1) Zhi observes that drawing a red face card from the full standard deck would w
ID: 3119524 • Letter: 1
Question
1) Zhi observes that drawing a red face card from the full standard deck would win the game. She knew the chances of drawing a face card that was red were not very high.
However, what is the probability of drawing a face card OR a red card? Enter your answer as a percent, rounded to the nearest whole number.
2) 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
Catch 1 45% $5
Catch 3 12% $10
Catch 3% $25
3) Kendra runs a local frozen yogurt shop. She found that her mean cost of a pint of frozen yogurt was $1.50, with a standard deviation of $0.10.
What are the low (A) __________ and high (B) __________ values three standard deviations from the mean?
4) Mark played on a softball team Wednesday and Friday nights. The team's season wrapped up, and Mark looked at their wins and losses. He drew a Venn diagram using two circles: Circle A represents all of their wins (on either day) and Circle B shows all of the games they played on Wednesday nights.
What proportion of their wins were on Wednesday?
Imagine this is a circle
A Middle B
9 4 6
Outside circle
1
5) Mark was looking at a player who has a 70% accuracy rating for free-throws. The player would either make the basket or not – each shot is independent of the previous shot, and it is reasonable that the probability stayed at 70% for this player. This particular player was fouled and given the chance to take two free-throws.
Using the geometric distribution, what is the probability that the player misses his first free-throw, but makes the free-throw on his second attempt?
6) Mark looked at the statistics for his favorite baseball player, Jose Bautista. Mark looked at seasons when Bautista played 100 or more games and found Bautista's probability of hitting a home run in a game is .173.
If Mark uses the normal approximation of the binomial distribution, what will be the variance of the number of home runs Bautista is projected to hit in 100 games?
7) 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 26-inch diameter at the tread tire would be __________?
8)
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 r2 will be (B) __________ (high or low). The explanatory variable is (C) __________.
9) Jesse was a veterinary student who worked with many different kinds of animals. He noticed that animals with high heartbeat rates, like rodents, tend to have shorter lives, while animals with low heartbeat rates, like tortoises, tend to live longer lives.
If we plotted data on heartbeat rate and lifespan, we would see a __________ association, which describes its direction.
10) 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.
11) At the clinic where Jesse worked, the weekly feed cost for a domestic shorthair cat was $2. The cat used for the study weighed ten pounds.
Using the equation = .3 + .12x for the regression line of weekly food cost on weight (weight is explanatory), the residual for the domestic shorthair cat is __________.
12) Shawna found a study of American women that had an equation to predict weight (in pounds) from height (in inches): = -260 + 6.6x. Shawna mom’s height was 68 inches and her weight was 179 pounds.
The value of the residual for Shawna mom's weight and height, using this regression equation, is __________.
13) 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 10 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 10 students, Jason and Myra will need to sample (A) __________ more students, to assume normality of the test statistic.
If Jason and Myra had sampled 30 students without replacement, the minimum population size would have to be (B) __________ to treat the observations as independent.
14) 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."
Match the correct null and alternative hypothesis.
Null hypothesis (H0)
Alternative hypothesis (HA)
A.
B.
C.
D.
E.
Submit My Answer
15) After several days of collecting responses to their study, Jason and Myra reached an important milestone, with 30 responses to their survey of college students’ sleeping habits.
Recognizing that they were both feeling a little burned-out by the demands of their busy lives, they spent part of the afternoon at their local ice skating rink. As they skated over the cool ice, Myra focused on the hundreds of people enjoying the clear, sunny day, and began to make an important connection. With a good-sized sample, she thought, it was time to establish a confidence level and significance level for their study.
Later in the evening, as they resumed their project in the library, Jason suggested they use a significance level of 0.02 and calculate a 98% confidence interval.
The critical t value used for calculating a 98% confidence interval, with 29 degrees of freedom, is __________.
16) 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.
17) “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: 6.5 hours
Subject #82: 7.5 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) __________.
18) “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) __________.
19) 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 7.3 hours of sleep each night, with a standard deviation of 2.60.
Using the formula and information from their sample, the t-statistic Jason and Myra calculate is __________.
20) “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 0.01, and a significance level of 5%, Jason and Myra should __________ the null hypothesis.
21) 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 statements about the distributions of counts and proportions are true?
OOPS
2 of your answers are incorrect.
HINT
When they note the color of the candy picked, it's either a certain color or not that certain color. What do we call it when a question type has two possible outcomes? According to the Central Limit Theorem, the mean of the sample proportions equals the population proportion and is normally distributed if the sample size is large enough.
How to: Select one option from each row.
True
False
The count of drawing a purple candy is not a binomial distribution.
The distribution of yellow candies can be modeled as approximately normal if they pick 10 times.
The count of drawing a red candy has a binomial distribution.
The sample proportion of green candies has a binomial distribution.
The distribution of green candies can be modeled as approximately normal is they pick over 75 times.
22) 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%.”
H0: p = 0.20
H0: p < 0.20
Randy Rejects H0
Type I Error
(A) __________
Randy Fails to Reject H0
Correct Decision
(B) __________
Fill in the missing blanks in the table with either Type I error, Type II error, or correct decision.
23) “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 the (A) __________, also known as alpha, or by increasing the (B) __________.
The power of a hypothesis test is the probability of not making a(n) (C) __________.
24) “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 94,” 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 __________.
25) 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?
OOPS
1 of your answers is incorrect.
HINT
Recall that, at a standard 5% significance level, 1 out of 20 hypothesis tests find significance just by chance. What could cause the increase in getting a false positive result? What are the conditions that must be met in order to perform a hypothesis test for population proportions?
A significance level of 5% is chosen after calculating the p-value.
Performing multiple tests on the same population.
The sample is collected using a random method and is representative of the population.
The expected number of successes and failures are greater than 10.
The sample size is less than 10% of the population.
The null and alternative hypothesis are stated before performing a hypothesis test.
26) The following night, Kathy, a medical laboratory technologist, wandered up to the unit to grab a few pieces of candy to sustain her during her night shift. Randy and Jeanette had just finished their z-test on the 502 pieces of candy, and Kathy was very interested to know what they had found. Randy walked her through the analysis they had done so far.
“This is a great start,” said Kathy, “but I actually work with this kind of testing all the time in my laboratory. We’re always looking at the proportion of different cell types in patient samples and comparing these to what we would expect to find, according to a specific hypothesis.”
“What would you have done differently?” asked Randy.
“The best type of test to use in these situations is the chi-square goodness-of-fit test because it allows you to compare the counts between multiple independent groups,” explained Kathy.
The chi-square statistic is found by summing all the squared difference of the (A) __________ values, minus the expected values, divided by the (B) __________ values.
27) 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
120
24%
Yellow
106
21%
Green
75
15%
Orange
111
22%
Purple
90
18%
“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 red is (A) __________, and its expected frequency is (B) __________.
28) Spencer and Tasha continued to review the list of public universities, looking for a group of at least three that they could attend, which offered their majors, had below-average tuition and fee costs, and were within driving distance of their home state. They finally narrowed their list to a group of six schools, three of which were located in Minnesota.
Spencer explained, “Let's consider which bordering states have the lowest average costs compared to the national average. In order to do this, we’re going to have to approach this statistical problem in a different way. In a situation where we we want to compare the means of multiple samples, we’re better off performing what’s known as an ANOVA test.”
The test statistic F is found by (A) __________ the variance between samples by the variance within samples. The (B) __________ the F value, the more likely it would be to reject the null hypothesis.
29) Spencer asked Tasha to show her the article. “Let’s see,” Spencer said. “According to this, the average tuition and fees costs for the 2012-2013 school year was normally distributed, with a mean of $28,936 and a standard deviation of $3,339.”
Tasha looked at Spencer quizzically. “I want to attend Parkview University, and they are charging $31,407. You’re taking statistics right now, Spencer. Based on the figures in the article, what percentage of schools are charging less tuition than Parkview University?”
Using the z-table below, to the nearest percentage point, the percent of private universities that charge less than Parkview University is __________%.
30) “We know the population standard deviation is $3,339, said Spencer. “So the next part of our analysis will be to calculate a z-test. The answer will help us determine if the average private university tuition and fee costs for our top 10 choices, which is $32,350, are significantly more expensive than the average claimed by the writer of the article.”
To determine if the average private university tuition and fee costs are more expensive than the claimed average of $29,056, the value of the z-test statistic is __________.
31) Spencer concluded her z-score calculation, and figured out that the p-value was 0.042, which indicated that they should reject the null hypothesis.
“I have another question, Spencer, or rather another idea,” said Tasha. “Suppose we look at attending a public university in a neighboring state. We may end up paying more in tuition and fees than we would if we were in-state students, but maybe it will be less expensive than a private university.”
Spencer said, “Tasha, that’s not only an alternative worth considering, but it’s also another way we could apply our knowledge of statistics. What does that article say is the cost of public university in our neighboring state?”
“It’s $21,032,” said Tasha. “And the average tuition and fees for all public universities is $21,706, with a standard deviation of $2,494.”
“Awesome, Tasha,” said Spencer. “I’ll take a random sample of tuition and fees for 25 public universities for out-of-state students, and from that I calculate a z-test statistic of -1.35.”
True
False
The count of drawing a purple candy is not a binomial distribution.
The distribution of yellow candies can be modeled as approximately normal if they pick 10 times.
The count of drawing a red candy has a binomial distribution.
The sample proportion of green candies has a binomial distribution.
The distribution of green candies can be modeled as approximately normal is they pick over 75 times.
Explanation / Answer
note : only one question allowed per submission
The probability of drawing a red card.
There are 26 red cards,
number of face cards = 12 ( 4 kings , 4 queens , 4 jacks )
6 of 12 are red ( diamonds or hearts)
So now
P(red or face ) = P(red) +P(face) - P(red interesection face)
=> 26+12 -6
=>32
in percentage =32/52 * 100 = 61.53 %
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