deviation of statistic Critical valuestatisticMargin of error
3Student’s t- distribution Developed by William GossetContinuous distributionUnimodal, symmetrical, bell-shaped density curveAbove the horizontal axisArea under the curve equals 1Based on degrees of freedomdf = n - 1
4How does the t-distributions compare to the standard normal distribution? Shorter & more spread outMore area under the tailsAs n increases, t-distributions become more like a standard normal distribution
5Standard error – when you substitute s for s. Formula:Standard deviation of statisticStandard error – when you substitute s for s.Critical valuestatisticMargin of error
6How to find t* Find these t* 90% confidence when n = 5 Can also use invT on the calculator!Need upper t* value with 5% is above – so 95% is belowinvT(p,df)Find these t*90% confidence when n = 595% confidence when n = 15t* =2.132t* =2.145
7Steps for doing a confidence interval: Assumptions –Calculate the intervalWrite a statement about the interval in the context of the problem.We are ________% confident that the true mean context is between ______ and ______.
8Assumptions for t-inference Have an SRS from population (or randomly assigned treatments)s unknownNormal (or approx. normal) distributionGivenLarge sample sizeCheck graph of dataUse only one of these methods to check normality
9Ex. 2) A medical researcher measured the pulse rate of a random sample of 20 adults and found a mean pulse rate of beats per minute with a standard deviation of 3.86 beats per minute. Assume pulse rate is normally distributed. Compute a 95% confidence interval for the true mean pulse rates of adults.We are 95% confident that the true mean pulse rate of adults is between &
10Ex. 3) Consumer Reports tested 14 randomly selected brands of vanilla yogurt and found the following numbers of calories per serving:Compute a 98% confidence interval for the average calorie content per serving of vanilla yogurt.We are 98% confident that the true mean calorie content per serving of vanilla yogurt is between calories & calories.
11Ex 3 continued) A diet guide claims that you will get 120 calories from a serving of vanilla yogurt. What does this evidence indicate?Note: confidence intervals tell us if something is NOT EQUAL – never less or greater than!Since 120 calories is not contained within the 98% confidence interval, the evidence suggest that the average calories per serving does not equal 120 calories.
12RobustCI & p-values deal with area in the tails – is the area changed greatly when there is skewnessAn inference procedure is ROBUST if the confidence level or p-value doesn’t change much if the normality assumption is violated.t-procedures can be used with some skewness, as long as there are no outliers.Larger n can have more skewness.Since there is more area in the tails in t-distributions, then, if a distribution has some skewness, the tail area is not greatly affected.
13Find a sample size:If a certain margin of error is wanted, then to find the sample size necessary for that margin of error use:Always round up to the nearest person!
14Ex 4) The heights of SHS male students is normally distributed with s = 2.5 inches. How large a sample is necessary to be accurate within inches with a 95% confidence interval?n = 43
15Some Cautions:The data MUST be a SRS from the population (or randomly assigned treatment)The formula is not correct for more complex sampling designs, i.e., stratified, etc.No way to correct for bias in data
16Cautions continued:Outliers can have a large effect on confidence intervalMust know s to do a z-interval – which is unrealistic in practice
18Steps for doing a hypothesis test “Since the p-value < (>) a, I reject (fail to reject) the H0. There is (is not) sufficient evidence to suggest that Ha (in context).”AssumptionsWrite hypotheses & define parameterCalculate the test statistic & p-valueWrite a statement in the context of the problem.H0: m = 12 vs Ha: m (<, >, or ≠) 12
19Assumptions for t-inference Have an SRS from population (or randomly assigned treatments)s unknownNormal (or approx. normal) distributionGivenLarge sample sizeCheck graph of dataUse only one of these methods to check normality
21Calculating p-values For z-test statistic – For t-test statistic – Use normalcdf(lb,ub)[using standard normal curve]For t-test statistic –Use tcdf(lb, ub, df)
22Example 1: Bottles of a popular cola are supposed to contain 300 mL of cola. There is some variation from bottle to bottle. An inspector, who suspects that the bottler is under-filling, measures the contents of six randomly selected bottles. Is there sufficient evidence that the bottler is under-filling the bottles? Use a = .1
23What are your hypothesis statements? Is there a key word? SRS?I have an SRS of bottlesNormal?How do you know?Since the boxplot is approximately symmetrical with no outliers, the sampling distribution is approximately normally distributedDo you know s?s is unknownWhat are your hypothesis statements? Is there a key word?H0: m = 300 where m is the true mean amountHa: m < 300 of cola in bottlesp-value =.0880a = .1Plug values into formula.Compare your p-value to a & make decisionSince p-value < a, I reject the null hypothesis.Write conclusion in context in terms of Ha.There is sufficient evidence to suggest that the true mean cola in the bottles is less than 300 mL.
24Example 3: The Wall Street Journal (January 27, 1994) reported that based on sales in a chain of Midwestern grocery stores, President’s Choice Chocolate Chip Cookies were selling at a mean rate of $1323 per week. Suppose a random sample of 30 weeks in 1995 in the same stores showed that the cookies were selling at the average rate of $1208 with standard deviation of $275. Does this indicate that the sales of the cookies is lower than the earlier figure?
25What is the potential error in context? Assume:Have an SRS of weeksDistribution of sales is approximately normal due to large sample sizes unknownH0: m = where m is the true mean cookie salesHa: m < per weekSince p-value < a of 0.05, I reject the null hypothesis. There is sufficient evidence to suggest that the sales of cookies are lower than the earlier figure.What is the potential error in context?What is a consequence of that error?
26Example 9: President’s Choice Chocolate Chip Cookies were selling at a mean rate of $1323 per week. Suppose a random sample of 30 weeks in 1995 in the same stores showed that the cookies were selling at the average rate of $1208 with standard deviation of $275. Compute a 90% confidence interval for the mean weekly sales rate.CI = ($ , $ )Based on this interval, is the mean weekly sales rate statistically less than the reported $1323?
27A special type of t-inference Matched Pairs TestA special type oft-inference
28Matched Pairs – two forms Pair individuals by certain characteristicsRandomly select treatment for individual AIndividual B is assigned to other treatmentAssignment of B is dependent on assignment of AIndividual persons or items receive both treatmentsOrder of treatments are randomly assigned or before & after measurements are takenThe two measures are dependent on the individual
29Is this an example of matched pairs? 1)A college wants to see if there’s a difference in time it took last year’s class to find a job after graduation and the time it took the class from five years ago to find work after graduation. Researchers take a random sample from both classes and measure the number of days between graduation and first day of employmentNo, there is no pairing of individuals, you have two independent samples
30Is this an example of matched pairs? 2) In a taste test, a researcher asks people in a random sample to taste a certain brand of spring water and rate it. Another random sample of people is asked to taste a different brand of water and rate it. The researcher wants to compare these samplesNo, there is no pairing of individuals, you have two independent samples – If you would have the same people taste both brands in random order, then it would be an example of matched pairs.
31Is this an example of matched pairs? 3) A pharmaceutical company wants to test its new weight-loss drug. Before giving the drug to a random sample, company researchers take a weight measurement on each person. After a month of using the drug, each person’s weight is measured again.Yes, you have two measurements that are dependent on each individual.
32A whale-watching company noticed that many customers wanted to know whether it was better to book an excursion in the morning or the afternoon. To test this question, the company collected the following data on 15 randomly selected days over the past month. (Note: days were not consecutive.)You may subtract either way – just be careful when writing HaDay123456789101112131415MorningAfter-noonSince you have two values for each day, they are dependent on the day – making this data matched pairsFirst, you must find the differences for each day.
33-1 -2 I subtracted: Morning – afternoon Day123456789101112131415MorningAfter-noonDifferences-1-2I subtracted:Morning – afternoonYou could subtract the other way!Assumptions:Have an SRS of days for whale-watchings unknownSince the normal probability plot is approximately linear, the distribution of difference is approximately normal.You need to state assumptions using the differences!Notice the granularity in this plot, it is still displays a nice linear relationship!
34Differences-1-212Is there sufficient evidence that more whales are sighted in the afternoon?Be careful writing your Ha!Think about how you subtracted: M-AIf afternoon is more should the differences be + or -?Don’t look at numbers!!!!If you subtract afternoon – morning; then Ha: mD>0H0: mD = 0Ha: mD < 0Where mD is the true mean difference in whale sightings from morning minus afternoonNotice we used mD for differences& it equals 0 since the null should be that there is NO difference.
35finishing the hypothesis test: Differences-1-212finishing the hypothesis test:Since p-value > a, I fail to reject H0. There is insufficient evidence to suggest that more whales are sighted in the afternoon than in the morning.In your calculator, perform a t-test using the differences (L3)Notice that if you subtracted A-M, then your test statistict = , but p-value would be the sameHow could I increase the power of this test?
37Remember:We will be interested in the difference of means, so we will use this to find standard error.
38mx-y =6 inches & sx-y =3.471 inches Suppose we have a population of adult men with a mean height of inches and standard deviation of 2.6 inches. We also have a population of adult women with a mean height of 65 inches and standard deviation of 2.3 inches. Assume heights are normally distributed.Describe the distribution of the difference in heights between males and females (male-female).Normal distribution withmx-y =6 inches & sx-y =3.471 inches
397165FemaleMale6Difference = male - females = 3.471
40What is the probability that the height of a randomly selected man is at most 5 inches taller than the height of a randomly selected woman?b) What is the 70th percentile for the difference (male-female) in heights of a randomly selected man & woman?P((xM-xF) < 5) = normalcdf(-∞,5,6,3.471) = .3866(xM-xF) = invNorm(.7,6,3.471) = 7.82
41Two-Sample Procedures with means When we compare, what are we interested in?The goal of these inference procedures is to compare the responses to two treatments or to compare the characteristics of two populations.We have INDEPENDENT samples from each treatment or population
42Assumptions:Have two SRS’s from the populations or two randomly assigned treatment groupsSamples are independentBoth distributions are approximately normallyHave large sample sizesGraph BOTH sets of datas’s unknown
43FormulasSince in real-life, we will NOT know both s’s, we will do t-procedures.
44Calculator does this automatically! Degrees of FreedomOption 1: use the smaller of the two values n1 – 1 and n2 – 1This will produce conservative results – higher p-values & lower confidence.Option 2: approximation used by technologyCalculator does this automatically!
46Pooled procedures: Used for two populations with the same variance When you pool, you average the two-sample variances to estimate the common population variance.DO NOT use on AP Exam!!!!!We do NOT know the variances of the population, so ALWAYS tell the calculator NO for pooling!
47Two competing headache remedies claim to give fast-acting relief Two competing headache remedies claim to give fast-acting relief. An experiment was performed to compare the mean lengths of time required for bodily absorption of brand A and brand B. Assume the absorption time is normally distributed. Twelve people were randomly selected and given an oral dosage of brand A. Another 12 were randomly selected and given an equal dosage of brand B. The length of time in minutes for the drugs to reach a specified level in the blood was recorded. The results follow: mean SD n Brand A Brand BDescribe the shape & standard error for sampling distribution of the differences in the mean speed of absorption. (answer on next screen)
48Describe the sampling distribution of the differences in the mean speed of absorption. Find a 95% confidence interval difference in mean lengths of time required for bodily absorption of each brand. (answer on next screen)Normal distribution with S.E. = 3.316
49Closest without going over Assumptions:Have 2 independent randomly assigned treatments Given the absorption rate is normally distributed s’s unknownState assumptions!Think “Price is Right”!Closest without going overFormula & calculationsFrom calculator df = 21.53, use t* for df = 21 & 95% confidence levelConclusion in contextWe are 95% confident that the true difference in mean lengths of time required for bodily absorption of each brand is between –5.685 minutes and minutes.
50Note: confidence interval statements Matched pairs – refer to “mean difference”Two-Sample – refer to “difference of means”
52Hypothesis Test:Since we usually assume H0 is true, then this equals 0 – so we can usually leave it out
53The length of time in minutes for the drugs to reach a specified level in the blood was recorded. The results follow:mean SD n Brand A Brand BIs there sufficient evidence that these drugs differ in the speed at which they enter the blood stream?
54State assumptions! Hypotheses & define variables! H0: mA= mB Ha:mA= mB Have 2 independent randomly assigned treatments Given the absorption rate is normally distributed s’s unknownState assumptions!Hypotheses & define variables!H0: mA= mBHa:mA= mBWhere mA is the true mean absorption time for Brand A & mB is the true mean absorption time for Brand BFormula & calculationsConclusion in contextSince p-value > a, I fail to reject H0. There is not sufficient evidence to suggest that these drugs differ in the speed at which they enter the blood stream.
55Suppose that the sample mean of Brand B is 16 Suppose that the sample mean of Brand B is 16.5, then is Brand B faster?No, I would still fail to reject the null hypothesis.
56Robustness:Two-sample procedures are more robust than one-sample proceduresBEST to have equal sample sizes! (but not necessary)