Statistical Inference: Hypothesis Testing and Confidence Intervals - Prof. Hal Sadofsky, Exams of Probability and Statistics

Download Statistical Inference: Hypothesis Testing and Confidence Intervals - Prof. Hal Sadofsky and more Exams Probability and Statistics in PDF only on Docsity! Solutions to practice final Part I: multiple choice. (1) (4 points) Suppose we are testing H0 : µ = µ0. We compute our z-statistic for some SRS, and find that z = 7− µ0 σ/ √ 10 = 1.96. (a) What is our sample mean? (i) 10 (ii) 7 (iii) 1.96 (iv) µ0 (v) 1.96. (b) What is our sample size? (i) 1.96 (ii) 7 (iii) √ 10 (iv) µ0 (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 true. (ii) .025 is the probability that the mean of some SRS satisfies x ≥ 7 if the null hypothesis is true. (iii) .025 is the probability that H0 is false. (iv) .025 is the probability that µ 6= µ0 if the null hypothesis is true. (v) .025 is the probability that the mean of some SRS satisfies x ≥ µ0 if the null hypothesis is true. (2) (4 points) Suppose you calculate a confidence interval for your businesses monthly electric bill from a SRS of size n. Your calculation tells you that µ is x± E with 95% confidence. (a) What can you do to decrease E? i. Raise σ. ii. Lower σ. iii. Lower n. iv. Raise n. v. Raise x. vi. Lower x. (b) µ in x̄± E with 95% confidence means: i. µ has a 95% probability of being between x̄− E and x̄ + E. ii. For 95% of samples, µ is between x̄− E and x̄ + E. iii. µ > x̄ + E 5% of the time. iv. For 95% of samples, µ = x̄. (3) (2 points) We carry out a one sample t test with 18 observations of the hypothesis H0 : µ = 8 against Ha : µ < 8. Our statistic has value t = −1.8. The P -value of the test is: (a) between .050 and .10 (b) less than .025 (c) between .025 and .05 (d) between .95 and .975. (4) (4 points) Let X be the sum of two random numbers between 0 and 1. The sum X can take any value between 0 and 2. The density curve of X is the triangle shown below. 1 0.5 1 1.5 2 0.2 0.4 0.6 0.8 1 The mean of X is (a) , and the probability that X is greater than .5 is (d) . (a) 1 (b) .5 (c) .75 (d) .875 (5) (2 points) A certain set of data has the following five-number summary −10 − 5 0 5 10 Which of the following statements is correct: (a) The mean of the data is 0 (b) The standard deviation of the data is 0 (c) About 50% of the data is less than 5 (d) About 75% of the data is between −5 and 10 (6) (3 points) You discover that your business’s utility bill is normally distributed with mean $1254 and standard deviation $272. What is the probability that the average bill for the next 6 months will be greater than $1500? (i) .0134 (ii) .0268 (iii) .9732 (iv) .3692 (7) (2 points) If data for a test of significance is significant at the .1 level then it will also be significant at the .05 level. (i) always (ii) sometimes (iii) never (8) (2 points) To test if taking aspirin regularly reduces heart attacks, a double- blind randomized comparative experiment assigned 15, 000 male doctors to take aspirin and another 10, 000 to take a placebo. After 5 years, 140 of the aspirin group and 180 of the control group had died of heart attacks. The pooled sample proportion is (a) 0.0093 (b) 0.0128 (c) 0.01365 (d) 0.018 (9) (4 points) You are planning a sample survey to determine what percent of voters will vote Republican in the next presidential election. If you want your sample to be accurate to within 2% with 95% confidence, how large a sample do you need? (a) 5023 (b) 79 (c) 2401 (d) 683 (10) (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 .721 is our pooled proportion of our sample who voted. z = .75− .69√ .721(1− .721)( 1 714 + 1 634 ) = 2.452. (c) Calculate your P -value. P (z ≥ 2.452) = .007. (d) State your conclusion. If the same proportion of the population votes in both Oregon and Wash- ington, then the probability of our samples showing this many more vot- ers in Oregon is .007. This is strong evidence that more people vote in Oregon. (16) (10 points) Based on a simple random sample of 250 credit card holders, a department store found that 185 card holders incurred a monthly interest charge on an unpaid balance. (a) Find a 90% confidence interval for the proportion of credit card holders who incur a monthly interest charge. Be sure to check that the procedure is safe to use. Our sample proportion is p̂ = 185/250 = .74. If p represents our popu- lation proportion, we get that p is .74± 1.645 √ .74(1− .74)/250 = .74± .0456 with 90% confidence. This uses the “large sample” procedure, which is OK if we have at least 15 failures and successes. We could also have used the “plus four” pro- cedure. For that we’d use p = 187/254 = .736. We’d then get p is .736± 1.645 √ .736(1− .736)/254 = .736± .0455. (b) How large should the sample size be if we want a 98% confidence interval with a margin of error of 0.03 (or less)? Our critical value for 98% is 2.326. So to get our sample size we find n so n ≥ (2.326 .03 )2.74(1− .74) = 1156.6. So we’d need to take n at least 1157. (17) (10 points) A study of the inheritance of speed and endurance in mice found a trade-off between these two characteristics, both of which help mice sur- vive. To test endurance, mice were made to swim in a bucket with a weight attached to their tails (they were rescued when exhausted). Here are the data on endurance in minutes for female and male mice. Group n Mean Std. Dev. Male 135 6.7 6.69 Female 162 11.4 26.09 (a) Give a 95% confidence interval for the mean endurance of female mice swimming. We use a one-sample t-statistic here with n = 162, so we’ll have 161 degrees of freedom. From table C, we can see our critical value is about t∗ = 1.983. So µ (our mean endurance for female mice) is 11.4± 1.98326.09√ 162 = 11.4± 4.065 with confidence 95%. The TI-83 will give µ in (7.352, 15.448) with 95% confidence. This is a slightly narrower range due to the TI-83 calculating the critical value more accurately than our estimate. (b) Give a 95% confidence interval for the mean difference in endurance times. We use a 2-sample t-statistic here. We are estimating the mean en- durance time for female mice minus the mean endurance time for male mice. This number will be 11.4− 6.7± 1.983 √ 26.092 162 + 6.692 135 = 4.7± 4.22 with 95% confidence. (Our critical value is an estimate for 134 degrees of freedom, which hap- pens to be the same estimate, at least to the first three decimal places.) If we ask the TI-83, we get (.4996, 8.9004). Again, this is a narrower (thus better) estimate than the one we did directly because the calculator does a more accurate job of determining the degrees of freedom, and thus the critical value. (18) (10 points) A Gallup Poll on energy use asked 512 adults if they favored increasing the use of nuclear power as an energy source. 225 said “Yes.” Does this poll give good evidence that fewer than half of adults favor increased use of nuclear power? As usual, state hypotheses, state your P -value, and state a conclusion. The conclusion should be stated in a way that non-statisticians can understand, and should explain what probability the P -value represents. (a) Let p be the proportion of adults favoring increased use of nuclear power. Let p0 = .5. (Let p̂ = .439 be our sample proportion.) We ask if p < p0 = .5. So H0 : p = p0 and Ha : p < p0 are our hypotheses. (b) Our test statistic is z = .439− .5√ .5(1− .5)/512 = −2.76. (c) Our P -value is P (Z ≤ −2.76) = .0029. (d) Conclusion: If half of all adults favor more nuclear power, then the probability of finding 225 or fewer in a sample of 512 who favor nuclear power is only .0029. This is strong evidence against the hypothesis that half of all adults favor more nuclear power, so we conclude that probably fewer than half of all adults favor more nuclear power.