Hypothesis Tests – 1 sample Means 1. Marriage. In 1960, census
Transcription
Hypothesis Tests – 1 sample Means 1. Marriage. In 1960, census
Hypothesis Tests – 1 sample Means 1. Marriage. In 1960, census results indicated that the age at which American men first married had a mean of 23.3 years. It is widely suspected that young people today are waiting longer to get married. We want to find out if the mean age of first marriage has increased during the past 40 years. a) Write appropriate hypotheses. b) We plan to test our hypothesis by selecting a random sample of 40 men who married for the first time last year. Do you think the necessary assumptions for inference are satisfied? Explain. c) Describe the approximate sampling distribution model for the mean age in such samples. d) The men in our sample married at an average age of 24.2 years with a standard deviation of 5.3 years. What's the P-value for this result? e) Explain (in context) what this P-value means. f) What's your conclusion? 2. Fuel economy. A company with a large fleet of cars hopes to keep gasoline costs down, and sets a goal of attaining a fleet average of at least 26 miles per gallon. To see if the goal is being met, they check the gasoline usage for 50 company trips chosen at random, finding a mean of 25.02 mpg and a standard deviation of 4.83 mpg. Is this strong evidence that they have failed to attain their fuel economy goal? a) Write appropriate hypotheses. b) Are the necessary assumptions to perform inference satisfied? c) Describe the sampling distribution model of mean fuel economy for samples like this. d) Find the P-value. e) Explain what the P-value means in this context. f) State an appropriate conclusion. 3. Cars. 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 in the United States; they believe it is 3000 pounds. To see if that estimate is correct, they checked a random sample of 91 cars. For that group the mean weight was 2919 pounds, with a standard deviation of 531.5 pounds. Is this strong evidence that the mean weight of all cars is not 3000 pounds? 4. Portable phones. A manufacturer claims that a new design for a portable phone has increased the range to 150 feet, allowing many customers to use the phone throughout their homes and yards. An independent testing laboratory found that a random sample of 44 of these phones worked over an average distance of 142 feet, with a standard deviation of 12 feet. Is there evidence that the manufacturer's claim is false? 5. Chips ahoy. In 1998, as an advertising campaign, the Nabisco Company announced a "1000 Chips Challenge," claiming that every 18-ounce bag of their Chips Ahoy cookies contained 1000 chocolate chips. Dedicated Statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies, and counted the chocolate chips. Some of their data are given below. (Chance, 12, no. 1 [1999]) 1219 1244 1214 1258 1087 1356 1200 1132 1419 1191 1121 1270 1325 1295 1345 1135 a) Check the assumptions and conditions for inference. Comment on any concerns you have. b) What does this evidence say about Nabisco's claim? 6. Yogurt. Consumer Reports tested 14 brands of vanilla yogurt and found the following numbers of calories per serving: 160 130 200 170 220 190 230 80 120 120 180 100 140 170 a) Check the assumptions and conditions for inference. b) A diet guide claims that you will get 120 calories from a serving of vanilla yogurt. What does this evidence indicate? 7. Maze. Psychology experiments sometimes involve testing the ability of rats to navigate mazes. The mazes are classified according to difficulty, as measured by the mean length of time it takes rats to find the food at the end. One researcher needs a maze that will take rats an average of about one minute to solve. He tests one maze on several rats, collecting the data at the right. a) Plot the data. Do you think Time (sec) the conditions for inference 38.4 57.6 are satisfied? Explain. 46.2 55.5 b) Test the hypothesis that the 62.5 49.5 mean completion time for 38.0 40.9 this maze is 60 seconds. 62.8 44.3 What is your conclusion? 33.9 93.8 c) Eliminate the outlier, and 50.4 47.9 test the hypothesis again. 35.0 69.2 What is your conclusion? 52.8 46.2 d) Do you think this maze 60.1 56.3 meets the "one-minute 55.1 average" requirement? Explain. 8. Braking. A tire manufacturer is considering a newly designed tread pattern for its all-weather tires. Tests have indicated that these tires will provide better gas mileage and longer treadlife. The last remaining test is for braking effectiveness. The company hopes the tire will allow a car traveling at 60 mph to come to a complete stop within an average of 125 feet after the brakes are applied. They will adopt the new tread pattern unless there is strong evidence that the tires do not meet this objective. The distances (in feet) for 10 stops on a test track were 129, 128, 130, 132, 135, 123, 102, 125, 128, and 130. Should the company adopt the new tread pattern? Test an appropriate hypothesis and state your conclusion. Explain how you dealt with the outlier, and why you made the recommendation you did. Answers: 1. a) H0: = 23.3; HA: > 23.3 b) We have a random sample that comprises less than 10% of the population. Population may not be normally distributed, as it would be easier to have a few much older men at their first marriage than some very young men. However, with a sample size of 40, y should be approximately Normal. We should check the histogram for severity of skewness and possible outliers. c) t39(23.3, s/ 40 ) d) 0.1414 e) If the average age at first marriage is still 23.3 years, there is a 14.5% chance of getting a sample mean of 24.2 years or older simply from natural sampling variation. f) We have not shown that the average age at first marriage has increased from the mean of 23.3 years. 2. a) H0: = 26; HA: < 26 b) We have a representative sample, less than 10% of all trips, and a large enough sample that skewness should not be a problem. c) t49(26, 4.83/ 50 ) d) 0.0789 e) If the average fuel economy is 26 mpg, the chance of obtaining a sample mean of 25.02 or less by natural sampling variation is 8%. f) Since 0.05 < P < 0.10, there is some evidence that the company may not be achieving the fuel economy goal. 3. No. Based on this large random sample, t = -1.45; P-value = 0.1494. Because the P-value is high, we fail to reject H0. These data do not provide evidence to support the claim that the average weight of all cars is not 3000 pounds. We retain the null hypothesis that the mean is 3000 pounds. 4. Yes. Based on this large random sample, t = -4.42, P-value = 3.29 X 10-5. Because the P-value is low, we reject H0. These data show that the range of these phones is not 150 feet; in fact, it is less. 5. a) Random sample; less than 10% of all Chips Ahoy bags; the nearly Normal condition seems reasonable from a Normal probability plot. The histogram is roughly unimodal and symmetric with no outliers. b) t = 10.105, P-value = 4.35 x 10-8. Since p-value < α, we reject that the population mean amount of chocolate chips in Chips Ahoy is 1000. 6. a) Random sample; this is less than 10% of all vanilla yogurt brands. Nearly Normal condition is reasonable by examining a Normal probability plot. The histogram is roughly unimodal (although somewhat uniform) and symmetric with no outliers. b) t = 3.165, P-value = 0.0075. Since p-value < α, we reject that the population mean amount of calories in vanilla yogurt is 120. 7. a) The Normal probability plot is relatively straight, with one outlier at 93.8 sec. Without the outlier, the conditions seem to be met. The histogram is roughly unimodal and symmetric with no other outliers. b) t = -2.63, P-value = 0.0160. With the outlier included, we might conclude that the mean completion time for the maze is not 60 seconds; in fact, it is less. c) t = -4.46, P-value = 0.0003. Because the P-value is so small, we reject H0. Without the outlier, we conclude that the average completion time for the maze is not 60 seconds; in fact, it is less. The outlier here did not change the conclusion. d) The maze does not meet the "one-minute average" requirement. Both tests rejected a null hypothesis of a mean of 60 seconds. 8. The data value of 102 feet is an outlier. When this is removed, the Normal probability plot is relatively straight. The Nearly Normal Condition seems satisfied. With the outlier removed, the histogram is roughly unimodal and symmetric with no other outliers. H0: = 125; HA: > 125 feet. With the outlier eliminated, x = 128.89, t = 3.29, Pvalue = 0.01. With a P-value this low, we reject H0. There is strong evidence to suggest that the tires will not bring the car to a complete stop within 125 feet. On the basis of these data, the company should not adopt the new tread pattern. Only 2 out of the 10 data values were less than the desired 125 feet, and 1 of these was an outlier.