Zhi plays a game in which she can purchase a ticket and it has several chances, or "catches" to win money. The table below shows the probability of winning at each stage, and how much money the ticket can win at each catch. Every time Zhi plays the game, her ticket is played through each catch, which means we can win money at each stage. Given the probabilities and payout values in this table, what is the expected value of Zhi's ticket? Catch 0 40% $1.00 Catch 1 45% $5.00 Catch 3 12% $10.00 Catch 4 3% $25.00 Zhi went to a blackjack table at the casino. At this table, the dealer is using one standard deck of 52 cards that has just been shuffled. Zhi knows that getting 21, a blackjack, is the best hand possible. She really likes when she is dealt an ace first. Before the second card is dealt Zhi wondered what her chances were of getting another ace if she did not get a face card. What is the probability that the second card is an Ace, given that it is not a face card? Enter your answer as a percent, rounded to the nearest whole number. Mark created a similar diagram from the previous season, and he wanted to look into the team's losses. In the Venn diagram, circle A shows wins and circle B shows games played on Wednesday nights. How many games did the team lose on a Wednesday? Mark went back to this session's results. Mark then looked at the games won and lost at the two locations where they played. He made a two-way table. What is the probability that the team lost a game if they played at Lincoln Park? Using the same table, what was the probability that the team played a game at Daryl Fields OR the team won a game? Another player had a 25% accuracy at making three-point shots. Mark thought it was reasonable that each attempt was independent and the probability stayed at 25% for this player. Using the geometric distribution, what is the probability that the player first makes a three-point goal on his third attempt? Switching over to baseball, Mark noticed that the probability that a certain player hit a home run in a single game is 0.165. Mark is interested in the variability of the number of home runs if this player plays 150 games. If Mark uses a binomial distribution to model the number of home runs, what is the variance for a total of 150 games? Tasha continued reading the study of the 20 top-selling cars and their weight and gas mileage. It told her that r = -0.82. For this data, the value of the coefficient of determination is __________. Tasha next considered how the speed of a car influences how much fuel it uses. She found a graph for four different compact cars at different speeds, from 20 to 70 mph. The value of r for this graph will be closest to (A) __________. The value of r 2 will be (B) __________ (high or low). The explanatory variable is (C) __________. David was interested about the performance of a sports car, but was confused about scatterplots. Tasha collected some performance data from the internet about the car's engine. Specifically, she plotted the horsepower against the engine speed (in 1000 revolutions per minute). The points fall closely on a line. Using the data values Tasha collected, the slope of the best-fit line is __________. By observing another set of data values, Tasha used a calculator for the tire diameter at the tread and circumference data and got an equation for the least-squares line: = 0.1 + 3.1x. Using this equation to predict circumference from diameter at the tread, the circumference for a 30-inch diameter at the tread tire would be __________. Tasha kept track of the miles she was able to drive versus the gallons of gas in her car's fuel tank. During this time, she always filled her tank completely when she got gas. After filling her tank eight times, she plotted the miles driven on the xaxis and the gallons on the y-axis. If she were to create a scatterplot of her data, the direction of the association would be (A) __________. The strength of the association would be (B) __________. Jesse put the heartbeat and lifespan data into a statistical software program. The software produced a scatterplot of the data, and Jesse noticed the downward trend as he read the graph from left to right. Then he used the software to calculate the coefficient of determination to be 0.64. The value of the correlation coefficient is __________. Simone checked the two anomalous data points and found that they were for reptiles that require more expensive food than the mammals. If Simone removes these two points and then recalculates r, the correlation coefficient (r= 0.7) will (A) __________ (increase, decrease, show no change)? (B) Would these points be considered influential? Jesse took two data points from the weight and feed cost data set and calculated a slope, or average rate of change. A hamster weighs half a pound and costs $2 per week to feed, while a Labrador retriever weighs 62.5 pounds and costs $10 per week to feed. Using weight as the explanatory variable, the slope of a line between these two points is __________. Jesse entered the weight (in pounds) and weekly food cost data (in $) into a statistics software package and found a regression equation of = .3 + .12x. This means that, on average, for each additional (A) __________, the weekly feed cost goes (B) __________ 12 cents. At the clinic where Jesse worked, the weekly feed cost for a rabbit was $2.20. The rabbit used for the study weighed nine pounds. Using the equation = .5 + .16x for the regression line of weekly food cost on weight (weight is explanatory), the residual for the rabbit is __________. Jesse used a regression equation for weight of an animal (explanatory) and weekly food cost (response). Using the regression equation = .3 + .12x, the predicted weekly food cost for a hedgehog weighing 1.3 pounds is __________. The study Shawna read about the weight and calories burned included results in the form of a scatterplot. Looking at the scatterplot, the form is (A) __________. The strength of this correlation can best be described as (B) __________ and r-squared is likely close to (C) __________. The study on calories burned per minute of walking, depending on weight, has a regression equation of = 1.1 + .03x. The predicted number of calories burned in a minute of walking for a 150 pound person is __________. Shawna looked at a study on mortality rates for men and women aged 65 to 74. When the researchers looked at calories of sweeteners consumed per day by women as the explanatory variable, they got a regression equation of = - 40.857 + 0.704x with r = 0.74. The percent of the change in mortality rate that is explained by the change in calories of sweeteners is __________. Shawna knew that a correlation between calories of sweeteners consumed and mortality rates does not, in itself, mean that more sweeteners leads to a higher chance of dying. It's possible that a __________ variable is actually causing a change in both variables. If calories from sweeteners are not the cause of early mortality, Shawna wondered what else might be a factor. Which of the the following methods is best suited, which is least suited to establish causation between the two variables? Order the methods from best suited to least suited. 1. 2. 3. a. retrospective cohort study b. controlled randomized clinical trial c. observational study Shawna read a scatterplot that displayed the relationship between the number of cars owned per household and the average number of insured citizens in neighborhoods across the country. The plot showed a strong positive correlation. Shawna recalled that correlation does not imply causation. In this example, Shawna saw that increasing the number of cars per household would not cause members of her community to purchase health insurance. The likely confounding variable that is causing an increase in both the number of cars and average number of insured citizens is __________. Shawna found a study of American women that had an equation to predict weight (in pounds) from height (in inches): = -260 + 6.6x. Shawna's height was 64 inches and her weight was 150 pounds. The value of the residual for Shawna's weight and height, using this regression equation, is __________. At the idea of surveying their fellow students about their sleeping habits, and comparing this to the general population, Myra began to perk up. \"I remember from my statistics course that we need to start by coming up with a hypothesis. I suspect that full-time college students are chronically sleep deprived, getting less than 8 hours of sleep.\" \"That's a good hypothesis, Myra.\" Jason thought for a minute about their sample. \"Let's call 15 of our friends randomly and see what they have to say about their sleeping habits. Remember that one of the conditions for performing a hypothesis test is that the observations are independent." If they have already sampled 15 students, Jason and Myra will need to sample (A) __________ more students, to assume normality of the test statistic. If Jason and Myra had sampled 40 students without replacement, the minimum population size would have to be (B) __________ to treat the observations as independent. A few days later, Myra and Jason had another late-night studying session to organize their thoughts about the hypothesis test they want to perform. After reviewing their statistics information further, Myra concluded, \"I think it would be meaningful to focus our efforts on examining if students are not getting an average of 8 hours of sleep." What is the null_____? and alternative hypothesis_______?. Myra and Jason continued their study and reviewed their data from, what was now, a sample of 101 students! The survey results show that their sample of students got an average of 6.5 hours of sleep each night with a standard deviation of 2.14. Using a 95% confidence level, they also found that t* = 1.984. A 95% confidence interval calculates that the average number of hours of sleep for working college students is between the lower value of (A) __________ hours and the upper value of (B) __________ hours. \"Myra,\" said Jason, \"this is where my knowledge of statistics is a little weaker than yours. I'm still not sure how we can use this sample to make a statement about the sleeping habits of an entire population of working college students.\" \"Look Jason,\" Myra replied. \"Here is the stack of survey responses. Pick out five of them at random, and then make a list of them on a sheet of paper.\" When Jason was finished, his list looked like this: Subject #12: 6 hours Subject #23: 6.5 hours Subject #49: 6 hours Subject #82: 7 hours Subject #98: 5 hours \"Let's consider these five as the population,\" said Myra. \"Now, all ten samples of the size n = 2 can be taken, and the sample means can be found. For example, the mean of the first two subjects is 6.25 hours of sleep. The Central Limit Theorem tells us that distributions with a finite mean and standard deviation, the distribution of the sample means is approximately normal for large sample size. \" Using the data from the 5 students above, the mean of these sample means is (A) __________, and the standard error of the sample mean is (B) __________. \"So now,\" continued Myra, \"We will proceed to do a one-sample t-test on a group. We'll use a new set of data values. Show me the list of values you get.\" Subject #15: 6.5 hours Subject #27: 5 hours Subject #48: 6 hours Subject #80: 7.5 hours Subject #91: 5.5 hours Jason learned that their t-distribution is similar to the normal distribution but has (A) __________ tails. The value for the degrees of freedom is (B) __________. Jason said, \"I think I'm beginning to understand this better, Myra. We know that the normal adult population gets, on average, 8 hours of sleep each night. Hypothesis testing helps us see if college students are different from the general population.\" \"Right,\" replied Myra. \"You're beginning to see the bigger picture. The t-test is how we'll know if we're dealing with a difference, or not.\" Jason and Myra tabulate that their sample of 101 students got an average of 7.1 hours of sleep each night, with a standard deviation of 2.48. Using the formula and information from their sample, the t-statistic Jason and Myra calculate is __________. \"There's one final thing we need to look at,\" Myra said. \"We don't know the population's standard deviation, although we do know the standard deviation of our sample.\" \"I think I remember this Myra,\" said Jason. \"I can't remember what it's called, but in statistics there's a test that will give us a number that will tell us about how likely it is that the relationship is due to pure chance.\" Myra smiled, knowing that she could count on Jason's grasp of statistics after all. \"That's the p-value, Jason. And once we know what that is, we'll be able to make a decision about our hypothesis.\" Given a p-value of 0.02, and a significance level of 3%, Jason and Myra should __________ the null hypothesis. As Myra and Jason wrapped up the basic statistical analysis of their project, they discussed if their project should consider if sleep deprivation varies across other age groups as well. Myra says \"Let's give our survey to an equal number of kids, teenagers, adults, and senior citizens. Then we'll use the new data to perform an ANOVA test.\" Later on, Myra and Jason decide to simultaneously study the number of hours of sleep across the age groups, as well as their primary residence location (urban, suburban and rural). In this case, they should use __________ as their statistical test. To begin the chi-square test, Kathy asked Randy and Jeanette to show her the data. They told Kathy that the company claims that the distribution of each color is 20%. %( Color Counts Red 123 23% Yellow 145 27% Green 64 12% Orange 129 24% Purple 75 14% ) \"We can't use the chi-square test in every instance because there are certain assumptions and conditions that must be met,\" explained Kathy. \"The first thing I look at is to see if all the expected frequencies are greater than 5.\" The observed frequency for the color yellow is (A) __________, and its expected frequency is (B) __________. Jeanette worked as a graveyard ward clerk at her local hospital. In addition to processing patient records, one of her assigned duties was to keep the candy bowl full on the ward's main desk counter, a perk that was often enjoyed by both patients' families and the hospital floor nursing and physician staff. Jeanette purchased several large bags of wrapped FruityTooty candy each week from a nearby warehouse store. Each bag contained a mixture of cherry, grape, apple, lemon and orange flavors. One night, Randy, the unit's charge nurse, stopped by the desk to drop off some records for Jeanette. Reaching into the bowl for a piece of candy, Randy remarked, \"I really love the green, apple-flavored ones, but there always seems to be a lot of cherry and grape flavors. Is that because the cherry candies are so popular, or does the bag contain less green ones than the other colors?\" It was a slow night on the floor, and Randy and Jeanette were intrigued by the question. They decided to randomly pick a candy from a new bag Jeanette had stored, note it's color, and place it back in the bag. Which of the following two statements are true? The count of drawing an orange candy has a binomial distribution. The distribution of the count of picking a red candy cannot be modeled as approximately normal is they pick candies over 100 times. The distribution of the count of purple candies cannot be approximated with a normal distribution. The count of drawing a green candy is not a binomial distribution. The sample proportion of red candies has a binomial distribution. The distribution of the count of picking a yellow candy can be modeled as approximately normal if they pick candies 15 times. Jeanette and Randy wondered if there were more red candies than what the company claimed. They counted 14 red, 8 green, 10 yellow, 10 purple and 6 orange candies in total from the bag. They decided to look at FruityTooty's website to see if the bags of candy were intentionally filled with more red candies. \"Hey Randy,\" said Jeanette. \"I found a question on the company's website where someone had asked the very same question.\" She showed Randy the answer from the FruityTooty's marketing department, claiming that each flavor was given an equal distribution in the bags. \"If this statement is true,\" said Randy, \"the total of each color should be 20%, which is not what we found in our simple sample.\" Determine the correct null hypothesis and alternative hypothesis. When Randy was studying for his bachelor's degree in nursing, he had taken a statistics course. Jeanette seemed eager to understand the meaning of these statistics a little better, and it was a slow evening, so he spent some time explaining to Jeanette about the null hypothesis and the alternative hypothesis. \"Let's look at the green candies.\" explained Randy, \"The hypothesis would be that there is a lower proportion of green candies than 20%, and the null hypothesis is what the company maintains, which is the proportion of colors is equal, and that green is 20%.\" A type I error would be made if the null hypothesis is (A) __________, but it is actually (B) __________. A type II error would be made if the null hypothesis is (C) __________, but it is actually (D) __________. \"So now,\" Jeanette asked, \"we are approaching the point where we can begin to test the hypothesis, right? And, we want to avoid making a type II error. How do we do that, Randy?\" Randy can increase the power of the test by increasing either alpha, or the (A) __________. Alpha, , is equal to the probability of making a(n) (B) __________. \"Now that we've established a confidence interval,\" Randy explained, \"we're now ready to test our hypothesis. Look at the counts we recorded, Jeanette. How many red candies did we find in our 400-count sample? And, what did the FruityTooty company claim would be the population proportion of red candies?\" \"There were 92,\" said Jeanette, \"and the company said the proportion would be 20%.\" \"Okay. Then we have enough information to compute the z-test statistic,\" said Randy. \"Here we go.\" Using the formula and data provided in the problem, the value of the z-test statistic is __________. Going back to their count from their purchased bags, and using the red candies as an example, Randy showed Jeanette how to calculate a 90% confidence interval. %( Color Counts Red 110 20% Yellow 151 27% Green 71 13% Orange 135 25% Purple 83 15% ) Using a 90% confidence interval for the population proportion, what are the lower and upper values of the interval? Lower Value - Upper Value A. 0.228 B. 0.166 C.0.172 D.0.241 While Jeanette was waiting for Randy to complete his computations, a thought occurred to her. \"Randy,\" she asked. \"wouldn't it make sense to run this test at least another 20 times? Wouldn't that make our analysis much more accurate?\" \"Well, it could, but if we did this as many as 20 times on this population, we could end up with making the wrong conclusion about the null hypothesis at least once or twice, which would be considered a false positive result,\" explained Randy. \"In other words, due to sampling error, we could end up with results that appear to be statistically significant, when they actually are not.\" Which of the following two choices could result in invalid hypothesis test results? The sample size is less than 10% of the population. A significance level of 5% is chosen after calculating the p-value. The expected number of successes and failures are greater than 10. The sample is collected using a random method and is representative of the population. The null and alternative hypothesis are stated before performing a hypothesis test. Performing multiple tests on the same population. To begin the chi-square test, Kathy asked Randy and Jeanette to show her the data. They told Kathy that the company claims that the distribution of each color is 20%. %( Color Counts Red 123 23% Yellow 145 27% Green 64 12% Orange 129 24% Purple 75 14% ) \"We can't use the chi-square test in every instance because there are certain assumptions and conditions that must be met,\" explained Kathy. \"The first thing I look at is to see if all the expected frequencies are greater than 5.\" The observed frequency for the color yellow is (A) __________, and its expected frequency is (B) __________