Stats exam 3
wider
A 2018 sample of 130 college students randomly selected from a university indicated that 91 were sexually active.The researcher calculated a 95% confidence interval for the data to be .62 to .78. If the researcher had created a 99.7% confidence interval instead, the interval would have been:
Event
Subset of all possible outcome to which a probability is assigned
P
Symbol for probability
P(A)
The proportion of time an event occurs over the long run
sampling distribution of the mean
The standard error of the mean is a name for the standard deviation of the:
decrease, remain the same, decrease
decide whether the following would cause an increase, decrease, or remain the same as a result of each of the following changes. a) the sample size is doubled from 400 to 800 b) the population size is doubled from 25million to 50million c) the level of confidence is lowered from 95% to 90%
10% (9,900/99,000)
given that out of 100,000 people, 1,000 have cancer. 800 of those who have cancer had a test show as malignant and 200 of the tests showed as benign. Out of the 99,000 of those who do not have cancer, 9,900 of their tests showed as malignant and 89,100 were shown as benign. What is the probability that a woman who does not have cancer gets a false positive (what is the probability of getting a positive test given that you do not have cancer)?
9,900 showed as malignant, 89,100 showed as benign (90% of 99,000 is 89,100) (remaining 10% is 9,900)
part c| we now know that the total of people who have cancer is 1,000 and 99,000 do not. We also know that out of the people who have cancer there was an 80% sensitivity, that means out of the 1,000 people 800 showed as malignant and the rest of the 20% were false positives meaning 200 showed as benign. If the specificity is 90%, then it correctly identifies 90% of the women who do not have cancer as not having cancer. Out of the women who do not have cancer, how many showed as malignant and how many showed as benign?
the sample mean
Which of the following values will always be within the upper and lower limits of the confidence interval?
0-1
Probability is a proportion and will always be between ____
Sample space
Total number of outcomes
Binary trial
Only two possible outcomes: yes/no, heads/tails, true/false, etc.
decrease
Other things being equal, as the sample size increases, the standard error of a statistic ______.
Experience
Personal probability is what an individual would expect to happen based on personal _____ Ex. Guessing the probability of getting a parking spot on Sunday based on own experience
Risk
Probability is solved the same way as ___
Inferences
Samples make ____ about the populatoon
1/32 (2^5)
What's the probability of flipping a coin 5x in a row and getting a head then a tail then a head then a tail then a head then a tail?
2N
When dealing with binary trials, how can we calculate the total possible outcomes?
true population proportion
if numerous large random samples or repetitions of the same size are taken from a population, the proportions from the various samples will have what apx. mean?
centered
sampling distributions are ___ over the true population parameter.
Simpson's Paradox
1973 admission rate at the university of Berkeley's graduate schools shows that more men were accepted into the school than women. They were sued by women for gender gap in admissions. Berkeley countered their graph with a graph showing that if the class is divided up by program, women are accepted into each program at a higher rate. What is this an example of?
Independent
2 events are ____ if probability of A is equal to probability of A given B p(A)=p(A|B)
12/52
A deck of 52 cards contains 12 royalty cards. If you randomly select a card from the deck, what is the probability of obtaining a royalty card?
b and c
A particular psychological test is used to measure need for achievement. The average test score for all university students in Oklahoma is 115. A university in southern Oklahoma estimates the mean test score for a sample of n students to be 110 and constructs a confidence interval based on their scores. Which of the following statements about the confidence interval is always true? A. The resulting interval will contain 115. B. The 95% confidence interval for n = 100 will be more narrow than the 95% confidence interval for n = 50. C. For n = 100, the 95% confidence interval will be wider than the 90% confidence interval.
22.74 to 24.06 ( 3/ sq rt (121) = 0.33 ; 23.4+/-2(0.33))
A random sample of 121 students from the University of Oklahoma had a sample mean ACT score of 23.4 with a sample standard deviation of 3.65. Construct a 95% confidence interval for the population mean ACT score of University of Oklahoma students.
3 (mu = mu sub x-bar)
A small town has 3500 families. The population mean (mu) number of children per family is = 3, with a population standard deviation (sigma) = 0.60. A sampling distribution of the mean for n= 100 is developed for this population. What is the mean of this sampling distribution?
Specificity
Ability to detect the absence of disease in people who do not have the disease
Specificity
Correctly identifying those who do not have the variable
Sensitivity
Correctly identifying those who have the variable
b (A: probability of occurrence + probability of not occurring = 1, probability of not having a girl (BBB)= 0.5x0.5x0.5=.125, 1-0.125= .875 chance of having atleast one occurrence (girl) ; B: (BBG)= 0.5x0.5x0.5= 0.125 ; C: would just be 0.5)
Given that the probability of having a boy or girl is p = 0.50, which probability is the smallest? a. The probability that at least one of a couple's 3 children is a girl. b. The probability that a couple's first girl occurs the third time around (so the couple has a boy and then another boy and then a girl). c. The probability that a couple's third child is a girl. d. The probabilities of the three other choices are all equal.
((49/50)(1/50)) 0.0196
Lyme disease is a disease carried by ticks, which can be transmitted to humans by tick bites. Suppose the probability of contracting the disease is 1/50 for each tick bite. What is the probability that you will not get the disease from your first tick bite and will get it from your second tick bite? (Assume the events are independent from one bite to the next.)
7.5% (800/10,700)
given that out of 100,000 people, 1,000 have cancer. 800 of those who have cancer had a test show as malignant and 200 of the tests showed as benign. Out of the 99,000 of those who do not have cancer, 9,900 of their tests showed as malignant and 89,100 were shown as benign. What is the probability that a woman who gets a positive test actually has cancer (what is the probability that a woman who has cancer given a positive test)?
Multiply
Joint (and) multiplication rule means that you must ___ p(A) to p(B|A)
128 (2^7)
If you flip a coin 7 times how many outcomes are possible?
Probability
= the number of times event occurs / sample space
c
A 95% confidence interval indicates that: Group of answer choices a)95% of the time, the interval will include the sample mean. b)95% of the possible population means will be included within the interval. c) 95% of the intervals constructed using this process based on samples from the population will include the population mean. d) 95% of the possible sample means will be included within the interval.
0.024 ( 6/50= 0.12 ; 10/50= 0.2 ; 0.2x0.12=0.024)
A refrigerator contains 50 pieces of fruit: 6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11 plums, and 8 mangos.Imagine that you put your hand in the refrigerator and pull out a piece of fruit. You decide you do not want to eat that fruit, so you put it back into the refrigerator and pull out another piece of fruit. What is the probability that you will pull a banana on the first try and an apple on the second try?
0.25 (at fault incidents from people with: low risk = 0 med risk = 45 high risk = 15 total 60 no fault incidents from people with: low risk = 225 med risk = 180 high risk = 35 total 440 low risk total = 225 med risk total = 225 high risk total = 50 total total = 500 15/60 = 0.25)
An insurance company writes policies for 500 newly-licensed drivers each year. Suppose 45% of these are low-risk drivers, 45% are moderate-risk, and 10% are high risk. The company has no way to know which group any individual driver falls in when it writes the policies. None of the low-risk drivers will have an at-fault accident in the next year, but 20% of the moderate-risk and 30% of the high-risk drivers will have an at-fault accident. Given that a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk? (Hint: use your percentages to create a table with risk level as your columns and fault/no-fault accidents as your rows, and 500 as your total number of people in the table.)
a ( 0.78, the sample proportion should be center of the confidence interval)
Beth and Kathy worked together to estimate the proportion of college students who say they drink alcohol of one kind or another.They collect a random sample of 50 college students and calculate a sample proportion of students who say they drink alcohol to be .78. They each create a 95% confidence interval for their finding. Beth's 95% confidence interval is .74 to .88, while Kathy's 95% confidence interval is .72 to .84. Which of the following statements is correct? a) Beth's confidence interval must be incorrect, because the sample proportion is not exactly in the middle of her confidence interval. b) Kathy's confidence interval must be incorrect, because the sample proportion is not exactly in the middle of her confidence interval. c) Neither Beth's nor Kathy's confidence interval can be correct, because the sample proportion is not one of the interval's end points. d) Both Beth's and Kathy's confidence interval can be correct, because they may have calculated different standard deviations which gave them different intervals.
they're all equally likely
If a fair die is rolled five times, which of the following ordered sequence of result, if any, is most likely to occur? A. 3 5 1 6 2 B. 4 2 6 1 5 C. 5 2 2 2 2
normal distribution
If sampling distributions of sample means are examined for samples of size 2, 16, and 50, you will notice that as n increases in size, the shape of the sampling distribution appears more like that of the:
independent
If the occurrence of one event does not influence the outcome of another event, then the two events are:
50%
If you flip a fair coin and get heads 5 times in a row, what is the probability of getting tails on the next flip?
sampling distribution
If you take all samples of a particular size from a selected population, find the mean of each sample, and then plot the means, what have you created?
0.9653 (1-(0.0332+0.0015))
In 2010 in the same country, the probability that a birth would result in twins was 0.0332, and the probability that a birth would result in triplets or more was 0.0015. What was the probability that a birth in 2010 resulted in a single child? (Assume that the probability of a birth resulting in four or more children is essentially zero.)
0.08 [sq rt. (0.37(1-0.37)/35)=0.08]
In a recent study of Vietnam veterans, researchers found that among the population of veterans, .37 have been divorced at least once. If I took all possible samples of size n= 35 from this population, calculated the sample proportion on each sample, and arranged the sample proportions into a frequency distribution, the standard deviation of this sampling distribution would equal ______.
would not apply to either
In which of the following situations would the gambler's fallacy not apply? A. When the events are not independent. B. When knowledge of one outcome affects the probability of the next one.
Yes .5=.5
Is getting tails on a first coin flip independent of getting tails on a second flip? P(T)=P(T|T on first flip)
caculated/observed
Relative frequency is what is actually ___. We went out into the world and we figured it out mathematically. Ex. Figuring out the probability of getting a parking spot on a Sunday by going outside and calculated the actual probability
0.025 ( 170/200=0.85 ; sq rt (0.85(1-0.85)/200))
Researchers were interested in the population proportion of undergraduate students at the University of Oklahoma who are from the state of Oklahoma. They took a random sample of 200 students and found that 170 of them are from Oklahoma. They calculated a 95% confidence interval for the population proportion p to be .8 to .9. What would you estimate the standard error of the proportion to be?
0.71 (table: those who tested positive that: have the disease = 32,200 don't have disease = 12900 total = 45100 those who tested negative that: have the disease= 2,800 don't have disease = 202100 total= 204900 total who have disease = 35000 total who don't= 215000 total total = 250,000 32200/45100=0.71)
Suppose Disease X occurs in 14% of a population with a total of 250,000 people. A test for Disease X has a sensitivity of 92% and a specificity of 94%. What is the probability that you actually have the disease given that your test results are positive (i.e. the test indicates that you have the disease)? Complete the table below to help you determine the answer.
a ( 12/75=0.16 and 4/25=0.16 ; 0.16=0.16)
Suppose that an organization is concerned about the number of its new employees who leave the company before they finish one year of work. In an effort to predict whether a new employee will leave or stay, the organization develops a standardized test and applies it to 100 new employees. After one year, they note what the test had predicted (stay or leave) and whether the employee actually stayed or left. They then compiled the data into the following table: predicted to stay that: actually stays = 63 actually leaves = 12 total=75 predicted to leave that: actually stays = 21 actually leaves = 4 total = 25 total that actually stayed= 84 total that actually left = 16 total total = 100 It is __________ that an employee predicted to leave will actually leave as it is for one who is predicted to stay to actually leave. a) equally likely b) less likely c) more likely
0.61 (mu sub p = p)
Suppose the population proportion of American citizens who are in favor of gun control is .61. If a sampling distribution of size n= 50 was created from this population, what would be the mean of this sampling distribution?
a
Suppose there is a population of test scores on a large, standardized exam for which the mean and standard deviation are unknown. Two different random samples of 50 data values are taken from the population. One sample has a larger sample standard deviation (SD) than the other. Each of the samples is used to construct a 95% confidence interval. How do you think these two confidence intervals would compare? a. The confidence interval based on the sample with the larger standard deviation would be wider. b. The two confidence intervals would have the same width because they are both 95% confidence intervals. c. The two samples would produce identical values for the lower and upper bounds of the two confidence intervals. d. The confidence interval based on the sample with the smaller standard deviation would be wider.
$247 and $253 ( 30/ sq rt 100 = 3 ; 250+/-3)
The amount of money college students spend each semester on textbooks is normally distributed with a mean of $250 and a standard deviation of $30. Suppose you take a random sample of 100 college students from this population. There would be a 68% chance that the sample mean (𝑥̅) amount spent on textbooks would be between:
d (mu sub x-bar = mu ; 8/sq rt (64) = 1 ; central limit theorem states that it will be a normal shape if n is greater than or equal to 30)
The amount of time it takes to take an exam has a negatively (left) skewed distribution with a population mean of 65 minutes and a standard deviation of 8 minutes. A sampling distribution of the mean for n = 64 will: Group of answer choices a) Have a mean of 0, a standard deviation of 1, and be negatively (left) skewed in shape. b) Have a mean of 65, a standard deviation of 1, and be negatively (left) skewed in shape. c) Have a mean of 65, a standard deviation of 8, and be normal in shape. d) Have a mean of 65, a standard deviation of 1, and be normal in shape.
96.6% (132,527/137,128)
The table below is based on records of accidents in 1988 compiled by the Department of Highway Safety and Motor Vehicles in the state of Florida. seatbelt worn with: nonfatal injury: 412,368 fatal injury: 510 total: 412,878 no seat belt worn with: nonfatal injury: 132,527 fatal injury: 4,601 total: 137,128 total nonfatal injuries: 544,895 total fatal injuries: 550,006 total total: 550,006 What is the probability of someone having a nonfatal accident given that no seatbelt was worn?
Expect
Theoretical probability is what we ____ to happen. Ex. The probability of getting heads when flipping a coin is .5
c
Which of the following is an example of the availability heuristic? Group of answer choices a) Believing that a warranty plan that completely covers 30% of the possible problems is a better deal than a plan that covers 30% of all possible problems. b) Thinking that events with more detailed descriptions are more likely than events with very general descriptions. c) Thinking that you won't get lung cancer if you smoke because you know many smokers and none of them have lung cancer. d) Believing that random events are self-correcting (i.e. if you are on a losing streak, believing that your luck is about to turn around)
samples
sampling distributions are composed of statistics from many, many ___
63.5, 0.5
the height of American adult women is distributed almost exactly as a normal distribution. The mean height of adult American women is 63.5 inches with a standard deviation of 2.5 inches. Imagine that all possible random samples of size 25 (n=25) are taken from the population of American adult women's heights. the means from each sample would then be graphed to form the sampling distribution of sample means. The mean of this sampling distribution is ____ and the standard deviation of this sampling distribution is ___
Probability
How do we determine which sample statistic is the best estimate of the population parameter?
# of outcomes classified as A / total # of outcomes
How p(a) is calculated
[first calculate the SD by taking the square root of the standard proportion (0.56) multiplied by 1 minus the standard proportion (1-.56) divided by the sample size (1600) to get an SD of 0.0124096736] 0.548-0.572, 0.535-0.585, 0.523-0.597, no, outside,2, is
IN case study 19.1, we learned that about 56% of American adults actually voted in the presidential election of 1992, whereas about 61% of a random sample claimed that they had voted. The size of the sample was not specified, but suppose it were based on 1600 American adults, a common size for such studies. Into what interval of values should the sample proportion fall 68%, 95%, and almost all of the time? (round to 3 decimal places) 68%: ___-___ 95%: ___-___ almost all: ___-____ is the observed value of 61% reasonable based on your answer in part a? ___, 61% is well ___ the expected range. Now suppose the sample had been of only 400 people. Compute a standardized score to correspond to the reported percentage of 61% (round to nearest integer.) ____ Comment on whether you believe people in the sample could all have been telling the truth, based on your result. It ___ possible that all the people sampled were telling the truth.
1/128 (2^7)
If I flip a coin 7 times, what's the probability of getting tails seven times in a row?
Not likely
If I sample from a given population, how likely am I to get the sample statistic that I did?
1.000 have cancer 99.000 do not, total 100.000 (1%x100,000; 99%x100,000)
part a| Use a base rate to calculate totals of who have cancer and don't have cancer. Imagine there are 100,000 women who have breast lumps, and you know that 1% of all breast lumps are malignant. create a chart and fill out the totals in the last column using this information. hint: base rate is 1% (99% is rest of population) population is 100,000
800 show malignant (80% of 1,000) 200 show benign (false negative to 20%)
part b| from part a, the total of people who have cancer is 1,000 and those who do not have cancer comes out to a total of 99,000. now use knowledge of specificity and sensitivity of test to fill in the rest of the table. If the sensitivity of the test is 80%, then it correctly identifies 80% of the women who have cancer as having Out of those who have cancer, how many of the tests showed as malignant and how many of the test show as benign?
anchoring
people's risk perception can be severely distorted when they are provided with a reference point. This point is called an anchor and the process of doing so is called ___. Most people tend to stay relatively close to the anchor. - an insurance agent may say that he is giving a particular health insurance for 200$ whereas the same product is being given by some other company for $250. Here $200 is the anchor that the agent is using. It is very likely that we will get attracted to this product because of the discount that the agent is giving.
baseline risk
risk of having "the trait" for a group used as a basis for comparison
narrower
sampling distributions get ___ with increasing sample size
bell-shaped
sampling distributions of size n>30 are typically _____
0.54 to 0.66 [180/300=0.6 which is the population proportion. so, the sq. rt of (0.6(1-0.6)/300)=0.0283 ; 0.6+/-2(0.0283) = 0.54 and 0.66] yes, can, 54%, 66%, 0.42 to 0.78 [0.6+/-2 (sq rt. of (0.6(1-0.6)/30)] no, cannot, 42, 78, above and below, smaller, smaller, larger
A university is contemplating switching from the quarter system to the semester system. The administration conducts a survey of a random sample of 300 students and finds that 180 of them prefer to remain on the quarter system. (a)Construct a 95% confidence interval for the true proportion of all students who would prefer to remain on the quarter system. (Round your answers to two decimal places.) __ to ___ b) Does the interval you computed in part (a) provide convincing evidence that the majority of students prefer to remain on the quarter system? Explain. (Round your answer to the nearest percent, when necessary.) ____(yes or no). We _____(can or cannot) be confident based on the 95% confidence interval of ___% to ___%, that the majority of students prefer to remain on the equal systems, since the interval is ___ (above, below, above and below) 50% c)Now suppose that only 30 students had been surveyed and that 18 said they preferred the quarter system. Compute a 95% confidence interval for the true proportion that prefers to remain on the quarter system. (Round your answers to two decimal places.) ___ to ___ Does the interval provide convincing evidence that the majority of students prefer to remain on the quarter system? (Round your answer to the nearest percent, when necessary.) __(yes or no) we ____(can or cannot) be confident based on the 95% confidence interval, of ___% to __%, that the majority of students prefer to remain on the quarter systems, since the interval is ___ (above, below, above and below) 50% d) Compare the sample proportions and the confidence intervals found in parts (a) and (c). Use these results to discuss the role sample size plays in helping make decisions from sample data. The larger sample size in part (a) yielded a ___ (smaller or larger) standard error, and thus a ___ (smaller or larger) confidence interval, than the smaller sample size in part c. In general, a ___(smaller or larger) sample size produces a smaller 95% confidence interval
Sensitivity
Ability to detect disease in people who actually have the disease
Observe
Empirical probability is what we ___ in practice. Ex. We know the probability of getting landing on heads when we flip a coin, but empirical probability would be what we actually get when we flip the coin
Mutual exclusivity
Due to _____, if A & B cannot occur at the same time, p(A and B)=0
Add
Due to the addition (or) rule, if A and B are mutually exclusive, then you can ask chances of getting A or B which would call for you to ___ A and B P(AorB)= p(A)+p(B)-p(AandB)
7.5%
Let's say that you know that 1% of breast lumps turn out to be malignant. You find a lump in your breast. Your doctor recommends an X-ray test designed to detect cancer and tells you that it is 80%accurate at detecting malignant lumps (I.e. it will misread a malignant lump as benign 20% of the time and give you a false negative). You are also told that it is 90% accurate at detecting benign lumps (I.e. it will misread a benign lump as malignant 10% of the time and give you a false positive) Your test comes back positive. What are the chances that your lump is truly malignant?
49/50 (1-(1/50))
Lyme disease is a disease carried by ticks, which can be transmitted to humans by tick bites. Suppose the probability of contracting the disease is 1/50 for each tick bite. (a) What is the probability that you will not get the disease when bitten once?
0.711 (725/1019), 0.014 [sq rt of (0.711(1-0.711)/1019)], 0.683 to 0.739, 68.3% to 73.9%
One of the questions asked in a certain poll was: "All in all, if you had your choice, would you want to be rich, or not?" (A previous question had asked respondents if they thought they were rich, and the 2% who said yes were not asked this question.) Of the 1,019 people asked this question, 725 said yes, they would want to be rich. (a) What is the sample proportion of respondents who said yes, they would want to be rich? (Round your answer to three decimal places.) ____ (b) Find the standard error to accompany the proportion you found in part (a). (Round your answer to three decimal places.) ____ (c)Use your answers from parts (a) and (b) to construct a 95% confidence interval for the population proportion who want to be rich. (Round your answers to three decimal places.) ___ to ___ (d)Interpret the confidence interval found in part (c) by writing a few sentences explaining what it means. (Round your answers to one decimal place, when necessary.) We are 95% confident that between ____% and ___% of the people who do not think they are rich, want to be rich.
0.111 to 0.149 [0.13+/-2(sq rt. (0.13(1-0.13)/1216))], 0.818 to 0.856 (1234/1474=0.8372 ; sq rt. (0.8372(1-0.8372)/1474)=0.00962 ; 0.8372+/-2(0.00962)]
Refer to News Story 2: "Research shows women harder hit by hangovers," and the accompanying Original Source 2: "Development and initial validation of the Hangover Symptoms Scale: Prevalence and correlates of hangover symptoms in college students." Table 3 of the journal article reports that 13% of the 1,216 college students in the study said that they had not experienced any hangover symptoms in the past year. (a) Assuming that the participants in this study are a representative sample of college students, find a 95% confidence interval for the proportion of college students who have not experienced any hangover symptoms in the past year. Use the formula in this chapter. (Round your answers to three decimal places.) ___ to ___ b)The journal article also reported that the study originally had 1,474 participants, but only 1,234 reported drinking any alcohol in the past year. (Only those who reported drinking in the past year were retained for the hangover symptom questions.) Use this information to find a 95% confidence interval for the proportion of all students who would report drinking any alcohol in the past year. (Round your answers to three decimal places.) ___ to ___
0.24
Say you have a population of 75,000. 10% of that population have a disease. there is a sensitivity of 85% and a specificity of 70%. What is the probability of having cancer given the test showed positive?
Simpson's Paradox
Statistical phenomenon when relationship between two variables appears to be in one direction, but when a third variable is included, the relationship reverses directions
0.20 (74/367), 0.021 (sq rt(0.20(1-0.20)/367)), .158 to .242(.20+/-2(0.021))
Suppose a an advertisement for a certain drug reports the results of a double-blind study in which 367 patients took the drug and 193 took a placebo. Headaches were reported as a side effect by 74 of those taking the drug. (a) What is the sample proportion of patients taking the drug who reported headaches? (Round your answer to two decimal places.) ___ b)What is the standard error accompanying the proportion computed in part (a)? (Round your answer to three decimal places.) ___ c)Construct a 95% confidence interval for the population proportion based on the information from parts (a) and (b). (Round your answer to three decimal places.) ___ to ___
52.6% to 61.4% (57+/-4.4), 52.6% to 61.4% (.57+/-2[sq rt(.57(1-.57)/506]), 88.5% to 93.5% (.91+/-2[sq. rt.(.91(1-.91)/506)]), 86.6% to 95.4% (.91+/-4.4)
Suppose a certain magazine surveyed 506 American Catholics, asking questions regarding the Pope. From this survey, it was reported that "the sampling error is ± 4.4%." (a)One question asked was, "Do you favor allowing women to be priests?" to which 57% of the respondents answered yes. Using the reported margin of error of 4.4%, calculate a 95% confidence interval for the response to this question (as a %). ___ % to __% b)Calculate a 95% confidence interval for the question in part (a), using the formula in this chapter rather than the reported margin of error. (Round your answers to one decimal place.) ___% to ___% c)Another question in the survey was, "Is it possible to disagree with the Pope and still be a good Catholic?" to which 91% of respondents said yes. Using the formula in this chapter, compute a 95% confidence interval for the true percentage who would answer yes to the question. (Round your answers to one decimal place, when necessary.) ___% to ___% now compute a 95% confidence interval using the reported margin of error 4.4% ___% to ___%
0.003 [take the square root of the sample proportion (.22) multiplied by 1 minus the sample proportion (1-0.22) divided by sample size (9,007)], almost all, less than, yes
Suppose a poll of 9,007 adults taken in 1996 found that 9% held a particular religious belief, whereas a poll of 1,000 adults taken in 2007 found that 22% held that belief. Assuming a proper random sample was used, verify that the sample proportion for the poll taken in 1996 almost certainly represents the population proportion to be within about 1% (round to nearest 3 decimal places). The standard deviation for the possible sample proportions is approximately ___. We know that ___ of the sample proportions are likely to be within 3 standard deviations of the true proportion. Since 3 SDs is ___ 1%, we can verify that the sample proportion for the poll almost certainly represents the population proportion to be within about 1%. Based on these results, would you conclude that the proportion of all adults who believe in this particular religious belief was higher in 2007 than it was in 1996? explain. ___, the value of 22% found in 2007 is well outside the range of the true proportion of 1996.
0.152 to 0.269 [41/195=.2103 ; sq rt of (.2103(1-.2103)/195) =0.0292 ; 0.2103+/-2(0.0292)]
Suppose an advertisement for a certain drug reports the results of a double-blind study in which 193 patients took the drug and 195 took a placebo. Headaches were reported as a side effect by 41 of those taking the placebo. (a) Compute a 95% confidence interval for the true population proportion that would get headaches after taking a placebo. (Round your answers to three decimal places.) ___ to ___
In this case the value of 0.65 is a relative frequency because it was her observed proportion correct over the long run.
Suppose another friend guessed the color of 1,000 cards and got 650 correct. The friend claims she has ESP and has a 0.65 probability of guessing correctly. Is the value of 0.65 a relative-frequency probability or a personal probability? Explain.
It is a personal probability because it is based neither on physical assumptions nor repeated observations, but on your friend's belief about their own guessing ability.
Suppose another friend has never tried the experiment but believes he has ESP and can guess correctly with probability 0.65. Is the value of 0.65 a relative-frequency probability or a personal probability? Explain.
0.261 (0.241+0.020)
Suppose in 2012, the probability that a randomly selected child in a country was living with his or her mother as the sole parent was 0.241 and with his or her father as the sole parent was 0.020. What was the probability that a child was living with just one parent?
0.111 (1-0.52=0.48 this is the chance of being a girl being born, 0.48x0.48x0.48)
Suppose probability that a birth results in a boy is about 0.52. You offer a bet to an unsuspecting friend. Each day you will call the local hospital and find out how many boys and how many girls were born the previous day. For each girl, you will give your friend $1 and for each boy your friend will give you $1 a) Suppose that on a given day there are three births. What is the probability (as a decimal) that you lose $3 on that day?
0.04 (1(probability of being boy) - 1(probability of being girl) 0.52-0.48)
Suppose probability that a birth results in a boy is about 0.52. You offer a bet to an unsuspecting friend. Each day you will call the local hospital and find out how many boys and how many girls were born the previous day. For each girl, you will give your friend $1 and for each boy your friend will give you $1 c)Notice that your net profit is $1 if a boy is born and −$1 if a girl is born. What is the expected value of your profit for each birth (in $)?
40$ (1000x expected profit from each birth, 1,000x0.04)
Suppose probability that a birth results in a boy is about 0.52. You offer a bet to an unsuspecting friend. Each day you will call the local hospital and find out how many boys and how many girls were born the previous day. For each girl, you will give your friend $1 and for each boy your friend will give you $1 d) knowing the answer from part c is 0.04 (expected value of your profit for each birth), how much can you expect to make after 1,000 births in dollars
relative frequency (it means that over the long run the portion of times luggage will be temporarily lost, delayed, damaged, or pilfered is 1/1000)
Suppose that "the probability that a piece of checked luggage will be lost, delayed, damaged, or pilfered on a flight with a particular airline is 1/1,000." Interpret that statement, using the appropriate interpretation of probability.
0.049 [the square root of the population proportion(0.62) x (1-the population proportion(0.62)) / the sample size (100)],2, less than, likely
Suppose that 33% of the students at a university favor the semester system, 62% favor the quarter system and 5% have no preference. Is a random sample of 100 students large enough to provide convincing evidence that the quarter system is favored? explain. The rule for sample proportions (because this is categorical) tells us that the sample proportions for this situation would be bell-shaped with a mean of 0.62 and a standard deviation (round to 3 decimal places) of ____. Therefore, 95% of the time, the sample proportion would be within ____ standard deviations of 0.62. So there is a probability of ____ ____ 2.5% that the sample proportion would fall below 0.50. Therefore, we can conclude that a random sample of 100 students is ___ to be large enough to provide convincing evidence that the quarter system is favored.
109, 2.5 [take the SD of the population (25) divided by the square root of the sample size (100)], 106.5 and 111.5 (1.9+/-2.5), 104 and 114 (1.9+/-(2x2.5)), 101.5 and 116.5 (109+/-(3x2.5))
Suppose the population of the IQ scores in the town or city where you live is bell-shaped, with a mean of 109 and a standard deviation of 25. Describe the frequency curve for possible sample means that would result from random samples of 100 IQ scores. Using the rule for sample means, we would expect the possible sample means to be bell-shaped, with a mean of ____ and a standard deviation of ____. Therefore, we would find them between: ___ and ___ about 68% of the time, between ___ and ___ 95% of the time, and about ___ and ___ 99.7% of the time.
211/212 (1-(1/212))
Suppose the probability of being born on a Friday the 13th is about 1/212. What is the probability of not being born on a Friday the 13th?
27.25, 27.5, 27.75, 28, 28.25, 28.5, 28.75 (SD-3, SD-2, SD-1, mean, SD+1, SD+2, SD+3)[knowing that the mean of the sampling distribution of the mean is equal to the mean of the population (28). take the mean and divide it by the square root of n (16) to get the SD.(0.25)]
Suppose you are interested in estimating the average number of miles per gallon of gasoline your car can get. You calculate the miles per gallon for each of the next nine times you fill the tank. Suppose, in truth, the values for your car are bell-shaped, with a mean of 28 miles per gallon and a standard deviation of 1. part a) Draw a bell curve of the possible sample means you are likely to get based on your sample of 16 observations. Include the intervals into which 68%, 95%, and almost all of the potential sample means will fall.
27.7, 27.8, 27.9, 28, 28.1, 28.2, 28.3 (SD-3, SD-2, SD-1, mean, SD+1, SD+2, SD+3)[knowing that the mean of the sampling distribution of the mean is equal to the mean of the population (28). take the mean and divide it by the square root of n (100) to get the SD.(0.1)] more, the same, smaller
Suppose you are interested in estimating the average number of miles per gallon of gasoline your car can get. You calculate the miles per gallon for each of the next nine times you fill the tank. Suppose, in truth, the values for your car are bell-shaped, with a mean of 28 miles per gallon and a standard deviation of 1. part b) Draw a bell curve of the possible sample means you are likely to get based on your sample of 100 observations. Include the intervals into which 68%, 95%, and almost all of the potential sample means will fall. compared to your answer on part A, part b (which has a larger sample size) now has a picture that is ___ compact. the mean is ____. The standard deviation is ____.
(To first find money on your third try, you must not find money on your first two. Since the probabilities are independent, the probability is (70%)(70%)(30%)) 14.7%
Suppose you routinely check coin-return slots in vending machines to see whether they have any money in them. You have found money about 30% of the time What is the probability that the next time you will find money is on the third try?
(Since the probability of not finding money by the third try is (70%)(70%)(70%) = 34.3% by rule 3, the probability must be) 65.7%
Suppose you routinely check coin-return slots in vending machines to see whether they have any money in them. You have found money about 30% of the time What is the probability that you will have found money by the third try?
(These events are complements, so the probability of not finding money is 100% − 30% = ) 70%.
Suppose you routinely check coin-return slots in vending machines to see whether they have any money in them. You have found money about 30% of the time. What is the probability that you do not find money the next time you check?
0.78, 0.109, 0.139, 0.17, 0.201, 0.231, 0.262 (SD-3, SD-2, SD-1, mean, SD+1, SD+2, SD+3) [square root of the population proportion x (1- population proportion) divided by the sample size][square root of (.17(1-.17)/150) add and subtract that number to the mean to get the SDs]
Suppose you want to estimate the proportion of students at your college who are left-handed. You decide to collect a random sample of 150 students and ask them which hand is dominant. Suppose the truth is that 0.17 or 17%, of the students are left-handed, and you take a random sample of 150 students. Use the rule for sample proportions to draw a bell curve consisting of the standard deviations and mean.
exists, is, is, likely, all, does
Suppose you want to estimate the proportion of students at your college who are left-handed. You decide to collect a random sample of 200 students and ask them which hand is dominant. Go through the conditions for which the rule for sample proportions applies and determine if the rule would apply to this situation. For condition 1: a population ____ with a fixed proportion with the specified trait. condition 2: a random sample of students ____ selected condition 3: the sample size ___ large enough because we are ___ to see at least ten with and ten without the specified trait. Since ___ of the three conditions are met, we conclude that the rule for sample proportions ___ apply to this situation.
1/2 (you have a 50/50 shot of guessing the right color out of two)
Suppose you wanted to test your extrasensory perception (ESP) ability using an ordinary deck of 52 cards, which has 26 red and 26 black cards. You have a friend shuffle the deck and draw cards at random, replacing the card and reshuffling after each guess. You attempt to guess the color of each card. What is the probability that you guess the color correctly by chance?
No (.14 doesn't equal .19 ((389/2799)=(144/757))
Table: history(a)= 213, history(b)=456, history(c)=113, history(d)=24, history(f)=9, psy(a)=300, psy(b)=500, psy(c)=200, psy(d)=12, psy(f)=10, math(a)=100, math(b)=131, math(c) =279, math(d)=43 math(f)= 20, bio(a)=144, bio(b)=130, bio(c)=100, bio(d)=9, bio(f)=6 Is being a biology major independent of making an A?
0.39 ((88/2799)+(1022/2799)-[(88/2799)x(12/88)])
Table: history(a)= 213, history(b)=456, history(c)=113, history(d)=24, history(f)=9, psy(a)=300, psy(b)=500, psy(c)=200, psy(d)=12, psy(f)=10, math(a)=100, math(b)=131, math(c) =279, math(d)=43 math(f)= 20, bio(a)=144, bio(b)=130, bio(c)=100, bio(d)=9, bio(f)=6 P(D or Psychology)?
1217/2799
Table: history(a)= 213, history(b)=456, history(c)=113, history(d)=24, history(f)=9, psy(a)=300, psy(b)=500, psy(c)=200, psy(d)=12, psy(f)=10, math(a)=100, math(b)=131, math(c) =279, math(d)=43 math(f)= 20, bio(a)=144, bio(b)=130, bio(c)=100, bio(d)=9, bio(f)=6 P(b)?
279/692
Table: history(a)= 213, history(b)=456, history(c)=113, history(d)=24, history(f)=9, psy(a)=300, psy(b)=500, psy(c)=200, psy(d)=12, psy(f)=10, math(a)=100, math(b)=131, math(c) =279, math(d)=43 math(f)= 20, bio(a)=144, bio(b)=130, bio(c)=100, bio(d)=9, bio(f)=6 What is the probability of being a math major given that you made a C? P(math|C)?
0.43 (473/1112)
There is a total of 473 males, 364 like trucks, the rest like dolls. There is a total of 639 females, 408 like dolls, the rest like trucks. Construct a table then calculate the probability of being male. P(male) (Hint the total number of trucks is 595, total number of dolls is 517, and total number of kids is 1112)
.54 (595/1112)
There is a total of 473 males, 364 like trucks, the rest like dolls. There is a total of 639 females, 408 like dolls, the rest like trucks. Construct a table then calculate the probability that you are female (Hint the total number of trucks is 595, total number of dolls is 517, and total number of kids is 1112)
3, 15, 225 (15^2)
The administration of a large university wants to use a random sample of students to measure student opinion of a new food service on campus. Administrators plan to use a continuous scale from 1-100, where 1 is complete dissatisfaction and 100 is complete satisfaction. They know from past experience with such questions that the standard deviation for the responses is going to be about 5, but they do not know what to expect for the mean. They want to be almost sure that the sample mean is within plus or minus 1 point of the true population mean value. How large will their random sample have to be? In order to be almost certain, we want our sample mean estimate to be within ___ standard deviations away from the mean, so the correct calculation for the necessary sample is ___ divided by the square root of n equals 1 (?/square root N=1) which, when we solve for n gives us a required sample size of n=___
0.39 (231/595)
There is a total of 473 males, 364 like trucks, the rest like dolls. There is a total of 639 females, 408 like dolls, the rest like trucks. Construct a table then calculate the probability that you are female, given that you like trucks. P(F|T) (Hint the total number of trucks is 595, total number of dolls is 517, and total number of kids is 1112)
.21 (109/517)
There is a total of 473 males, 364 like trucks, the rest like dolls. There is a total of 639 females, 408 like dolls, the rest like trucks. Construct a table then calculate the probability that you are male, given that you like dolls p(M|D) (Hint the total number of trucks is 595, total number of dolls is 517, and total number of kids is 1112)
0.37 ((517/1112)x(408/517))
There is a total of 473 males, 364 like trucks, the rest like dolls. There is a total of 639 females, 408 like dolls, the rest like trucks. Construct a table then calculate the probability that you like dolls AND are female p(dolls)xp(females|dolls) (Hint the total number of trucks is 595, total number of dolls is 517, and total number of kids is 1112)
population standard deviation, square root of the sample size, smaller
The rule for sample means says that the standard deviation of the frequency curve for the samples will be the ______ divided by the _____. So the larger the sample size the ___ the standard deviation of this frequency curve, and hence the better the estimate.
Variability
There is ____ in all samples, even if from the same population.
0.23 (109/473)
There is a total of 473 males, 364 like trucks, the rest like dolls. There is a total of 639 females, 408 like dolls, the rest like trucks. Construct a table then calculate p(D|M) probability of liking dolls given that you are male (Hint the total number of trucks is 595, total number of dolls is 517, and total number of kids is 1112)
.36 (231/639)
There is a total of 473 males, 364 like trucks, the rest like dolls. There is a total of 639 females, 408 like dolls, the rest like trucks. Construct a table then calculate the probability that you like trucks given that you are female. P(T|F) (Hint the total number of trucks is 595, total number of dolls is 517, and total number of kids is 1112)
No, .54 doesn't equal .77 (595/1112 = 364/473)
There is a total of 473 males, 364 like trucks, the rest like dolls. There is a total of 639 females, 408 like dolls, the rest like trucks. Construct a table then calculate whether liking trucks and being male is independent from each other. P(T)=P(T|M) (Hint the total number of trucks is 595, total number of dolls is 517, and total number of kids is 1112)
Yes ((517/1112)x(0/517)=0)
There is a total of 473 males, 364 like trucks, the rest like dolls. There is a total of 639 females, 408 like dolls, the rest like trucks. Construct a table. Is the probability of liking dolls and trucks mutually exclusive p(dolls and trucks) = p(dolls) * p(trucks|dolls) (Hint the total number of trucks is 595, total number of dolls is 517, and total number of kids is 1112)
.63 ((595/1112)+(473/1112)-[(595/1112)x(364/595)])
There is a total of 473 males, 364 like trucks, the rest like dolls. There is a total of 639 females, 408 like dolls, the rest like trucks. Construct a table. What is the probability of liking trucks OR being male p(trucks or male)=p(trucks)+p(males)-p(trucks and males) (Hint the total number of trucks is 595, total number of dolls is 517, and total number of kids is 1112)
68%, 95%, 100% (usually 99.7%)
Use the Empirical Rule to specify what level of confidence (as a %) would accompany each of the following confidence intervals. (Round your answers to the nearest percent.) a) sample proportion +/- 1 (SEP) b) sample proportion +/-2 (SEP) c) sample proportion +/- 3 (SEP)
133 and 137 lbs (12/sqrt 36 gives us an SD of 2. 135+/-2)
Weight is a measure that tends to be normally distributed. Suppose the mean weight of all women at a large university is 135 pounds, with a standard deviation of 12 pounds. If you took a random sample of 36 university women, there would be a 68% chance that the sample mean (x-bar) weight would be between: ___ and ___ lbs
does, population, 0.360 to 0.420 [calculating the SD of sample proportion, so take the square root of the sample proportion (.39) multiply it by one minus the sample proportion (1-0.39) and divide it by the sample size (1027) to get the SD (0.15219907). Take the population proportion (0.39) +/- (0.15219907) to get the interval (0.360-0.420)], yes, would, outside
according to the Sacramento Bee, "a 1997-98 survey of 1,027 Americans conducted by the National Sleep Foundation found that23% of adults say they have fallen asleep at the wheel in the last year. Conditions 2 and 3 needed to apply the rule for Sample Proportions are met because this result is based on a large random sample of adults. Explain how condition 1 is also met. There ___ exist an actual ___ of adults, a fixed proportion of whom have fallen asleep at the wheel in the last year. The article also said that (based on the same survey) "37% of adults report being so sleepy during the day that it interferes with their daytime activities." If, in truth, 39% of all adults have this problem, find the interval in which about 95% of all sample proportions should fall, based on samples of size 1,027. (round to nearest 3 decimal places) ___ to ____ does the result of this survey fall into that interval, yes or no? Suppose a survey based on a random sample of 1,027 college students was conducted and 21% reported being so sleepy during the day that it interferes with their daytime activities. Would it be reasonable to conclude that the population proportion of college students who have this problem differs from the proportion of all adults who have the problem? explain. it ___ be reasonable to conclude that they differ; 21% is well ___ the interval computed in part b.
