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Confidence Intervals with Means Chapter 9 Formula: Standard deviation of Critical value statistic Confidence Interval : x z * n statistic Margin of error Student’s t- distribution • Developed by William Gosset • Continuous distribution • Unimodal, symmetrical, bell-shaped density curve • Above the horizontal axis • Area under the curve equals 1 • Based on degrees of freedom df = n - 1 How does the t-distributions compare to the standard normal distribution? • Shorter & more spread out • More area under the tails • As n increases, t-distributions become more like a standard normal distribution Formula: Standard deviation of Standard error – Critical value statistic when you substitute s for . Confidence Interval : s x t * n statistic Margin of error How to find t* Can also use invT on the calculator! Need upper t* value with 5% is above – so 95% is below invT(p,df) Find these t* 90% confidence when n = 5 95% confidence when n = 15 t* =2.132 t* =2.145 Steps for doing a confidence interval: 1) Assumptions – 2) Calculate the interval 3) Write a statement about the interval in the context of the problem. We are ________% confident that the true mean context is between ______ and ______. Assumptions for t-inference • Have an SRS from population (or randomly assigned treatments) • unknown • Normal (or approx. normal) distribution – Given – Large sample size – Check graph of data Use only one of these methods to check normality Ex. 2) A medical researcher measured the pulse rate of a random sample of 20 adults and found a mean pulse rate of 72.69 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 70.883 & 74.497. Ex. 3) Consumer Reports tested 14 randomly selected brands of vanilla yogurt and found the following numbers of calories per serving: 160 200 220 230 120 180 140 130 170 190 80 120 100 170 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 126.16 calories & 189.56 calories. Ex 3 continued) A diet guide claims that you will get 120 calories from a serving Note: confidence intervals tell us of vanilla yogurt. What does this if something is NOT EQUAL – evidence indicate? 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. CI & p-values deal with area in the tails Robust – is the area changed greatly when there is skewness • An inference procedure is ROBUST if the confidence level or p-value doesn’t change much if the normality assumption is violated. Since there is more area in the tails in tdistributions,can then, a distribution has • t-procedures beif used with some some skewness, tail area not skewness, as long the as there areisno greatly affected. outliers. • Larger n can have more skewness. Find a sample size: • If a certain margin of error is wanted, then to find the sample size necessary for that margin of error use: m z * n Always round up to the nearest person! Ex 4) The heights of SHS male students is normally distributed with = 2.5 inches. How large a sample is necessary to be accurate within + .75 inches with a 95% confidence interval? n = 43 Some 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 Cautions continued: • Outliers can have a large effect on confidence interval • Must know to do a z-interval – which is unrealistic in practice Hypothesis Tests One Sample Means Steps for doing a hypothesis test “Since the p-value < (>) a, I reject 1) Assumptions (fail to reject) the H0. There is (is not) sufficient evidence to suggest thathypotheses Ha (in context).” 2) Write & define parameter H0: m = 12 vs Ha: m (<, >, or ≠) 12 3) Calculate the test statistic & p-value 4) Write a statement in the context of the problem. Assumptions for t-inference • Have an SRS from population (or randomly assigned treatments) • unknown • Normal (or approx. normal) distribution – Given – Large sample size – Check graph of data Use only one of these methods to check normality Formulas: unknown: statistic - parameter test statistic standard deviation of statistic t= x m s n Calculating p-values • For z-test statistic – – Use normalcdf(lb,ub) – [using standard normal curve] • For t-test statistic – – Use tcdf(lb, ub, df) Example 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 = .1 299.4 297.7 298.9 300.2 297 301 • I have an SRS of bottles SRS? Normal? •Since the boxplot is approximately symmetrical with no outliers, the sampling distribution is approximatelyHow do you know? normally distributed Do you know ? What are your H0: m = 300 where m is the true mean amount hypothesis statements? Is Ha: m < 300 of cola in bottles there a key word? 299 .03 300 t 1.576 p-value =.0880 a = .1 1.503 Plug p-value values to Compare your 6 into decision formula. a & make Since p-value < a, I reject the null hypothesis. Writethat conclusion in There is sufficient evidence to suggest the true context in terms of Ha. mean cola in the bottles is less than 300 mL. • is unknown Example 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? Assume: •Have an SRS of weeks •Distribution of sales is approximately normal due to large sample size • unknown H0: m = 1323 where m is the true mean cookie sales error in context? Ha: m < 1323What is the per potential week What is a consequence of that error? 1208 1323 t 2.29 p value .0147 275 30 Since 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. Example 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 = ($1122.70, $1293.30) Based on this interval, is the mean weekly sales rate statistically less than the reported $1323? Matched Pairs Test A special type of t-inference Matched Pairs – two forms • Pair individuals by certain characteristics • Randomly select treatment for individual A • Individual B is assigned to other treatment • Assignment of B is dependent on assignment of A • Individual persons or items receive both treatments • Order of treatments are randomly assigned or before & after measurements are taken • The two measures are dependent on the individual Is 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 employment No, there is no pairing of individuals, you have two independent samples Is 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 samples No, 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. Is 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. A 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 You may subtract either company thewhen following data on 15 way – collected just be careful writing Hadays over the past randomly selected month. (Note: days were not consecutive.) Day 1 2 Morning 8 9 3 4 5 6 7 8 9 10 11 12 13 14 15 7 9 10 13 10 8 2 5 7 7 6 8 7 After8 10 9 8 9 11 8 noon Since you have two values for 10 4 7 8 9 6 6 9 First, you must find the differences for each day. each day, they are dependent on the day – making this data matched pairs Day 1 2 3 4 5 6 7 Morning 8 9 7 9 10 13 10 Afternoon 8 10 9 8 9 11 8 9 10 11 8 2 5 1 2 1 3 1 4 1 5 7 7 6 8 7 8 10 4 7 8 9 6 6 9 I subtracted: Differen Morning – afternoon - 0 -1 -2 1 1 2 2 -2 -2 -2 0 2 ces 1 2 2 You could subtract the other Assumptions: way! • Have an SRS of days for whale-watching You need to state assumptions using the • unknown differences! •Since the normal probability plot is approximately linear, the distribution of difference is approximately Notice the granularity in this normal. plot, it is still displays a nice linear relationship! Differences 0 -1 -2 1 1 2 2 -2 -2 -2 -1 -2 0 2 Is there sufficient evidence that more whales are sighted in the afternoon? H0: mD = 0 Ha: mD < 0 Be careful writing your Ha! Think about how you– If you subtract afternoon subtracted: M-A Hdifferences mD>0should Notice morning; we mthen a:more D foris Ifused afternoon & it equals since the nullbeshould the0 differences + or -? be that there NOat difference. Don’t islook numbers!!!! Where mD is the true mean difference in whale sightings from morning minus afternoon -2 Differences 0 -1 -2 1 1 2 2 -2 finishing the hypothesis test: x m .4 0 t .945 s 1.639 n 15 p .1803 df 14 a .05 -2 -2 -1 -2 0 2 In your calculator, perform t-test Notice athat if the youusing subtracted differences (L3) A-M, then your test statistic t = + .945, but pvalue would be the same Since p-value > a, I fail to reject H0. There is How could I insufficient evidence to suggest that more whales increase theare sighted in the afternoon than in the morning. power of this test? -2 Two-Sample Inference Procedures with Means Remember: m m m x y x y x y 2 2 x y We will be intereste d in the differen ce of means, so we will use this to find standard error. Suppose we have a population of adult men with a mean height of 71 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 (malefemale). Normal distribution with mx-y =6 inches & x-y =3.471 inches Female 65 Male 71 Difference = male - female 6 = 3.471 a) What 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? P((xM-xF) < 5) = normalcdf(-∞,5,6,3.471) = .3866 b) What is the 70th percentile for the difference (male-female) in heights of a randomly selected man & woman? (xM-xF) = invNorm(.7,6,3.471) = 7.82 • Two-Sample Procedures with means When we compare, The goal of these inferencewhat are we procedures is to compare theinterested in? responses to two treatments or to compare the characteristics of two populations. • We have INDEPENDENT samples from each treatment or population Assumptions: • Have two SRS’s from the populations or two randomly assigned treatment groups • Samples are independent • Both distributions are approximately normally – Have large sample sizes – Graph BOTH sets of data • ’s unknown Formulas Since in real-life, we will NOT know both ’s, we will do t-procedures. Degrees of Freedom Option 1: use the smaller of the two values n1 – 1 and n2 – 1 This will produce conservative results – higher p-values & lower confidence. Calculator Option 2: approximation used bydoes this automatically! technology s s 2 2 1 2 1 2 2 n n df 1 s 1 s n 1 n n 1 n 1 2 2 1 2 1 2 2 Confidence Called intervals: standard error CI statistic critical value SD of statistic s s x x t * n n 1 2 2 1 2 1 2 2 Pooled procedures: • Used for two populations with the same variance • When you pool, you average the twosample 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! Two competing headache remedies claim to give fastacting 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 20.1 8.7 12 Brand B 18.9 7.5 12 Describe the shape & standard error for sampling distribution of the differences in the mean speed of absorption. (answer on next screen) Describe the sampling distribution of the differences in the mean speed of absorption. Normal distribution with S.E. = 3.316 Find a 95% confidence interval difference in mean lengths of time required for bodily absorption of each brand. (answer on next screen) Assumptions: State assumptions! Thinkrandomly “Price assigned is Right”! Have 2 independent treatments Given the absorption rate is normally distributed ’s unknown Closest without going over & calculations s12 s22 Formula x1 x2 t * df 21.53 n1 n2 2 2 8.7 7.5 20.1 18.9 2.080 (5.685,8.085) 12 12 From calculator df = Conclusion in context We are 95% confident that the true difference in mean 21.53, use t* for df = lengths of time required for bodily absorption of each 21 & 95% confidence brand is between –5.685 minutes and 8.085 minutes. level Note: confidence interval statements • Matched pairs – refer to “mean difference” • Two-Sample – refer to “difference of means” Hypothesis Statements: H0: m1 - m2 = 0 H0: m1 = m2 Ha: m1 - m2 < 0 Ha: m1< m2 Ha: m1 - m2 > 0 Ha: m1> m2 Ha: m1 - m2 ≠ 0 Ha: m1 ≠ m2 Be sure to define BOTH m1 and m2! Hypothesis Test: Test statistic Since we usually assume H0 is true, statistic parameter then this equals 0 – can usually SDsoofwestatistic leave it out x x m m t 1 2 1 2 2 1 2 1 2 s s n n 2 The length of time in minutes for the drugs to reach a specified level in the blood was recorded. The results follow: Brand A Brand B mean 20.1 18.9 SD 8.7 7.5 n 12 12 Is there sufficient evidence that these drugs differ in the speed at which they enter the blood stream? Have 2 independent randomly assigned treatments State assumptions! Given the absorption rate is normally distributed ’s unknown H0: mA= mB Hypotheses & define variables! Where mA is the true mean absorption time for Brand A & mB is the true mean absorption time for Brand B Ha:mA= mB x1 x2 20.1 18.9 t .361& calculations Formula s12 s22 8.7 2 7.52 n1 n2 12 12 Conclusion in context p value .7210 df 21.53 α .05 Since 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. Suppose that the sample mean of Brand B is 16.5, then is Brand B faster? t x1 x2 s12 s22 n1 n2 20.1 16.5 8.7 2 7.52 12 12 1.085 p value .2896 df 21.53 α .05 No, I would still fail to reject the null hypothesis. Robustness: • Two-sample procedures are more robust than one-sample procedures • BEST to have equal sample sizes! (but not necessary)