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## Confidence Intervals with Means

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**Confidence Intervals with Means**Chapter 9**Formula:**Standard deviation of statistic Critical value 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 statistic Standard error – when you substitute s for s. Critical value 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:**• Assumptions – • Calculate the interval • 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) • s 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.**Note: confidence intervals tell us if something is NOT EQUAL**– never less or greater than! Ex 3 continued) A diet guide claims that you will get 120 calories from a serving of vanilla yogurt. What does this evidence indicate? 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.**Robust**CI & p-values deal with area in the tails – 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. • 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.**If a certain margin of error is wanted, then to find the**sample size necessary for that margin of error use: Find a sample size: Always round up to the nearest person!**Ex 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 + .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 s to do a z-interval – which is unrealistic in practice**Steps for doing a hypothesis test**• Assumptions • Write hypotheses & define parameter • Calculate the test statistic & p-value • Write a statement in the context of the problem. “Since the p-value < (>) a, I reject (fail to reject) the H0. There is (is not) sufficient evidence to suggest that Ha (in context).” H0: m = 12 vs Ha: m (<, >, or ≠) 12**Assumptions for t-inference**• Have an SRS from population (or randomly assigned treatments) • s unknown • Normal (or approx. normal) distribution • Given • Large sample size • Check graph of data Use only one of these methods to check normality**Formulas:**s unknown: m t =**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**SRS?**• I have an SRS of bottles Normal? How do you know? • Since the boxplot is approximately symmetrical with no outliers, the sampling distribution is approximately normally distributed Do you know s? • sis unknown What are your hypothesis statements? Is there a key word? H0: m = 300 where m is the true mean amount Ha: m < 300 of cola in bottles p-value =.0880 a = .1 Plug values into formula. Compare your p-value to a & make decision Since 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.**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 • s unknown • H0: m = 1323 where m is the true mean cookie sales • Ha: m< 1323 per week • 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. What is the potential error in context? What is a consequence of that error?**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**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 Matched Pairs – two forms**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 Is this an example of matched pairs? No, there is no pairing of individuals, you have two independent samples**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 Is this an example of matched pairs? 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.**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. Is this an example of matched pairs? 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 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 Ha Since you have two values for each day, they are dependenton the day – making this data matched pairs First, you must find the differences for each day.**I subtracted:**Morning – afternoon You could subtract the other way! • Assumptions: • Have an SRS of days for whale-watching • s unknown • Since 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!**Is there sufficient evidence that more whales are sighted in**the afternoon? Be careful writing your Ha! Think about how you subtracted: M-A If afternoon is more should the differences be + or -? Don’t look at numbers!!!! If you subtract afternoon – morning; then Ha: mD>0 H0: mD = 0 Ha: mD < 0 Where mD is the true mean difference in whale sightings from morning minus afternoon Notice we used mD for differences & it equals 0 since the null should be that there is NO difference.**finishing 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 statistic t = + .945, but p-value would be the same How could I increase the power of this test?**Remember:**We will be interested in the difference 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 (male-female). Normal distribution with mx-y =6 inches & sx-y =3.471 inches**Female**Male 65 71 Difference = male - female 6 s = 3.471**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? 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**Two-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**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’s unknown**Formulas**Since in real-life, we will NOT know both s’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. Option 2: approximation used by technology Calculator does this automatically!**Confidence intervals:**Called standard error**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!!!!! Pooled procedures: We do NOT know the variances of the population, so ALWAYS tell the calculator NO for pooling!**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 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. 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**Assumptions:**Have 2 independent randomly assigned treatments Given the absorption rate is normally distributed s’s unknown State assumptions! Think “Price is Right”! Closest without going over Formula & calculations From calculator df = 21.53, use t* for df = 21 & 95% confidence level Conclusion in context We 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 8.085 minutes.**Note: confidence interval statements**• Matched pairs – refer to “mean difference” • Two-Sample – refer to “difference of means”