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University of Toronto Scarborough
STAB22 Final Examination
April 2009
For this examination, you are allowed two handwritten letter-sized sheets of notes (both sides) prepared by you, a non-programmable, non-communicating calculator, and writing implements.
This question paper has 24 numbered pages. Before you start, check to see that you have all the pages. You should also have a Scantron sheet on which to enter your answers, and a set of statistical tables. If any of this is missing, speak to an invigilator.
This examination is multiple choice. Each question has equal weight, and there is no penalty for guessing. To ensure that you receive credit for your work on the exam, fill in the bubbles on the Scantron sheet for your correct student number (under “Identification”), your last name, and as much of your first name as fits.
Mark in each case the best answer out of the alternatives given (which means the numerically closest answer if the answer is a number and the answer you obtained is not given.)
If you need paper for rough work, use the back of the sheets of this question paper. The question paper will be collected at the end of the examination, but any writing on it will not be read or marked.
Before you begin, two more things:
- Check that the colour printed on your Scantron sheet matches the colour of your question paper. If it does not, get a new Scantron from an invigilator.
- Complete the signature sheet, but sign it only when the in- vigilator collects it. The signature sheet shows that you were present at the exam.
The correct alternative has an asterisk next to it.
- A simple random sample of 70 coffee drinkers was taken. Each sampled coffee drinker was asked to taste two unmarked cups of coffee, one of which is actually Brand A and the other is Brand B, and was asked which one they preferred. 44 coffee drinkers preferred Brand A, and the other 26 preferred Brand B. Use this information to answer this question and the next one. The people who commissioned the survey are trying to find out whether a majority of all coffee-drinkers (that is, more than 50% of them) prefer brand A. Carry out a suitable test of significance to assess the evidence. What is the P-value of your test?
(a) between 0.05 and 0. (b) between 0.025 and 0. (c) greater than 0. (d) * between 0.01 and 0. (e) less than 0. Null p = 0.50, alternative p > 0 .50; ˆp = 44/70 = 0.629; standard error, using the null value of p, 0.50, is
(0.50)(1 − 0 .50)/70 = 0.060, so z = (0. 629 − 0 .5)/ 0 .060 = 2.15. One-sided P-value is 0.0158.
- In the situation of Question 1, it turned out that the coffee drinkers had always been given an unmarked cup containing Brand A coffee first, and Brand B coffee second. How do you react to this knowledge?
(a) The P-value was small, so there will still be good evidence that Brand A is preferred. (b) The coffee drinkers received their coffee in unmarked cups, so it doesn’t matter which brand is actually tasted first. (c) Brand A must have an advantage by being tasted first, so there cannot be a significant difference between brands A and B. (d) The P-value was large, so there will still be no evidence that Brand A is preferred. (e) * A better approach would have been to toss a coin to decide whether each drinker gets Brand A first or Brand B first.
If the test gives a small P-value (as it did), this could be because brand A really is preferred, or because the brand given first, whichever it was, is preferred, and we have no way of telling which is the reason for the small P-value. So (a) and (c) could be true, or could be false. (d) was actually false anyway, but if you thought the P-value from the previous question was large, (d) here could also be either true or false. (b) is a red herring: the fact that the cup is unmarked doesn’t stop Brand A from being tasted first, so if being first is the reason for the difference, whether the cup is unmarked or not will not matter. That leaves (e), and we already know that randomization is a good thing.
- A university financial aid office took a simple random sample of students to see how many of them were employed the previous summer. Of the 750 men sampled, 703 had been employed the previous summer; of the 650 women sampled, 592 had been employed the previous summer. Use this information for this question and the next one. Test whether the proportion of all male students employed last summer is different from the proportion of all female students employed last summer. What is the P-value of your test?
(a) * between 0.05 and 0. (b) between 0.01 and 0. (c) between 0.025 and 0. (d) less than 0.
- A simple random sample of 30 observations is taken from some population. The sample mean is 120. You wish to test the null hypothesis μ = 116 against the alternative μ 6 = 116, but you have only the output below.
N Mean SE Mean 95% CI 30 120.000 1.826 (116.422, 123.578)
N Mean SE Mean 99% CI 30 120.000 1.826 (115.297, 124.703)
What can you say about the P-value of your test?
(a) It is less than 0.01. (b) It is greater than 0.95. (c) It is less than 0.05. (d) It is greater than 0.05. (e) * It is less than 0.05 but greater than 0.01.
116 is outside the 95% confidence interval, so the two-sided P-value would be less than 0. (an implausible value); 116 is inside the 99% interval, so the P-value is greater than 0. (plausible).
- It is desired to estimate the mean of a population using a 95% confidence interval. The population is known to have a highly skewed shape. A sample of size 45 is proposed. What do you think of this choice of sample size?
(a) * the sample size may not be large enough to allow the Central Limit Theorem to apply (b) the sample size is bigger than 30, so the calculation should be accurate (c) the sampling distribution of the sample mean has exactly a normal shape regardless of the shape of the population (d) the normal distribution will be an accurate approximation to the t distribution in this case (e) The Law of Large Numbers says that the sample mean and population mean will be close, so there is no need to calculate a confidence interval
Don’t get hung up on a “magic” sample size of 30: larger is better, and if the population is highly skewed, larger is necessary for a normal approximation to be good. So (b) and (d) are wrong. The Central Limit Theorem is only an approximation, never the exact truth, so (c) is no good (replace “exactly” by “approximately” and you would be all right). In (e), the Law of Large Numbers does indeed say they will be close, but not how close; the normal approx, if it applies, does say how close. That leaves (a), which is right on the money.
- A student’s mark on a Psychology exam has a normal distribution with mean 65 and SD 10. The same student’s mark on a Chemistry exam has a normal distribution with mean 60 and SD 15. The two marks are independent of each other. Use this information for this question and the next two. What is the probability that the student scores over 80 on both exams?
(a) between 0.01 and 0. (b) between 0.20 and 0. (c) between 0.10 and 0. (d) * less than 0.
(e) greater than 0.
Two parts to this: use the normal distribution to find the chance of the student getting 80 on each exam, and then use the laws of probability to figure out “both”. On the psychology exam, a score of 80 goes with a z of (80 − 65)/10 = 1.5, so, from Table A, the probability of a score over 80 is 0.0668. Likewise, on the chemistry exam, z = (80 − 60)/15 = 1.33, and a prob of 0.0912. For independent events (as here), the probability of both is gotten by multiplying each probability together: (0.0668)(0.0912) = 0.0061. You can guess that the answer would be pretty small because getting over 80 on even one of the exams would be hard enough without having to get over 80 on both.
- What is the probability that the student scores over 80 on exactly one of the two exams?
(a) between 0.01 and 0. (b) between 0.20 and 0. (c) * between 0.10 and 0. (d) less than 0. (e) greater than 0. Use 0.0668 and 0.0912 from the previous question. The student has to get (over 80 on psych and not on chem) or (less than 80 on psych and over 80 on chem). Figure out the probability of each one by multiplying, then add the results: 0.0668(1 − 0 .0912) + (1 − 0 .0668)(0.0912) = 0.1458. You can’t use the binomial here because the chances of getting over 80 on the two exams are different (if they were the same, you could).
- What is the probability that the student’s mean mark in the two courses is over 80?
(a) between 0.10 and 0. (b) greater than 0. (c) * between 0.01 and 0. (d) between 0.20 and 0. (e) less than 0.
The total score has mean 65 + 60 = 125 and SD
102 + 15^2 = 18.03 (remember rules for variances and SDs). For the student’s mean mark to be above 80, the total on the two exams has to be above 160, so z = (160 − 125)/ 18 .03 = 1.94, and the probability of being above this is 0.0261. (It’s easier to get a mean score above 80 than it is to get over 80 on both exams because a high enough score on one exam will allow you to have a score below 80 on the other and still get an average above 80, such as 84 and 78.)
- The time it takes (for any student) to complete a STAB22 final exam is a random variable having a normal distribution with mean 160 minutes and standard deviation 15 minutes. Use this information for this question and the next one. Anne and Bob are two friends writing this exam. What is the probability that Anne will complete the exam at least ten minutes before Bob completes the exam? (Assume that all students start the exam at the same time and that their completion times are independent.)
(a) this probability is greater than 0.45 but less than 0. (b) this probability is less than 0. (c) this probability is greater than 0.35 but less than 0. (d) * this probability is greater than 0.01 but less than 0.
numbers). Thus the probability that at least one does is 1 − 0 .0282 = 0.9718. (Even though the chance of reaching a live person is pretty small when you only call once, making 10 calls gives you a decent chance of catching somebody; the mean number of live people you encounter is 3.)
- An experiment was conducted to compare a new drug with a standard drug with respect to mean recovery time. It is known that the mean recovery time for the standard drug is 26 days. Following the experiment, which involved 65 patients, a 95% confidence interval estimate was constructed for the mean recovery time (in days) for patients on the new drug. The 95% confidence interval was found to be from 24.6 to 27.8. Which of the following statements is a valid conclusion based on this information?
(a) The experimenter should reject the claim that the new drug is the same as the standard drug with respect to mean recovery time. (b) There is evidence, at the 5% level of significance, that the new drug is better than the standard drug. (c) * There is insufficient evidence, at the 1% level of significance, to reject the claim that there is no difference between the new drug and the standard drug with respect to mean recovery time. (d) There is sufficient evidence, at the 5% level of significance, to reject the claim that there is no difference between the new drug and the standard drug with respect to mean recovery time. (e) None of the other statements is a valid conclusion based on the information from this study.
The confidence interval contains 26, the mean recovery time for the standard drug, so there is not evidence (at the 0.05 level, at least) that the new drug is different from the older one. This rules out (a), (b) and (d). If the results are not significant at the 0.05 level, the P-value is bigger than 0.05, which means it must be bigger than 0.01 as well, and so the results are not significant at the 0.01 level either. This means that (c) is true. (Assessment of significance from a confidence interval is appropriate for a two-sided test, so that, if 26 had been outside the confidence interval, we would have had to think more carefully about whether (b) is true.)
- Following the analysis of a well-designed completely randomized experiment it was reported that the observed effect was “statistically significant”. Which of the following statements best explains the meaning of the phrase “statistically significant”?
(a) The observed result made sense to the experimenter since it was what was hoped would happen. (b) The observed effect happened because the experiment was properly designed and carried out without bias. (c) The experimenter carefully employed the basic principles of experimental design in conducting the study. (d) * The observed effect was sufficiently large so that it would rarely occur simply by chance. (e) The laws of probability say that this observed result would be expected to happen by chance.
“Statistically significant” means that the result was unlikely to happen by chance if there was actually no effect (and therefore we conclude that there is an effect). This is what (d) says. (b) and (c) are true but irrelevant: a properly designed experiment is a given, before you can start talking about statistical significance. (a) is not how science works (or should work), and (e) is missing a “not”.
- During a recent year, the average cost of making a movie was $58 million. A simple random sample of 15 action movies was taken during that year; the sampled action movies cost a mean of $62.5 million to make, with a sample standard deviation of $9.5 million. Is this evidence that action movies cost more to make, on average, than movies in general? Use α = 0.05.
(a) * The P-value is small, so we can conclude that action movies cost more to make on average. (b) The P-value is small, so we have evidence that action movies cost the same to make as movies generally. (c) The P-value is not small, so we cannot conclude that action movies cost more to make on average. (d) The P-value is not small, so we have evidence that action movies cost the same to make as movies generally.
We only have a sample SD, so we have to use a t-test, one-sided since we are looking for evidence of “more”. The test statistic is t = (62. 5 − 58)/(9. 5 /
- = 1.83, with 14 − 1 = 13 df. In Table D, the P-value is between 0.025 and 0.05, so we do have evidence at α = 0. 05 that the action movies cost more to make on average, which is (a). (b) and (d) are wrong because we don’t have evidence in favour of “the same”, only against, and in (c) the P-value is smaller than 0.05.
- The histogram below shows the scores of 77 players in a golf tournament. Use the histogram for this question and the one following.
What is the median score for these golfers?
(a) * between 289 and 292 (b) between 286 and 289 (c) cannot tell from a histogram (d) between 289 and 292 (e) greater than 292
The median score is the (77 + 1)/2 = 39th smallest (or largest) one. 12 of the golfers scored between 283 and 286, 19 scored between 286 and 289, and 18 scored between 289 and 292. (You might be off by one reading the frequencies from the scale, but that doesn’t matter.) So 12 + 19 = 31 of the golfers scored less than 289; this is less than 39 of them, so we need to take the next group: 12 + 19 + 18 = 49 which is more than 39, so the median is in the last group we added, between 289 and 292.
- From the histogram in Question 18, how does the mean score compare with the median?
(a) cannot estimate mean from a histogram (b) the mean score is about the same as the median (c) * the mean score is bigger than the median
- In an experiment, the soles of boys’ shoes are made of two different synthetic materials, A and B. To see which material is better (lasts longer), measurements were made on the amount of wear of the soles of shoes worn by 10 boys. Each boy wore a special pair of shoes - the left sole was made with material A and the right sole with material B. Based on this information, which one of the following statements is true?
(a) The decision as to whether the left or the right sole is to be made with material A or B should be determined by using a table of the standard normal distribution. (b) The type of material is the response variable, with two levels, A and B. (c) * None of the above other statements is true. (d) This is an example of a completely randomized design. (e) Increasing the sample size (number of replications) is one way to eliminate the need for any randomization in assigning the materials.
All of the statements apart from (c) are false: in (a), a table of random digits ought to be used (or you could toss a coin). In (b), the response is how long the sole lasts; the material type is the factor (or is an explanatory variable). This is a matched pairs design (the two materials are compared on the same boy), so (d) is false, and in (e), randomization is always required, regardless of the sample size.
- The amount of money spent by a customer at a discount store is a random variable with mean $ and standard deviation $30. What is the approximate probability that a randomly selected group of 50 shoppers will spend a total of more than $5300? Choose the number closest to the answer from the options given below.
(a) 0. (b) *0. (c) 0. (d) 0. (e) 0.
The easiest way to tackle this is via the sample mean (and using the ideas of §5.2). If the n = 50 shoppers spend a total of more than $5300, the sample mean is bigger than 5300 /50 = 106. According to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normal with mean 100 and SD 30/
(remember, we are talking about a sample mean, so divide by
n). To find the prob of the sample mean being bigger than 106, find z = (106 − 100)/(30/
- = 1.41, and the prob of being more than this is 1 − 0 .9207 = 0.0793, which is closer to 0.10 than any of the other options. Or you can work directly with the total, using the rules for means and SDs from Chapter 4. The sample total has mean 50(100) = 5000, and variance 50(30^2 ) = 45000, so the SD of the sample total is
45000 = 212.13. Again we have to use the Central Limit Theorem to claim that the sampling distribution of the sample total is approximately normal (the CLT works for means and totals). For the sample total being bigger than 5300, z = (5300 − 5000)/ 212 .13 = 1 .41, which is the same z as above, and therefore the probability is the same, as you would guess.
- According to a report, the mean salary for mayors in all Canadian cities is $120,000. A newspaper believes that this figure is not correct, and wishes to assess the evidence against it. A reporter at the newspaper takes a simple random sample of 10 Canadian cities, and finds that the mean salary for mayors in those cities is $132,200. Use this information for this question and the following one. What would a suitable null hypothesis for a test of significance say?
(a) The sample mean salary is equal to $132,200. (b) * The mean salary of all mayors is equal to $120,000. (c) The mean salary of all mayors is not equal to $120,000. (d) The sample mean salary is not equal to $132,200. (e) The mean salary of all mayors is greater than $120,000.
A null hypothesis makes a claim that the population mean is equal to some value, which rules out (c), (d) and (e). The value in (a) is the sample mean, so the right answer must be (b).
- Using the information in Question 24, what would a suitable alternative hypothesis for a test of signif- icance say?
(a) * The mean salary of all mayors is not equal to $120,000. (b) The sample mean salary is not equal to $132,000. (c) The mean salary of all mayors is equal to $120,000. (d) The sample mean salary is equal to $132,200. (e) The mean salary of all mayors is greater than $120,000.
An alternative hypothesis makes the claim that the population mean is in some way different from some value. That rules out (c) and (d). (b) uses the sample mean. The newspaper believes only that the null-hypothesis $120,000 is “not correct”, so a two-sided test is called for, as in (a). (e) would be correct if we were doing a particular one-sided test.
- A fair 10-sided die has faces numbered from 1 to 10. The probability of rolling a 1 on any roll is 0.1. If the die is rolled 12 times, what is the value of r such that the probability of rolling r 1’s or less is 0.8891?
(a) there is no value of r that works (b) 2. (c) * 2 (d) 5 (e) 1
The number of 1’s rolled has a binomial distribution with n = 12 and p = 0.1 (12 independent rolls, and the chance of rolling a 1 is always the same). So look at Table C and cast your eye down the n = 12, p = 0.1 section until you have gone far enough to make the probabilities add up to 0.8891 (since you want r or less). The first two add up to 0.6590, and the first three add up to 0.8891, so you need to go as far as 2 (k in the table). In the days before computers were used for role-playing games, people used to have real 10-sided dice. They were 10-sided prisms: if you looked at them end-on, you saw a regular 10-sided decagon, but if you looked at them side-on, they looked just like a rectangle. You rolled them sideways, and got the score of the face that was (horizontal and) uppermost.
- In a population, 45% of people have type O blood, 40% have type A, 11% have type B, and 4% have type AB. Use this information for this question and the next one. Consider an accident victim with type B blood. She can only receive a transfusion from a person with type B or type O blood. There are three people from this population who are willing to donate blood, but they don’t know what their blood type is. Assuming that the blood types of the three individuals are independent, what is the probability that at least one of them will be a suitable donor? Choose the closest answer from the following options.
The regression equation is CR = 0.0294 + 0.00235 PVTY
Predictor Coef SE Coef T P Constant 0.029387 0.001670 17.59 0. PVTY 0.0023498 0.0002356 9.97 0.
S = 0.00896965 R-Sq = 32.8% R-Sq(adj) = 32.5%
Regression Analysis: CR versus HSC
The regression equation is CR = 0.0952 - 0.000630 HSC
Predictor Coef SE Coef T P Constant 0.09521 0.01155 8.24 0. HSC -0.0006303 0.0001442 -4.37 0.
S = 0.0104614 R-Sq = omitted R-Sq(adj) = omitted
Use this information for this question and the next one.
Which of the following statements is true?
(a) The linear regression of CR on PVTY explains more than 50% of the variability in CR. (b) * Using the regression of CR on PVTY, the residual for an observation with PVTY = 3.7 and CR = 0.05 is greater than 0.005. (c) The linear regression of CR on HSC explains more than 25% of the variability in CR. (d) None of the other four statements is true. (e) The MINITAB output for the regression analysis of CR versus PVTY above indicates the need for a higher order (curved) model rather than a linear model.
Judge each of the four statements individually. You can stop as soon as you reach a true one, though it’s probably sensible to check the others as well in case you made a mistake. In (a), the percent of variability explained by the regression is R-squared, the square of the correlation. Here, the correlation between CR and PVTY is 0.573, and the square of this is less than 0.50, so (a) is false. In (b), the regression equation for predicting CR from PVTY is 0.0294 + 0. 00235 P V T Y , so when PVTY is 3.7, the predicted CR is 0.0294 + (0.00235)(3.7) = 0.0381. Residual is observed minus predicted, 0. 05 − 0 .0381 = 0.0119, which is less than 0.05, so this statement is true. If you’re pressed for time, go ahead and mark B and move on to the next question, otherwise... (c) is the same idea as (a). Find the correlation between CR and HSC, which is − 0 .293, and square it to get 0.086. This is safely less than 0.25, so the statement is false. For (e), look for any evidence that a curve would be better. The best place to look is the plot of residuals vs. fitted values (which you interpret the same way as a plot of residuals vs. x): there is no discernible pattern at all, let alone a curved pattern, so there is no evidence that a curve would be better. In short, (b) is true and the other statements are false.
- Using the information in Question 29 above, calculate the slope of the regression line for predicting PVTY from HSC). Choose the closest value from the options below.
(a) − 0. 5 (b) * − 0. 3 (c) − 0. 2 (d) − 0. 4 (e) − 0. 1
The slope of the line for predicting PVTY from HSC is rSDP V T Y /SDHSC , where r is the correlation between PVTY and HSC (− 0 .572). We don’t know the standard deviations, but we do know two other slopes: predicting CR from PVTY and predicting CR from HSC. So we know that:
0. 00235 = 0 .573(SDCR/SDP V T Y )
− 0. 000630 = − 0 .293(SDCR/SDHSC )
If we knew SDP V T Y /SDHSC , we’d be able to find the slope we want. A little thought reveals that dividing the second equation by the first will get that (because SDCR will cancel, and everything else is numbers). So we get
(− 0. 000630 / 0 .00235) = (− 0. 293 / 0 .573)(SDP V T Y /SDHSC )
(careful with the zeroes!) and rearranging,
SDP V T Y /SDHSC = (− 0. 000630 / 0 .00235)/(− 0. 293 / 0 .573) = 0. 5243.
This is correctly positive since it is two positive things divided by each other. The last step is to mutiply by the correlation: the slope is (− 0 .572)(0.5243) = − 0 .2999. Yes, this was a difficult question.
- A car magazine commissions a study to determine whether single drivers do more driving for pleasure than married drivers. The magazine collects samples of 35 single drivers and 35 married drivers, and records how many kilometres per week of pleasure driving each driver does. Some Minitab output from an analysis is shown below:
If John breaks the balloon on the first throw, he will win 100 − 15 = 85 (the prize minus the cost of making one throw. The probability of doing this is 0.1. If John breaks the balloon on the second throw (which means that he missed the first time), he will win 100 − 2(15) = 70, with probability (0.9)(0.1) = 0.09. If he breaks the balloon on the third throw, having missed the first two times, he wins 100 − 3(15) = 55 with probability (0.9)^2 (0.1) = 0.081. The rest of the time, he misses with all three throws and then quits. In this case, he loses 3(15) = 45, that is, wins −$45. The probability of this is 1 − 0. 1 − 0. 09 − 0 .081 = 0.729. Then go through and multiply winnings by probabilities to get a mean for X of
85(0.1) + 70(0.09) + 55(0.081) − 45(0.729) = − 13. 55.
Even though there are 3 different ways to win, the most likely outcome is that he will lose, and therefore the mean comes out negative.
- In a study of the effects of college student employment on academic performance, the researchers analyzed the GPAs of a random sample of students who were employed (denoted Emp) and a random sample of students who were not employed (denoted NotEmp). Some MINITAB output obtained from this study is given below. In the questions below, μEmp denotes the population mean GPA of all students employed and μN otEmp denotes the population mean GPA of all students not employed.
Descriptive Statistics: Emp, NotEmp
Variable N N* Mean StDev SE Mean Minimum Q1 Median Q3 Maximum Emp 55 0 2.8734 0.4294 0.0579 1.9257 2.5987 2.8761 3.2340 3. NotEmp 65 0 3.0224 0.2717 0.0337 2.3223 2.8569 3.0286 3.2633 3.
Use this information for this question and the following two questions. Which of the following statements is true?
(a) * Look at the sample of employed students. At least 25% of these students have a GPA equal to or below 2.6000. (b) None of the other four statements is true. (c) Look at the distribution of the GPAs in the sample of students who were not employed. This distribution is right skewed.
(d) Look at the sample of students who were employed. In this sample, more than 15 students have a GPA of 3.25 or higher. (e) Look at the maximum observed GPA of the sample of employed students. According to the
- 5 × IQR criterion, this value is an outlier.
In (a), for 25% less than a certain value, we need to look at Q1. Here Q1 is 2.5987, so 25% of the GPAs are less than 2.5987, and because 2.6 is slightly more than Q1, slightly more than 25% of the GPAs are less than 2.6. (Or exactly 25% if there are no data values between 2.5987 and 2.6). So (a) is true. You could mark A and move on to the next question, or check the others: Skipping (b), you can use the right-hand normal quantile plot to assess normality (and thereby symmetry/skewness). I’d say the plot is pretty straight, which would imply the data are normally distributed and thus symmetric. If there is any curve, it is lower in the middle than you would expect for a straight line: this means that the lower values are more spaced out than you would expect from a normal distribution and the higher ones are closer together. This would mean that the values are skewed to the left. Either thought process is good, but right-skewness is definitely wrong. In (d), Q3 says that 25% of employed students have a GPA above 3.234. 25% of 55 is 13.75. So there must be 13 or fewer students with a GPA above 3.234, and fewer than this with a GPA above 3.25. So (d) is false. In (e), use the criterion to figure out which upper-end values would be outliers. 1. 5 × IQR is 1.5(3. 2340 − 2 .5987) = 0.953, and Q3 plus this is 3.2340 + 0.953 = 4.187. The highest observed value is just above 3.6, nowhere near as big as this, so the highest observed value is not an outlier.
- Using the information in Question 33 above, which of the following statements is true?
(a) Consider the 90% confidence interval for μEmp, the population mean GPA of students who were employed. The margin of error of this confidence interval is greater than 0.11. (b) Based on the on the information given in Question 33, we see that t-procedures cannot be applied for this data set. (c) Consider the t-test for testing the null hypothesis μEmp = 3.0 against the alternative hypothesis μEmp < 3 .0. The P-value of this test is less than 0.01. (d) * None of the other four statements is true. (e) Consider the t-statistic for testing the null hypothesis H 0 : μEmp = 3.0 against the alternative hypothesis Ha : μEmp < 3 .0. This test statistic is between − 2 .00 and 2.00.
Same idea again: evaluate each statement and stop as soon as you find a true one (or go on and check the others). Another approach is to check the easiest one to check first (so that, with luck, you don’t even have to check the hardest one). In (a), go ahead and figure out the margin of error of the CI, using t procedures (since the population SD is not known). For the employed students: df = 55 − 1 = 54, so use 50 in Table D, and t∗^ = 1.676, which is a smidgen bigger than z∗^ would have been. The margin is 1 .676(0.4294)/
55 = 0.097. This is not greater than 0.11, so (a) is false. In (b), t procedures are good if the sample size is big enough to overcome any non-normality in the population. Here, (i) 55 is a decently big sample, and (ii) in the normal quantile plot, there is no serious evidence of non-normality, so the t procedure should be (doubly) OK. (b) is therefore false. In (c), work out the test statistic and then get the P-value from it. The null hypothesis says μ = 3, so t = (2. 8734 − 3)/(0. 4294 /
- = − 2 .186. We are on the correct side, so look up 2.186 without the minus sign in the 50 df line of Table D (closest). The one-sided P-value is between 0.01 and 0.02. The P-value is not less than 0.01, so this one is also false. (Or do it
(b) * 0. (c) 0. (d) 0. (e) 0.
A type II error means that the alternative hypothesis is true, but we do not decide to reject the null. Here, that means that the second distribution is the true one, but we got a value of X that was 0 or 1 (not 2 or bigger). The probability of this is 0.10 + 0.30 = 0.40.
- When would you prefer to use t procedures (confidence interval, test of significance) instead of z procedures?
(a) When you want to obtain a smaller confidence interval for the same sample size. (b) When you have a small sample. (c) * When the population standard deviation is not known. (d) When the population distribution is approximately normal. (e) When you have a large sample.
The only distinction between t and z is that you do a t when you don’t know the population SD. Some texts assert that it has to do with the sample size as well, but when you have a large sample it doesn’t matter much which test you do, and in this course we’ve said it only matters whether you know the population SD or not. So (b) is defensible, but not the best answer. The other alternatives are wrong; for example, if the population distribution is approx normal, you could use either t or z depending on whether you know the population SD or not.
- The costs of major surgery can vary substantially from one place to another. A study of the costs involved in a particular surgery was done in California and Montana. The 95% confidence intervals for the mean costs in the two states were reported to be from $5826.76 to $6173.24 in Montana and from $6061.41 to $6338.59 in California. No other information was given in the report. Assume that these intervals were calculated based on the normal distribution (i. e. assume that both the sample sizes were very large). Calculate the upper limit of the 95% confidence interval for the difference between the two population means (i.e. mean cost in California minus mean cost in Montana). Choose the closest answer from the following options.
(a) * $ (b) $ (c) $ (d) $ (e) $
“Assume that both sample sizes were very large” means to use the bottom line of the t table, Table D. That means taking t∗^ = 1.96. We don’t have the sample SDs, but we can work out what we need from the individual CI’s given, because the margin of error is half the length of the confidence interval (we go up and down from the sample mean). Label California as 1 and Montana as 2. Then:
t∗s 1 /
n 1 = (6338. 59 − 6061 .41)/2 = 138. 59 t∗s 2 /
n 2 = (6173. 24 − 5826 .76)/2 = 173. 24
Since we are taking t∗^ = 1.96, divide both of these by 1.96 to get s 1 /
n 1 = 70.71 and s 2 /
n 2 = 88.39. So that’s what we have. What we need is the difference in sample means, and the margin of error is calculated as t∗
s^21 /n 1 + s^22 /n 2. The sample means are the midpoints of the given intervals, 6200 and 6000, while the things inside the square root are the squares of what we found above: for example, s^21 /n 1 = (s 1 /
n 1 )^2. So those are 4999.8 and 7812.4. Finally, the margin of error of your two-sample interval is 1. 96
4999 .8 + 7812.4 = 222, and so the upper limit is 6200 − 6000 + 222 = 422. This one was difficult, but it has in common with all the difficult questions that you have to figure out what you want, and then try to get it from what you have. (Was it worth struggling through all the above for one mark? Your call.)
- A 90% confidence interval for a population mean, based on a sample of size 30, is from 40 to 65. This confidence interval turned out to be too big. Which of the following is a way of making the confidence interval shorter?
(a) * Make the sample size bigger. (b) Use the median instead of the mean. (c) Make the sample size smaller. (d) Choose a higher level of confidence.
Choosing a higher level of confidence like 99%, or taking a smaller sample size, will make the interval longer. The confidence intervals we have seen do not involve medians at all. So that leaves (a).
- A magazine is considering the launch of an online edition. In a small survey, the magazine contacted a random sample of 500 current subscribers and asked whether they would be interested in an online edition. Some of these subscribers showed an interest and some did not. Based on the information from this sample, the investigators tested the null hypothesis H 0 : p = 0.3 against Ha : p > 0 .3 , where p is the population proportion of subscribers who will be interested in an online edition. The value of the Z-statistic for testing H 0 : p = 0.3 against Ha : p > 0 .3 was 5.86. If we were interested in testing H 0 : p = 0.4 against Ha : p > 0 .4 based on the same sample used above, what could we say about the value of the Z-statistic for this test? This value is:
(a) between 1.5 and 3.0. (b) between 3.0 and 4.5. (c) greater than 4.5. (d) * between 0.5 and 1. (e) less than 0.5.
Start with what we have, and figure out how to get what we want. What we have is the z-statistic for testing p = 0.3. This is
z =
pˆ − 0. 3 √ (0.3)(0.7)/ 500
We don’t know ˆp, but we know everything else, so we can work it out:
pˆ = 5. 86
This makes sense: the test statistic was (very) positive, so ˆp should be quite a bit bigger than 0.3. Then go back and figure out z when H 0 : p = 0.4:
z = (0. 4201 − 0 .4)/
