140 2 6 1 1 1 1 4 2 4 5 1 1 1 1 1 3 5 1 1 1 1 1 2 6 2 1 1 2 4 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 3 5 1 1 1 1 1 4 5 1 1 1 1 1 6 5 1 1 1 1 1 2 5 1 1 1 1 1 6 5 1 1 1 1 1 3 5 1 1 1 1 1 7 6 1 1 1 1 1 2 1 5 1 1 1 1 1 3 5 1 1 1 1 1 2 5 1 1 1 1 1 3 5 1 1 1 1 1 2 5 2 3 2 1 1 4 6 1 1 1 1 1 1 4 6 1 1 1 1 1 1 5 5 1 1 1 1 1 5 5 1 1 1 1 2 2 5 1 1 1 1 1 2 6 1 1 1 1 1 1 1 5 1 1 1 1 1 3 5 1 1 1 1 1 1 5 2 1 2 1 1 1 5 1 2 1 1 1 4 5 1 1 1 1 1 10 5 1 1 1 1 1 1 5 1 1 1 1 1 6 5 1 1 1 1 1 2 5 2 1 1 1 1 5 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 7 5 1 1 1 1 1 3 5 1 1 1 1 1 3 5 1 1 1 1 1 2 5 1 1 1 1 1 1 5 1 1 1 1 1 6 5 1 1 1 1 1 1 5 1 1 1 1 1 2 5 1 1 1 1 1 5 5 1 1 1 1 1 11 5 1 1 1 1 1 1 5 1 1 1 1 1 1 5 1 1 1 1 1 46 1 1 46 1 1 18 5 1 1 1 1 1 8 5 1 1 1 1 1 5 5 1 1 1 1 1 5 5 2 2 2 2 2 3 5 3 3 3 3 3 6 5 3 3 3 3 3 1 5 2 2 2 2 2 3 5 2 2 2 2 2 19 5 2 2 2 2 2 2 5 1 1 1 1 1 5 5 1 1 1 1 1 8 5 2 2 2 2 2 13 5 1 1 1 1 1 8 5 2 2 2 2 2 10 5 1 1 1 1 1 8 5 3 3 3 3 3 6 5 2 2 2 2 2 4 5 1 1 1 1 1 9 5 1 1 1 1 1 2 5 4 4 4 4 4 5 5 1 1 1 1 1 6 5 1 1 1 1 1 2 5 1 1 1 1 1 5 5 1 1 1 1 1 6 5 1 1 1 1 1 8 5 1 1 1 1 1 7 5 1 1 1 1 1 2 2 1 1 3 2 1 1 3 2 1 1 2 2 1 1 2 5 1 1 1 1 1 7 5 1 1 1 1 1 13 5 1 1 1 1 1 14 5 1 1 1 1 1 3 5 1 1 1 1 1 1 5 1 1 1 1 1 6 5 1 2 2 2 2 3 5 1 1 1 1 1 6 5 1 1 1 1 1 1 5 1 1 1 1 1 1 5 1 1 1 1 1 2 5 1 1 1 1 1 5 5 1 1 1 1 1 1 5 1 1 1 1 2 2 5 1 1 1 1 1 2 5 1 1 1 1 1 6 5 1 1 1 1 1 3 5 1 1 1 1 1 2 5 1 1 1 1 1 1 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 1 5 1 2 1 1 1 1 5 1 1 1 1 1 5 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 16 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 6 5 1 1 1 1 1 2 5 1 1 1 1 1 7 5 1 1 1 1 1 1 5 1 1 1 1 1 1 5 1 1 1 1 1 10 5 1 1 1 1 1 2 5 1 1 1 1 1 2 5 1 1 1 1 1 3 5 1 1 1 1 1 1 5 1 1 1 1 1 1 5 1 1 1 1 1 1 5 1 1 1 1 1 1 5 1 1 1 1 1 1 5 2 2 1 2 2 3 2 1 1 Which of the following is not a property of the binomial distribution? The individual trials are dependent on each other. There are exactly two possible outcomes for each trial. The probability of success is the same for each trial. The number of trials is fixed. The mean, variance, and standard deviation of the number of successes can be calculated knowing only the number of trials and the probability of success for a given trial. All of the other answers is a property of the binomial distribution. A multiple choice exam has 15 questions. Each question has 4 possible answers. Only one answer is correct per question. What is the probability that by just guessing, a student will get at most 7 correct? 0.9827 0.0393 0.0173 0.0566 50--50 If the random variable $X$ has a binomial distribution with $n = 10$ and $p = .5$, what is the variance of $X$? 2.5 5 0.25 1.58 null hypothesis Which of the following is true of a normal distribution? Each of the other answers is a property of a normal distribution. The graph is a symmetric, bell--shaped curve. The graph is centered at the mean $\mu$. The total area under of the region between the curve and the horizontal axis is one. If the mean $\mu$ and the variance ${\sigma}^2$ are known, the standard normal table may be used to obtain relevant probabilities for the corresponding normal distribution. The entire graph is always strictly above the horizontal axis. The area of the region under the standard normal curve between $-2.0$ and $-1.0$ is 0.1359 0.0228 0.4772 0.3413 None of the other answers is correct. Let $Z$ be standard normal. Then $P( Z > -1.65) =$ 0.9505 0.0495 0.9010 0.0165 \$11,111.11 Let$ Z $be standard normal. The value of$ z $such that$ P(Z > z) = 0.0217 $is approximately 2.02$ -2.02 $0.5382 0.7643$ \mu $Let$ Z $be standard normal. The value of$ z $such that$ P(-z < Z < z) = .8500 $is approximately 1.44 1.03 1.04 1.00$ {\sigma}^2 $For the standard normal distribution, if the$ z $value is equal to 0, this implies that the raw score is the same as the mean the raw score is 0 the raw score does not exist the raw score is extremely large the raw score is not so raw If$ X $is a normally distributed random variable with a mean of 15 and a variance of 9, then$ P(x < 18) $is 0.8413 0.0 0.3413 0.1587 unknown, to say the least If$ X $is a normally distributed random variable with a mean of 15 and a variance of 9, then$ P(X = 18) $is 0.0 0.8413 0.3413 0.1587 negative If IQ scores are normally distributed with a mean of 100 and a standard deviation of 20, then the probability of a person having an IQ score of at least 130 is 0.0668 0.1332 0.9332 does not exist 0.0 The manager of a video rental store believes that the total annual revenue of the store can be approximated by a normal distribution with a mean of \$250,000 and a standard deviation of \$30,000. She knows she needs \$180,000 to break even. What is the probability that the store will make a profit next year? 0.9901 0.0099 0.9599 0.0401 0.0 The sample statistic $\overline{x}$ is a point estimate of the population mean the sample median the population mode the population variance nothing The mean score of international students at a certain university on the TOEFL English language exam is normally distributed with a mean of 490 and a standard deviation of 80. Suppose groups of 30 students are studied. The mean and standard deviation for the distribution of sample means will be, respectively, 490, 14.61 490, 8/30 16.33, 80 490,213.33 binomially distributed, uniformly distributed Let $X$ be a random variable with mean 4.83 and standard deviation 2.79. A random sample of size 6 is selected. The standard error of the mean is 1.139 2.790 7.784 1.297 unknown, probably forever A random sample of size 7 was taken from a population of size 40. The observed values were $$3 \qquad 8 \qquad 13 \qquad 15 \qquad 17 \qquad 22 \qquad 27 .$$ A point estimate for the population mean is 15 116 105 2.625 7/40 not random since the population only has size 40 which is not evenly divisble by 7 The Central Limit Theorem applies to any probability distribution only discrete distributions only continuous distributions only sem--discrete distributions only normal distributions When considering sampling distributions, if the population from which we sample is normally distributed, then the distribution of the sample means is normally distributed is binomially distributed has a $T$ distribution has an unknown distribution is not distributed fairly As the sample size increases, the length of the confidence interval for the population mean should decrease increase stay the same decrease and then increase Who knows? If we change the confidence level from 98\% to 95\% when constructing a confidence interval for the mean, we can expect the width of the interval to decrease increase stay the same decrease and then increase Who knows? A 90\% confidence interval for a population mean indicates that we are 90\% certain that the population mean will fall within the interval we are 90\% certain that the interval will contain all possible sample means with the same sample size taken from the given distribution we are 90\% certain that the population mean will be the same as the sample mean used in constructing the interval we are 90\% confident there are no free lunches we are never 90\% confident when it comes to statistics A 99\% confidence interval is to be constructed for the population mean from a random sample of size 22. If the population standard deviation is unknown, the table value to be used in the computation is 2.831 2.518 2.330 2.580 2.576 Who knows? A 99\% confidence interval is to be constructed for the population mean from a random sample of size 22. If the population standard deviation is known, the table value to be used in the computation is 2.576 2.831 2.518 2.330 2.580 Who knows? The heights in inches of the students on a campus are assumed to have a normal distribution with a standard deviation of 4 inches. A random sample of 49 students was taken with a mean of 68 inches. A 95\% confidence interval for the population mean $\mu$ is 66.88 inches to 69.12 inches 67.08 inches to 68.94 inches 63.42 inches to 72.48 inches 64.24 inches to 71.76 inches cannot be measured in inches A random sample was taken from a normal population. The mean of the sample is 30, an estimate for the standard deviation of the mean is 5, and the $t$--value from the table is 1.725. The lower limit for the corresponding confidence interval is 21.375 28.071 26.673 20.357 unknown since the confidence level is not specified directly In hypothesis testing, the level of significance ($\alpha$) is the probability of rejecting a true null hypothesis failing to reject a true null hypothesis failing to reject a false null hypothesis rejecting a false null hypothesis constructing a confidence interval The collection of all elements under study and about which one is trying to draw conclusions is a population a random sample a biased sample the data set unknown stranger than anything Statistical inference is the drawing of conclusions from data the presentation and summarization of data a subset of the population a quality which characterizes a population silly, just silly The major objective of statistical analysis is to make $\dots$ about a $\dots$ based on information obtained from a $\dots$. inferences, population, sample inferences, sample, population hypotheses, statistic, sample hypotheses, parameter, population wishes, better life, fortune cookie A random variable is always a variable whose value is determined by the outcome of some chance experiment a variable that assumes its values only at isolated points a variable that, prior to the experiment, can conceivably assume any value in some interval of real numbers None of the other answers is correct. At least two of the other answers is correct. An example of a continuous random variable is the length of a telephone conversation the interest paid on six--month money market certificates per week the number of dots on the top face of a fair die dress size discrete To test the theory that the country is pleased with the present economic conditions, a booth is set up on Wall Street and persons who pass by the booth are asked their opinion. The results of this survey might be suspect because the sample isn't random, therefore, probably not representative people are reluctant to talk to strangers the sample isn't large enough there are no numbers in the raw data well, just because Use the following sample of the ages of forest rangers in Virginia to estimate the true mean age of all forest rangers in Virginia. \aligned & 35 \qquad 38 \qquad 36 \\ & 37 \qquad 34 \qquad 33 \\ & 40 \qquad 35 \qquad 37 \endaligned 36.111 35 or 37 34.000 35.500 34.222 The range of the sample $3, 1, 2, 8, 6, 5$ is 7 2 6 8 not defined Lorenzo Watts is a statistician for the Fuquay Tyle Utility Company. Long years of experience have shown that at the main power plant in Tyleville, the amount of actual work time the company linemen put in during an eight--hour day has a mean of four hours and a variance of 0.64. What is the standard deviation of the hours worked? 0.8 0.4096 0.32 0.4 unknown The reason the standard deviation is more easily interpreted than the variance is the standard deviation is expressed in the same units as the original data the standard deviation is a measure of average deviation the standard deviation is normally distributed the standard deviation takes the mean into account the standard deviation is equal to the sample mean Bernie Ooly is a telephone salesman for the Everlast Cemetary Company. It is his experience that 10\% of his calls lead to a sale, and that each call is independent of all other calls. What is the probability that Bernie will make exactly 3 sales in 10 calls? 0.0574 0.0128 0.9426 0.2431 0.9872 Five fair coins are tossed. The probability of obtaining two or fewer heads is 0.5000 0.1875 0.3125 0.1563 can't happen since 5 is not evenly divisible by 2 If the random variable $Z$ has a standard normal distribution, then $P(Z \le -1.74)$ is 0.0409 $-0.0409$ 0.9591 $-0.9591$ $\dfrac{\sqrt{2}}{2}$ If $Z$ is standard normal, then the value of $z$ such that $P(Z \ge z) = .516$ is $-0.04$ 0.04 0.484 1.04 $-0.484$ Dr. I. C. Trout, a professional iththyologist, is studying a population of catfish with his assistant, Etta Bass. Trout and Bass have established that the mean weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Additionally, weights of catfish are normally distributed. What is the probability that a catfish will weigh more than five pounds? 0.0122 0.4878 0.9878 0.9556 can't happen; catfish don't come that big Refer to the previous question. If a sample of 64 fish is taken from the population of catfish, what would the standard error of the mean weight equal? 0.100 0.013 0.080 0.112 0.800 Given that the random variable $X$ is normally distributed with a mean of 3 and a standard deviation of 5, find $E(X)$. 3 5 5/3 3/2 $\sigma$ As the size of the sample is increased, the variance of the distribution of the sample mean is decreased is increased will stay the same disappears approaches $\mu$ A point estimator is a statistic an interval extimator a random sample a parameter very big Consider the following sample of data $$700 \qquad 500 \qquad 340 \qquad 280 \qquad 600 .$$ What is a point estimate for the mean of the population from which the data were taken? 484 500 605 340 unknown Refer to the previous problem. The sample median is 500 484 605 340 unknown As the confidence interval increases, the width of the interval increases remains the same decreases waivers ceases to exist A random sample of five observations provided a mean of 0.32 and a standard deviation of 0.0043. In constructing a 95\% confidence interval for the mean of the population from which the data were selected, the appropriate table value to use in the formula is 2.776 2.571 3.132 1.96 not available Audie O'Fyle is testing whether a certain brand of audio tape provides less than one decibel of tape hiss. His hypotheses are \aligned H_0 &: \mu_0 = 1 \\ H_1 &: \mu < 1 \endaligned If Audie concludes that the tape does provide at least one decibel of hiss, when this is not true, Audie has made a Type I error Type II error Type III error tragic, tragic mistake hissless tape Which of the following could be called a null hypothesis? $H_0 : \mu = 10$ $H_0 : \overline{x} = 3200$ $H_0 : \mu = 1.5$ $H_1 : \mu > 150$ More than one of the other answers could be a null hypothesis. Acceptance of a false null hypothesis is a Type II error a Type I error a Type III error a significance level a correct decision One of the most feared predators in the New River is the Giant Guppy. Although it is known that the Giant Guppy grows to a mean length of 21 feet, an RU student in marine biology believed the guppies near Radford are smaller due to unusual feeding habits (which we won't go into). To test his claim, he decided to tackle randomly chosen guppies, measure them, and, of course, then set them free. Because this is a difficult, costly, and dangerous, not to mention silly, operation, only three guppies were sampled. Their measured lengths were 22, 20, and 17 feet. Test the hypothesis that Radford Guppies are smaller than the average Giant Guppy. Use a significance level of 1\%. \vskip .05in \item{ a. } Null hypothesis: \vskip .5in \item{ b. } Alternative hypothesis: \vskip .5in \item{ c. } Test statistic: \vskip .5in \item{ d. } Appropriate table: \vskip .5in \item{ e. } Table value for test statistic: \vskip .5in \item{ f. } Rejection region: \vskip .5in \item{ g. } Data value of the test statistic: \vskip .5in \item{ h. } Conclusion: \vskip .5in \item{ i. } If your conclusion is wrong, to what type of error are you subject? \vskip .5in \item{ j. } What are the practical consequences of the error committed in part i)? \vskip .5in 1 One of the most feared predators in the New River is the Giant Guppy. Although it is known that the Giant Guppy grows to a mean length of 16 feet, an RU student in marine biology believed the guppies near Radford are larger due to unusual feeding habits (which we won't go into). To test his claim, he decided to tackle randomly chosen guppies, measure them, and, of course, then set them free. Because this is a difficult, costly, and dangerous, not to mention silly, operation, only three guppies were sampled. Their measured lengths were 21, 20, and 15 feet. Test the hypothesis that Radford Guppies are larger than the average Giant Guppy. Use a significance level of 5\%. \vskip .05in \item{ a. } Null hypothesis: \vskip .5in \item{ b. } Alternative hypothesis: \vskip .5in \item{ c. } Test statistic: \vskip .5in \item{ d. } Appropriate table: \vskip .5in \item{ e. } Table value for test statistic: \vskip .5in \item{ f. } Rejection region: \vskip .5in \item{ g. } Data value of the test statistic: \vskip .5in \item{ h. } Conclusion: \vskip .5in \item{ i. } If your conclusion is wrong, to what type of error are you subject? \vskip .5in \item{ j. } What are the practical consequences of the error committed in part i)? \vskip .5in 1 The number of letters per word for words used in the crossword puzzle in a typical issue of the {\it Tartan} has an interesting distribution. The following table shows $x$ the number of letters in a word and $n(x)$ the number of words having this number of letters in a recent issue. \aligned \quad x & \quad \, 2 \quad \, \, \, 3 \quad \, \, \, 4 \quad \, \, 5 \quad 6 \quad 7 \\ n(x) & \quad 11 \quad 43 \quad 14 \quad 4 \quad 4 \quad 2 \endaligned Estimate the number of letters per word expected in a {\it Tartan} crossword puzzle. 3.4 4.0 2.8 3.2 4.5 A sample of underweight babies was fed a special diet and the following weight gains (lbs) were observed at the end of three months: $$6.7 \quad 2.7 \quad 2.5 \quad 3.6 \quad 3.4 \quad 4.1 \quad 4.8 \quad 5.9 \quad 8.3 .$$ The mean and standard deviation are: 4.67, 1.95 4.67, 3.82 3.82, 4.67 1.95, 4.67 4.67, 1.84 The heights in centimeters of 5 students are: $$165, \quad 175, \quad 176, \quad 159, \quad 170 .$$ The sample median and sample mean are respectively: 170, 169 170, 170 169, 170 176, 169 176, 176 If most of the measurements in a large data set are of approximately the same magnitude except for a few measurements that are quite a bit larger, how would the mean and median of the data set compare and what shape would a histogram of the data set have? the mean would be larger than the median and the histogram would be skewed with a long right tail the mean would be smaller than the median and the histogram would be skewed with a long left tail the mean would be larger than the median and the histogram would be skewed with a long left tail the mean would be smaller than the median and the histogram would be skewed with a long right tail the mean would be equal to the median and the histogram would be symmetrical In measuring the center of the data from a skewed distribution, the median would be preferred over the mean for most purposes because: the mean may be too heavily influenced by the larger observations and this gives too high an indication of the center the median is the most frequent number while the mean is most likely the median is less than the mean and smaller numbers are always appropriate for the center the mean measures the spread in the data the median measures the arithmetic average of the data excluding outliers A sample of 99 distances has a mean of 24 feet and a median of 24.5 feet. Unfortunately, it has just been discovered that an observation which was erroneously recorded as 30'' actually had a value of 35.'' If we make this correction to the data, then: the median remains the same, but the mean is increased the mean remains the same, but the median is increased the mean and median remain the same the mean and median are both increased we do not know how the mean and median are affected without further calculations; but the variance is increased Which of the following is false? The numbers 1, 5, 9 have a smaller standard deviation than 101, 105, 109 The numbers 3, 3, 3 have a standard deviation of 0 The numbers 3, 4, 5 have the same standard deviation as 1003, 1004, 1005 The standard deviation is a measure of spread around the center of the data The standard deviation can only be computed for interval or ratio scaled data In a statistical test for the equality of a mean, such as $\text{ Null }: \, {\mu}_0 = 10$ , if $\alpha = 0.05$, 5\% of the time we will say that there is a real difference when there is no difference 95\% of the time we will make an incorrect inference 5\% of the time we will say that there is no real difference when there is a difference 95\% of the time the null hypothesis will be correct 5\% of the time we will make a correct inference DDT is an insecticide that accumulates up the food chain. Predator birds can be contaminated with quite high levels of the chemical by eating many lightly contaminated prey. One effect of DDT upon birds is to inhibit the production of the enzyme carbonic anhydrate which controls calcium metabolism. It is believed that this causes egg shells to be thinner and weaker than normal and makes the eggs more prone to breakage. (This is one of reasons why the condor in California is near extinction.) An experiment was conducted where 16 sparrow hawks were fed a mixture of 3 ppm dieldrin and 15 ppm DDT (a combination often found in contaminated prey). The first egg laid by each bird was measured and the mean shell thickness was found to be 0.19 mm with a standard deviation of 0.01 mm. A normal egg shell has a mean thickness of 0.2 mm. The null and alternate hypotheses are: $\text{ Null }: {\mu}_0 = 0.2 \qquad \text{ Alternative }: {\mu}_0 < 0.2$ $\text{ Null }: {\mu}_0 < 0.2 \qquad \text{ Alternative }: {\mu}_0 = 0.2$ $\text{ Null }: {\mu}_0 = 0.2 \qquad \text{ Alternative }: {\mu}_0 < 0.2$ $\text{ Null }: {\mu}_0 = 0.19 \qquad \text{ Alternative }: {\mu}_0 = 0.2$ $\text{ Null }: {\mu}_0 = 0.2 \qquad \text{ Alternative }: {\mu}_0 = 0.2$ Same question. The value of the data test statistic is: $-4.00$ $-1.00$ 0.01 1.96 1.75 It is important to detect a decrease in the average thickness to .18 mm because then the eggs are so fragile that few survive. What sample size would be needed to be 80\% sure of detecting this decrease at $\alpha = 0.05$? 27 8 128 34 101 The average time it takes for a person to experience pain relief from aspirin is 25 minutes. A new ingredient is added to help speed up relief. Let ${\mu}_0$ denote the average time to obtain pain relief with the new product. An experiment is conducted to verify if the new product is better. What are the null and alternative hypotheses? $\text{ Null }: {\mu}_0 = 25 \text{ vs } \text{ Alternative }: {\mu}_0 < 25$ $\text{ Null }: {\mu}_0 = 25 \text{ vs } \text{ Alternative }: {\mu}_0 = 25$ $\text{ Null }: {\mu}_0 < 25 \text{ vs } \text{ Alternative }: {\mu}_0 = 25$ $\text{ Null }: {\mu}_0 < 25 \text{ vs } \text{ Alternative }: {\mu}_0 > 25$ $\text{ Null }: {\mu}_0 = 25 \text{ vs } \text{ Alternative }: {\mu}_0 > 25$ In order to study the amounts owed to the city, a city clerk takes a random sample of 16 files from a cabinet containing a large number of delinquent accounts and finds the average amount owed to the city to be \$230 with a sample standard deviation of \$36. It has been claimed that the true mean amount owed on accounts of this type is greater than \$250. If it is appropriate to assume that the amount owed is a normally distributed random variable, the value of the test statistic appropriate for testing the claim is:$ -2.22  -3.33  -1.96  -0.55  -2.1314 $An appropriate 95\% confidence interval for$ {\mu}_0 $has been calculated as$(-0.73,1.92)$based on$ n = 15 $observations from a normally distributed population with a a standard deviation of 2. The hypotheses of interest are$ \text{ Null }: {\mu}_0 = 0 \text{ vs } \text{ Alternative }: {\mu}_0 \ne 0 $. Based on this confidence interval, we should not reject the null hypothesis at the$ \alpha = 0.05 $level of significance we should reject the null hypothesis at the$ \alpha = 0.05 $level of significance we should reject the null hypothesis at the$ \alpha = 0.10 $level of significance we should not reject the null hypothesis at the$ \alpha = 0.10 $level of significance we cannot perform the required test since we do not know the value of the test statistic The Roanoke Times claims that the time of travel from downtown to the Roanoke Airport via bus has an average of$ {\mu}_0 = 27 $minutes. A person who often takes this bus believes that$ {\mu}_0 $is greater than 27 minutes. A sample of six ride--times taken to test the hypothesis of interest gave$ \overline{x} = 27.5 $minutes and a standard deviation$ s = 2.43 $minutes. The value of the test statistic for testing this hypothesis is:$ 0.504  -0.532  0.460  -0.504  -0.460 $A 95\% confidence interval for$ {\mu}_0 $is calculated to be$ (1.7 , 3.5) $. It is now decided to test the hypothesis$ \text{ Null }: {\mu}_0 = 0 \text{ vs } \text{ Alternative }: {\mu}_0 \ne 0 $at the$ \alpha = 0.05 $level, using the same data as was used to construct the confidence interval. we would reject the null hypothesis at level$ \alpha = 0.05 $we cannot test the hypothesis without the original data we cannot test the hypothesis at the$ \alpha = 0.05 $level since the$ \alpha = 0.05 $test is connected to the 97.5\% confidence interval. we can only make the connection between hypothesis testing and confidence intervals if the sample sizes are large we would accept the null hypothesis at level$ \alpha = 0.05 $. We want to test$ \text{ Null }: {\mu}_0 = 1.5 \text{ vs } \text{ alternative }: {\mu}_0 \ne 1.5 \text{ at } \alpha = .05 $. A 95\% confidence interval for$ {\mu}_0 $calculated from a given random sample is$ (1.4, 3.6) $. Based on this finding we: fail to reject the null hypothesis reject the null hypothesis cannot make any decision at all because the value of the test statistic is not available cannot make any decision at all because the distribution of the population is unknown cannot make any decision at all because$ (1.4,3.6) $is only a 95\% confidence interval for$ {\mu}_0 $A random sample of 100 observations is to be drawn from a population with a mean of 40 and a standard deviation of 25. The probability that the mean of the sample will exceed 45 is: 0.0228 0.4772 0.4207 0.0793 not possible to compute, based on the information provided Cans of salmon have a nominal net weight of 250 grams. However, due to variation in the canning process, the actual net weight has an approximate normal distribution with a mean of 255 grams and a standard deviation of 10 grams. According to Consumer Affairs, a sample of 16 tins should have less than a 5\% chance that the mean weight is less than 250 grams. What is the actual probability that a sample of 16 tins will have a mean weight less than 250 grams? .0228 .1915 .3085 .4772 .0500 The Central Limit Theorem is important in Statistics because: it enables reasonably accurate probabilities to be determined for events involving the sample average when the sample size is large regardless of the distribution of the variable it tells us that large samples do not need to be selected it guarantees that, when it applies, the samples that are drawn are always randomly selected it tells us that if several samples have produced sample averages which seem to be different than expected, the next sample average will likely be close to its expected value it is the basis for much of the theory that has been developed in the area of discrete random variables and their probability distributions A random sample of 15 people is taken from a population in which 40\% favor a particular political stand. What is the probability that exactly 6 individuals in the sample favor this political stand? 0.2066 0.4000 0.5000 0.4000 0.0041 It has been estimated that about 30\% of frozen chickens contain enough salmonella bacteria to cause illness if improperly cooked. A consumer purchases 12 frozen chickens. What is the probability that the consumer will have more than 6 contaminated chickens? .039 .961 .118 .882 .079 Which of the following is not an assumption of the Binomial Distribution? the probability of success is equal to .5 in all trials all trials must be identical all trials must be independent each trial must be classified as a success or a failure the number of successes in the trials is counted The probability that a certain machine will produce a defective item is 0.20. If a random sample of 6 items is taken from the output of this machine, what is the probability that there will be 5 or more defectives in the sample? .0016 .0001 .0154 .0015 .2458 Seventeen people have been exposed to a particular disease. Each one independently has a 40\% chance of contracting the disease. A hospital has the capacity to handle 10 cases of the disease. What is the probability that the hospital's capacity will be exceeded? .035 .965 .989 .011 .736 There are 10 patients on the Neo-Natal Ward of a local hospital who are monitored by 2 staff members. If the probability (at any one time) of a patient requiring emergency attention by a staff member is .3, assuming the patients to be behave independently, what is the probability at any one time that there will not be sufficient staff to attend all emergencies? .6172 .3828 .3000 .0900 .9100 Newsweek in 1989 reported that 60\% of young children have blood lead levels that could impair their neurological development. Assuming that a class in a school is a random sample from the population of all children at risk, the probability that at least 5 children out of 10 in a sample taken from a school may have a blood level that may impair development is: about .84 about .25 about .20 about .16 about .64 The number of children in a family is an example of a continuous random variable. False True The Central Limit Theorem guarantees that the population we sample from is normal whenever the sample size is sufficiently large. False True As the degrees of freedom increases to infinity, the$ T $distribution approaches the normal distribution with mean one and variance two. False True Confidence intervals for$ {\mu}_x $are valid only if the sample is taken from a normal distribution. False True As the sample size increases, the variance of the distribution of$ \overline{X} $decreases increases remains the same flucuates back and forth flits hither and yon The Council on Alcohol and Drug Abuse wants to determine the percentage of young people in the United States in the 35--40 year old age group who have a serious problem with alcohol. A sample of 1000 people in this group yielded 250 showing signs of alcohol abuse. A point estimate for the desired proportion is 0.250 250 2.50 0.500 not enough information is given to answer this question Ashley and Shannon were late to Statistics class because they had stopped to pick some poppies. In order to not get yelled at by their other class mates for interrupting class, they collected a sample of 23 poppies, counted the number of petals on each, and brought the following information to class: $$\overline{X} \, = 9.16 \, \, \text{ and } \, \, S \, = \, 1.3 .$$ What is a point estimate of the average number of petals on a poppy leaf? 9.16 1.3$ {1.3}^2  9.16 \, {}_{-}^{+} \, 0.271 $none of the other answers is even vaguely familiar to me A machine is producing metal pieces that are cylindrical in shape. A sample of pieces is taken and the diameters, measured in inches, are recorded as follows. $$1.01 \qquad \qquad 1.04 \qquad \qquad 0.99$$ $$0.97 \qquad \qquad 0.99 \qquad \qquad 1.01$$ $$1.03 \qquad \qquad 0.98 \qquad \qquad 1.03$$ A point estimate for the mean diameter of metal pieces from this machine is 1.0056 inches 2.0056 inches 0.0005 inches 0.0224 inches Inches? What's an inch? A sample of size 16 gives$ \overline{X} = 7864 $and$ S^2 = 552 $. A point estimate of the standard error of the mean is 5.87 138.0 552.0 34.5 none of the other answers is correct The standard error of the mean is the standard deviation of the distribution of the sample mean an error often made on Statistics exams an error or mistake in the calculation of the sample mean the arithmetic mean of the distribution of the sample mean does not exist unless the population is at least approximately normal A 95\% confidence interval for a population mean was calculated from a random sample of 50 observations, and turned out to be$ 24 \, {}_{-}^{+} \, 7 $({\it i.e.}, 18 to 32). For practical purposes we can afford to draw only one sample and conclude We are 95\% confident that$ \mu $lies between 18 and 32. We are confident that about 95\% of the population means are between 18 and 32. We are confident that 95\% of our observations in our sample are between 18 and 32. We are 95\% confident that the sample mean will fall between 18 and 32. We are confident of nothing; the only certainties are death and taxes. To construct a 93.56\% confidence interval on$ \mu $when$ \sigma $is known and the sampled population is normal, the table value needed for substitution in the formula is 1.85 0.0644$ -1.52 $1.52 No such confidence interval exists. Use the following information to answer the next three questions. The travel time for a business woman between Radford and Roanoke is uniformly distributed between 40 and 90 minutes. What is the probability she will finish her trip in 80 minutes or less? 0.8 0.02 0.2 1.0 none of the other answers is correct The probability her trip will take longer than 60 minutes is 0.6 0.4 0.02 1.00 none of the other answers is correct The probability her trip will take exactly 50 minutes is 0.0 1.0 0.02 0.06 none of the other answers is correct All of the following statements about the normal distribution are true except the variance is always 1 mean = median = mode the graph of the distribution is a bell--shaped curve the mean may be unequal to 0 the distribution is symmetric about$ \mu $Let$ X $denote the score obtained by a particular gymnast on her floor routine. Assume that$ X $is normally distributed with mean 9.0 and standard deviation 0.25. The probability that the score for a given performance falls between 8.75 and 9.25 is approximately equal to 0.68 0.95 0.99 0.75 \$2.50 The value of a standard normal random variable tells us the number of standard deviations from the mean the mean of any normal distribution the variance of any normal distribution whether the random variable is discrete or continuous that for every light on Broadway, there's a broken heart in Topeka Suppose $Z$ is standard normal. Find $p(Z = 2.6)$. 0.0 0.9953 0.0047 $-.0047$ $\dfrac{{\sigma}_x}{\sqrt{n}}$ Suppose $Z$ is standard normal. Find the value of $L$ for which $p(Z \ge L) = 0.1251$. 1.15 $-1.15$ 0.5517 0.8749 not enough information is given to answer this question Use the following information to answer the next three questions. Let $X$ denote the amount of radiation that can be absorbed by an individual before death ensues. Assume that $X$ is normally distributed with a mean of 500 roentgens and a standard deviation of 150 roentgens. Find $p(X \le 560)$. 0.6554 0.3446 0.5 0.4 Roentgens? What's a roentgen? How many standard deviations from the mean is an individual who can absorb 560 roentgens of radiation before death? 0.4 1.0 0.2 $-0.4$ Roentgens? What's a roentgen? Above what dosage level will only 2.5\% of those exposed survive? 794 975 1.96 650 Roentgens? What's a roentgen? How are statistics and parameters related? A sample statistic estimates a population parameter. usually by birth, but sometimes by marriage A population statistic estimates a sample parameter. A population parameter estimates a sample statistic. A sample parameter estimates a population statistic. For the standard normal probability distribution, the area to the left of the mean is 0.5 greater than 0.5 $-0.5$ 1 none of the other answers is correct If a $X$ value is to the left of the mean, the corresponding $Z$ value is negative positive zero non--existent none of the other answers is correct Suppose $Z$ is standard normal. Find $p(-1.50 \le Z \le 1.90)$. 0.9045 0.0381 $-0.0381$ 0.4 none of the other answers is correct Suppose $Z$ is standard normal. Find $p(-2.0 \le Z \le -1.0)$. 0.1359 0.8185 0.1469 1.0000 none of the other answers is correct Suppose $Z$ is standard normal. Find $p( 2.0 \le Z \le 2.5)$. 0.0166 0.9710 0.5000 4.500 none of the other answers is correct Suppose $Z$ is standard normal. Find $p(-2.54 \le Z \le 2.54)$. 0.9890 0.4945 0.0000 0.5400 none of the other answers is correct Suppose $Z$ is standard normal. Find $p(2.32 \le Z \le 3.05)$. 0.0091 0.4989 0.9887 0.9909 none of the other answers is correct Suppose $Z$ is standard normal. If the area between zero and $L$ is 0.4115, then $L$ is 1.35 2.70 1.00 0.2077 none of the other answers is correct Suppose $Z$ is standard normal. If the area to the right of $L$ is 0.8413, then $L$ is $-1.0$ 1.0 2.0 $-2.0$ none of the other answers is correct Suppose $Z$ is standard normal. If the area to the right of $L$ is 0.0668, then $L$ is 1.50 0.17 2.00 1.00 none of the other answers is correct A standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1 which is the standard by which all future normal distributions should be judged with a mean of 1 and a standard deviation of 0 with any mean and any standard deviation with a mean of 0 and any standard deviation A normal probability distribution is a continuous probability distribution is a discrete probability distribution can be either continuous or discrete must have a mean of 0 none of the other answers is correct Use the following information to answer the next three questions. The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 miles and a standard deviation of 5,000 miles. What is the probability that a randomly selected tire will have a life of at least 30,000 miles? 0.9772 0.4772 0.0228 none of the other answers is correct Mile? What's a mile? What is the probability that a randomly selected tire will have a life of at least 47,500 miles? 0.0668 0.4332 0.9332 50--50 none of the other answers is correct What percentage of the tires will have a life between 34,000 and 46,000 miles? 76.98\% 38.49\% 50--50 3\% none of the other answers is correct Use the following information to answer the next four questions. A volunteer ambulance service handles 0 to 5 service calls on any given day. The actual probability distribution for the number of service calls is as follows. $$x: \qquad 0 \qquad 1 \qquad 2 \qquad 3 \qquad 4 \qquad 5$$ $$f(x): \quad .10 \quad .15 \quad .30 \quad .20 \quad .15 \quad .10$$ What is the expected number of service calls per day? 2.45 2.50 6.25 4.0 none of the other answers is correct What is the standard deviation of the number of service calls per day? 1.43 2.05 2.45 2.92 none of the other answers is correct What is the probability of at least three service calls in a given day? 0.45 0.20 0.35 0.75 none of the other answers is correct What is the probability of fewer than two service calls in a given day? 0.25 0.30 0.55 0.75 none of the other answers is correct Use the following information to answer the next two questions. Twenty--nine percent of lawyers and judges are women ({\it Statistical Abstract of the United States}, 1997). In a jurisdiction with 30 lawyers and judges, what is the expected number of women? 8.7 15.0 2.9 17.4 none of the other answers is correct In the same jurisdiction, what is the standard deviation of the number of women? 2.5 1 or 2 6.2 more than 10 none of the other answers is correct Use the following information to answer the next three questions. According to the WTA and the ATP Tour, forty percent of women tennis players use Wilson rackets. Suppose ten women players are chosen at random. What is the probability exactly one of them uses a Wilson racket? 0.1209 0.1673 .0464 0.5000 none of the other answers is correct What is the probability exactly two use Wilson rackets? 0.2150 0.1673 0.3823 0.16 50--50 What is the probability none of them use a Wilson racket? 0.0060 0.0404 0.9940 50--50 Wilson rackets? What a Wilson racket? Use the following information to answer the next three questions. Many utility companies have begun to promote energy conservation by offering discount rates to consumers who keep their usage below certain established subsidy standards. A recent EPA reports indicates that 70\% of the consumers of one area have reduced their usage enough to qualify for discounted rates. Suppose six consumers from the region are chosen randomly. What is the probability all six will qualify for discounted rates? 0.1176 1.0000 0.8824 0.0102 none of the other answers is correct What is the probability at least four qualify for discounted rates? 0.7443 0.2557 0.5798 0.4202 none of the other answers is correct What is the probability fewer than two qualify for discounted rates? 0.0109 0.0705 0.0007 0.7443 none of the other answers is correct Use your TI calculator to construct a frequency histogram for this set of data. Let it determine the number of classes and class boundaries. What is the number of classes it uses? 7 5 3 9 1 What is the frequency of the third class? 7 8 4 5 6 What are the boundaries of the second class? 18.03--18.87 17.20--18.03 18.87--19.70 17--22 There is no well to tell. Which class is the modal class? 3 7 5 1 There is no modal class. Which of the following {\it best} describes the shape of the histogram? Somewhat symmetric with a well-defined mean Skewed left Extremely variable Bimodal It makes no sense to talk about shape. Which of these is not a reason that Statistics is of interest? Statistics prove with 100\% certainty what we want to know about a population. It is often hard to express our requirements without the use of numbers. Numerical quantification can add specificity to vague statements. Most statistical studies are performed using samples of data from the population of interest. Statistics allows us to make inferences about important population parameters. One of the aims of inferential statistics is to make inferences about samples from a population by analyzing the entire population. False True