Stats 250 Practice Exam 2 + Answers - 1 .

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Unformatted text preview: Stats 250 W12 Exam 2 Solutions 1. Age at first marriage – A researcher investigating changing trends in marriage wants to determine if the average age of first marriage is higher for men than for women. He takes a random sample of 30 recently married heterosexual couples from a list of newspaper wedding announcements and records the ages of the man and woman for each couple. An excerpt of the data is shown below: Couple 1 2 … 30 Mean Standard Deviation Male age 27.3 28.5 30.7 28.2 2.8 Female age 27.6 27.8 32.4 27.6 3.2 Difference (M – F) ‐0.3 0.7 ‐1.7 0.65 1.5 a. State the appropriate hypotheses to test the researcher's theory and complete the definition of the corresponding parameter. The research has decided to use a 5% significance level. [3] H0: ____ d = 0 _________________ Ha: ________ d > 0 _____________________ where the parameter represents the population mean difference ___ in age at first marriage (m – f) for all couples ________ Paired Samples Test Mean Pair 1 m ‐ f Std. Deviation .65 1.50 Paired Differences 95% Confidence Interval of the Difference Std. Error Mean Lower Upper t df Sig. (2‐tailed) .2746 .09 1.21 2.37 29 .024 b. Use the SPSS output to complete the statement by filling in the blank and circling the appropriate word: [2] The observed sample mean difference in ages is ____ 2.37 ________ standard errors (Circle one) ABOVE BELOW the hypothesized population mean difference in ages. c. If there really is no difference in the average age of first marriage for men and women in the population of all such couples, what is the expected value of the test statistic? [1] Final answer _______ 0 _______ d. Provide a complete sketch of the p‐value; include a label and values on the x axis, a full label for the distribution, the shaded area for the p‐value, and the actual value of the p‐value [4] p‐value = ____0.024/2 = 0.012____________ e. Circle the appropriate statistical decision: Reject H0 [1] f. What type of error could the researcher have made? Fail to reject H0 [1] (circle one) Type I Error Type II Error Page 1 Cannot be determined Stats 250 W12 Exam 2 Solutions 2. Caffeine Consumption – Concerned about caffeine consumption by young adults, a doctor decides to investigate the caffeine consumption of college students and high school students. She randomly samples 72 college students (group 1) and 64 high school students (group 2) to investigate if the mean caffeine consumption (in mg) differs between college students and high school students. It has been a few years since she took her statistics course, so she uses Google to find a few commands in SPSS and obtains the output below: a. The researcher wishes to estimate the difference in population mean caffeine consumption (college – high school) using a 95% confidence interval. She would like to use the pooled procedure. She notes that the two sample standard deviations are somewhat similar, but would like to be further assured by examining Levene’s Test results. i. Clearly state the null hypothesis for Levene’s Test. [1] H0: ________ ____or equal POPULATION variances ____________ ii. Explain how Levene’s test results support her decision to pool (include numerical values). [1] The p‐value for Levene’s test of 0.400 is larger than the general significance level for such a test of 0.10, so we fail to reject the null hypothesis of equal population variances. b. In order to create the pooled confidence interval, an estimate of the common population standard deviation must be obtained. What is the value of this estimate? (Show all work). [3] . . , . ≅ . → . Final answer ___ 55.6184____ c. Consider the following statement written by the researcher in her report. It is not quite correct. Change ONE word in the following statement to make it correct: [1] With 95% confidence, population we would estimate the difference in sample mean caffeine consumption (College – High School) to be between ‐0.01 mg to 37.79 mg. Page 2 Stats 250 W12 Exam 2 Solutions 3. Waiting at Wal‐Mart – According to Wal‐Mart, check‐out time for their customers (during the 4 to 8 pm time period) has been uniformly distributed between 4 and 11 minutes. This rectangular shaped distribution has a mean of 7.5 minutes and a standard deviation of 2.02 minutes. a. A summer intern working at Wal‐Mart has been given the task to gather some data to assess if this model for check‐out time is still reasonable. So he plans to take a random sample of 36 customers checking out between 4 and 8 pm, record each check‐out time, and compute the average check‐out time. i. [3] What is the approximate model (or distribution) for the average check‐out time based on a random sample of 36 customers? Provide all details. The sample mean will be approximately N(7.5, 2.02/√ ) or N(7.5, 0.3367). ii. What is the probability the average check‐out time based on a random sample of 36 customers will be more than 8.5 minutes? Show all work. [2] Z = (8.5 – 7.5)/0.3367 = 2.97; the area to the left of z=2.97 is 0.9985; so our probability is 1 – 0.9985 = 0.0015. Final answer _____ 0.0015______ b. Suppose the intern was running out of time to complete the project and instead decided he would take a random sample of 9 customers. Explain how the distribution of the sample average in part (a) (i) above would be different by completing the following statements. [3 points] i. The mean of the distribution would: (circle one) increase decrease stay the same ii. The standard deviation of the distribution would: (circle one) increase decrease stay the same iii. The shape of the distribution would: (circle one) change stay the same 4. Alpha – Environmentalists concerned about the impact of high‐frequency radio transmissions on birds found that there was no evidence of a higher mortality rate among hatchlings in nests near cell towers. They based this conclusion on a test using = 0.05. Would they have made the same decision at = 0.10? [2] Circle one: Yes No Can’t Tell Briefly explain: We only know that the p‐value is more than 0.05, but we don’t know how much more; in particular, we don’t know how the p‐value compares to 0.10. Page 3 Stats 250 W12 Exam 2 Solutions 5. Plastic Hardness – Companies that manufacture plastic parts measure the hardness of their products in Brinell units. Different parts will have different minimal hardness criteria depending on the function of that particular part. Since measuring the hardness of a part (the Brinell score) involves stressing each part to the point of failure or breakage, companies often prefer to take a small random sample of parts. The population mean Brinell score for a particular plastic part in a new toy needs to be estimated using a 95% confidence interval. A random sample of 16 parts has been selected from last week’s stable production run and the hardness (the Brinell score) has been obtained for each part. a. Before constructing the 95% confidence interval, the manager of the quality control group creates the graph you see at the right, along with a few summary measures. Clearly state the assumption that is assessed by looking at this graph. [2] The assumption is that the population of all hardness scores (or Brinell scores) has a normal model. Mean = 229.5 Std Error = 4.13 b. The sample mean Brinell score was 229.5 points. What was the standard deviation for these 16 observations? Show your work. [2] The standard error of the mean = 4.13 = √ √ ; so s = 4.13(4) = 16.52 points. Final answer __ s = 16.52 points __ c. Provide the 95% confidence interval for estimating the population mean Brinell score. Show your work. [3] sample mean ± t*(standard error) 229.5 ± (2.13)(4.13) 229.5 ± 8.7969 note: t* = 2.13 was found using df = 15 and 95% confidence. Final answer: 95% Confidence Interval: ____ 220.7031____ to ____ 238.2969____ d. For this particular plastic part, the minimal level for acceptable hardness is 222 points. Based on the data, does it appear that this production run of plastic parts have acceptable hardness on average? Circle your answer and provide a brief explanation with numerical support for your answer. [2] Circle one: Yes No because . . . the value of 222 is the minimal level, we want to be assured that the average hardness will be at or above 222; but the interval includes 222 and values below 222. Page 4 Stats 250 W12 Exam 2 Solutions 6. Herbs may ease Migraines – A homeopathic preparation of ginger and feverfew offered some relief for migraine sufferers in a recent study. In the study, a group of 16 migraine sufferers treated themselves with the remedy as soon as they felt a migraine developing. Of the 16 subjects, 10 were pain free within three hours. The researcher used these results to test the claim (at a 5% significance level) that a majority of all migraine sufferers using this remedy would be pain free within three hours. Let p be the population proportion of all migraine sufferers using this remedy would be pain free within three hours. a. State the appropriate hypotheses: H0: __ p = 0.50______ Ha: __ p > 0.50______ [2] b. Provide the observed test statistic value: _____ X = 10 ______ [1] c. The p‐value for the test is 0.227. What distribution was used to find this p‐value? Provide all details. [2] Final answer _Binomial distribution with n = 16 and p = 0.50 or written Bin(16, 0.50)_ d. At a 5% significance level, the results (circle one) ARE ARE NOT statistically significant. [1] e. It has been recommended to repeat this study using a larger sample size. For the new study design (compared to the original study design), the probability of correctly concluding that a majority of all migraine sufferers using this remedy would be pain free within three hours … [2] Circle one: would increase would decrease would stay the same The statistical name for this probability is: ______power (of the test) ____________________ f. Besides the original study design having a small sample of 16 subjects, what is another major issue regarding this study? [1] Circle one: non‐response bias response bias it is just an observational study 7. What is that p‐value? A school district was interested in assessing if the drop‐out rate for their high school students is less than the national level of 10%, that is, testing Ho: p = 0.10 versus Ha: p < 0.10 at a 5% significance level. A large random sample of high school students from the district was selected and ˆ the sample proportion that dropped out was p = 0.12. [2] Which of the following is the only reasonable for the p‐value? Circle one: 0.02 0.05 0.10 Page 5 0.12 0.75 Stats 250 W12 Exam 2 Solutions 8. Voter Appeal – In studying his campaign plans, Mr. Lehman wishes to learn if there is any difference in his voter appeal for male voters as opposed to female voters. His campaign manager takes two independent random samples and provides the following results: Gender 1 = Male Voter 2 = Female Voter If the election were held today, would you vote for Mr. Lehman? Yes No Total 78 122 200 94 106 200 The campaign manager uses these results to test if there is a difference (at the 10% level) in the likelihood of voting for Mr. Lehman for the two populations of male voters and female voters, namely, to test: H0: p1 = p2 versus Ha: p1 ≠ p2 a. If the null hypothesis is true, there is just one common population proportion. Give an estimate of that common population proportion. Show all your work. [2] = (78 + 94)/(200+200) = 172/400 = 0.43 Final answer: ____ 0.43 ____________ b. The researcher remembers there is a data condition that needs checking before conducting the test. Provide that check (include numerical support). = (200)(0.43) = 86, = (200)(1 – 0.43) = 114, [2] = (200)(0.43) = 86, and = (200)(1 – 0.43) = 114 are all at least 10. Note, the common is used to check this condition in hypothesis testing. c. The resulting test statistic value is ‐1.62. Provide the corresponding p‐value. Show all work. [2] The area to the left of ‐1.62 is 0.0526, since we have a two‐sided alternative, our p‐value is 2(0.0526) = 0.1052. p‐value: _____ 0.1052 ___________ d. Which of the following is the appropriate statistical decision and conclusion at the 5% level? Check one. [2] ____ Reject H0 and conclude there is not a significant difference in the likelihood of voting for Mr. Lehman for the two populations of male voters and female voters. ____ Reject H0 and conclude there is a significant difference in the likelihood of voting for Mr. Lehman for the two populations of male voters and female voters. _ X _ ail to reject H0 and conclude there is not a significant difference in the likelihood of voting for F Mr. Lehman for the two populations of male voters and female voters. ____ Fail to reject H0 and conclude there is a significant difference in the likelihood of voting for Mr. Lehman for the two populations of male voters and female voters. Page 6 Stats 250 W12 Exam 2 Solutions 9. On Time Orders – A catalog sales company promises to deliver orders placed on the Internet within 3 days. Follow‐up calls to a large random sample of recent customers gave a 95% confidence interval for the proportion of all orders that arrive on time to be (82%, 92%). a. What proportion of sampled orders arrived on time? [2] Circle one: 82% 92% 87% 10% 5% Can’t be determined since we do not know the sample size n b. The margin of error of this interval is: [2] Circle one: 82% 92% 87% 10% 5% Can’t be determined since we do not know the sample size n c. The researcher felt that the confidence interval was too wide to provide a precise estimate of the population proportion. What could the researcher have done to produce a narrower (more precise) confidence interval estimate of the population proportion of all orders arriving on time? Circle your answer. [2] i. Increase the confidence level to 99%. ii. Increase the sample size. iii. Both i and ii. iv. None of the above. 10. Name that Scenario – Below are a number of research problems. For each, determine if the most appropriate statistical analysis technique is conducting a hypothesis test (HT) or constructing a confidence interval (CI). Then also give the symbol for the corresponding parameter that is of interest. [6 points] a. Police set up an auto checkpoint at which drivers are stopped and their cards inspected for safety problems. They want to estimate the percentage of all cars on the road that may be unsafe. Circle one: HT CI Parameter of interest: ______ p _________ b. A company has recently started an exercise break program for its workers to see if it will improve job satisfaction on average, as measured by a questionnaire with higher scores indicating more satisfied. Scores for 10 randomly selected workers before and after the implementation of the exercise program were recorded. Circle one: HT CI Parameter of interest: _____ d _______ c. Researcher Colinda would like to determine if the proportion of all adults favoring the death penalty is the higher than that for all teenagers. She asks a random sample of adults (group 1) whether or not they favor the death penalty. She also asks an independent random sample of teens (group 2) whether or not they favor the death penalty. Circle one: HT CI Parameter of interest: ___ p1 – p2 ______ Page 7 Stats 250 W12 Exam 2 Solutions 11. Carbon Monoxide in LA – The Environmental Protection Agency (EPA) conducted a test to assess if the population average carbon monoxide level in the downtown Los Angeles area is higher than 4.9 parts per million (ppm). The hypotheses tested were: H0: = 4.9 versus Ha: > 4.9 and a 5% significance level was used. The results from a random sample of 25 carbon monoxide readings gave a sample mean of 5.40, a sample standard deviation of 1.03, a test statistic value of 2.43 with a corresponding p‐value of 0.011. Below are possible statements for including in the EPA report. You have been asked to determine which statements are appropriate to include (True) and which are not appropriate to include (False). Clearly circle your answer for each statement. [1 point each] a. The value of 5% represents the tolerable risk of concluding the population average carbon monoxide level is greater than 4.9 ppm, when in fact, it is not. True False b. The power of the test is the probability that the EPA correctly concludes the population average carbon monoxide level is greater than 4.9 ppm. True False c. The standard normal distribution was used to calculate the p‐value of 0.011. True False d. The p‐value of 0.011 is the probability that the null hypothesis is true. True False e. Based on the data, the carbon monoxide level in all areas in downtown Los Angeles is greater than 4.9. True False 12. Memory Experiment – A researcher conducts an experiment on human memory and recruits 10 people to participate in her study. She performs the experiment and analyzes the results. She obtains a p‐value of 0.13. Which of the following is a reasonable interpretation of her results? Her significance level was 10%. [2] Circle one: i. This proves that her experimental treatment has no effect on memory. ii. She should reject the null hypothesis. iii. There could be a treatment effect, but the sample size was too small to detect it. Page 8 ...
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