Statistics Exam: Hypothesis Testing and Confidence Intervals, Exams of Probability and Statistics

Download Statistics Exam: Hypothesis Testing and Confidence Intervals and more Exams Probability and Statistics in PDF only on Docsity! Statistics 150: Introduction to Statistics Exam 3 - Chapter 7.3, 7.4, 8, 9.1, 9.2 April 27, 2001 Name: ____________________________________ Section 1 Directions: This section contains six computational problems worth a total of 85 points. For each problem, you must present all your work, as well as the final answer, in order to receive full or partial credit. 1. A company that manufactures automobile brake pads developed a new kind of pads made of carbon and special metallic compounds. According to the company, this type of pads lasts 10% longer than the regular carbon-based pads. Historical data showed that the mean lifetime of the regular pads was 57,000 mi. with a standard deviation of 11,000 mi. Test data based on 35 common vehicles revealed that the mean lifetime of the new pads was 61,000 mi. Test the hypothesis that the new pads last longer than the regular ones at α = .01. (a) State the null and alternative hypotheses. (5 pts.) H0: µ 57000 ≤ H1: µ 57000 > (b) Show the supporting work. (8 pts.) z = (61000 – 57000)/(11000/351/2) = 2.151 z.01 = 2.326 ⇒ z > 2.326 (c) State the conclusion (reject or retain H0) and interpretation. (5 pts.) Retain H0 There is no sufficient evidence to suggest that the new pads last longer than the regular ones. ⇒ 2. Refer to Problem 1 above. Assuming that the true mean lifetime of the new pads is 62,000 mi., compute the probability that the test will result in a Type II error. (8 pts.) 0x = 57000 + 2.326(11000/351/2) = 61324.8 β = P( X < 61324.8) = P(( X – 62000)/(11000/351/2) < (61324.8 – 62000)/(11000/351/2)) = P(Z < –0.36) = .5 – .1406 = .3594 Page 1 of 4 3. Twist-off caps are very common for bottled drinks these days. However, many consumers are concerned about the cleanliness of the contents of the bottles (i.e., bottles can easily be opened before a customer purchases them). A major beer company intends to switch back to using regular caps if more than 75% of all consumers are concerned about the cleanliness of the contents. The company conducted a survey with a random sample of 250 consumers and found that 195 of them were concerned about the issue. (a) Construct a 95% confidence interval for the proportion of the people who are concerned about the cleanliness of the contents. (8 pts.) p̂ =195/250 = .780 .780 1.960(.780(.220)/250)± 1/2 = .780± .051 = (.729, .831) (b) Based on your answer in (a) above, would you conclude that the company should stop using twist-off caps? Justify your answer. (5 pts.) No, because the confidence interval contains .75. 4. An experiment was conducted to examine whether the short-term-memory capacity would differ in the morning and in the afternoon. Seven volunteering participants took a digit-span test (participants were given a series of random digits and were asked to recite the digits backwards) at 10:00 a.m. and 4:00 p.m. The obtained data and the results of the analysis using MINITAB are presented below. Mean Numbers of Digits Correctly Recited Time of day 1 2 3 4 5 6 7 x s Morning 4.3 4.4 2.9 6.1 3.3 6.7 5.2 4.70 1.39 Afternoon 4.1 4.2 3.0 5.7 3.1 6.4 5.0 4.50 1.28 Difference 0.2 0.2 –0.1 0.4 0.2 0.3 0.2 0.20 0.15 Paired T-Test and Confidence Interval Paired T for Morning - Aftnoon N Mean StDev SE Mean Morning 7 4.700 1.394 0.527 Aftnoon 7 4.500 1.275 0.482 Difference 7 0.2000 0.1528 0.0577 95% CI for mean difference: (0.0587, 0.3413) T-Test of mean difference = 0 (vs not = 0): T-Value = 3.46 P-Value = 0.013 (a) State the null and alternative hypotheses. (5 pts.) H0: µMORNING=µAFTERNOON H1: µMORNING≠ µAFTERNOON (b) With α = .05, should the null hypothesis be rejected? Justify your answer. (Note: Answer this question in reference to the confidence interval shown in the output.) (5 pts.) Yes, because the confidence interval does not contain zero. Page 2 of 4