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Do you believe the mayor of Toronto smoked crack?. Yes No This is not relevant to how great of a mayor he is. Chapter 17 &18. Inference About Means. When the conditions are met, we are ready to find the confidence interval for the population mean, μ . The confidence interval is
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Do you believe the mayor of Toronto smoked crack? • Yes • No • This is not relevant to how great of a mayor he is.
Chapter 17 &18 Inference About Means
When the conditions are met, we are ready to find the confidence interval for the population mean, μ. The confidence interval is where the standard error of the mean is The critical value depends on the particular confidence level, C, that you specify and on the number of degrees of freedom, n – 1, which we get from the sample size. A Confidence Interval for Means? (cont.) One-sample t-interval for the mean
Chips Problem • Students investigating the packaging potato chips purchased 6 bags of chips from Kroger marked with a net weight of 28.1 grams. • They weighed the contents of each bag, recording the weight as follows: • 29.2, 28.2, 29.1, 28.5, 28.8, 28.6
Data: 29.2, 28.2, 29.1, 28.5, 28.8, 28.6 • Find the • Sample mean • Sample standard deviation • Create • 95% confidence interval for the mean weight
This data is on the weight of a bag of potato chips. Interpret the 95% CI • 95% of all bags of chips will have a mean weight that falls within the interval • 95% of the chips will be contained in the interval • The interval contains the true mean weight of the contents of a bag of chips 95% of the time • We are 95% confident that the interval contains the true mean weight of the contents of a bag of chips.
Comment on the company’s state net weight of 28.1g • Since the interval is ABOVE the stated weight of 28.1 grams, there is evidence that the company is filling the bags to MORE than the stated weight, ON AVERGAGE. • Since the interval is BELOW the stated weight of 28.1 grams, there is evidence that the company is filling the bags to LESS than the stated weight, ON AVERGAGE. • Since the interval CONTAINS the stated weight of 28.1 grams, there is evidence that the company is filling the bags to the stated weight, ON AVERGAGE.
A practical sampling distribution model for means When the conditions are met, the standardized sample mean follows a Student’s t-model with n – 1 degrees of freedom. We estimate the standard error with The t-distribution and our sample
Test Rules • If |t| > t*, then reject H0 • If |t| < t*,then fail to reject H0 • If α > p-value, then reject H0 • If α < p-value, then fail to reject H0
Find p-values • Online Program • http://www.tutor-homework.com/statistics_tables/statistics_tables.html • TI-83/TI-84/TI-89 • http://www.stat.osu.edu/~stated/resources/calc_how.php
Using your software, find the following p-values • What is the p-value for t≥2.61 with 4 degrees of freedom? • What is the p-value for |t|>1.81 with 21 degrees of freedom? • What is the p-value for |t|<1.53 with 21 degrees of freedom?
Chocolate Chip Problem • A company announced a ‘1000 Chips Trial’ claiming that every 18-ounce bag of its cookies contained at least 1,000 chocolate chips. • Students purchased random bags of cookies from different stores and counted the number of chips in each bag. Data were recorded to test the claim.
Data – Create a 95% CI 1022 1142 1120 1269 1276 1228 1202 1317 1325 1491
Test the hypothesis that the average number of chips in a bag is greater than 1,000. What would you conclude? • The average number of chips is 1,000. • We have evidence that the average number of chips is greater than 1,000. • We fail to find evidence that the average number of chips is different from 1,000.
Test the hypothesis AGAIN using the t-score method with a significance level of 0.01. What would you conclude? • The average number of chips is 1,000. • We have evidence that the average number of chips is greater than 1,000. • We fail to find evidence that the average number of chips is different from 1,000.
The company claims at least 1000 Chips in EVERY bag. What would you conclude? • The company’s claim is true • The company’s claim is false • We cannot test this claim
A practical sampling distribution model for means When the conditions are met, the standardized sample mean follows a Student’s t-model with n – 1 degrees of freedom. We estimate the standard error with The t-distribution and our sample
Vehicle weight problem • One of the important factors in auto safety is the weight of the vehicle. • Insurance companies are interested in knowing the average weight of cars currently licensed. They believe it is 3,000 lbs. (i.e. hypothesize). • To test this belief, they checked a random sample of 91 cars and found: • Mean weight 2,855lbs. • SD 531.5lbs • Is this strong evidence that the mean weight of all cars is NOT 3,000lbs.?
Is this strong evidence that the mean weight of all cars is not 3000lbs, assume a significance level of 0.05? • Yes, there is sufficient evidence the mean is different from 3000 • No, there is sufficient evidence the mean is different from 3000 • Yes, there is NOT sufficient evidence the mean is different from 3000 • No, there is NOT sufficient evidence the mean is different from 3000
Test the hypothesis that the mean completion time for this maze is 60 sec., assume a significance level of 0.05. • Reject null, there is sufficient evidence to suggest the mean time is NOT 60 sec. • Reject null, there is NOT sufficient evidence to suggest the mean time is NOT 60 sec. • Fail to reject null, there is sufficient evidence to suggest the mean time is NOT 60 sec. • Fail to reject null, there is NOT sufficient evidence to suggest the mean time is NOT 60 sec.
Eliminate the outlier then, test the hypothesis that the mean completion time for this maze is 60 sec • Reject null, there is sufficient evidence to suggest the mean time is NOT 60 sec. • Reject null, there is NOT sufficient evidence to suggest the mean time is NOT 60 sec. • Fail to reject null, there is sufficient evidence to suggest the mean time is NOT 60 sec. • Fail to reject null, there is NOT sufficient evidence to suggest the mean time is NOT 60 sec.
Do you think THIS maze meets the “one-minute average” requirement? • There is NOT evidence that the mean time required for rats to complete the maze is different from 60s. The maze meets the requirements. • There is evidence that the mean time required for rats to complete the maze is different from 60s. The maze meets the requirements. • There is evidence that the mean time required for rats to complete the maze is different from 60s. The maze DOES NOT meet the requirements. • There is NOT evidence that the mean time required for rats to complete the maze is different from 60s. The maze DOES NOT meet the requirements.
Upcoming in Class • Homework #11 due Sunday • Quiz #6 is next Wednesday • Exam #2 is in two weeks • Last week of class, work on data project.