Statistical Inference: Confidence Intervals and Sampling Distributions, Exercises of Probability and Statistics

Download Statistical Inference: Confidence Intervals and Sampling Distributions and more Exercises Probability and Statistics in PDF only on Docsity! STA 2023 Practice Questions Exam 2 Chapter 7- sec 9.2 Formulas Given on the test: Memorize these Formulas: General Format for Confidence Interval: estimator +/- (t or z) est. standard error General Format of Test Statistic: # ( ) oestimator from H t or z estimate of stderr   Determining sample size for estimating proportions and means. Practice Problems on next page. Case parameter estimator standard error Estimate of standard error Sampling Distribution one mean  x n  n s t (n-1) one prop. p p̂ n pp )1(  CI: ˆ ˆ(1 )p p n  ST: n pp )1( 00  z 1. You take a random sample from some population and form a 96% confidence interval for the population mean, Which quantity is guaranteed to be in the interval you form? a) 0 b)  c) x d) .96 2. Suppose you conduct a significance test for the population proportionand your p-value is 0.184. Given a 0.10 level of significance, which of the following should be your conclusion? a) accept HO b) accept HA c) Fail to reject HA d) Fail to reject HO e) Reject Ho 3. Decreasing the sample size, while holding the confidence level the same, will do what to the length of your confidence interval? a) make it bigger b) make it smaller c) it will stay the same d) cannot be determined from the given information 4. Decreasing the confidence level, while holding the sample size the same, will do what to the length of your confidence interval? a) make it bigger b) make it smaller c) it will stay the same d) cannot be determined from the given information 5. If you increase the sample size and confidence level at the same time, what will happen to the length of your confidence interval? a) make it bigger b) make it smaller c) it will stay the same d) cannot be determined from the given information 6. Which of the following is a property of the Sampling Distribution of x ? a) if you increase your sample size, x will always get closer to the population mean. b) the standard deviation of the sample mean is the same as the standard deviation from the original population  c) the mean of the sampling distribution of x is the population mean. d) x always has a Normal distribution. 7. Which of the following is true about p-values? a) a p-value must be between 0 and 1. b) if a p-value is greater than .01 you will never reject HO. c) p-values have a N(0,1) distribution d) None of the above are true. Questions 20-23 Researchers are concerned about the impact of students working while they are enrolled in classes, and they’d like to know if students work too much and therefore are spending less time on their classes than they should be. First, the researchers need to find out, on average, how many hours a week students are working. They know from previous studies that the standard deviation of this variable is about 5 hours. 20. A survey of 200 students provides a sample mean of 7.10 hours worked. What is a 95% confidence interval based on this sample? a) (6.10, 8.10) b) (6.41, 7.79) c) (6.57, 7.63) d) (7.10, 8.48) 21. Suppose that this confidence interval was (6.82, 7.38). Which of these is a valid interpretation of this confidence interval? a) There is a 95% probability that the true average number of hours worked by all UF students is between 6.82 and 7.38 hours. b) There is a 95% probability that a randomly selected student worked between 6.82 and 7.38 hours. c) We are 95% confident that the average number of hours worked by students in our sample is between 6.82 and 7.38 hours. d) We are 95% confident that the average number of hours worked by all UF students is between 6.82 and 7.38 hours. 22. We have 95% confidence in our interval, instead of 100%, because we need to account for the fact that: a) the sample may not be truly random. b) we have a sample, and not the whole population. c) the distribution of hours worked may be skewed d) all of the above 23. The researchers are not satisfied with their confidence interval and want to do another study to find a shorter confidence interval. What should they change to ensure they find a shorter confidence interval? a) They should increase their confidence level and increase their sample size. b) They should increase their confidence level but decrease their sample size. c) They should decrease their confidence level but increase their sample size. d) They should decrease their confidence level and decrease their sample size. 24. Suppose our p-value is .044. What will our conclusion be at alpha levels of .10, .05, and .01? a) We will reject Ho at alpha=.10, but not at alpha=.05 b) We will reject Ho at alpha=.10 or .05, but not at alpha=.01 c) We will reject Ho at alpha=.10, .05, or .01 d) We will not reject Ho at alpha=.10, .05, or .01 25. For each of the following situations, can we use the Z table to compute probabilities (T/F): _____ a. Weights of adults are approximately Normally distributed with mean 150 lbs and stdev 25 lbs. We want to know the probability that a randomly selected person weights more than 200 pounds. _____ b. Weights of adults are approximately Normally distributed with mean 150 lbs and stdev 25 lbs. We want to know the probability that the average weight of 10 randomly selected people is more than 200 pounds. _____ c. Weights of adults are approximately Normally distributed with mean 150 lbs and stdev 25 lbs. We want to know the probability that the average weight of 50 randomly selected people is more than 200 pounds. _____ d. Salaries at a large corporation have mean of $40,000 and stdev of $20,000. We want to know the probability that a randomly selected employee makes more than $50,000. _____ e. Salaries at a large corporation have mean of $40,000 and stdev of $20,000. We want to know the probability that the average of ten randomly selected employees is more than $50,000. _____ f. Salaries at a large corporation have mean of $40,000 and stdev of $20,000. We want to know the probability that the average of fifty randomly selected employees is more than $50,000. _____ g. A club has 50 members, 10 of which think the president should be deposed. What is the probability that, if we select 20 members at random, 18% or more in our sample think the president should be deposed? _____ h. A club has 5000 members, 1000 of which think the president should be deposed. What is the probability that, if we select 91 members at random, 18% or more in our sample think the president should be deposed? Questions 26-27 Recent studies have shown that 20% of Americans fit the medical definition of obese. A random sample of 100 Americans is selected and the number of obese in the sample is determined. 26. What is the sampling distribution of the sample proportion? a) p̂ ~ N(10, 0.2) b) p̂ ~N(2, 1.27) c) p̂ ~N(0.2, 0.04) d) Can not be determined 27. What is the probability that the sample proportion is greater than 0.24? a) 0.1841 b) 0.1587 c) 0.8413 d) 1.0 28. An auto insurance company has 32,000 clients, and 5% of their clients submitted a claim in the past year. We will take a sample 3,200 clients, and determine how many of them have submitted a claim in the past year. What is the sampling distribution of p̂ ? a) p̂ ~ N(3200, 0.2) b) p̂ ~N(160, 152) c) p̂ ~N(0.05,0.003852) d) Can not be determined Questions 29- 30 Suppose 20 donors come to a blood drive. Assume that the blood donors are not related in any way, so that we can consider them independent. The probability that the donor has type-O blood is 0.06, which is constant from donor to donor. Let X = the number of donors that have type-O blood. 29. For a sample of 100 donors, what is the sampling distribution of the sample proportion? a) p̂ ~ Binomial (100, 0.06) b) p̂ ~Normal (0.06, 0.0237) c) p̂ ~Normal(6, 2.37) d) Can not be determined 30. For a sample of 300 donors, what is the sampling distribution of the sample proportion? a) p̂ ~ Binomial (200, 0.06) b) p̂ ~Normal (12, 3.359) c) p̂ ~Normal(0.06, 0.013711) d) Can not be determined 31. For the sample of 300 donors, what is the probability that the sample proportion is greater than 0.10? a) 0.0019 b) 0.181 c) 0.819 d) 0.991 32. The executives at Sandbachian, Inc. having recently solved their widget crises, have another major problem with one of their products. Many cities are sending complaints that their manhole covers are defective and people are falling into the sewers. Sandbachian, Inc. is pretty sure that only 4% of their manhole covers are defective, but they would like to do a study to confirm this number. They are hoping to construct a 95% confidence interval to get within 0.01 of the true proportion of defective manhole covers. How many manhole covers need to be tested? a) 8 b) 1476 c) 9604 d) 9605 43. You would like to estimate the proportion of "regular users of vitamins" in a large population. In order to find a confidence interval for the proportion, a) we must assume that we have a random sample from a normal population b) we must assume that we have a random sample from a binomial population where np> 15 and n(1-p)> 15 c) we must assume that the population is normal (but we do not require a random sample because of the Central Limit Theorem). d) we do not need to assume that the population is normal nor that the sample is random (because of the Central Limit Theorem). e) We do not need to assume anything. 44. A survey was conducted to get an estimate of the proportion of smokers among the graduate students. Report says 38% of them are smokers. Chatterjee doubts the result and thinks that the actual proportion is much less than this. Choose the correct choice of null and alternative hypothesis Chatterjee wants to test. a) Ho: p=.38 versus Ha: p .38. b) Ho: p=.38 versus Ha: p > .38. c) Ho: p=.38 versus Ha: p< .38. d) None of the above. 45. A political poll of Americans was conducted to investigate their opinions on gun control. Each person was asked if they were in favor or gun control or not in favor of gun control - no respondents were removed from the results. The survey found that 25% of people contacted were not in favor of gun control laws. These results were accurate to within 3 percentage points, with 95% confidence. Which of the following is NOT correct? a) The 95% confidence interval is approximately from (22% to 28%). b) We are 95% confidence that the true proportion of people not in favor of gun control is within 3 percentage points of 25%. c) In approximately 95% of polls on this issue, the confidence interval will include the sample proportion. d) If another poll of similar size were taken, the percentage of people IN FAVOR of gun control would likely range from 72% to 78%. 46. Suppose we are interested in finding a 95% confidence interval for the proportion p of UF undergraduate students who are from the state of Florida. We take a random sample of 20 students, and we find that 17 of them are from Florida. Which of the following is the small- sample confidence interval for p, using 95% confidence? a) (.694, 1.000) b) (.629, .954) c) (.850, .930) d) (.688, 1.000) 47. Which of the following statements about small-sample and large-sample confidence intervals for proportions are true? I. The large-sample confidence interval formula for proportions is valid if np ≥ 15 and n(1-p) ≥ 15. II. Large-sample confidence intervals always contain the true parameter value, whereas small-sample confidence intervals may not. III. We form small-sample confidence intervals by using the large-sample formula after adding 4 successes and 4 failures. a) I and III only b) II only c) I only d) I, II, and III Questions 48-50: Suppose we are interested in finding a 95% confidence interval for the mean SAT Verbal score of students at a certain high school. Five students are sampled, and their SAT Verbal scores are 560, 500, 470, 660, and 640. 48. What is the standard error of the sample mean? a) 16.71 b) 37.36 c) 83.55 d) 113.2 49. What is the 95% confidence interval for the population mean? a) (462.3, 669.7) b) (469.9, 662.1) c) (486.3, 645.7) d) (492.8, 639.2) 50. The method used to calculate the confidence interval in the previous question assumes which one of the following? a) The sample mean equals the population mean. b) The sample standard deviation does not depend on the sample drawn. c) The population has an approximately normal distribution. d) The degrees of freedom df ≥ 30. 51. A sample of size 45 is drawn from a slightly skewed distribution. What is the approximate shape of the sampling distribution? a) Skewed Distribution b) Binomial Distribution c) Normal Distribution d) Uniform Distribution Questions 52-53 We know that 65% of all Americans prefer chocolate over vanilla ice cream. Suppose that 1000 people were randomly selected. 52. The standard error of the sample proportion is a) 0.03567 b) 0.01508 c) 0.01798 d) 0.3785 53. The Sampling Distribution of the sample proportion is a) Binomial ( 1000, 0.65) b) Normal( 0.65, 0.01508) c) Normal(10000,0.65) d) None of the above 54. What is the probability that our sample will have more than 70% of people prefer chocolate ice cream? a) 0.9995 b) 0.0005 c) 0.70 d) none of the above 55. We are doing an experiment where we record the number of heads when we get when we flip an unbiased coin many times. For what sample sizes below would the sampling distribution of the sample proportion be approximately normally distributed? a) 5 b) 28 c) 50 d) All of the above e) None of the above 56. For a test with the null hypothesis Ho: p = 0.5 vs. the alternative Ha: p > 0.5, the null hypothesis was not rejected at level alpha=.05. Das wants to perform the same test at level alpha=.025. What will be his conclusion? a) Reject H0. b) Fail to Reject H0. c) No conclusion can be made. d) Reject Ha. 57. The null hypothesis Ho: p=.5 against the alternative Ha: p>.5 was rejected at level alpha=0.01. Nate wants to know what the test will result at level alpha=0.10. What will be his conclusion? e) Reject H0. f) Fail to Reject H0. g) No conclusion can be made. h) Reject Ha. 10. D. The sampling distribution of x is the distribution of values taken by x in all possible samples of the same size from the same population. 11. B. Because we infer conclusions about the population from data on selected Individuals (all sample). 12. a. F. In a very large number of samples, 95% of the confidence intervals would contain the population mean. If the endpoints of the CI are given, use the term confidence, not probability. b. T. The definition of confidence interval. We are 95% confidence that the unknown  lies between (1.15, 4.20). c. F. The center of each interval is at x , and therefore varies from sample to sample. So, when 100 intervals calculated the same way, we can expect 100 of them to capture their own sample mean. Not only 95% of them. d. F. This sentence states that individuals (all American households) is in that interval. This is wrong. CI made statements about not individuals. e. T. In a very large number of samples, 95% of the confidence intervals would contain the population mean. f. T. The center of each interval is at x , and therefore varies from sample to sample. So, when 100 intervals calculated the same way, we can expect 100 of them to capture the sample mean. 13. C. Use the rule : p-value <alpha, reject H 0 . Our usual alpha levels are .10, .05, and .01. We reject H 0 at all these levels, so III is true. II is not true because there is not an interval in HT. I is true because the definition of the p-value is the probability that you would see a result this extreme if the null were true. This p-value is so low that the probability of getting a sample like this if H 0 were true is unlikely. 14. B. A parameter is a number that describes the population. A statistic is a number that describes the sample. 15. B This problem is a question about the sampling distribution of the sample means. The amount of money earned in tips is a quantitative variable. The sample mean has a Normal distribution with mean equal to 10 and standard error equal to 2.5 35 . Draw the picture. 13 10 7.09 2.5 35 z    The probability greater than 7.09 is a very small number –almost zero. 16. C. A parameter is a number that describes the population. So here, the parameter should be the average number of jelly beans in all packages made, which is 375. 17. A. z= 6/8 375373 =-.61. Look the table A, the probability of being less than -.61 is .2709. 18. C. Since the number of jelly beans follows the normal distribution, we can use the z table. 19. C. According the central limit theory, when n is large, the sampling distribution of the sample mean x is approximately normal. That is, x ~       n  , . 20. B. The formula for the confidence interval for a population mean is: x  t s n .However, n is large, so we can use the z instead of the t. x  z s n . x =7.1. For 95% confidence, z = 1.96. So the confidence interval is 7.1 1.96* 200 5 =7.1 .69=(6.41 , 7.79) 21. D. The definition of confidence interval. We are 95% confident that the unknown population mean work hours lies between 6.82 and 7.38. A is wrong because it was the term probability when the numbers are given. B is wrong because it talks about individuals rather than the population mean. C is wrong because of it estimates the average “in our sample”. A CI estimates the average in our population. 22. B. The estimate ( x in this case) is our guess for the value of the unknown parameter (). So, we need to calculate the margin of error shows how accurate we believe our guess is, based on the variability of the estimate. That’s why we have 95% confidence in our interval, instead of 100%. 23. C. From the conclusion of question 4 and 5, we know that the confidence interval will become narrower when the size increases and the confident level decreases. 24. B. We will reject the null when the p-value is smaller than the significance level. The p-value of this test is 0.044, which is smaller than the levels at .10, .05, but larger than .01. So we reject the H 0 when  =0.10 and .05, but fail to reject the null when  =.01. 25. a. T. Since the population has a normal distribution, we can use the Normal table for the probability that one person is more than 200 lbs. b. T. Since the population has a normal distribution, the sampling distribution of x is normal. So, we can use the z table. c. T. Since the population has a normal distribution, the sampling distribution of x is also normal. So, we can use the z table. d. F. The distribution is not Normal because the 68,95,99.7% rule does not apply. The sample size is quite small (1), so the CLT does not apply. So, we can’t use the z table. e. F. The sample size is quite small (10), So the CLT does not apply. So, we can’t use the z table. f. T. According to the CLT, when we draw an SRS of size n from any population with mean  and finite standard deviation  . When n is large, the sampling distribution of the sample mean x is approximately normal. x ~N(  , n  ). Here, the sample size n is large, so we can apply the CLT. Therefore, we can use the Z table to find the probability. g. F. p̂ ~N( p, n pp )1(  ) when values of n, p satisfying np 15 and n(1-p) 15. However, np =20* 50 10 =4<15, therefore, you can’t use Normal table here to find this probability. NOTE – we learned in class that you can make CI for this data if you add 2 successes and 2 failures. The trick only works for CI for p – not for significance tests, or finding this type of probability, or doing problems about means. h. T. p̂ ~N( p, n pp )1(  ) when values of n, p satisfying np 15 and n(1-p) 15. np =91* 5000 1000 =18.2. n(1-p)=91*(4000/5000)=72.8. So we can use the z table here. 26. C p̂ ~N( p, n pp )1(  ) when values of n, p satisfying np 15 and n(1-p) 15. np =0.2*100=20> 15 and n(1-p)=100*0.8 = 80  15. So, p̂ ~N( 0.2, 0.2(0.8) 100 )  p̂ ~N( 0.2, 0.04 ) 27. B Use the sampling distribution of the sample proportion that you used above and the z- score. 0.24 0.20 1.0 0.04 z    Look up 1.00 in the table. 0.8413 is listed in the table. This is the proportion less than, we want the proportion greater than so we take 1-0.8413=0.1587. 28. C p̂ ~N( p, n pp )1(  ) when values of n, p satisfying np 15 and n(1-p) 15. np =0.05*3200=160 > 15 and n(1-p)=3200*0.95=3040  15. So, p̂ ~N( 0.05, 0.05(0.95) 3200 )  p̂ ~N( 0.05, 0.003852) 45. C is the incorrect statement. The confidence interval is suppose to estimate the population proportion –not the sample proportion. “A” is just giving the confidence interval that is o.k. “B” is talking about estimating the population proportion with the confidence interval that is correct. “D” is estimating the complement of “not in favor of gun control” –“in favor of gun control”. 46. B ˆnp = 17 > 15, but ˆ(1 )n p = 3 < 15. Therefore, we can compute the confidence interval using the large sample formula if we add 2 successes and 2 failures. Then   0829. 420 )2083)(.7917(.ˆ1ˆ 791. 420 217 ˆ          n pp se p and the resulting 95% confidence interval is   ).954.,629(.083.96.1792.)(96.1ˆ  sep 47. C The only correct statement is the first one --The large-sample confidence interval formula for proportions is valid if np ≥ 15 and n(1-p) ≥ 15. The large sample confidence interval only contain the true value a certain percentage of the time. A 95% CI will contain the value 95% of the time. You add 2 successes and 2 failures. 48. B First, we use a calculator to find that the sample standard deviation s = 83.55. Then .36.37 5 55.83  n s se 49. A The 95% confidence interval for the population mean is  .025. setx  In this particular problem, we have 36.370.566  sex Using df = n – 1 = 4, we look up (in a table) that t.025 = 2.776. Then our confidence interval is     ).7.669,3.462(36.37776.20.566025.  setx 50. C Assumptions for the confidence interval for the mean are as follows: data is quantitative, random sample, data comes from a normal distribution. Only statement (c) is true. 51. C According to the Central Limit Theorem for large n the sampling distribution of sample mean is Normal. 52. B The standard error is √ (0.65 * 0.35)/ 1000= 0.01508 53. B p̂ ~N( p, n pp )1(  ) when values of n, p satisfying np 15 and n(1-p) 15. n, p satisfying 1000*0.65 65 and 1000*0.35 15. So, p̂ ~N( 0.65, 0.65(1 0.65) 1000  )  p̂ ~N( 0.65,0.01508) 54. B The Z- value for this (0.7 – 0.65)/ 0.01508 = 3.32 Now P( Z> 3.32) = 1 – P(Z≤3.32) = 0.0005 55. C p = 0.5 = the probability of getting heads when you flip an unbiased(fair) coin You need to have np> 15 and np> 15. This happens when n = 50. (50*.5=25 and 50*(1-.5) = 25) 56. B The hypothesis was not rejected at level alpha=.05.So p value was higher than 0.05 and so higher than 0.025 as well. So the test will again fail to reject the null hypothesis at level =0.025. 57. A The hypothesis was rejected at level=0.01.So, p value was less than 0.01 and so less than 0.10 as well. Hence the test will again reject the null hypothesis at level=0.10. 58. C The hypothesis was rejected at level=0.10.So p value was less than 0.10.But that might be more than 0.05 or might be less than 0.05 which we don’t know from above information. Therefore we don’t know what will happen for the test at level=0.05. 59. D They want to show that more whales turn away than usual with the extra sounds emitted. 60. D Solution: p-hat is 24/52=0.4615. 067937.0 52 )4.01(4.0)1( 0      n pp se . Thus, 905.0 067937.0 4.04615.0ˆ 0 0      se pp z The probability shaded greater than 0.905 is (1-0.8186) =0.1814. p-value = 0.1814. p-value is not less than alpha So, we fail to reject Ho. 61. B The assumptions of the hypothesis test for a proportion are the data must be categorical, data must come from a random sample, np> 15 and n(1-p)> 15.