Solved Final Exam - Introduction to Methods of Probability and Statistics | MATH 243, Exams of Probability and Statistics

Download Solved Final Exam - Introduction to Methods of Probability and Statistics | MATH 243 and more Exams Probability and Statistics in PDF only on Docsity! Math 243 Final Exam NAME: March 20th 2006 Student ID: Part I: Multiple Choice. 1A. (6 points) Suppose we are testing H0 : µ = 10. We compute our z-statistic for some SRS, and find that z = 8− 10 σ/ √ 14 = −1.96. (a) What is our sample mean? (i) 10 (ii) 14 (iii) 8 (iv) µ (v) -1.96. (b) What is our sample size? (i) -1.96 (ii) 14 (iii) √ 14 (iv) 8 (v) 10. (c) We now calculate P (Z ≤ −1.96) = .025. Which of the following is true? (i) .025 is the probability that H0 is false. (ii) .025 is the probability that the mean of an SRS of size 14 satisfies x ≤ 10 if H0 is true. (iii) .025 is the probability that H0 is true. (iv) .025 is the probability that µ < 10 if the null hypothesis is true. (v) .025 is the probability that the mean of an SRS of size 14 satisfies x ≤ 8 if H0 is true. 1B. (4 points) Suppose you calculate a confidence interval for some normally distributed random variable using a sample of size 50. Your calculation tells you that µ is x ± 4.6 with 95% confidence. If you want an estimate with error 2.3 at 95% confidence, what sample size should you use? (i) 13 (ii) 25 (iii) 100 (iv) 150 (v) 200 (vi) 250 1C. (4 points) We carry out a one sample t test with 23 observations of the hypothesis H0 : µ = µ0 against Ha : µ > µ0. Our statistic has value t = 2.3. The P -value of the test is: (a) between .050 and .10 (b) between .01 and .02 (c) between .025 and .05 (d) between .8 and .9 (e) between .95 and .975. 1D. (4 points) Let X be a random number between −2 and 4. The density curve of X is shown below. 1 -2 -1 1 2 3 4 0.05 0.1 0.15 0.2 0.25 0.3 The mean of X is 1 , and the probability that X < 0 is (a) . (a) 1/3 (b) 1/2 (c) 2/3 (d) 3/4 1E. (2 points) A February 24 Zogby poll of 2665 people states that 51% of Amer- icans believe that tax cuts will lead to greater government income. Fill in the blanks below with terms from this list: population, parameter, 2665, 51%, 49% (a) The sample size: 2665 . (b) The percentage of U.S. adults who believe tax cuts will lead to greater government income: parameter . (c) The percentage of adults polled who believe tax cuts will lead to greater government income: 51% . 1F. (2 points) If data for a test of significance is significant at the .1 level then it will also be significant at the .01 level. (i) always (ii) sometimes (iii) never 1G. (2 points) The following question asks you to compare the t-distributions to the normal distribution. (a) The normal distribution is symmetrical while the t-distributions are slightly skewed. (b) The proportion of area beyond a specific value of t is less than the proportion of area beyond the corresponding value of z. (c) The greater the df, the more the t-distributions resemble the standard normal distribution. (d) All of the above. (e) None of the above. 1H. (4 points) A store has weekly sales distributed N(40, 000, 5, 000). What is the probability that the average sales for the next 4 weeks will be under $34,000? (i) .0082 (ii) .0138 (iii) .9836 (iv) .3692 (v) .0164 1I. (8 points) In the following situations, determine which one of the following statistical procedures is appropriate. Fill in each blank with the letter of the correct procedure. Procedures: A. one sample z test B. one sample t test Our calculator gives that p, our proportion is between .2457 and .3643. with confidence 99%. (b) How large should the sample size be if we want a 98% confidence interval with a margin of error of 0.02 (or less)? Our critical value for 98% is 2.326. So we calculate 2.326 .02 2 .25 = 3381.4. So we would use n = 3392. If we use our estimate of p = .305, we get z∗ m p(1 − p) = 2867.1, so we can use 2868, which is obviously better. 7. (10 points) 46 mice were trained to run a maze. After the training period they were divided into two groups of 22 and 24 each. The first group was given water laced with caffeine, and the other was given ordinary water. After one week of this, the rats were timed in the mazes again. Group n Mean maze time Std. Dev. Caffeine 22 178 49 No caffeine 24 164 26 (a) Give a 95% confidence interval for the mean time in the maze for the caffeine treated mice. We are estimating µ1 - mean time in the maze for mice given caffeine. We use the t distribution with 21 degrees of freedom. From table C, t∗ = 2.08. So our confidence interval is µ1 is in 178± 2.08 · 49/ √ 22 = 178± 21.729 with 95% confidence. (b) Give a 95% confidence interval for the difference of the mean time for the caffeine treated mice minus the mean time for the non-caffeine treated mice. µ2 is the mean time in the maze for mice who aren’t given caffeine. We want to estimate µ1 − µ2. Again we use 21 degrees of freedom (since 22 is the smaller sample size). Also t∗ is still 2.08. So our confidence interval is µ1 − µ2 is 178− 164± 2.08 √ 492 22 + 262 24 = 14± 24.373 with 95% confidence. 8. (10 points) A February 20 Zogby Poll showed 416 of 1039 approved of Pres- ident Bush’s job performance. Zogby released a similar poll of 1004 adults on March 6th showing 382 out of 1004 adults approving of Bush’s job per- formance. Do hypothesis testing to determine whether there is statistically significant evidence that Bush’s approval rating fell between February 20 and March 6. Let p1 be the proportion of adults approving of Bush’s job performance on February 20th, p2 the proportion of adults approving of Bush’s job perfor- mance on March 6. p̂1 = 416/1039 = .400, p̂2 = 382/1004 = .380. The pooled sample proportion is p̂ = 798 2043 = .391. • H0 : p1 = p2 Ha : p1 > p2. • Our test statistic is z = .4− .38√ .391(1− .391)( 1 1039 + 1 1004 ) = .926 • Our P -value is about .177. • .177 > .05 so this result is not statistically significant. We do not re- ject H0, and there is not statistically significant evidence that Bush’s approval rating fell during this period.