Chapter 8 & 10 Questions, Chapter 6 & 7 Statistics Questions, Statistics Probability Exam

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The complement of​ "at least​ one" is​ _______.

"none"

"At least​ one" is equivalent to​ _______.

"one or more"

>continued... When you bet that the outcome is an odd number, the payoff odds are​ 1:1. How much profit do you make if you bet ​$16 and​ win? How much profit should you make on the $16 bet if you could somehow convince the casino to change its payoff odds so that they are the same as the actual odds against​ winning?

$16 & $18.46

Assume that all​ grade-point averages are to be standardized on a scale between 0 and 4. How many​ grade-point averages must be obtained so that the sample mean is within 0.014 of the population​ mean? Assume that a 95​% confidence level is desired. If using the range rule of​ thumb, σ can be estimated as range/4=4−0/4=1.

((1.96*1)/0.014)^2 = 19600 No, because fairly large.

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 18 subjects had a mean wake time of 102.0min. After​ treatment, the 18 subjects had a mean wake time of 79.4min and a standard deviation of 21.2min. Assume that the 18 sample values appear to be from a normally distributed population and construct a 95​% confidence interval estimate of the mean wake time for a population with drug treatments

(menu 6, 6, 2) 68.9, 89.9 The confidence interval does not include the mean wake time of 102.0min before the​ treatment, so the means before and after the treatment are different. This result suggests that the drug treatment has a significant effect.

A data set includes 106 body temperatures of healthy adult humans having a mean of 98.7°F and a standard deviation of 0.66°F. Construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body​ temperature?

(menu 6, 6, 2) 98.532, 98.868. This suggests that the mean body temperature could very possibly be 98.6°F.

Which of the following values cannot be​ probabilities?

-0.45, 5/3, sqrt2, 1.24

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find the bone density test scores that can be used as cutoff values separating the lowest 12​% and highest 12​%, indicating levels that are too low or too​ high, respectively.

-1.17, 1.17

In a certain instant lottery game, the chance of a win is stated as 5​%. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive.

0.05

For a certain casino slot machine, the odds in favor of a win are given as 11 to 89. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive.

0.11 because 11/(11+89)=11/100

In a certain weather forecast, the chances of a thunderstorm are stated as "3 in 20​." Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive.

0.15 because 3/20

Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the probability that a randomly selected thermometer reads between −1.99 and −0.88.

0.1661

A genetic experiment with peas resulted in one sample of offspring that consisted of 421 green peas and 159 yellow peas. a. Construct a 95​% confidence interval to estimate of the percentage of yellow peas. b. It was expected that​ 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?

0.238<p<0.310 b. No

Fourteen of the 50 digital video recorders ​(DVRs) in an inventory are known to be defective. What is the probability that a randomly selected item is​defective?

0.28 because 14/50=0.28

Assuming boys and girls are equally​ likely, find the probability of a couple having a baby girl when their fourth child is​ born, given that the first three children were all girls.

0.5

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were​ born, and 272 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born.

0.744<p<0.856 (Yes, the proportion of girls is significantly different from 0.5)

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 300 babies were​ born, and 270 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective?

0.855<p<0.945 (yes)

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than 3.65.

0.9999

A classic counting problem is to determine the number of different ways that the letters of "misspell" can be arranged. 8!/(2!2!)

10080

How many ways can you make change for a​ quarter?

12

Here are summary statistics for randomly selected weights of newborn​ girls: n=205​, x=29.8​hg, s=7.9hg. Construct a confidence interval estimate of the mean. Use a 90​% confidence level. Are these results very different from the confidence interval 28.1hg<μ<31.7hg with only 12 sample​ values, x=29.9​hg, and s=3.5​hg?

28.9, 30.7 No, because the confidence interval limits are similar.

Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 98% confidence level. What does the confidence interval tell about the population of all college students in the state? 3.5, 2.9, 3.7, 4.8, 3.1, 4.1, 3.2, 4.4, 4.5, 4.4, 4.1, 3.9, 3.6, 3.7, 3.5

3.46, 4.21 No

Listed below are student evaluation ratings of​ courses, where a rating of 5 is for​ "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 95​% confidence level. What does the confidence interval tell about the population of all college students in the​ state? 4.0​,2.9​,3.9​,4.6​,3.1,4.4​,3.7​,4.5,4.1​,4.1​,4.2​,3.6,3.2​,4.0​,3.8.

3.59, 4.15 No

Listed below are student evaluation ratings of​ courses, where a rating of 5 is for​ "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 95​% confidence level. What does the confidence interval tell about the population of all college students in the​ state? 3.7​, 3.0​, 3.9​, 4.6​, 3.1​, 4.0​, 3.8​, 4.6​, 4.5​, 4.2​, 4.6​, 3.8​, 3.5​, 3.9​, 4.0

3.67, 4.23 No

Listed below are student evaluation ratings of​ courses, where a rating of 5 is for​ "excellent." the ratings were obtained at one university in a state. construct a confidence interval using a 90​% confidence level. what does the confidence interval tell about the population of all college students in the​ state? 3.93.13.74.43.24.13.34.44.3​,4.44.5 3.93.2​4.2,4.

3.74, 4.13 No

Listed below are student evaluation ratings of​ courses, where a rating of 5 is for​ "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 90​% confidence level. What does the confidence interval tell about the population of all college students in the​ state?4.0​,3.2​,3.9​,4.8​,2.9​,4.5​,3.8​,4.8​,4.4​,4.4​,4.6​,4.1​,3.6​,4.3​,3.9

3.84, 4.32 The results tell nothing about the population of all college students in the​ state, since the sample is from only one university.

Here are summary statistics for randomly selected weights of newborn​ girls: n=203​, x=32.1hg, s=6.7hg. Construct a confidence interval estimate of the mean. Use a 99​% confidence level. Are these results very different from the confidence interval 30.6hg<μ<34.2hg with only 15 sample​ values, -x-=32.4hg, and s=2.3hg?

30.9, 33.3 No, because the confidence interval limits are similar.

Assume that all​ grade-point averages are to be standardized on a scale between 0 and 6. How many​ grade-point averages must be obtained so that the sample mean is within 0.005 of the population​ mean? Assume that a 95​% confidence level is desired. If using the range rule of​ thumb, σ can be estimated as range/4=6−0/4=1.5. Does the sample size seem​ practical?

345,732 No, because fairly large.

The data given to the right includes data from 42 candies, and 6 of them are red. The company that makes the candy claims that 32​% of its candies are red. Use the sample data to construct a 90​% confidence interval estimate of the percentage of red candies.

5.4%<p<23.3% (no)

Find the probability that when a couple has three children, at least one of them is a boy. ​(Assume that boys and girls are equally​ likely.)

7/8

Assume that the sample is a simple random sample obtained from a normally distributed population of flight delays at an airport. Use the table below to find the minimum sample size needed to be 95​% confident that the sample standard deviation is within 5% of the population standard deviation.

768. The computed minimum sample size is not likely correct.

How many different ways can the letters of "happiness​" be​ arranged? 9!/(2!2!)

90720

Which of the following is NOT a principle of​ probability?

All events are equally likely in any probability procedure.

In horse​ racing, a trifecta is a bet that the first three finishers in a race are​ selected, and they are selected in the correct order.

Because the order of the first three finishers does make a​ difference, the trifecta involves permutations.

Listed below are speeds​ (mi/h) measured from traffic on a busy highway. This simple random sample was obtained at​ 3:30 P.M. on a weekday. Use the sample data to construct a 95​% confidence interval estimate of the population standard deviation.

CI estimate = 3.2, 7.6. No. The confidence interval is an estimate of the standard deviation of the population of speeds at​ 3:30 on a​ weekday, not other times.

Twelve different video games showing substance use were observed and the duration of times of game play​ (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. Use the sample data to construct an 80​% confidence interval estimate of σ​, the standard deviation of the duration times of game play.

CI estimate = 320.1, 563.3

The​ _____________ distribution is used to develop confidence interval estimates of variances or standard deviations.

Chi-square

Determine whether the value is a discrete random​ variable, continuous random​ variable, or not a random variable.

Continuous random variable: Amount of rainfall in City B, the amount of snowfall in December in City A, and the time it takes to fly from City A to City B. Discrete random variable: The number of people in a restaurant and the number of free-throw attempts before the first shot is made. Not a random variable: The gender of college students.

If A denotes some​ event, what does -A- denote? If P(A)=0.992​, what is the value of P(-A-​)?

Event -A-denotes the complement of event​ A, meaning that -A- consists of all outcomes in which event A does not occur.

Express the confidence interval 0.222 < p < 0.666 in the form ^p +- E.

Find point estimate (0.666+0.222)/2 = 0.444 AND the margin of error (0.666-0.222)/2 = 0.222 so it is 0.444 +- 0.222

Which of the following is not a commonly used​ practice?

If the distribution of the sample means is normally​ distributed, and n>​30, then the population distribution is normally distributed.

A researcher collects a simple random sample of​ grade-point averages of statistics​ students, and she calculates the mean of this sample. Under what conditions can that sample mean be treated as a value from a population having a normal​ distribution?

If the population of grade-point averages has a normal distribution & the sample has more than 30 grade-point averages.

The accompanying table describes results from groups of 10 births from 10 different sets of parents. The random variable x represents the number of girls among 10 children. Use the range rule of thumb to determine whether 1 girl in 10 births is a significantly low number of girls.

Max=8.4 girls Min=1.7 girls Yes, 1 girl is a significantly low number of​ girls, because 1 girl is below the range of values that are not significant.

Refer to the accompanying​ table, which describes the number of adults in groups of five who reported sleepwalking. Find the mean and standard deviation for the numbers of sleepwalkers in groups of five.

Mean=1.4 Standard deviation=1.0

Refer to the accompanying​ table, which describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children. Find the mean and standard deviation for the number of girls in 8 births.

Mean=4.0 girls. Standard deviation=1.4 girls.

A combination lock uses three numbers between 1 and 87 with​ repetition, and they must be selected in the correct sequence.

No, because the multiplication counting rule would be used to determine the total number of combinations.

Nine different senators from the current U.S. Congress are randomly selected without replacement and whether or not they've served over 2 terms is recorded.

No, because the trials of the experiment are not independent and the probability of success differs from trial to trial.

Weights of golden retriever dogs are normally distributed. Samples of weights of golden retriever​ dogs, each of size n=​15, are randomly collected and the sample means are found. Is it correct to conclude that the sample means cannot be treated as being from a normal distribution because the sample size is too​ small?

No; the original population is normally​ distributed, so the sample means will be normally distributed for any sample size.

Annual incomes are known to have a distribution that is skewed to the right instead of being normally distributed. Assume that we collect a large (n>​30) random sample of annual incomes. Can the distribution of incomes in that sample be approximated by a normal distribution because the sample is​ large?

No; the sample means will be normally​ distributed, but the sample of incomes will be skewed to the right.

Drive-thru Orders: 338, 261, 241, 131, 40, 54, 32, 12. If one order is​ selected, find the probability of getting an order that is not accurate.

P=0.124

A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the​ drug, 14 subjects had a mean wake time of 93.4min and a standard deviation of 42.7min. Assume that the 14 sample values appear to be from a normally distributed population and construct a 90​% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments.

ONINE CALC: ignore the mean. CI estimate = 32.56, 63.43. No, the confidence interval does not indicate whether the treatment is effective.

Which of the following is NOT a requirement of the Permutations​ Rule, nPr=n!/(n−r)!​, for items that are all​ different?

Order is not taken into account​ (rearrangements of the same items are considered to be the​ same).

When a man observed a sobriety checkpoint conducted by a police​ department, he saw 661 drivers were screened and 3 were arrested for driving while intoxicated. Based on those​ results, we can estimate that P(W)=0.00454​, where W denotes the event of screening a driver and getting someone who is intoxicated. What does P(-W-) denote, and what is its​ value?

P(-W-) denotes the probability of screening a driver and finding that he or she is not intoxicated. P(-W-)=0.99546.

Assume that a procedure yields a binomial distribution with n=7 trials and a probability of success of p=0.90. Use a binomial probability table to find the probability that the number of successes x is exactly 4.

P(4)=0.023

Confusion of the inverse occurs when we incorrectly believe​ _______.

P(B|A)=P(A|B)

In the game of blackjack played with one​ deck, a player is initially dealt 2 different cards from the 52 different cards in the deck. A winning​ "blackjack" hand is won by getting 1 of the 4 aces and 1 of 16 other cards worth 10 points.

P(blackjack hand)=32/663. The percentage is 4.83%.

In a survey of consumers aged 12 and​ older, respondents were asked how many cell phones were in use by the household.​ (No two respondents were from the same​ household.) Among the​ respondents, 202 answered​ "none," 280 said​ "one," 374 said​ "two," 135 said​ "three," and 91 responded with four or more. A survey respondent is selected at random. Find the probability that​ his/her household has four or more cell phones in use. Is it unlikely for a household to have four or more cell phones in​ use?

P(four or more cell phones)=0.084. No​, because the probability of a respondent with four or more cell phones in use is greater than 0.05.

To reduce laboratory​ costs, water samples from four public swimming pools are combined for one test for the presence of bacteria. Further testing is done only if the combined sample tests positive. Based on past​ results, there is a 0.007 probability of finding bacteria in a public swimming area.

P(positive test result)=0.028. The probability is quite​ low, indicating that further testing of the individual samples will be a rarely necessary event.

Testing for a disease can be made more efficient by combining samples. If the samples from two people are combined and the mixture tests​ negative, then both samples are negative. On the other​ hand, one positive sample will always test​ positive, no matter how many negative samples it is mixed with.

P(positive test result)=0.19. The probability is not​ low, so further testing of the individual samples will frequently be a necessary event.

When three basketball players are about to have a​ free-throw competition, they often draw names out of a hat to randomly select the order in which they shoot.

P(shoot free throws in alphabetical order)=1/6

A modified roulette wheel has 28 slots. One slot is​ 0, another is​ 00, and the others are numbered 1 through 26​, respectively. You are placing a bet that the outcome is an odd number.​ (In roulette, 0 and 00 are neither odd nor​ even.)

P(winning)=13/28 Actual odds against winning= 15:13.

A survey showed that 80​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If 9 adults are randomly​ selected, find the probability that no more than 1 of them need correction for their eyesight. Is 1 a significantly low number of adults requiring eyesight​ correction?

P=0.000. Yes, because the probability of this occurring is small.

Among 500 randomly selected drivers in the 20−24 age​ bracket, 5 were in a car crash in the last year. If a driver in that age bracket is randomly​ selected, what is the approximate probability that he or she will be in a car crash during the next​ year? Is it unlikely for a driver in that age bracket to be involved in a car crash during a​ year? Is the resulting value high enough to be of concern to those in the 20−24 age​ bracket?

P=0.01. Yes, it is unlikely. No, the probability is not high enough to be of concern.

Drive-thru Orders: 340, 272, 246, 134, 30, 54, 35, 14. If three different orders are​ selected, find the probability that they are all from restaurant C.

P=0.0156

The accompanying table shows the results from a test for a certain disease. Find the probability of selecting a subject with a negative test​ result, given that the subject has the disease. 312, 2, 10, 1111.

P=0.031. The subject would not receive treatment and could spread the disease.

If a procedure meets all of the conditions of a binomial distribution except the number of trials is not​ fixed, then the geometric distribution can be used. The probability of getting the first success on the xth trial is given by P(x)=p(1−p)^x−1​, where p is the probability of success on any one trial. Subjects are randomly selected for a health survey. The probability that someone is a universal donor​ (with group O and type Rh negative​ blood) is 0.11.

P=0.0547

Assume that when adults with smartphones are randomly​ selected, 53​% use them in meetings or classes. If 12 adult smartphone users are randomly​ selected, find the probability that fewer than 5 of them use their smartphones in meetings or classes.

P=0.1411

Assume that random guesses are made for 8 multiple-choice questions on a test with 5 choices for each​ question, so that there are n=8 ​trials, each with probability of success​ (correct) given by p=0.20. Find the probability of no correct answers.

P=0.168

To the right are the outcomes that are possible when a couple has three children. Assume that boys and girls are equally​ likely, so that the eight simple events are equally likely. Find the probability that when a couple has three​ children, there is exactly 1 girl.

P=0.375 because 3/8. The table looks like 1st - 2nd - 3rd boy-boy-boy

Drive-thru Orders: 311, 278, 233, 132, 38, 53, 35, 16. If one order is​ selected, find the probability of getting an order from restaurant B or C or an order that is not accurate.

P=0.596

Drive-thru Orders: 331, 271, 250, 143, 35, 54, 40, 16. If one order is​ selected, find the probability of getting food that is not from Restaurant A.

P=0.679 because total of food not from restaurant A dived by total of all orders = 774/1140.

The table below displays results from experiments with polygraph instruments. Find the positive predictive value for the test. That​ is, find the probability that the subject​ lied, given that the test yields a positive result. 11, 42, 33, 8.

P=0.792

In a study of helicopter usage and patient​ survival, among the 41,499 patients transported by​ helicopter, 204 of them left the treatment center against medical​ advice, and the other 41,295 did not leave against medical advice. If 40 of the subjects transported by helicopter are randomly selected without​ replacement, what is the probability that none of them left the treatment center against medical​ advice?

P=0.821

The data represent the results for a test for a certain disease. Assume one individual from the group is randomly selected. Find the probability of getting someone who tests negative​, given that he or she did not have the disease. 134, 20, 10, 136.

P=0.872.

Among 6194 cases of heart pacemaker​ malfunctions, 256 were found to be caused by​ firmware, which is software programmed into the device. If the firmware is tested in 3 different pacemakers randomly selected from this batch of 6194 and the entire batch is accepted if there are no​ failures, what is the probability that the firmware in the entire batch will be​ accepted?

P=0.881 and it will likely result in the entire batch being accepted.

In a certain​ country, the true probability of a baby being a boy is 0.534. Among the next four randomly selected births in the​ country, what is the probability that at least one of them is a girl​?

P=0.919

Based on a​ poll, 67​% of Internet users are more careful about personal information when using a public​ Wi-Fi hotspot. What is the probability that among four randomly selected Internet​ users, at least one is more careful about personal information when using a public​ Wi-Fi hotspot?

P=0.988. It is very possible that the result is not valid because the sample may not be representative of the people who use public​ Wi-Fi.

A certain group of women has a 0.51​% rate of​ red/green color blindness. If a woman is randomly​ selected, what is the probability that she does not have​ red/green color​ blindness?

P=0.9949

A Social Security number consists of nine digits in a particular​ order, and repetition of digits is allowed. After seeing the last four digits printed on a​ receipt, if you randomly select the other​ digits, what is the probability of getting the correct Social Security number of the person who was given the​ receipt?

P=1/100000

A moving company has a truck filled for deliveries to five different sites. If the order of the deliveries is randomly​ selected, what is the probability that it is the shortest​ route? 5!

P=1/120

Winning the jackpot in a particular lottery requires that you select the correct five numbers between 1 and 42 and, in a separate​ drawing, you must also select the correct single number between 1 and 24. Combination

P=1/20416032

In a small private​ school, 3 students are randomly selected from 12 available students. What is the probability that they are the three youngest​ students? Combination

P=1/220

A presidential candidate plans to begin her campaign by visiting the capitals in 4 of 42 states. What is the probability that she selects the route of four specific​ capitals? Permutation

P=1/2686320

If you know the names of the remaining seven students in the spelling​ bee, what is the probability of randomly selecting an order and getting the order that is used in the spelling​ bee? 7!

P=1/5040

A main goal in statistics is to interpret and understand the meaning of statistical values. The​ _______ can be very helpful in understanding the meaning of the mean and standard deviation.

Range Rule of Thumb

​If, under a given​ assumption, the probability of a particular observed event is extremely​ small, we conclude that the assumption is probably not correct. This represents the​ _______.

Rare Event Rule

Refer to the accompanying data set of mean​ drive-through service times at dinner in seconds at two fast food restaurants. Construct a 99% confidence interval estimate of the mean​ drive-through service time for Restaurant X at​ dinner; then do the same for Restaurant Y.

Restaurant X = 155.2, 203.1 Restaurant Y = 133.8, 171.4. The confidence interval estimates for the two restaurants overlap​, so there does not appear to be a significant difference between the mean dinner times at the two restaurants.

Three randomly selected households are surveyed. The numbers of people in the households are 2​, 4​, and 9. Assume that samples of size n=2 are randomly selected with replacement from the population of 2, 4​, and 9.

Sample proportion (number of even values over n)=0,0.5,1. Probabilities=1/9, 4/9, 4/9. The mean of the sample proportions=0.667. The population proportion=0.667.

Which of the following is NOT needed to determine the minimum sample size required to estimate a population​ proportion?

Standard Deviation

Which of the following is NOT a requirement of the Combinations​ Rule, nCr=n!/r!(n−r)!​, for items that are all​ different?

That order is taken into account​ (consider rearrangements of the same items to be different​ sequences).

Find the area of the shaded region. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. #102-#132

The area of the shaded region is 0.4305

Find the area of the shaded region. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. #97

The area of the shaded region is 0.5793

Find the area of the shaded region. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. #95-#125

The area of the shaded region is 0.5827

Find the area of the shaded region. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. #105

The area of the shaded region is 0.6305

In a state pick 4 lottery​ game, a bettor selects four numbers between 0 and 9 and any selected number can be used more than once. Winning the top prize requires that the selected numbers match those and are drawn in the same order.

The combination and permutations rules do not apply because repetition is allowed and numbers are selected with replacement. The multiplication counting rule applies to this problem.

Which of the following is NOT a property of the sampling distribution of the sample​ mean?

The distribution of the sample mean tends to be skewed to the right or left.

Find the indicated IQ score. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. x w/ 0.3949 area

The indicated IQ score is 104 (change the z score to positive because the known area is to the right)

Find the indicated IQ score. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. x w/ 0.55 area to the left

The indicated IQ score, x, is 101.9 (calculator with area then use the formula x=meu+(z*sigma))

Find the indicated IQ score. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. x w/ an area of 0.6 to the right

The indicated IQ score, x, is 96.2 (always change the z score sign)

Which of the following is NOT a requirement for using the normal distribution as an approximation to the binomial​ distribution?

The sample is the result of conducting several dependent trials of an experiment in which the probability of success is p.

Which of the following is NOT a property of the​ chi-square distribution?

The mean of the chi-square distribution is 0

Assume that the sample is a simple random sample obtained from a normally distributed population of IQ scores of statistics professors. Use the table below to find the minimum sample size needed to be 95​% confident that the sample standard deviation s is within 20​% of σ.

The minimum sample size needed is 48. Yes the sample size is practical, because the sample size is small enough for most applications.

A thief steals an ATM card and must randomly guess the correct five​-digit pin code from a 7​-key keypad. Repetition of digits is allowed. 7*7*7*7*7

The number of possible codes is 16807. The probability that the correct code is given on the first try is 1/16807.

When testing for current in a cable with seven color-coded wires, the author used a meter to test five wires at a time. How many different tests are required for every possible pairing of five wires? Combination

The number of tests required is 21.

Assume that military aircraft use ejection seats designed for men weighing between 133.7 lb and 202 lb. If​ women's weights are normally distributed with a mean of 171.6 lb and a standard deviation of 46.2 lb, what percentage of women have weights that are within those​ limits? Are many women excluded with those​ specifications?

The percentage of women that have weights between those limits is 53.86%. Yes, the percentage of women who are​ excluded, which is the complement of the probability found​ previously, shows that about half of women are excluded.

Assume a population of 40​, 45​, 47​, and 53. Assume that samples of size n=2 are randomly selected with replacement from the population. Listed below are the sixteen different samples.

The population median is not equal to the mean of the sample medians​ (it is also not half or double the mean of the sample​ medians). The sample medians do not target the population​ median, so sample medians are biased​ estimators, because the mean of the sample medians does not equal the population median.

Which of the following is NOT a requirement of constructing a confidence interval estimate for a population​ variance?

The population must be skewed to the right.

Based on a​ survey, when 1009 consumers were asked if they are comfortable with drones delivering their​ purchases, 42% said yes. The probability of randomly selecting 30 of the 1009 consumers and getting exactly 24 who are comfortable with the drones is represented as 0+. What does 0+ indicate?

The probability 0+ indicates that the probability is a very small positive value. It indicates that the event is​ possible, but very unlikely.

Drug Screening: 40, 9, 20, 33 (add all numbers together and put the # you're trying to find over that total).

The probability of a false positive result is 0.196. The person tested would suffer because he or she would be suspected of using drugs when in reality he or she does not use drugs.

In a genetics experiment on​ peas, one sample of offspring contained 398 green peas and 314 yellow peas. Based on those​ results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the value of 3/4 that was​ expected?

The probability of getting a green pea is approximately 0.559. No, this probability is not reasonably close to 3/4.

Polygraph Tests: 15, 44, 28, 11 (add all numbers together and put the # you're trying to find over that total).

The probability that a randomly selected polygraph test subject was not lying is 0.439. Is the result close to the probability, rounded to three decimal​ places, of 0.398 for a negative test​ result? Yes, because there is less than a 0.050 absolute difference between the probability of a true response and the probability of a negative test result.

Let event A=subject is telling the truth and event B=polygraph test indicates that the subject is lying. Use your own words to translate the notation P(B|A) into a verbal statement.

The probability that the polygraph indicates lying given that the subject is actually telling the truth.

An elevator has a placard stating that the maximum capacity is 1932 lb—12 passengers.​ So, 12 adult male passengers can have a mean weight of up to 1932/12=161 pounds. If the elevator is loaded with 12 adult male​ passengers, find the probability that it is overloaded because they have a mean weight greater than 161 lb.​

The probability the elevator is overloaded is 0.7703 because z = (161-167)/(28/sqrt12) = -0.74 = calc552 >> 0.2297 >> 1-0.2297. It is indeed overloaded.

Is the random variable given in the accompanying table discrete or​ continuous? Explain. 0-4: 0.063, 0.250, 0.375, 0.250, 0.063.

The random variable given in the accompanying table is discrete because there are a finite number of values.

If the order of the items selected​ matters, then we have a​ _______. 0-3: 0.125, 0.375, 0.375, 0.125.

The random variable is​ x, which is the number of girls in three births. The possible values of x are​ 0, 1,​ 2, and 3. The values of the random value x are numerical.

​_____________ is the distribution of all values of the statistic when all possible samples of the same size n are taken from the same population.

The sampling distribution of a statistic

_____________ is the distribution of sample​ proportions, with all samples having the same sample size n taken from the same population.

The sampling distribution of the proportion

Which of the following is not a requirement of the binomial probability​ distribution?

The trials must be dependent.

With a short time remaining in the​ day, a delivery driver has time to make deliveries at 6 locations among the 8 locations remaining. How many different routes are​ possible? Permutation

There are 20160 possible different routes.

If radio station call letters must begin with either K or W and must include either two or three additional​ letters, how many different possibilities are​ there?

There are 36504 different possibilities.

A weather forecasting website indicated that there was a 25​% chance of rain in a certain region. Based on that​ report, which of the following is the most reasonable​ interpretation?

There is a 0.25 probability that it will rain somewhere in the region at some point during the day.

Which of the following is NOT one of the three methods for finding binomial probabilities that is found in the chapter on discrete probability​ distributions?

Use a simulation

The YSORT method of sex selection, developed by the Genetics & IVF Institute, was designed to increase the likelihood that a baby will be a boy.

Yes, because the procedure satisfies all the criteria for a binomial distribution.

Surveying 100 college students and asking if they like pirates or ninjas better, recording Pirate or Ninja.

Yes, it results in a binomial distribution because all 4 requirements are satisfied.

Seven hundred different voters in a region with two major political parties are randomly selected from the population of 2.7 million registered voters.

Yes, the result is a binomial distribution.

When conducting research on color blindness in​ males, a researcher forms random groups with five males in each group. The random variable x is the number of males in the group who have a form of color blindness. 0-5: 0.659, 0.287, 0.048, 0.005, 0.001, 0.000.

Yes, the table shows a probability distribution. Mean=0.4 males. Standard deviation=0.6 males.

Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder. 0-5: 0.026, 0.147, 0.327, 0.327, 0.147, 0.026.

Yes, the table shows a probability distribution. Mean=2.5 children. Standard deviation=1.1 children.

A Gallup poll of 1236 adults showed that​ 12% of the respondents believe that it is bad luck to walk under a ladder. Consider the probability that among 30 randomly selected people from the 1236 who were​ polled, there are at least 2 who have that belief.

Yes. Although the selections are not​ independent, they can be treated as being independent by applying the​ 5% guideline.

Express the confidence interval 0.222<p<0.444 in the form ^p±E.

^p+-E = 0.333 +- 0.111

A drug is used to help prevent blood clots in certain patients. In clinical​ trials, among 4394 patients treated with the​ drug, 147 developed the adverse reaction of nausea. Construct a 90​% confidence interval for the proportion of adverse reactions.

a) 0.033 b) 0.004 c) 0.029 < p < 0.037 d) One has 90​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

A genetic experiment with peas resulted in one sample of offspring that consisted of 438 green peas and 161 yellow peas. a. Construct a 90​% confidence interval to estimate of the percentage of yellow peas. b. It was expected that​ 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?

a) 0.239<p<0.299b) No, the confidence interval includes 0.25, so the true percentage could easily equal 25%

In a study of cell phone use and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 2501 subjects randomly selected from an online group involved with ears. 1082 surveys were returned. Construct a 99​% confidence interval for the proportion of returned surveys.

a) 0.433 b) 0.026 c) 0.407 < p < 0.459 d) One has 99​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll, n=910 and x=512 who said​ "yes." Use a 90% confidence level.

a) 0.563 b) 0.027 c) 0.536 < p < 0.590 d) One has 90​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll, n=1023 and x=531 who said​ "yes." Use a 90% confidence level.

a) ^p = x/n = 0.519 b)E = 1.645 sqrt(0.519(1-0.519))/1023 = 0.026 c) All requirements are satisfied. The confidence interval is 0.493 < p < 0.545. d) would actually contain

Assume that females have pulse rates that are normally distributed with a mean of μ=74.0 beats per minute and a standard deviation of σ=12.5 beats per minute.

a. (convert pulse rate to z-score using z=(x-μ)/σ ) then use that in the calculator (menu, 5, 5, 2). Therefore, the probability that a randomly selected adult​ female's pulse rate is less than 81 beats per minute is 0.7123.

A genetics experiment involves a population of fruit flies consisting of 2 males named Barry and Carlos and 2 females named Diana and Erin.

a. 0 -> 1/4 0.5 -> 1/2 1 -> 1/4 b. The mean of the sampling distribution is 0.5. c. Yes, the sample mean is equal to the population proportion of males. These values are always​ equal, because proportion is an unbiased estimator.

Assume that 26.7​% of people have sleepwalked. Assume that in a random sample of 1542 adults, 456 have sleepwalked.

a. 0.0059 b. Yes, because less, 0.05. c. Since the result of 456 adults that have sleepwalked is significantly​ high, it is strong evidence against the assumed rate of 26.7​%.

Based on a smartphone​ survey, assume that 42​% of adults with smartphones use them in theaters. In a separate survey of 284 adults with​ smartphones, it is found that 98 use them in theaters.

a. 0.0062 b. Yes, because the probability of this event is less than the probability cutoff that corresponds to a significant event, which is 0.05.

In a survey of 1428 ​people, 968 people said they voted in a recent presidential election. Voting records show that 65​% of eligible voters actually did vote.

a. 0.0140 b. Some people are being less than honest because P(x≥968​) is less than​ 5%

In a survey of 1370 people, 913 people said they voted in a recent presidential election. Voting records show that 64​% of eligible voters actually did vote.

a. 0.0222 because subtract from 1 b. Some people are being less than honest because P(x>913) is less than 5%

A​ gender-selection technique is designed to increase the likelihood that a baby will be a girl. In the results of the​ gender-selection technique, 880 births consisted of 442 baby girls and 438 baby boys. In analyzing these​ results, assume that boys and girls are equally likely.

a. 0.0266 b. 1-0.5396=0.4597 No, because it is not far. c. The result from part​ (b) is more​ relevant, because one wants the probability of a result that is at least as extreme as the one obtained. d. No​, because the probability of having 505 or more girls in 969 births is not ​unlikely, and​thus, is attributable to random chance.

Based on a smartphone​ survey, assume that 51​% of adults with smartphones use them in theaters. In a separate survey of 296 adults with​ smartphones, it is found that 150 use them in theaters.

a. 0.4787 b. Not significantly low.

In a​ state's Pick 3 lottery​ game, you pay $1.36 to select a sequence of three digits​ (from 0 to​ 9), such as 922. If you select the same sequence of three digits that are​ drawn, you win and collect $423.49.

a. 1000 different selections are possible. b. P(winning)=0.001 c. Net profit=$422.13 d. Expected value=$-0.94 e. Neither bet is better because both games have the same expected value.

TInterval (13.046,22.15) -x-=17.598 Sx=16.01712719 n=50

a. 13.05, 22.15 (menu 6, 6, 2) b. same as -x- so 17.60 Mbps & margin of error = high minus low in (a) so it's 4.55 Mbps. c. Because the sample size of 50 is greater than​ 30, the distribution of sample means can be treated as a normal distribution.

The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be 90​% confident that his estimate is within seven percentage points of the true population​ percentage?

a. 139 b. 95 c. ​No, a sample of students at the nearest college is a convenience​ sample, not a simple random​ sample, so it is very possible that the results would not be representative of the population of adults.

In a study of government financial aid for college​ students, it becomes necessary to estimate the percentage of​ full-time college students who earn a​ bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use 0.03 E and 0.99%

a. 1842 b. 1769 c. No, using the additional survey information from part (b) only slightly reduces the sample size.

Gum & Money: 33, 16, 15, 30. A student given four quarters is more likely to have spent the money than a student given a​ $1 bill.

a. P(spent the money with quarters)= 0.673 b. P(spent the money with one dollar)= 0.333

In a study of government financial aid for college​ students, it becomes necessary to estimate the percentage of​ full-time college students who earn a​ bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.02 margin of error and use a confidence level of 95​%.

a. 2401 b. 2305 c. No, slightly.

The pulse rates of 154 randomly selected adult males vary from a low of 44bpm to a high of 116bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want 95​% confidence that the sample mean is within 2bpm of the population mean.

a. 312 (sigma = range/4) b. 102 c. Sample size in part (a) is larger so the result from part (b) is likely to be better because it uses a better estimate of sigma.

A clinical test on humans of a new drug is normally done in three phases. Phase I is conducted with a relatively small number of healthy volunteers. For​ example, a phase I test of a specific drug involved only 6 subjects. Assume that we want to treat 6 healthy humans with this new drug and we have 11 suitable volunteers available.

a. 332,640 (permutation) b. There are 462 different treatment groups possible. c. P(selecting the 6 youngest subjects)=1/462

The pulse rates of 151 randomly selected adult males vary from a low of 44 bpm to a high of 112 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want 98​% confidence that the sample mean is within 2bpm of the population mean.

a. 391 b. 202 c. The result from part​ (a) is larger than the result from part​ (b). The result from part (b) is likely to be better because it uses a better estimate of sigma .

A survey found that​ women's heights are normally distributed with mean 62.6 in. and standard deviation 2.8 in. The survey also found that​ men's heights are normally distributed with mean 67.7 in. and standard deviation 3.2 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 63 in.

a. 7.05% men meet the height requirement (don't change the signs) Since most men do not meet the height​ requirement, it is likely that most of the characters are women. b. The new height requirements are a minimum of 62.4 in and a maximum of 67.7 in.

A survey found that​ women's heights are normally distributed with mean 63.5 in and standard deviation 2.5 in. A branch of the military requires​ women's heights to be between 58 in and 80 in.

a. 98.61% No, because only a small percentage of women are not allowed to join this branch of the military because of their height. b. All women are eligible except the shortest 1% and the tallest 2% which is at least 57.7 in and at most 68.6 in. (change the highest z score sign).

Gum & Money: 25, 17, 13, 34. A student given four quarters is more likely to have spent the money.

a. P(spent the money with quarters)=0.595 b. P(kept the money with quarters)=0.405

The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8 cm.

a. Find the probability that an individual distance is greater than 214.80 cm. 1-0.8775=0.1225 b. Find the probability that the mean for 25 randomly selected distances is greater than 203.70 cm. (σ/sqrtN) = 8/sqrt25 = 1.6 >> (203.7-205.5)/1.6 = -1.125 >> 1-0.1302 = 0.8697 c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30? The normal distribution can be used because the original population has a normal distribution.

A ski gondola carries skiers to the top of a mountain. Assume that weights of skiers are normally distributed with a mean of 199lb and a standard deviation of 40lb. The gondola has a stated capacity of 25 passengers, and the gondola is rated for a load limit of 3750lb.

a. Given that the gondola is rated for a load limit of 3750lb, what is the maximum mean weight of the passengers if the gondola is filled to the stated capacity of 25 passengers? 3750/25=150 b. If the gondola is filled with 25 randomly selected​ skiers, what is the probability that their mean weight exceeds the value from part​ (a)? 40/sqrt25 = 8 >>> (150-199)/8 = -6.13. Thus, if the gondola is filled with 25 randomly selected​ skiers, the probability that their mean weight exceeds the value from part​ (a) is 1.0. c. If the weight assumptions were revised so that the new capacity became 20 passengers and the gondola is filled with 20 randomly selected​ skiers, what is the probability that their mean weight exceeds 187.5​lb, which is the maximum mean weight that does not cause the total load to exceed 3750lb? 40/sqrt20=8.9443 >> 187.5-199/8.9443=-1.29. Thus, if the gondola is filled with 20 randomly selected​ skiers, the probability that their mean weight exceeds 187.5lb is 0.9015. d. The gondola will be overloaded if the mean weight of the passengers is above the maximum allowed mean weight. Not safe enough.

In a survey of 3013 adults aged 57 through 85​ years, it was found that 83.1% of them used at least one prescription medication.

a. How many of the 3013 subjects used at least one prescription​ medication? 2504 b. Construct a​ 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication. 82%<p<84.2% c. What do the results tell us about the proportion of college students who use at least one prescription​ medication? The results tell us nothing about the proportion of college students who use at least one prescription medication.

Assume that females have pulse rates that are normally distributed with a mean of μ=76.0 beats per minute and a standard deviation of σ=12.5 beats per minute.

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is between 70 beats per minute and 82 beats per minute. p=0.3688 b. If 16 adult females are randomly​ selected, find the probability that they have pulse rates with a mean between 70 beats per minute and 82 beats per minute. p=0.9451 c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30? Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Assume that there is a 5​% rate of disk drive failure in a year.

a. P(w 2 the catastrophe can be avoided)=0.9975 b. P(w 3 the catastrophe can be avoided)=0.999875

Drive-thru Orders: 338, 274, 239, 134, 40, 59, 31, 15. If two orders are​ selected, find the probability that they are both accurate.

a. P=0.7598 and the events are independent. b. P=0.7597 and the events are not independent.

Assume that females have pulse rates that are normally distributed with a mean of μ=75.0 beats per minute and a standard deviation of σ=12.5 beats per minute.

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 79 beats per minute. (79-72)/12.5 then calculator >> p=0.6255 b. If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 79 beats per minute. (σ/sqrtN) 12.5/sqrt25 = 2.5... so then you do z=(79-75)/2.5 = 1.6 THEN you find the probability with the calculator (552) which is 0.9452 c. Does the population have pulse rates that are normally​ distributed? YES. Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8612g and a standard deviation of 0.0515g. A sample of these candies came from a package containing 440 candies, and the package label stated that the net weight is 375.7g.​ (If every package has 440 candies, the mean weight of the candies must exceed 375.7440=0.8539g for the net contents to weigh at least 375.7g.)

a. If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8539g. p=0.5564 b. If 440 candies are randomly​ selected, find the probability that their mean weight is at least 0.8539g. p=0.9985 c. Yes, because the probability of getting a sample mean of 0.8539g or greater when 440 candies are selected is not exceptionally small.

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150lb and 201lb. The new population of pilots has normally distributed weights with a mean of 160lb and a standard deviation of 30.4lb.

a. If a pilot is randomly​ selected, find the probability that his weight is between 150lb and 201lb. p=0.5402 b. If 37 different pilots are randomly​ selected, find the probability that their mean weight is between 150lb and 201lb. p=0.9773 c. When redesigning the ejection​ seat, which probability is more​ relevant? Part (a) because single

The accompanying table describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children.

a. P(6 girls in 8 births):0.109 b. P(6 or more girls in 8 births)=0.132 c. The result from part​ b, since it is the probability of the given or more extreme result. d. No, since the appropriate probability is greater than​ 0.05, it is not a significantly high number.

Based on a​ poll, 60​% of adults believe in reincarnation. Assume that 7 adults are randomly​ selected, and find the indicated probability.

a. P(6 of 7)=0.131 b. P(all)=0.028 c. P(at least 6)=0.159 d. No, because the probability that 6 or more of the selected adults believe in reincarnation is greater than 0.05.

Multiple-choice questions each have four possible answers (a, b, c, d), one of which is correct. Assume that you guess the answers to three such questions.

a. P(WCC)=3/64 b. P(CCW)=3/64 P(CWC)=3/64 c. P(two correct answers when three guesses are made)=9/64

Hospitals typically require backup generators to provide electricity in the event of a power outage. Assume that emergency backup generators fail 35​% of the times when they are needed. A hospital has two backup generators so that power is available if one of them fails during a power outage.

a. P(both fail)=0.1225 b. P(working generator)=0.8775. No, it is not high enough because both generators fail about 12% of the time.

You want to obtain cash by using an​ ATM, but​ it's dark and you​ can't see your card when you insert it. The card must be inserted with the front side up and the printing configured so that the beginning of your name enters first.

a. P(inserted correctly)=1/4 b. P(incorrect on first attempt but correct on second)=3/16 c. Trick question

Based on a​ poll, among adults who regret getting​ tattoos, 19​% say that they were too young when they got their tattoos. Assume that seven adults who regret getting tattoos are randomly​ selected, and find the indicated probability.

a. P(none)=0.2288 b. P(one)=0.3756 c. P(0 or 1)= 0.6044 d. No, because the probability that at most 1 of the selected adults say that they were too young is greater than 0.05.

Gum & Money: 28, 16, 17, 25. A student given a​ $1 bill is more likely to have kept the money.

a. P(spent the money with one dollar)= 0.405 b. P(kept the money with one dollar)= 0.595

Assume that females have pulse rates that are normally distributed with a mean of μ=72.0 beats per minute and a standard deviation of σ=12.5 beats per minute.

a. Rate less then 79 bpm p=0.7123 b. If 16 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 79 beats per minute. p=0.9875 c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30? Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

A corporation must appoint a​ president, chief executive officer​ (CEO), chief operating officer​ (COO), and chief financial officer​ (CFO). It must also appoint a planning committee with three different members. There are 13 qualified​ candidates, and officers can also serve on the committee.

a. There are 17160 ways to appoint the officers (permutation). b. There are 286 ways to appoint the committee. c. P(getting the three youngest of the qualified candidates)=1/286

TInterval (13.046,22.15) -x-=17.598 Sx=16.01712719 n=50

a. df= 49 b. t(a/2)=2.01 c. The number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values.

Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby. This method is designed to increase the likelihood that each baby will be a​ girl, but assume that the method has no​ effect, so the probability of a girl is 0.5. Assume that the groups consist of 17 couples.

a. mean=8.5 standard deviation=2.1 b. Sig. low=4.3 girls Sig high=12.7 girls c. The result of 15 girls is significantly high, because it is greater than 12.7 girls. A result of 15 girls would suggest that the method is effective.

The lengths of pregnancies are normally distributed with a mean of 269 days and a standard deviation of 15 days.

a. p=0.0038 (1-0.9962) b. 243 days

A​ gender-selection technique is designed to increase the likelihood that a baby will be a girl. In the results of the​ gender-selection technique, 969 births consisted of 505 baby girls and 464 baby boys. In analyzing these​ results, assume that boys and girls are equally likely.

a. p=0.0108 b. p=0.0994 No, not far. c. Part b d. No, not unlikely.

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 120lb and 161lb. The new population of pilots has normally distributed weights with a mean of 128lb and a standard deviation of 25.2 lb.

a. p=0.5294 b. p=0.9638 c. Part (a) because single

Before every​ flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The aircraft can carry 39 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6,591lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than 6,591 lb39=169lb. What is the probability that the aircraft is​ overloaded?

a. p=0.758 b. The pilot needs to take action.

A boat capsized and sank in a lake. Based on an assumption of a mean weight of 135​lb, the boat was rated to carry 60 passengers​ (so the load limit was 8,100lb). After the boat​ sank, the assumed mean weight for similar boats was changed from 135lb to 175lb.

a. p=1 b. 40.4/sqrt14 = 10.797354 >> z = 175-178.9/10.7974 = -0.36 = 1-0.3590 = p=0.6410 c. Because there is a high probability of​ overloading, the new ratings do not appear to be safe when the boat is loaded with 13 passengers.

When using the​ _______ always be careful to avoid​ double-counting outcomes.

addition rule

Which word is associated with multiplication when computing​ probabilities?

and

In a probability​ histogram, there is a correspondence between​ _______.

area and probability

Which of the following groups of terms can be used interchangeably when working with normal​ distributions?

areas, probability, and relative frequencies

The conditional probability of B given A can be found by​ _______.

assuming that event A has​ occurred, and then calculating the probability that event B will occur

A​ _______ is any event combining two or more simple events.

compound event

A​ _______ probability of an event is a probability obtained with knowledge that some other event has already occurred.

conditional

A​ _______ random variable has infinitely many values associated with measurements.

continuous

Nicotine in menthol cigarettes 99​% confidence; n=21​, s=0.21 mg.

df = 20 X^2L = 7.434 (the value right to the left of it) X^2R = 39.997 (all the way to the right on the table) so the confidence interval estimate of sigma o is 0.15 & 0.34

Platelet Counts of Women 80​% confidence; n=24​, s=65.1.

df = 23 FIND 1-given percentage (which is alpha)(and TWO TAILED) and plug that into online calculator for both values... X^2L = 14.848 & X^R = 32.007. CI estimate = 55.2, 81.0

Platelet Counts of Women 95​% confidence; n=24​, s=65.8.

df = 23 X^2L = 11.689 & X^2R = 38.076 CI estimate = 51.1, 92.3

Nicotine in menthol cigarettes 98​% confidence; n=26​, s=0.21 mg.

df = 25 X^2L = 11.524 (left next to 13.120) X^2R = 44.314 (all the way to the right but not the last one) so the CI estimate of sigma is 0.16, 0.31

A​ _______ random variable has either a finite or a countable number of values.

discrete

Events that are​ _______ cannot occur at the same time.

disjoint

The classical approach to probability requires that the outcomes are​ _______.

equally likely

The​ _______ of a discrete random variable represents the mean value of the outcomes.

expected value

Find the critical value za/2 that corresponds to the confidence level 84%

find alpha (1-.84) and then divide by 2. 1-(a/2)=0.92 so find z(0.08)... menu553(0.92,0,1) = 1.41

Two events A and B are​ _______ if the occurrence of one does not affect the probability of the occurrence of the other.

independent

Here are summary statistics for randomly selected weights of newborn​ girls: n=289​, x=28.4hg, s=7.5hg. The confidence level is 90​%.

menu 5, 5, 6: t(a/2) = 1.65

Pulse rates of women are normally distributed with a mean of 77.5 beats per minute and a standard deviation of 11.6 beats per minute. Answer the following questions.

meu = 0 & sigma = 1 The z scores are numbers without units of measurement.

Based on a​ survey, assume that 36​% of consumers are comfortable having drones deliver their purchases. Suppose that we want to find the probability that when four consumers are randomly​ selected, exactly two of them are comfortable with delivery by drones.

n=4 x=2 p=0.36 q=0.64

If np≥5 and nq≥​5, estimate P(at least 10) with n=13 and p=0.5 by using the normal distribution as an approximation to the binomial​ distribution; if np<5 or nq<​5, then state that the normal approximation is not suitable.

np = (13*0.5) = 6.5 & nq = (13*0.5) = 6.5 μ=6.5 & sigma=sqrtnpq=1.803 x<10-0.5=9.5 z=(9.5-6.5)/1.803=1.66 P(z<1.66)=0.952 = 1-0.952 = 0.048

Use a normal approximation to find the probability of the indicated number of voters. In this​ case, assume that 167 eligible voters aged​ 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged​ 18-24, 22% of them voted. Probability that fewer than 39 voted.

np=167*0.22=36.74 nq=167*0.78=130.26 meu=np=36.74 sigma=sqrtnpq=5.353 x=39 so z=(38.5-36.74)/5.353=0.33 so the area left of z=0.33 is 0.6293 MINUS 0.0005 equals 0.6288

In the binomial probability​ formula, the variable x represents the​ _______.

number of successes

An elevator has a placard stating that the maximum capacity is 2490lb—15 passengers.​ So, 15 adult male passengers can have a mean weight of up to 2490/15=166 pounds. If the elevator is loaded with 15 adult male​ passengers, find the probability that it is overloaded because they have a mean weight greater than 166 lb.​ (Assume that weights of males are normally distributed with a mean of 176 lb and a standard deviation of 27 lb​.)

p(overloaded)=0.9243. No, there is a good chance that 15 randomly selected adult male passengers will exceed the elevator capacity.

Assume that adults have IQ scores that are normally distributed with a mean of 99.6 and a standard deviation of 24.5. Find the probability that a randomly selected adult has an IQ greater than 146.1.

p=0.0288 (1-0.9712)

Use a normal approximation to find the probability of the indicated number of voters. In this​ case, assume that 198 eligible voters aged​ 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged​ 18-24, 22% of them voted. Probability that exactly 48 voted

p=0.0512

Assume that adults have IQ scores that are normally distributed with a mean of μ=105 and a standard deviation σ=15. Find the probability that a randomly selected adult has an IQ between 95 and 115.

p=0.4950

Use a normal approximation to find the probability of the indicated number of voters. In this​ case, assume that 170 eligible voters aged​ 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged​ 18-24, 22% of them voted. Probability that exactly 39 voted.

p=0.6517-0.5793=0.0724

Assume that adults have IQ scores that are normally distributed with a mean of μ=105 and a standard deviation σ=20. Find the probability that a randomly selected adult has an IQ less than 125.

p=0.8413 (convert to z score then calc552)

If the order of the items selected​ matters, then we have a​ _______.

permutation problem

A​ _______ is a single value used to approximate a population parameter.

point estimate

A​ _______ variable is a variable that has a single numerical​ value, determined by​ chance, for each outcome of a procedure.

random

P(A)+P(-A-) ​= 1 is one way to express the​_______.

rule of complementary events

The​ _______ is the best point estimate of the population mean.

sample mean

The​ _______ for a procedure consists of all possible simple events or all outcomes that cannot be broken down any further.

sample space

The best point estimate of the population variance σ2 is the​ _____________.

sample variance, s^2

The confidence level is 95​%, σ is not​ known, and the histogram of 63 player salaries​ (in thousands of​ dollars) of football players on a team is as shown.

t(a/2) = 2

The confidence level is 99​%, σ is not​ known, and the histogram of 58 player salaries​ (in thousands of​ dollars) of football players on a team is as shown.

t(a/2) menu, 5, 5, 6, but erase the negative. = 2.66

For a sequence of two events in which the first event can occur m ways and the second event can occur n​ ways, the events together can occur a total of​ (m)(n) ways. This is called​ _______.

the fundamental counting rule

As a procedure is repeated again and​ again, the relative frequency of an event tends to approach the actual probability. This is known as​ _______.

the law of large numbers

P(A or B) indicates​_______.

the probability that in a single​ trial, event A​ occurs, event B​ occurs, or they both occur.

Which is NOT a criterion for distinguishing between results that could easily occur by chance and those results that are highly​ unusual?

the sample size is less than​ 5% of the size of the population

A picture of line segments branching out from one starting point illustrating the possible outcomes of a procedure is called a​ _______.

tree diagram

A critical​ value, zα​, denotes the​ _______.

z dash score with an area of alpha to its right.

The confidence level is 90​%, σ=3815 thousand​ dollars, and the histogram of 56 player salaries​ (in thousands of​ dollars) of football players on a team is as shown.

z(a/2) = 1.65

The confidence level is 95​%, σ=3593 thousand​ dollars, and the histogram of 57 player salaries​ (in thousands of​ dollars) of football players on a team is as shown.

z(a/2) = 1.96 because of table

Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. A=0.9162

z=1.28 (calculator: menu, 6, 5, 3)

Find the critical value za/2 that corresponds to the confidence level 87%

za/2 = 1.51

​Claim: The mean pulse rate​ (in beats per​ minute) of adult males is equal to 69bpm. For a random sample of 133 adult​ males, the mean pulse rate is 68.4bpm and the standard deviation is 11.3bpm.

-0.61 because of the second mean equation on chart (the claim number is the second number on the top).

Claim: Fewer than 91​% of adults have a cell phone. In a reputable poll of 1057 ​adults, 86​% said that they have a cell phone.

-5.68 because of Z score equation from chart.

The claim is that for 12 AM body​ temperatures, the mean is μ<98.6°F. The sample size is n=9mand the test statistic is t=−2.366.

0.023

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 48.0 and 58.0 minutes. Find the probability that a given class period runs between 51.5 and 51.75 minutes.

0.025 because 0.25*0.1=0.025

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 48.0 and 53.0 minutes. Find the probability that a given class period runs between 51.5 and 51.75 minutes.

0.05 because 0.25*0.2=0.05

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 8 minutes. Find the probability that a randomly selected passenger has a waiting time less than 0.75 minutes.

0.094

The claim is that for a smartphone​ carrier's data speeds at​ airports, the mean is μ=11.00 Mbps. The sample size is n=27 and the test statistic is t=−1.167.

0.254 (online calculator)

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 6 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.

0.623

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. z=0.54

0.7054 (calculator: menu, 5, 5, 2)

Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1. z= between -0.94 and 1.23

0.7171

Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1. z=-1.09

0.8621 because 1-0.1378=0.8621

​Claim: The standard deviation of pulse rates of adult males is less than 12bpm. For a random sample of 160 adult​ males, the pulse rates have a standard deviation of 11.5bpm.

146.03 because of standard deviation equation on chart (the claim goes on the bottom).

Match these values of r with the accompanying​ scatterplots: 0.996​, 0.774​, 0.392​, −0.774​, and −0.996.

1=-0.774, 2=-0.996, 3=0.392, 4=0.774, 5=0.996

Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 617 randomly selected adults showed that 64​% of them would erase all of their personal information online if they could.

6.96 because of Z score equation from chart.

Suppose IQ scores were obtained for 20 randomly selected sets of twins. The 20 pairs of measurements yield x=100.07​, y=99.75​, r=0.950​, P-value=​0.000, and y=−24.86+1.25x​, where x represents the IQ score of the twin born first. Find the best predicted value of y given that the twin born first has an IQ of 96​? Use a significance level of 0.05.

95.14

Which of the following is NOT true when testing a claim about a​ proportion?

A conclusion based on a confidence interval estimate will be the same as a conclusion based on a hypothesis test.

Which of the following is NOT a true statement about error in hypothesis​ testing?

A type I error is making the mistake of rejecting the null hypothesis when it is actually false.

Listed below are numbers of Internet users per 100 people and numbers of scientific award winners per 10 million people for different countries. Construct a​ scatterplot, find the value of the linear correlation coefficient​ r, and find the​ P-value of r.

A, r=0.794, p=0 & p/=0, t=2.61, p=0.059, GREATER than.is NOT.

Twelve different video games showing alcohol use were observed. The duration times of alcohol use were​ recorded, with the times​ (seconds) listed below. What requirements must be satisfied to test the claim that the sample is from a population with a mean greater than 80sec?

A. Either the population is normally​ distributed, or n>​30, or both. & B. The sample observations must be a simple random sample. No. The sample size is not greater than​ 30, the sample does not appear to be from a normally distributed​ population, and there is not enough information given to determine whether the sample is a simple random sample.

Listed below are amounts of bills for dinner and the amounts of the tips that were left. Construct a​ scatterplot, find the value of the linear correlation coefficient​ r, and find the​ P-value of r. Determine whether there is sufficient evidence to support a claim of linear correlation between the two variables. alpha = 0.01

C, r=0.966, p=0 & p/=0, t=7.47, p=0.002, less than or equal to..is, 1.

Twelve different video games showing drug use were observed. The duration times of drug use were​ recorded, with the times​ (seconds) listed below. Assume that these sample data are used with a 0.10 significance level in a test of the claim that the population mean is greater than 85sec. If we want to construct a confidence interval to be used for testing that​ claim, what confidence level should be used for a confidence​ interval? If the confidence interval is found to be 19.7sec<μ<192.6 sec, what should we conclude about the​ claim?

Confidence level = 80%. The given confidence interval contains the value of 85​sec, so there is not sufficient evidence to support the claim that the mean is greater than 85 sec.

Which of the following is NOT one of the three common errors involving​ correlation?

Correlation does not imply causality.

Which of the following is not equivalent to the other​ three?

Dependent variable

Which of the following is NOT a requirement in determining whether there is a linear correlation between two​ variables?

If r > 1, then there is a positive linear correlation.

Which of the following is NOT a criterion for making a decision in a hypothesis​ test?

If the​ P-value is less than​ 0.05, the decision is to reject the null hypothesis.​ Otherwise, we fail to reject the null hypothesis.

Which of the following is NOT true for a hypothesis test for correlation.

If |r|>critical ​value, we should fail to reject the null hypothesis and conclude that there is not sufficient evidence to support the claim of a linear correlation.

If we find that there is a linear correlation between the concentration of carbon dioxide in our atmosphere and the global​ temperature, does that indicate that changes in the concentration of carbon dioxide cause changes in the global​ temperature?

No. Does not imply one is the cause of the other.

Consider a drug that is used to help prevent blood clots in certain patients. In clinical​ trials, among 5881 patients treated with this​ drug, 163 developed the adverse reaction of nausea. Use a 0.01 significance level to test the claim that 3​% of users develop nausea.

Null: p=0.03 Alternative: p=/0.03 Test Statistic: -1.03 P-value: 0.303 Fail to reject, not sufficient evidence to warrant. Since the rate of nausea appears to be relatively low, it is not a problematic adverse reaction.

In a study of the accuracy of fast food​ drive-through orders, one restaurant had 39 orders that were not accurate among 306 orders observed. Use a 0.10 significance level to test the claim that the rate of inaccurate orders is equal to​ 10%.

Null: p=0.1 Alternative: p=/0.1 Test Statistic: (on chart) 1.60 P-value 0.110 (from online calculator using test stat as z score). Fail to reject..not sufficient evidence to warrant. Since there is not sufficient evidence to reject the claim that the rate of inaccurate orders is equal to​ 10%, it is plausible that the inaccuracy rate is ​10%. This rate would be too high​, so the restaurant should work to lower the rate.

Which of the following is NOT true of using the binomial probability distribution to test claims about a​ proportion?

One requirement of this method is that np>5 and nq>5.

Identify the type I error and the type II error that correspond to the given hypothesis. The percentage of households with more than 1 pet is equal to 65%.

Reject - actually then fail - different

A bottle contains a label stating that it contains pills with 500 mg of vitamin​ C, and another bottle contains a label stating that it contains pills with 325 mg of aspirin. When testing claims about the mean contents of the​ pills, which would have more serious​ implications: rejection of the vitamin C claim or rejection of the aspirin​ claim?

Rejection of the claim about aspirin is more serious because the wrong aspirin dosage could cause more serious adverse reactions than a wrong vitamin C dosage. It would be wise to use a smaller significance level for testing the claim about the aspirin.

Which of the following is not a characteristic of the t test?

The Student t distribution has a mean of t=0 and a standard deviation of s=1.

Which of the following is NOT true about the tails in a​ distribution?

The inequality symbol in the alternative hypothesis points away from the critical region.

Which of the following is NOT a property of the linear correlation coefficient​ r?

The linear correlation coefficient r is robust. That is, a single outlier will not affect the value of r.

Which of the following is not a requirement for regression analysis?

The method for regression analysis line is not robust. It is seriously affected by a small departure from a normal distribution.

Forty different video games showing drug use were observed. The duration times of drug use (in seconds) were recorded. When using this sample for a t test of the claim that the population mean is greater than 95sec, what does df​ denote, and what is its​value?

The number of degrees of freedom -- 39

Which of the following is not a requirement for testing a claim about a population with σ not​ known?

The population mean is equal to 1.

​Claim: Most adults would not erase all of their personal information online if they could. A software firm survey of 529 randomly selected adults showed that 0.5​% of them would erase all of their personal information online if they could.

The results are significantly low, so there is sufficient evidence to support the claim that most adults would not erase all of their personal information online if they could.

​Claim: The mean respiration rate (in breaths per minute) of students in a large statistics class is less than 29. A simple random sample of the students has a mean respiration rate of 12.5.

The sample is not unusual if the claim is true. The sample is unusual if the claim is false.​ Therefore, there is sufficient evidence to support the claim.

Which of the following is NOT a requirement for testing a claim about a population mean with σ known?

The sample mean is greater than 30.

What is the relationship between the linear correlation coefficient r and the slope b1 of a regression​ line?

The value of r will always have the same sign as the value of b1.

Which of the following is NOT true about​ P-values in hypothesis​ testing?

The​ P-value separates the critical region from the values that do not lead to rejection of the null hypothesis.

The claim is that for a smartphone​ carrier's data speeds at​ airports, the mean is μ=16.00 Mbps. The sample size is n=20 and the test statistic is t=1.725.

Two tailed so P-value = 0.101

Identify the type I error and the type II error that corresponds to the given hypothesis. The proportion of people who write with their left hand is equal to 0.18.

Type I Error: Reject the claim that the proportion of people who write with their left hand is 0.18 when the proportion is actually 0.18. Type II Error: Fail to reject the claim that the proportion of people who write with their left hand is 0.18 when the proportion is actually different from 0.18.

When making predictions based on regression lines, which of the following is not listed as a consideration?

Use the regression line for predictions only if the data go far beyond the scope of the available sample data.

Which of the following statements about correlation is true?

We say that there is a positive correlation between x and y if the​ x-values increase as the corresponding​ y-values increase.

Original​ claim: More than 41​% of adults would erase all of their personal information online if they could. The hypothesis test results in a​ P-value of 0.2292.

a) Fail to reject Ho because the P-value is greater than alpha. b) There is not sufficient evidence.

Original​ claim: The standard deviation of pulse rates of a certain group of adult males is more than 12bpm. The hypothesis test results in a​ P-value of 0.3193.

a) Fail to reject because greater b) There is not

Original​ claim: The mean pulse rate​ (in beats per​ minute) of a certain group of adult males is 68bpm. The hypothesis test results in a​ P-value of 0.0094.

a) Reject Ho because less than b) There is

Original​ claim: The mean pulse rate​ (in beats per​ minute) of a certain group of adult males is 73bpm. The hypothesis test results in a​ P-value of 0.0818.

a) Reject Ho because less than or equal to b) There is sufficient evidence

Original​ claim: More than 43​% of adults would erase all of their personal information online if they could. The hypothesis test results in a​ P-value of 0.0186.

a) Reject Ho because the P-value is less than or equal to alpha b) There is sufficient evidence.

In a certain​ survey, 514 people chose to respond to this​ question: "Should passwords be replaced with biometric security​ (fingerprints, etc)?" Among the​ respondents, 53​% said​ "yes." We want to test the claim that more than half of the population believes that passwords should be replaced with biometric security.

a) The sample observations are not a random​ sample, so a test about a population proportion using the normal approximating method cannot be used. b) This statement means that if the​ P-value is very​ low, the null hypothesis should be rejected. c) This statement seems to suggest that with a high​ P-value, the null hypothesis has been proven or is​ supported, but this conclusion cannot be made. d) Choosing this specific of a significance level could give the impression that the significance level was chosen specifically to reach a desired conclusion.

A certain drug is used to treat asthma. In a clinical trial of the​ drug, 29 of 288 treated subjects experienced headaches​ (based on data from the​ manufacturer). The accompanying calculator display shows results from a test of the claim that less than 11​% of treated subjects experienced headaches.

a) left-tailed b) z=-0.50 c) P-value=0.3069 d) p=0.11 Fail to reject the null hypo because is greater. e) There is not sufficient evidence to support the claim.

A poll of 2,081 randomly selected adults showed that 93​% of them own cell phones. The technology display below results from a test of the claim that 92​% of adults own cell phones.

a) two-tailed b) test statistic (equal to Z-value) is 1.57 c) 0.115 d) p=0.92 Fail to reject because greater. e) There is not sufficient evidence to warrant rejection of the claim that 92% of adults own a cell phone.

The test statistic of z=−3.46 is obtained when testing the claim that p<0.72.

a. -2.33 (critical value calculator online) b. Reject Ho, there is sufficient evidence to support the claim that p<0.72

The test statistic of z=−1.68 is obtained when testing the claim that p=3/5.

a. 1.96, -1.96 b. Fail to reject H0.There is not sufficient evidence to warrant rejection of the claim that p=3/5.

a. What is a​ residual? b. In what sense is the regression line the straight line that​ "best" fits the points in a​ scatterplot?

a. A residual is a value of y−y​ hat, which is the difference between an observed value of y and a predicted value of y. b. The regression line has the property that the sum of squares of the residuals is the lowest possible sum.

Use the given data set to complete parts​ (a) through​ (c) below.​ (Use α=​0.05.) (10,9.14)(8,8.14)(13,8.74)

a. B b. r=0.817 (online calculator) There is sufficient evidence to support the claim of linear correlation. c. The scatterplot reveals a distinct pattern that is not a straight-line pattern.

The test statistic of z=2.30 is obtained when testing the claim that p>0.7.

a. This is a right-tailed test. b. 0.011 because it's the area to the right after menu 6,5,2 c. Reject Ho. There is sufficient evidence to support the claim that p>0.7

The test statistic of z=0.60 is obtained when testing the claim that p>0.8.

a. This is a right-tailed test. b. 0.274 menu 6,5,2 then 1 minus that c. Fail to reject Ho. There is not sufficient evidence to support the claim that p>0.8

​Claim: The mean pulse rate​ (in beats per​ minute) of adult males is equal to 68.5bpm. For a random sample of 164adult​ males, the mean pulse rate is 69.8bpm and the standard deviation is 11.4bpm.

a. meu=68.5 b. Ho: meu=68.5 H1: meu is not equal to 68.5

​Claim: A minority of adults would erase all of their personal information online if they could. A software firm survey of 599 randomly selected adults showed that 37​% of them would erase all of their personal information online if they could.

a. p<0.5 b. Ho: p=0.5 H1: p<0.5

The test statistic of z=2.36 is obtained when testing the claim that p>0.4.

a. right-tailed b. 0.009 c. Reject Ho. There is.

​Claim: The standard deviation of pulse rates of adult males is less than 10bpm. For a random sample of 155 adult​ males, the pulse rates have a standard deviation of 9.8bpm.

a. sigma<10 b. Ho: sigma=10 H1: sigma<10

The test statistic of z=1.49 is obtained when testing the claim that p≠0.642.

a. two-tailed b. 0.136 menu 6,5,2 then 1 minus that and then multiply that by two c. Fail to reject Ho. There is not.

a. Using the pairs of values for all 10 ​points, find the equation of the regression line. b. After removing the point with coordinates (8,3)​, use the pairs of values for the remaining 9 points and find the equation of the regression line. c. Compare the results from parts​ (a) and​ (b).

a. y hat=6.797+-0.422x b. y hat=6 c. The removal of the point has a significant impact on the regression line.

Different hotels in a certain area are randomly​ selected, and their ratings and prices were obtained online. Using​ technology, with x representing the ratings and y representing​ price, we find that the regression equation has a slope of 130 and a​ y-intercept of −363.

a. y with hat = -363 + (130)x b. The symbol y hat represents the predicted value of price.

The test statistic of z=−1.86 is obtained when testing the claim that p<0.26.

a. z=-1.65 b. If test statistics is in the critical region, reject Ho.

A​ __________ exists between two variables when the values of one variable are somehow associated with the values of the other variable.

correlation

A​ _____________ is a procedure for testing a claim about a property of a population.

hypothesis test

Paired sample data may include one or more​ ___________, which are points that strongly affect the graph of the regression line.

influential points

A straight line satisfies the​ __________________ if the sum of the squares of the residuals is the smallest sum possible.

least-squares property

The​ ______________ measures the strength of the linear correlation between the paired quantitative​ x- and​ y-values in a sample.

linear correlation coefficient r

In working with two variables related by a regression​ equation, the​ _________________ in a variable is the amount that it changes when the other variable changes by exactly one unit.

marginal change

The​ _________ hypothesis is a statement that the value of a population parameter is equal to some claimed value.

null

In a​ scatterplot, a(n) ______________ is a point lying far away from the other data points.

outlier

Listed below are amounts of court income and salaries paid to the town justices. All amounts are in thousands of dollars. Construct a​ scatterplot, find the value of the linear correlation coefficient​ r, and find the​ P-value using α=0.05.

p=0 & p/=0, A, r=0.868, t=4.625, p=0.002, LESS & IS, It does appear that justices might profit by levying larger fines.

A data set includes weights of garbage discarded in one week from 62 different households. The paired weights of paper and glass were used to obtain the results shown to the right. Is there sufficient evidence to support the claim that there is a linear correlation between weights of discarded paper and​ glass? Watch your decimal places!

p=0, p/=0, r=0.125 (not 1 but the other # in the chart), There are two critical values at r=+-0.250 (from the table link with the number closest to n), Because the absolute value of the test statistic is less than or equal to (in between) the positive critical​ value, there is NOT sufficient evidence to support the claim

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6966 subjects randomly selected from an online group involved with ears. There were 1337 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than​ 20%.

p=0.2, p<0.2, -1.69, 0.046, Because the​ P-value is greater than the significance​ level, fail to reject the null hypothesis. There is insufficient evidence to support the claim that the return rate is less than​ 20%.

Suppose 239 subjects are treated with a drug that is used to treat pain and 50 of them developed nausea. Use a 0.10 significance level to test the claim that more than 20​% of users develop nausea.

p=0.2, p>0.2, 0.36, 0.359, Fail to reject..not sufficient evidence to warrant.

Suppose that in a random selection of 100 colored​ candies, 23​% of them are blue. The candy company claims that the percentage of blue candies is equal to 29​%. Use a 0.10 significance level to test that claim.

p=0.29, p=/0.29, -1.32, 0.187, Fail to reject..not sufficient evidence to warrant.

Assume that adults were randomly selected for a poll. They were asked if they​ "favor or oppose using federal tax dollars to fund medical research using stem cells obtained from human​ embryos." Of those​ polled, 481 were in​ favor, 396 were​ opposed, and 117 were unsure. A politician claims that people​ don't really understand the stem cell issue and their responses to such questions are random responses equivalent to a coin toss. Exclude the 117 subjects who said that they were​ unsure, and use a 0.10 significance level to test the claim that the proportion of subjects who respond in favor is equal to 0.5.

p=0.5, p=/0.5, 2.87, 0.004, reject Ho..there is sufficient evidence to warrant rejection, the result suggests that the politician is wrong in claiming that the responses are random guesses.

In a study of 809 randomly selected medical malpractice​ lawsuits, it was found that 507 of them were dropped or dismissed. Use a 0.01 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed.

p=0.5, p>0.5, 7.21, 0.000, Reject null because less than, there is sufficient evidence to warrant rejection.

The​ _________ of a hypothesis test is the probability (1−β​) of rejecting a false null hypothesis.

power

Given a collection of paired sample​ data, the ____________________ yhat=b0+b1x algebraically describes the relationship between the two​ variables, x and y.

regression equation

For a pair of sample​ x- and​ y-values, the​ ______________ is the difference between the observed sample value of y and the​ y-value that is predicted by using the regression equation.

residual

A​ ______________ is a scatterplot of the​ (x,y) values after each of the​ y-coordinate values has been replaced by the residual value y−yhat.

residual plot

When determining whether there is a correlation between two​ variables, one should use a​ ____________ to explore the data visually.

scatterplot

The​ ___________ is a value used in making a decision about the null hypothesis and is found by converting the sample statistic to a score with the assumption that the null hypothesis is true.

test statistic

Park officials make predictions of times to the next eruption of a particular​ geyser, and collect data for the errors​ (minutes) in those predictions. The display from technology available below results from using the prediction errors to test the claim that the mean prediction error is equal to zero.

u=0, u=/0, -7.63, 0.000, Reject..sufficient..is not..some error.

In a test of the effectiveness of garlic for lowering​ cholesterol, 64 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes​ (before minus​ after) in their levels of LDL cholesterol​ (in mg/dL) have a mean of 0.4 and a standard deviation of 1.83.

u=0, u>0, 1.75, 0.042, Reject..sufficient..is greater than.

A data set includes data from 500 random tornadoes. The display from technology available below results from using the tornado lengths​ (miles) to test the claim that the mean tornado length is greater than 2.6 miles.

u=2.6, u>2.6, 2.61, 0.005, Reject..sufficient

The display provided from technology available below results from using data for a smartphone​ carrier's data speeds at airports to test the claim that they are from a population having a mean less than 4.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level.

u=4, u<4, test statistic = -3.70, P-value = 0.000, Reject Ho..there is sufficient evidence to support.

A data set includes data from student evaluations of courses. The summary statistics are n=86​, x=4.09​, s=0.55. Use a 0.10 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25.

u=4.25, u=/4.25, -2.70, 0.008, Reject..sufficient..is not.

A data set lists earthquake depths. The summary statistics are n=500​, x=5.84km, s=4.77km. Use a 0.01 significance level to test the claim of a seismologist that these earthquakes are from a population with a mean equal to 5.00.

u=5, u=/5, 3.94, 0.000, Reject...sufficient...is not.

A data set about speed dating includes​ "like" ratings of male dates made by the female dates. The summary statistics are n=197​, x=5.81​, s=2.02. Use a 0.05 significance level to test the claim that the population mean of such ratings is less than 6.00.

u=6, u<6, -1.32, 0.094, Fail to reject..not sufficient..less than.

Use the given data to find the equation of the regression line. Examine the scatterplot and identify a characteristic of the data that is ignored by the regression line. (10,12.92)(12,14.11)(8,10.51)

y hat = 2.40+0.9x (up and then curves down) Identify a characteristic of the data that is ignored by the regression line...The data has a pattern that is not a straight line.

Find the regression​ equation, letting overhead width be the predictor​ (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 1.6 cm. Can the prediction be​ correct?

y hat=-268.5+56.9x... -177.5 kg...The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data.

Listed below are systolic blood pressure measurements​ (in mm​ Hg) obtained from the same woman. Find the regression​ equation, letting the right arm blood pressure be the predictor​ (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 85 mm Hg.

y hat=55.7+1.1x Given that the systolic blood pressure in the right arm is 85 mm​ Hg, the best predicted systolic blood pressure in the left arm is 154.8 mm Hg (the mean of the left arm values).


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