Stats HW and Quizzes for Exam 2
In the population, people's salary follow a positively skewed distribution, with μ= 70,000 and σ = 20,000. 1. If we randomly pick a person, what's the probability that this person's salary is over 72,000? 2. If we randomly pick 400 people, what's the probability that the average salary of these 400 people is over 72,000?
(1) we dont know because we dont have a normal distribution (2) SE= 20,000/√400 = 1000. So Z-score=2. Probs= 2.28%
Context: In essence, we can try to estimate the population mean SAT score for all those who received coaching. If this mean is about the same as the mean for all those who did not receive coaching, then it suggests SAT coaching might not be very useful. In the present scenario, this high school teacher collected a sample of 30 students received coaching. The mean SAT score of this sample is X bar =503. Suppose the population standard deviation is σ=110. Question: The researcher plans to calculate a 68% confidence interval for the population mean. What is the confidence level in this case? (A) 68% (C) 100% (B) 32% (D) 0% (E) 100% or 0%
(A) 68%
Context: In essence, we can try to estimate the population mean SAT score for all those who received coaching. If this mean is about the same as the mean for all those who did not receive coaching, then it suggests SAT coaching might not be very useful. In the present scenario, this high school teacher collected a sample of 30 students received coaching. The mean SAT score of this sample is X bar =503. Suppose the population standard deviation is σ=110. Question: The length/width of a confidence interval (CI) is defined as: (upper end - lower end). Compare the 68%, 95%, and 70% confidence intervals you just calculated in Q11, Q14, and Q15. Which interval is shortest? (B) 95% CI (E) there is not enough information (D) the width is the same for the 3 intervals (C) 70% CI (A) 68% CI
(A) 68% CI
Sarah and Eric want to estimate the mean of a population. Sarah plans to collect a sample of size n = 100 and compute a 90% confidence interval for µ. Eric plans to collect a sample of size n = 200 and compute an 80% confidence interval for µ. In the long run, whose confidence interval will cover µ more often? (D) all of the above are possible (B) Eric's interval (C) Sarah's interval and Eric's interval will cover µ equally often in the long run (E) there is not enough information to answer this question (A) Sarah's interval
(A) Sarah's interval
Context: In essence, we can try to estimate the population mean SAT score for all those who received coaching. If this mean is about the same as the mean for all those who did not receive coaching, then it suggests SAT coaching might not be very useful. In the present scenario, this high school teacher collected a sample of 30 students received coaching. The mean SAT score of this sample is X bar =503. Suppose the population standard deviation is σ=110. Question: If I computed a 90% CI, without any serious calculation, how would the width of the interval change? (A) longer than the 68% CI, but shorter than the 95% CI (B) shorter than the 68% CI, but longer than the 95% CI (D) it is impossible to predict the length the 90% CI (C) sometimes longer than the 68% CI, but sometimes shorter than the 68% CI
(A) longer than the 68% CI, but shorter than the 95% CI
Because normal distribution is symmetric, _____________ (A) scores above the mean are distributed the same as scores below the mean. (B) extreme scores are less likely to appear. (C) there are infinitely many normal distributions. (D) all of the above
(A) scores above the mean are distributed the same as scores below the mean.
Context: In essence, we can try to estimate the population mean SAT score for all those who received coaching. If this mean is about the same as the mean for all those who did not receive coaching, then it suggests SAT coaching might not be very useful. In the present scenario, this high school teacher collected a sample of 30 students received coaching. The mean SAT score of this sample is X bar =503. Suppose the population standard deviation is σ=110. Question:Does the confidence interval you just calculated in Q11 (483.12,523.08) above cover the population mean µ? (A). it may or may not cover µ, but we do not know because µ is unknown. (C). it covers µ with 68% probability (D). it is very likely to cover µ, but the specific probability is unknown (B). it covers µ with 100% probability
(A). it may or may not cover µ, but we do not know because µ is unknown.
20. Let us use W to denote people's weight. Suppose for all the UCM students, W ~ N(150, 20). We then calculate the Z-score for each UCM student, and thus we have a new data set about the Z-scores of all the UCM students. What is the distribution of those Z-scores? (A) N(150, 20) (B) N(0, 1) (C) will be a normal distribution, but mean and standard deviation are unknown (D) N(-3, 3) (E) the z-scores will not follow a normal distribution
(B) N(0, 1)
Context: In essence, we can try to estimate the population mean SAT score for all those who received coaching. If this mean is about the same as the mean for all those who did not receive coaching, then it suggests SAT coaching might not be very useful. In the present scenario, this high school teacher collected a sample of 30 students received coaching. The mean SAT score of this sample is X bar =503. Suppose the population standard deviation is σ=110. Question: Do we care whether the confidence interval we just computed in Q11 (483.12,523.08) covers µ? (A) Yes, we care. If this particular interval does not cover µ, the statistical inference we just did was a failure. (B) No. We do not care how well this particular interval does, but the overall performance of all the possible intervals in the long term. (C) There is 68% probability that the confidence interval in Q11 covers µ. So we have to just settle for this 68% probability.
(B) No. We do not care how well this particular interval does, but the overall performance of all the possible intervals in the long term.
Which of the following is a wrong method for point estimation? (B) use sample median to estimate population median (C) use sample variance to estimate population variance (A) use sample mean to estimate population mean (D) all of the above are correct methods for point estimation
(B) use sample median to estimate population median
SAT scores have a standard deviation σ=110 in the population. California Department of Education is interested in the average SAT score of all the students in the Central Valley region. The state government hired two companies to independently infer this mean SAT score. Company A collected a sample of 900 students and obtained a mean SAT score of 522. Company B collected a sample of 1600 students and obtained a mean SAT score of 530. If both Companies A and B form a 90% confidence interval for the population mean SAT score, whose confidence interval is narrower? Hint: this question does not need any serious calculation. (E) there is not enough information (B). company B (A). company A (C). the same (D). sometimes company A, sometimes company B
(B). company B
Tom and Jerry want to estimate the mean for a population. Tom plans to collect a sample of size n = 100 and calculate a 90% confidence interval for µ. Jerry plans to collect a sample of size n = 200 and calculate a 90% confidence interval for µ. In the long run, whose confidence interval will cover µ more often? (C) Tom's interval and Jerry's interval will cover µ equally often in the long run (A) Tom's interval (E) there is not enough information to answer this question (B) Jerry's interval (D) all of the above are possible
(C) Tom's interval and Jerry's interval will cover µ equally often in the long run
Suppose individual X scores in the population follow a normal distribution N(38, 20). A researcher draws numerous samples of sample size n = 100 from the population, and in each sample she calculates the sample mean. Then 68% of these sample means should approximately fall within ____________________ (C) 38 and 44 (D) 36 and 40 (A) 34 and 40 (B) 34 and 38
(D) 36 and 40
ACT scores follow N(20.8, 4.8). SAT scores follow N(500, 110). Tom took ACT and he did better than 75% of other ACT test takers. Which of the following is correct? A) Tom's ACT score is about 22.1. (B) Tom is on the 25th percentile. (C) Tom's z-score is 0.75. (D) If Tom took SAT instead, he should have got about 574. (E) Tom's z-score is 0.25.
(D) If Tom took SAT instead, he should have got about 574.
Which of the following is INCORRECT? (A) The normal distribution is a family of distributions (B) There are infinitely many normal distributions (C) To identify a normal distribution, we need to know two things about this distribution (D) If a normal distribution is perfectly symmetric, it is called standard normal distribution.
(D) If a normal distribution is perfectly symmetric, it is called standard normal distribution.
Knowing that data follow a normal distribution allows us to (A) calculate the probability of obtaining a score greater than some specified value. (B) understand how data points in the population ended up being in my sample (C) calculate what range of values are likely or unlikely to occur (D) all of the above
(D) all of the above
For a normal distribution N(µ, σ), the mean can take on any value and the standard deviation can take on any positive value. Therefore, (A) normal distribution is also called standard normal distribution (B) N(2, 5) and N(3, 6) are two different normal distributions (C) there are an infinite number of possible normal distributions (D) in some situations data may approximately follow N(-5, -2) (E) B and C only (F) all of the above
(E) B and C only
Jerry collected a sample of size n = 50, and calculated a 70% confidence interval for µ as [104, 126]. What is the probability that µ is between 104 and 126? (A) 100% (B) 0% (E) either A or B (D) 30% (C) 70%
(E) either A or B
Jerry collected a sample of size n = 50, and formed a 70% confidence interval for µ as [104, 126]. What is the probability that µ is between 104 and 126? (A) 100% (B) 0% (C) 70% (D) 30% (E) either A or B
(E) either A or B
Suppose for all the sales managers in this country, their salaries have a mean of 135,000 and a standard deviation of 20,000. What % of sales managers have a salary of 155,000 or higher? (A) 16% (B) 15.87% (C) 34.13% (D) 68% (E) there is not enough information
(E) there is not enough information
Suppose weights of men follow N(160, 20) in the population. Which is more likely to appear? (I) a man weights above 175 lbs (II) total weight of 10 random men exceeds 1750 lbs (I) & (II) have exactly the same probability (II) is more likely to appear (I) is more likely to appear (I) & (II) have similar probability
(I) is more likely to appear
Consider the sampling distribution of sample means. With a large sample size and a small population variance, all the possible sample means (a) will be close to the population mean. (c) will slightly overestimate the population mean. (b) will slightly underestimate the population mean. (d) will equal the population mean
(a) will be close to the population mean.
Tom collected a sample of size n = 100, and formed a 90% confidence interval for µ as [4, 6]. Suppose later he find out that µ = 7 in the population. Which of the following statements is correct? (a) Tom must have done something wrong in collecting data. (b) Tom's interval was too wide. (c) Tom must have made a mistake in calculation. (d) It does not necessarily mean Tom made a mistake. Sometimes this can happen in practice.
(d) It does not necessarily mean Tom made a mistake. Sometimes this can happen in practice.
Tom collected a sample of size n = 100, and got a 90% confidence interval for µ as [4, 6]. Suppose later he finds out that µ = 7 in the population. Which of the following statements is correct? (d) It does not necessarily mean Tom made a mistake. Sometimes this can happen in practice. (b) Tom's interval was too wide. (a) Tom must have done something wrong in collecting data. (c) Tom must have made a mistake in calculation.
(d) It does not necessarily mean Tom made a mistake. Sometimes this can happen in practice.
Given a particular sample, we compute a 95% confidence interval for µ as [112, 118]. Which of the following is INCORRECT? (d) We cannot say there is 95% probability that µ falls within [112, 118], because µ is not a random variable but a fixed value. (a) The probability 95% is about all the confidence intervals based on all the possible samples, not about this particular one [112, 118]. (e) The probability that µ falls within [112, 118] is 95%. (c) If this interval covers µ, then µ is somewhere between 112 and 118. (b) We don't know whether this interval covers µ because µ is unknown.
(e) The probability that µ falls within [112, 118] is 95%.
Standard error ___ as sample size ___
decreases, increases
What is the standard error?
-just a standard deviation BUT of the sampling distribution of sample means
Given a particular sample, we compute a 90% confidence interval for µ as [15, 20]. Which of the following is correct? -90% of the time, µ is in [15, 20] -µ must be in [15, 20] -we don't know whether the confidence interval in this sample covers µ, and we don't care whether the interval covers µ either -if we use statistical methods properly, the confidence interval we get in this sample will cover µ -if we use statistical methods properly, there is 90% probability that the confidence interval we get in this sample will cover µ
-we don't know whether the confidence interval in this sample covers µ, and we don't care whether the interval covers µ either
The population has 50,000 athletes. We randomly pick a sample of 2,000 athletes from the population, and calculate their average height. Based on the average height of these 2,000 people, we have _______ sampling distribution(s) of average heights.
0
Suppose the distribution of family incomes of all the US households is positively skewed with mean 71,000 and standard deviation 15,000. -We randomly select 2000 families. There is __________% probability that the average income of this group of 2000 families is below 70,000.
0.14
X ~ N(12, 0.11) A supermarket sells this type of cola in boxes, and each box has 36 cans. We consider each box as a random sample of size n = 36. If we randomly pick a box from the supermarket, there is ________% probability that the average weight of the box of 36 cans exceeds 12.05 oz.
0.32
A supermarket sells peaches in boxes, and each box has 16 peaches. The peaches vary in size, but all the boxes have the same label that says the net weight of the box is 64 oz and the price is $7. Suppose workers at this supermarket never weight the boxes, but just randomly put 16 peaches in each box. Some customers worry that selling peaches in this way is unfair, because they may get a box that weights much less than 64 oz but they pay the same price. The supermarket manager tries to dissuade their concern. Suppose weight of individual peaches in the population follow X ~ N(4, 0.2). If we randomly pick a box, what is the probability that the weight of the box is less than 62 oz (ie, 2 oz less than what the label says)?
0.62
Suppose the weight of adults follow a normal distribution with mean 160 lbs and standard deviation 20. A certain scenic boat tour company operates on a small boat that has a capacity of 10 people only. To ensure safety, every passenger is required to have their weight measured before getting on the boat. If the total weight of the 10 passengers on the boat exceeds 1750 lbs, the boat is overweight and not allowed to operate. Assuming tourists all randomly come to the boat, how often will the boat be overweight? We want to know the probability of the total weight of the 10 people in the boat exceeding 1750 lb. This is equivalent to the probability of the average weight exceeding 175 lb. The probability is ________%.
0.89
The population has 20,000 athletes. We randomly put these athletes on buses, and each bus has 50 people. Within each bus, we calculate the average height of the 50 athletes in that particular bus. Because there are 400 buses, we have 400 average heights. Based on these 400 average heights, we have _______ sampling distribution(s) of average heights.
1
Marathon runners in the population. In the population σ=20 (suppose μ is unknown) Randomly put every 50 runners in a bus. In Bus #23, the mean weight is x̄ =150 1. Help the bus driver #23 calculate a 95% confidence interval for μ 2. Does this 95% CI include μ? 3. What's the probability that μ is the 95% CI from Q2?
1. [144,156] 2. We don't know because μ is unknown 3. Either 100% or 0%, but no other possibilities
Context: In essence, we can try to estimate the population mean SAT score for all those who received coaching. If this mean is about the same as the mean for all those who did not receive coaching, then it suggests SAT coaching might not be very useful. In the present scenario, this high school teacher collected a sample of 30 students received coaching. The mean SAT score of this sample is X bar =503. Suppose the population standard deviation is σ=110. Question: To compute a 70% confidence interval for the population mean, the Z-score we should use for the interval is ______________
1.04
Suppose the weight of adults follow a normal distribution with mean 160 lbs and standard deviation 30. A small private plane can carry 10 passengers only, and it cannot operate if the total weight of the 10 passengers exceeds 1800 lbs. Suppose passengers of this plane all randomly come from the population of adults. The probability that the total weight of 10 passengers exceeds 1800 lbs is __________%.
1.7
Assume that the population of heights of men (in inches) has a normal distribution N(69.5, 2.4). If we draw a random sample of 100 men from the population, what is the probability that the mean height of the sample is less than 69.0 inches?
1.88
N(µ = 140, σ = 32). An LDL level below 100 is considered as optimal. If we randomly pick a man from the general public, what is the probability that this man has an optimal cholesterol level?
10.56
N(582, 150)If I random select a sample of 100 farms, what is the probability that the sample mean is greater than 600 acres?
11.51
In a normal distribution, what % of data have a z-score that is larger than 1.20?
11.51%
Scores of a certain test follow a normal distribution whose µ and σ are unknown. We know exactly 50% of the scores are below 120, and the other 50% are above 120. Then what is the mean of this normal distribution?
120
ACT scores follow N(20.8, 4.8). SAT scores follow N(500, 110). Jack's SAT score is 380. Jack is on the _________ percentile.
13.79 or 13.57 or 14
N(µ = 140, σ = 32). If an LDL level between 160 and 190 means the man should start taking preventive actions, what percent of men need preventive actions?
20.49
What is the median of ACT scores in the population? ACT scores follow N(20.8, 4.8). SAT scores follow N(500, 110).
20.8
N(µ = 140, σ = 32). If we randomly pick a man from the general public, what is the probablity that this man's LDL score is below 100 or above 180? Note: the answer is the total of the two areas of interest
21.12
A population has a mean (m) of 24.12 and a standard deviation (s) of 4. Assume that a sampling distribution of sample means has been constructed, based on repeated samples of n = 100 from this population. a. What would be the value of the mean of the sampling distribution? b. What would be the value of the standard error of the mean?
24.12, .40
Suppose the weight of adults follow a normal distribution with mean 160 lbs and standard deviation 30. A small private plane can carry 10 passengers only, and it cannot operate if the total weight of the 10 passengers exceeds 1800 lbs. Suppose passengers of this plane all randomly come from the population of adults. The probability that an individual passenger's weight exceeds 180 lbs is __________%.
25.14
Individual X scores in the population have a standard deviation of 15. For all the possible samples with sample size n = 25, what is the standard error of those sample means?
3
Assume that cans of cola are filled so that the actual amounts are normally distributed with a mean of 12.00 oz and a standard deviation of 0.11 oz. We use X to denote the weight of a can of cola, so X ~ N(µ = 12, σ = 0.11). If we randomly pick a can of cola from the production line, what is the probability that this can weights less than 11.80 oz?
3.44
ACT sores ~ N(20.8, 4.8) SAT sores ~ N(500, 110) June's SAT is 760. What would she have got if she took ACT instead?
32.13
X ~ N(12, 0.11) If we randomly pick a can of cola from the production line, there is _________% probability that the can of cola we get is heavier than 12.05 oz.
32.64
I have collected a sample of data (2,2,2,6,8) What's my best guess of the population mean? Do I know how much off this particular guess is?
4
ACT scores follow N(20.8, 4.8). SAT scores follow N(500, 110). Amy took SAT and her score is on the 30th percentile. What is Amy's SAT score?
442.8 or 441.7
Context: In essence, we can try to estimate the population mean SAT score for all those who received coaching. If this mean is about the same as the mean for all those who did not receive coaching, then it suggests SAT coaching might not be very useful. In the present scenario, this high school teacher collected a sample of 30 students received coaching. The mean SAT score of this sample is X bar =503. Suppose the population standard deviation is σ=110. Question: Compute a 95% confidence interval for the population mean: [ _______ , 542.36].
462.83
N(µ = 140, σ = 32). If an LDL level above 190 is considered as dangerous, what percent of men have a dangerous cholesterol level?
5.94
N(µ = 140, σ = 32). If we randomly pick a man from the general public, what is the probability that this man's LDL score is below 90?
5.94
The population has 50,000 athletes. We randomly put these athletes on buses, and each bus has 50 people. So there are 1,000 buses. Within each bus, we calculate the average height of the 50 athletes in that particular bus. In the sampling distribution of average heights, each data point is based on ________ athletes.
50
ACT scores follow N(20.8, 4.8). SAT scores follow N(500, 110). What is the mode of SAT scores in the population?
500
Context: In essence, we can try to estimate the population mean SAT score for all those who received coaching. If this mean is about the same as the mean for all those who did not receive coaching, then it suggests SAT coaching might not be very useful. In the present scenario, this high school teacher collected a sample of 30 students received coaching. The mean SAT score of this sample is X bar =503. Suppose the population standard deviation is σ=110. Question: The SAT sample mean for these 30 students mentioned above is X¯=503. Based on this sample, a 68% confidence interval for the population SAT mean is [483.12, ____________]
523.08
ACT scores follow N(20.8, 4.8). SAT scores follow N(500, 110). Joe's took SAT and his score is on the 74th percentile. What is Joe's SAT score?
570.4 or 571.5
A population has a mean (m) of 615 and a standard deviation (s) of 90. Assume that a sampling distribution of sample means has been constructed, based on repeated samples of n = 400 from this population. a. What would be the value of the mean of the sampling distribution? b. What would be the value of the standard error of the mean?
615, 4.50
N(µ = 140, σ = 32). If an LDL level below 160 is considered as acceptable, what percent of men have an acceptable cholesterol level?
73.57
N(µ = 140, σ = 32). If an LDL level below 115 is considered as healthy, what percent of men have a no-so-healthy cholesterol level (that is, above 115)?
78.23
In a normal distribution, what percent of data are above Z = -0.84 (negative 0.84)?
79.95%
Suppose SE=5, x̄= 20 Calculate 50%, 80%, 90%, and 99% CI, which is wider?
99%?
The standard error in the sampling distribution of sample means is A) just standard deviation B) also called sampling deviation C) sampling error D) standard deviation of individual X scores
A) just standard deviation
Suppose the distribution of family incomes of all the US households is positively skewed with mean 71,000 and standard deviation 15,000. -If I draw numerous samples from the US and each sample has 2000 families. For each sample I calculate the average income of the families in that particular sample. What would the sampling distribution of average family incomes look like?
Approximately normal
Suppose the income of all the American families is like this: (positively skewed) Draw numerous samples. Each sample has 1,000 families. For each sample I calculate its average income. The distribution of all these income averages should be:
Approximately normal
To obtain the sampling distribution of sample means, we need to _____ A) have a large sample B) have numerous samples C) draw a sample; then make a histogram for the sample D) use a normal distribution to calculate paobabitlies
B) have numerous samples
Why can we not use sample median, mode to detect the populations mean/median/mode?
Because in the long run, we will still make mistakes
The sampling distribution of X̅ _____ A) estimates the population mean B) is the mean of the distribution over all the possible samples C) describes how X̅ changes over all the possible samples D) is the distribution of data in a sample
C)describes how X̅ changes over all the possible samples
Which case is more likely to appear? (I) a can of cola weights above 12.05 oz (II) the average weight of 36 cans of cola is above 12.05 oz
Case (I) is more likely to appear (based on previous questions)
Suppose the weights of adults follow N(160,20). Which is more likely to appear? A) a person weights above 175 lbs B) total weight of 10 people is above 175 lbs C) the two cases are equally likely
Case A: 22.66% Case B: 0.89% Case A is more likely to appear.
Which of the following is NOT part of the Central Limit Theorem? A) The mean of X̅ approximately equals the mean of X B) The standard error of X̅ approximately equals the standard deviation of X divided by √n C) The sampling distribution of X̅ is approximately normal regardless of what distribution X has D) all of the above are part of the Central Limit Theorem
D) all of the above are part of the Central Limit Theorem
For Point Estimation, which one is true : A) sample median -> population median B) sample median -> population mean C) sample mode -> population mode D) sample mean -> population mean
D) sample mean -> population mean
If we draw a sample of size N =1,000,000 from a population, we will have a sampling distribution of sample means because the sample size is so large. T or F
False
If we draw a sample with sample size 500,000 from a population, we will have a sampling distribution of sample means because the sample size is so large. T or F
False
If we draw numerous random samples of sample size n = 50 from a population, we will have numerous sampling distributions of sample means. T or F
False
If we draw numerous random samples of size N =50 from a population, we will have numerous sampling distributions of sample means. T or F
False
In a random sample, the selection of any one unit affects the selection of any other unit. True or False
False
Suppose scores of a certain personality test follow a negatively skewed distribution in the population. We randomly draw a sample of 1000 people from the population. Because the sample size is large and we use random sampling, based on the central limit theorem, these 1000 test scores in our sample should approximately follow a normal distribution. T or F
False
The mean of a large, properly drawn sample will be exactly the same as the population mean. T or F
False
We can only calculate Z-scores in a normal distribution. True or False?
False, for any distribution we can calculate Z-scores
What is the confidence interval for the mean?
It's an interval or range of values within which the true mean of the population is believed to be located
ACT sores ~ N(20.8, 4.8) SAT sores ~ N(500, 110) Johnny's ACT = 26 June's SAT = 760 Who did better?
June
Texas has roughly 225,000 farms, more than any other state in the US. The actual mean farm size is µ = 582 acres, and the standard deviation is σ = 150. Suppose the size of these farms follows a normal distribution. For random samples of n = 100 farms, what is the sampling distribution of sample means?
N(582, 15)
Can statistics guarantee that our guess is free of mistakes in a sample?
No. It only guarantees on average no mistakes in the long run
Suppose the income of all the American families is like this: (peak left) Randomly draw a sample of 5,000 families, within this sample, the distribution of incomes should be:
Positively skewed
ACT scores follow N(20.8, 4.8). SAT scores follow N(500, 110). If we draw the two normal curves in the same figure, what would the two curves look like?
SAT's curve is flatter and wider; ACT's curve is taller and narrower ACT's curve is on the left; SAT's curve is on the right
Assume that your class (a very large class) took an exam last week. The exam scores have a mean of 85 and standard deviation of 5. Your instructor told you that 30% of the students had a score of 90 or above. Which of the following is most likely to be correct? (A) 50% of the scores are below 85. (B) 20% of the students have scores between 85 and 90. (C) The scores do not follow a normal distribution. (D) Such a set of scores cannot appear in reality.
The scores do not follow a normal distribution.
According to the Central Limit Theorem, what is the relationship between the standard deviation of the population (σ) and the standard error?
The standard error is equal to σ divided by the square root of the sample size
Which is correct about the normal distribution? There are infinitely many normal distributions The larger the mean, the flatter the normal distribution becomes Standard normal distribution is denoted as N(1, 1) Some normal distributions are positively skewed
There are infinitely many normal distributions
In a certain test, scores follow N(µ = 100, σ = 30). Which of the following statements is correct? If we randomly pick a student, the probability of seeing this student's score below 70 is 34.13% Tom got 130 on this test. He is better than 84.13% of the test-takers. The median of the data is 50 These test scores follow a standard normal distribution
Tom got 130 on this test. He is better than 84.13% of the test-takers.
In a random sample, every case in the population has an equal chance of being selected. T or F
True
Scores on a science exam follow a negatively skewed distribution with μ = 100 and σ = 15. What % of scores are above 130?
We don't know. We cannot use the normal distribution table to find % because this is not a normal distribution.
X follows a bimodal distribution in the population, and the population mean is µ = 30. We draw numerous samples using sample size n = 1000 from the population. How often will X¯>30 in those samples? it's unknown because we don't have a normal distribution fairly less than 50% of the time fairly more than 50% of the time about 50% of the time
about 50% of the time
Variable Y follows a bimodal distribution in the population. If we randomly collect a sample of size n = 100,000, what's the data distribution in that sample? unimodal normal approximately normal bimodal
bimodal
To obtain the sampling distribution of sample means, we need to __________________________.
draw many samples, calculate the mean of each sample, and make a histogram of the means
To obtain the sampling distribution of sample means, we need to ___________
draw numerous samples
In a random sample, every unit in the population has an ____ chance of being selected.
equal
The mean of a population = 54.72, and the mean of a sample from that population = 54.92. Assuming the difference between the two values is due to chance, we can refer to the difference as sampling ____.
error
The purpose of constructing a confidence interval for the mean is to _______the true value of the population mean, based upon the mean of a _______ .
estimate, sample
Standard error ___ as sample size ___
increases, decreases
X follows a bimodal distribution in the population, and the population mean is µ = 30. We draw a sample using sample size n = 5000 from the population. How often will X > 30 in that sample? fairly more than 50% of the time it's unknown because we don't have a normal distribution about 50% of the time fairly less than 50% of the tim
it's unknown because we don't have a normal distribution
Statistics is about the _____
long term
A confidence interval for the mean is calculated by adding and subtracting a value to and from the sample______.
mean
According to the Central Limit Theorem, the mean of a sampling distribution of sample means will equal the _____ of the population from which the samples were drawn.
mean
A sampling distribution of sample means is based on taking repeated samples of size (n) from the same population and plotting the ___ of different samples.
means
In the Central Limit Theorem, what does n represent?
n represents the number of cases in each sample
Sample size: n=1000 vs n=2000, which X bar is closer to µ?
n=2000, they are more reliable than values in the 1000 curve
The shape of a sampling distribution of sample means will approach the shape of a ________ curve.
normal
Suppose the distribution of family incomes of all the US households is positively skewed with mean 71,000 and standard deviation 15,000. -I randomly pick 2000 families in the US and make a histogram of the incomes of these 2000 families. What would the histogram look like?
positively skewed
Consider the following process. What will we get at the end of this process? Step 1: draw a sample Step 2: calculate the mode of this sample Step 3: repeat Steps 1 & 2 numerous times Step 4: make a histogram for all the sample mode values we have
sampling distribution of sample modes
Consider the following process. What will we get at the end of this process? Step 1: draw a sample Step 2: calculate the standard deviation of this sample Step 3: repeat Steps 1 & 2 numerous times Step 4: make a histogram for all the sample standard deviation values we have
sampling distribution of sample standard deviations
Consider the following process. What will we get at the end of this process? Step 1: draw a sample Step 2: calculate the variance of this sample Step 3: repeat Steps 1 & 2 numerous times Step 4: make a histogram for all the sample variance values we have
sampling distribution of sample variance
The difference between a sample statistic and a population parameter that is due to chance is referred as ____
sampling error
According to the Central Limit Theorem, and given a sampling distribution of sample means, the standard error of the mean will equal the ____ of the population divided by the ____ of the sample size.
standard deviation, square root
The standard deviation in the sampling distribution of sample means is usually called __________
standard error of mean
The sampling distribution of sample means consists of ___________
the distribution of sample means for all the possible samples
If I randomly pick a family in the US, what is the probability that this family's income is below 70,000?
there is not enough information
Is the sample mean more stable than individual scores? If, so does the mean have extreme values?
yes it is more stable, and no, the mean is less likely to have extreme values
