STA 270
The use of topical painkiller ointment or gel rather than pills for pain relief was approved just within the last few years in the US for prescription use only.1 Insurance records show that the average copayment for a month's supply of topical painkiller ointment for regular users is $30. A sample of size n=75 regular users found a sample mean copayment of $27.90. Identify each of 30 and 27.90 as a parameter or a statistic
30---> parameter "mu" 27.90----? statistic "x^bar"
In this exercise, we are conducting many hypothesis tests to test a claim. Assume that the null hypothesis is true. If 100 tests are conducted using a significance level of 5%, approximately how many of the tests will incorrectly find significance?
5
In estimating the mean score on a fitness exam, we use an original sample of size n=30 and a bootstrap distribution containing 5000 bootstrap samples to obtain a 95% confidence interval of 67 to 73. A change in this process is described below. If all else stays the same, which of the following confidence intervals (A, B, or C) is the most likely result after the change: Using the data to find a 99% confidence interval
66 to 74 bc we now include more numbers
In estimating the mean score on a fitness exam, we use an original sample of size n=30 and a bootstrap distribution containing 5000 bootstrap samples to obtain a 95% confidence interval of 67 to 73. A change in this process is described below. If all else stays the same, which of the following confidence intervals (A, B, or C) is the most likely result after the change: Using an original sample of size n=45
67.5 to 72.7, larger sample size make the data more accurate
Percentile^=102-3.34⋅Years. predict the cognitive percentile for someone who has played football for 8 years and for someone who has played football for 13 years.
8 years = 75.28 13 years= 58.58 For every additional year playing football, the predicted cognitive percentile goes down 3.34.
State whether the following claim is one of association and causation, association only, or neither association nor causation. Cat owners tend to be more educated than dog owners.
Association
A random sample of maple trees in a forest is used to estimate the mean base circumference of all maple trees in the forest. Give notation for the quantity that is being estimated. Give notation for the quantity that gives the best estimate.
Mu and x^bar
The population is all trees in a forest. We walk through the forest and pick out trees that appear to be representative of all the trees in the forest. State whether or not the sampling method described produces a random sample from the given population.
Not Random
State the null and alternative hypotheses for the statistical test described below. Testing to see if there is evidence that a proportion is greater than 0.3
Null: p=.3 Alternative: p>.3
The following describes a sample. The information given includes the five number summary, the sample size, and the largest and smallest data values in the tails of the distribution. Five number summary: (16,43,53,57,72) n=120 Tails: 16,21,29,31,32,...,65,66,66,67,72 find the outliers using the IQR method
Q1=43 and Q3=57 so the interquartile range is IQR=57-43=14. We compute Q1-1.5(IQR)=43-1.5(14)=22, and Q3+1.5(IQR)=57+1.5(14)=78. anything below 22 is an outlier and anything above 78 is too
IQR
Q3-Q1
Variable Mean SE MeanStDev Minimum Q1 Median Q3 Maximum 28.766 0.476 3.369 21.300 26.375 29.400 31.150 35.100 range, IQR, The location of the 40th percentile and 90th percentile is between:
Range= 13.8 IQR= 4.775 40%= 26.375 to 29.400 90%=31.150 to 35.100
Z-scores
Sample z= x- mean/standard deviation Population z=x- population parameter/ standard deviation
In the situation below, indicate whether it makes more sense to use a relatively large significance level (such as α=0.10) or a relatively small significance level (such as α=0.01). Testing to see if a well-known company is lying in its advertising. If there is evidence that the company is lying, the Federal Trade Commission will file a lawsuit against them.
a small significance level
A random sample of n=755 US cell phone users age 18 and older in May 2011 found that the average number of text messages sent or received per day is 41.5 messages,1 with standard error about 6.1. (a) Use the information from the sample to give the best estimate of the population parameter. (b) Find a 95% confidence interval for the mean number of text messages.
a) 41.5 b) 29.3 to 53.7
The website fox6now.com held an online poll in June 2015 asking "What do you think about the concept of having an everyday uniform for work, like Steve Jobs did?" Of the people who answered the question, 24% said they loved the idea, 58% said they hated the idea, and 18% said that they already wore a uniform to work. (a) Are the people who answered the poll likely to be representative of all adult workers? b) Is it reasonable to generalize this result and estimate that 24% of all adult workers would like to wear a uniform to work?
a) NO b)NO
Scientists studying lion attacks on humans in Tanzania1 found that 95 lion attacks happened between 6 pm and 10 pm within either five days before a full moon or five days after a full moon. Of these, 71 happened during the five days after the full moon while the other 24 happened during the five days before the full moon. Does this sample of lion attacks provide evidence that attacks are more likely after a full moon? In other words, is there evidence that attacks are not equally split between the two five-day periods? (Note that this is a test for a single proportion since the data come from one sample.) (a) State the null and alternative hypotheses. (b) Use StatKey or other technology to find the p-value. (c) What is the conclusion that can be drawn?
a) Null: p=.5 alternative: p>.5 b) p value= 0 C)Reject H0, attacks are more likely after a full moon.
27.5% of US adults are college graduates. a) Use StatKey or other technology to generate a sampling distribution for the sample proportion of college graduates using a sample size of n=50. Generate at least 1000 sample proportions. Give the center of the sampling distribution and give the standard error. b) Repeat part (a) using a sample size of n=500. use confidence interval for proportion
a) center= .272 Standard error= .062 b)center=.275 standard error=.020
In July 2015, a poll asked a random sample of 1236 registered voters in Iowa whether they agree or disagree that the world needs to do more to combat climate change.1 The results show that 65% agree, while 25% disagree and 10% don't know. (a) Is the sample likely to be representative of all registered voters in Iowa? (b) Is it reasonable to generalize this result and estimate that 65% of all registered voters in Iowa agree that the world needs to do more to combat climate change?
a)YES b)YES
Figure 1 shows a scatterplot of the acidity (pH) for a sample of n=53 Florida lakes vs the average mercury level (ppm) found in fish taken from each lake. There appears to be a negative trend in the scatterplot, and we wish to test whether there is significant evidence of a negative association between pH and mercury levels. (a) What are the null and alternative hypotheses? (b) For these data, a statistical software package produces the following output: r=-0.575 p-value=0.000017. What is the conclusion of the test? c) Is this convincing evidence that low pH causes the average mercury level in fish to increase?
a)null: rho=0 alternative: rho<0 b) reject null There is very strong evidence of a negative association between pH and mercury levels in fish. c)no
Is the following an experiment or an observational study? To examine whether planting trees reduces air pollution, we find a sample of city blocks with similar levels of air pollution and we then plant trees in half of the blocks in the sample. After waiting an appropriate amount of time, we measure air pollution levels.
experiment, since they are manipulating where to and not plant trees
A somewhat surprising fact about coffee is that the longer it is roasted, the less caffeine it has. Thus an "extra bold" dark roast coffee actually has less caffeine than a light roast coffee. What is the explanatory variable and what is the response variable?
explanatory: roast amount Response: less caffeine negative correlation
Year,HotDogs 2015,62 2014,61 2013,69 2012,68 2011,62 2010,54 2009,68 2008,59 2007,66 2006,54 2005,49 2004,54 2003,45 2002,50 find regression line, correlation, interpret the slope, & Predict the winning number of hot dogs in 2016
hotdogs^= -2647+1.35(year) ----->The slope is the expected change in winning number of hot dogs for a 1 year increase in time. r= .726 predicted # of hot dogs for 2016= 81
A regression line to predict weight in pounds from height in inches is Weight^=-200+5(Height). We can interpret the slope as:
if the height is 1 inch more, then the predicted weight increases by 5 pounds
If the P-value is less than or equal to alpha then.... if the p value is greater than alpha then....
less than or equal to= reject null greater than= fail to reject null
From its founding through 2012, the Rock and Roll Hall of Fame has inducted 303 groups or individuals, and 206 of the inductees have been performers while the rest have been related to the world of music in some way other than as a performer. The full dataset is available in RockandRoll. Give the correct notation for the proportion of inductees who have been performers? What proportion of inductees have been performers? If we took many samples of size 50 from the population of all inductees and recorded the proportion who were performers for each sample, what shape do we expect the distribution of sample proportions to have? Where do we expect it to be centered?
p .680 bell shaped, centered at .680
The null and alternative hypotheses for a population proportion, as well as the sample results, are given. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: H0:p=0.5 vs Ha:p<0.5 statistic p^= 38/100= .38
p-value= . 014
Random samples of people in Canada and people in Sweden are used to estimate the difference between the two countries in the proportion of people who have seen a hockey game (at any level) in the past year. Give notation for the quantity that is being estimated Give notation for the quantity that gives the best estimate
p1-p2 p^1-p^2
Give the relevant proportion using correct notation. Of all 1,672,395 members of the high school class of 2014 who took the SAT (Scholastic Aptitude Test), 793,986 were minority students.1
p= .475
Assume that the sampling distribution is symmetric and bell-shaped.x^bar=60 and the standard error is 2 Indicate the parameter being estimated. Use the information to give a 95% confidence interval.
parameter= mu CI= 56 to 64
r=-0.45 and the margin of error for 95% confidence is 0.02 . Indicate the parameter being estimated. Use the information to give a 95% confidence interval.
parameter= rho CI= -.47 to -.43
Two variables are defined, a regression equation is given, and one data point is given. Hgt=height in inches Age=age in years of a child Hgt^=24.4+2.71(Age) The data point is a child 12 years old who is 60 inches tall.
predicted value= 56.92 residual= 3.08 slope interpretation: Given a one year increase in Age, expected change in Hgt is 2.71 inches.
negative correlation
r<0 neg association
Use StatKey or other technology to find the correlation for the following data. X,Y 3,1 5,2 1,1.5 7,4 6,2.5 use two quantitative variables
r=.821
no correlation
r=0 no association
positive correlation
r>0 positive association
State the conclusion of the test based on a p-value of 0.0002, if we use a 5% significance level.
reject null
Type 1 error (false positive)
rejection of a true null hypothesis if null is true, alpha is the chance of getting a type one error. if a type one error is serious choose a small significance level BUT if a type one error isn't so bad choose a bigger significance level
Null and alternative hypotheses for a test are given below. Give the notation (x¯, for example) for a sample statistic we might record for each simulated sample to create the randomization distribution. H0: p1=p2 versus Ha: p1≠p2
sample p^1-p^2
Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technology to estimate the standard error from a bootstrap distribution generated from the sample. Then use the standard error to give a 95% confidence interval for the proportion of the population to agree with the statement. StatKey tip: Use "CI for Single Proportion" and then "Edit Data" to enter the sample information. In a random sample of 250 people, 180 agree. Estimate the standard error. Find the 95% confidence interval.
standard error= .028 95% CI is .66 to .78
confidence interval
stat - 2(SE) or stat+ 2(SE)
indicate whether the following five number summary corresponds most likely to a distribution that is skewed to the left, skewed to the right, or symmetric. (16,25,36,47,56)
symmetric
Give the correct notation for the quantity described and give its value. Mean number of cell phone calls made or received per day by cell phone users. In a survey of 1917 cell phone users, the mean was 13.10 phone calls a day.
x^bar= 13.10
residual
y- y^= Actual value - predicted value
"For many people, being left alone with their thoughts is a most undesirable activity,." says a psychologist involved in a study examining reactions to solitude.1 In the study, 146 college students were asked to hand over their cell phones and sit alone, thinking, for about 10 minutes. Afterward, 76 of the participants rated the experience as unpleasant. Use this information to estimate the proportion of all college students who would find it unpleasant to sit alone with their thoughts. (This reaction is not limited to college students: in a follow-up study involving adults ages 18 to 77, a similar outcome was reported.) 1.Give notation for the quantity that is being estimated. 2.Give notation for the quantity that gives the best estimate. 3.Give the value for the quantity that gives the best estimate. 4.Give a 95% confidence interval for the quantity being estimated, given that the margin of error for the estimate is 9%.
1. p 2. p^ 3. .521 4. .431 to .611
Use the 95% rule and the fact that the summary statistics come from a distribution that is symmetric and bell-shaped to find an interval that is expected to contain about 95% of the data values. A bell-shaped distribution with mean 1050 and standard deviation 9.
1032 to 1068 1050- 9(2) 1050+9(2)
The following gives information about the proportion of a sample that agree with a certain statement. Use StatKey or other technology to find a confidence interval at the given confidence level for the proportion of the population to agree, using percentiles from a bootstrap distribution. StatKey tip: Use "CI for Single Proportion" and then "Edit Data" to enter the sample information. Find a 90% confidence interval if 112 agree and 288 disagree in a random sample of 400 people.
112/400 agree 90% CI is .245 to .315
Perhaps the most popular fighter since the turn of the decade, Ronda Rousey is famous for defeating her opponents quickly. The five number summary for the times of her first 12 UFC (Ultimate Fighting Championship) fights, in seconds, is (14,25,44,64,289).(a) Only three of her fights have lasted more than a minute, at 289, 267, and 66 seconds, respectively. Use the IQR method to see which, if any, of these values are high outliers. How many high outliers are there?
2 high outliers 287 and 267
Find and interpret the z-score for the data value given. The value 235 in a dataset with mean 175 and standard deviation 26
2.31 standard deviations above the mean
Give the correct notation for the quantity described and give its value. Proportion of US adults who own a cell phone. In a survey of 1006 US adults in 2014, 90% said they had a cell phone. Give the correct notation for the quantity described.
^ p= notation= .9
Type 2 error (false negative)
Believing that something is not real when it is.
Indicate whether we should trust the results of the following study. Is the method of data collection biased? Ask a random sample of students at the library on a Friday night "How many hours a week do you study?" to collect data to estimate the average number of hours a week that all college students study.
Biased
This exercise provides information about data in a survey of students. The survey included 51 students who smoke and 309 who don't. Find p^, the proportion who smoke. Round your answer to three decimal places.
^ p=.142
the regression line: Wins^=34.85+0.070(Runs). find explanatory variable, interpret slope. The San Francisco Giants won 88 games while scoring 665 runs in 2014. Predict the number of games won by San Francisco using the regression line.
Explanatory variable: runs Slope: The expected change in number of wins given one more run. predicted wins in 2014= 81.4 residual= 88-81.4= 6.6
The variable AudienceScore in the HollywoodMovies dataset gives the audience rating on the Rotten Tomatoes website of movies that came out of Hollywood between 2007 and 2013. Use StatKey or other technology to find the mean, the standard deviation, and the five number summary for the data in this variable. ( one quantitative variable) find mean, SD, and five number summary
Mean: 61.3 SD:16.6 19, 49,61,74,96
State whether the data are best described as a population or a sample. The U.S. Department of Transportation announces that of the 250 million registered passenger vehicles in the US, 2.1% are electro-gas hybrids.
Population
Give the relevant proportion using correct notation. A recent headline states that "45% Think Children of Illegal Immigrants Should Be Able to Attend Public School". The report gives the results of a survey of 1000 randomly selected likely voters in the US.1
^ p=.45
regression line
^ y = a+ bx line of best fit a= intercept b= slope x= stat
It is believed that sunlight offers some protection against multiple sclerosis (MS) since the disease is rare near the equator and more prevalent at high latitudes. What is it about sunlight that offers this protection? To find out, researchers1 injected mice with proteins that induce a condition in mice comparable to MS in humans. The control mice got only the injection, while a second group of mice were exposed to UV light before and after the injection, and a third group of mice received vitamin D supplements before and after the injection. In the test comparing UV light to the control group, evidence was found that the mice exposed to UV suppressed the MS-like disease significantly better than the control mice. In the test comparing mice getting vitamin D supplements to the control group, the mice given the vitamin D did not fare significantly better than the control group. If the p-values for the two tests are 1) 0.424 and 2) 0.002, which p-value goes with which test?
UV test group= .002 vitamin D group = .424
X,Y 10,111 20,100 30,100 40,99 50,82 60,77 find the regression line
Y^= 117.3-.64x
A regression line is given along with one of the data points used to create the line. Y^=90-3X; One data point is X=16, Y=33 Find the predicted value of Y for that data point and find the residual.
Y^=90-3X----> 90- 3(16)= 42 predicted = 42 actual = 33 33-42 residual =-9
A recent study1 examined the relationship of football and concussions on hippocampus volume in the brain. The study included three groups with n=25 in each group: healthy controls who had never played football, football players with no history of concussions, and football players with a history of concussions. In this exercise, we test for evidence that average brain size is larger in football players who have never had a concussion (FBNoConcuss) than in football players with a history of concussions (FBConcuss). The data are in FootballBrain where the variable Hipp measures brain size as the volume of the hippocampus (in ml) for each subject. Let group 1 be the football players with no concussion and group 2 be the football players with a history of concussions. (a) State the null and alternative hypotheses (b) Use StatKey or other technology to find the value of the relevant sample statistic. (c) Use StatKey or other technology to find the p-value. (d) Does it appear that this difference in brain size is just due to random chance?
a) m1=m2 or m1>m2 b) x1-x2=724.6 c)0 d)no
How many ants will climb on a piece of a peanut butter sandwich left on the ground near an ant hill? To study this, a student in Australia left a piece of a sandwich for several minutes, then covered it with a jar and counted the number of ants. He did this eight times, and the results are shown in Table 1. (In fact, he also conducted an experiment to see if there is a difference in number of ants based on the sandwich filling.)1 use bootstrap CI for mean (a) Find the mean and standard deviation of the sample. (b) What is the population parameter of interest? (c) What is the best point estimate for that parameter? (d) A bootstrap distribution of 5000 bootstrap statistics gives a standard error of 4.85. Use the standard error to find a 95% confidence interval for the parameter defined in part (b).
a) mean = 34 standard error= 14.63 b) parameter= mu c) 34 bc x^bar is 34 d) 95 % CI is 25 to 45
Next time you see an elderly man, check out his nose and ears! While most parts of the human body stop growing as we reach adulthood, studies show that noses and ears continue to grow larger throughout our lifetime. In one study1 examining noses, researchers report "Age significantly influenced all analyzed measurements:" including volume, surface area, height, and width of noses. In a test to see whether males, on average, have bigger noses than females, the study indicates that "p<0.01." Let group 1 be males and group 2 be females. (a) State the hypotheses. (b) Give the formal decision using a 5% significance level.
a) null: m1=m2 alternative= m1>m2 b)reject null bc p-value=.01
A survey was conducted in the United Kingdom, where respondents were asked if they had a university degree. One question asked, "In the last 20 years the proportion of the world population living in extreme poverty has...", and three choices were provided: 1.)"increased" 2.) "remained more or less the same" and 3.) "decreased". Of 373 university degree holders, 45 responded with the correct answer: decreased; of 639 non-degree respondents, 57 responded with the correct answer1. We would like to test if the percent of correct answers is significantly different between degree holders and non-degree holders. Let group 1 be the degree holders and let group 2 be the non-degree holders. (a) What are the null and alternative hypotheses? (b) Using technology, construct a randomization distribution and compute the p-value. (c) State the conclusion in context.
a) null:p1=p2 alternative: p1 does not equal p2 b) .083 *2= .16 c)We do not have evidence that the proportion giving the correct answer is significantly different between degree holders and non-degree holders.
Female primates visibly display their fertile window, often with red or pink coloration. Do humans also do this? A study1 looked at whether human females are more likely to wear red or pink during their fertile window (days 6-14 of their cycle). They collected data on 24 female undergraduates at the University of British Columbia, and asked each how many days it had been since her last period, and observed the color of her shirt. Of the 10 females in their fertile window, 4 were wearing red or pink shirts. Of the 14 females not in their fertile window, only 1 was wearing a red or pink shirt. Let group 1 be females wearing a red or pink shirt in the fertile group; and group 2 be females wearing a red or pink shirt in the not fertile group. (a) State the null and alternative hypotheses. (b) Calculate the relevant sample statistic, p1^-p2^, for the difference in proportion wearing a pink or red shirt between the fertile and not fertile groups. (c) For the 1000 statistics obtained from the simulated randomization samples, only 6 different values of the statistic p1^-p2^ are possible. The table below shows the number of times each difference occurred among the 1000 randomizations. Calculate the p-value.
a) p1=p2 b)p1>p2 c) 68+5/1000=.073
The data in CarDepreciation contains information on both New price and Depreciation for a sample of 20 automobile models. use bootstrap CI for slope/correlation (a) Find the correlation between New price and Depreciation from the original sample of 20 automobiles. (b) Use StatKey or other technology to create a bootstrap distribution of correlations and report the standard error. (c) Using the standard error, create a 95% confidence interval for the correlation between New price and Depreciation of automobile models.
a) r=.126 b) standard error=.24 c) 95% CI is -.3 to .66
Original sample: 85, 74, 80, 98, 88 Do the values given constitute a possible bootstrap sample from the original sample? (a) 80, 88, 74, 85, 98 (b) 98, 85, 88, 74, 80. (d) 80, 88, 98, 85, 69, 74
a) yes b) yes d)no
Information on a sample of 362 college students is collected. The complete dataset is available at StudentSurvey. We see that 27 of the 193 males in the sample smoke while 16 of the 169 females in the sample smoke. (a) What is the best estimate for the difference in the proportion of smokers, using male proportion minus female proportion? (b) Find a 99% confidence interval for the difference in proportions.
a).045 b)99% CI is -.046 to .134
Overeating for just four weeks can increase fat mass and weight over two years later, a Swedish study shows.1 Researchers recruited 18 healthy and normal-weight people with an average age of 26. For a four-week period, participants increased calorie intake by 70% (mostly by eating fast food) and limited daily activity to a maximum of 5000 steps per day (considered sedentary). Not surprisingly, weight and body fat of the participants went up significantly during the study and then decreased after the study ended. Participants are believed to have returned to the diet and lifestyle they had before the experiment. However, two and a half years after the experiment, the mean weight gain for participants was 6.8 lbs with a standard error of 1.2 lbs. A control group that did not binge had no change in weight. (a) Give a 95% confidence interval for the parameter. (b) Give the margin of error.
a)4.4 to 9.2 b) margin error = 2.4
The phrase "You dirty rat" does rats a disservice. In a recent study,1 rats showed compassion that surprised scientists. Twenty-three of the 30 rats in the study freed another trapped rat in their cage, even when chocolate served as a distraction and even when the rats would then have to share the chocolate with their freed companion. (Rats, it turns out, love chocolate.) Rats did not open the cage when it was empty or when there was a stuffed animal inside, only when a fellow rat was trapped. We wish to use the sample to estimate the proportion of rats to show empathy in this way. The data are available in the dataset CompassionateRats. (a) Give the correct notation of the relevant parameter. (b) Give the best estimate of the parameter in part (a). (c) Use StatKey or other technology to create a bootstrap distribution. What is the standard error? (d) Use the standard error to find a 95% confidence interval for the proportion of rats likely to show empathy.
a)p= proportion b) 23/30= .767 c).078 d) 95% CI is .60 to .90
A headline in June 2015 proclaims "Infections can lower IQ."1 The headline is based on a study in which scientists gave an IQ test to Danish men at age 19. They also analyzed the hospital records of the men and found that 35% of them had been in a hospital with an infection such as an STD or a urinary tract infection. The average IQ score was lower for the men who had an infection than for the men who hadn't. (a) What are the cases in this study? (b) Is the explanatory variable categorical or quantitative? (c) Is the response variable categorical or quantitative? (d) Does the headline imply causation? (e) Is the study an experiment or an observational study? (f) Is it appropriate to conclude causation in this case?
a. Danish men b. categorical c. quantitative d. yes e. observational f. no
It is well-known that lack of sleep impairs concentration and alertness, and this might be due partly to late night food consumption. A 2015 study1 took 44 people aged 21 to 50 and gave them unlimited access to food and drink during the day, but allowed them only 4 hours of sleep per night for three consecutive nights. On the fourth night, all participants again had to stay up until 4 am, but this time participants were randomized into two groups; one group was only given access to water from 10 pm until their bedtime at 4 am while the other group still had unlimited access to food and drink for all hours. The group forced to fast from 10 pm on performed significantly better on tests of reaction time and had fewer attention lapses than the group with access to late night food. (a) Is the explanatory variable categorical or quantitative? (b) Are the response variables categorical or quantitative? (c) Is this an observational study or a randomized experiment? (d) Can we conclude that eating late at night worsens some of the typical effects of sleep deprivation (reaction time and attention lapses)? (e) Are there likely to be confounding variables?
a. categorical b. quantitative c. experiment d. yes e. no
randomized comparative experiment
an experiment that uses both comparison of two or more treatments and random assignment of subjects to treatments
Where is null hypothesis centered?
around the null
State whether the following claim is one of association and causation, association only, or neither association nor causation. Cell phone radiation leads to deaths in honey bees.
association with causation
State whether the following claim is one of association and causation, association only, or neither association nor causation. Daily exercise improves mental performance.
association with causation
Construct an interval estimate for the given parameter using the given sample statistic and margin of error. for μ1-μ2, using x¯1-x¯2=4 with margin of error 11.
confidence interval -7 to 15
The null and alternative hypotheses for a test are given, as well as some information about the actual sample and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. H0:μ1=μ2 vs Ha:μ1≠μ2
centered at 0 bc of null
A random sample of n=461 smartphone users in the US in January 2015 found that 355 of them have downloaded an app. Give notation for the parameter of interest. Give the notation for the quantity we use to make the estimate
p p^= .770
2(SE)
margin of error
symmetric skew
mean = median
skewed right
mean is greater than median
five number summary 9,10,13,16,18,19,20,23,24,28,30
mean: 19.1 SD:6.9 9,14.5,19,23.5,30
skewed left
median is greater than mean
five number summary
minimum, Q1, median, Q3, maximum
For a random sample of households in the US, we record annual household income, whether the location is east or west of the Mississippi River, and number of children. We are interested in determining whether there is a difference in average household income between those east of the Mississippi and those west of the Mississippi. (a) State the null and alternative hypotheses. (b) What statistic(s) from the sample would we use to estimate the difference?
null: m1=m2 Alternative: m1 does not equal m2 sample x^bar 1- x^bar 2
State the null and alternative hypotheses for the statistical test described below Testing to see if average sales are higher in stores where customers are approached by salespeople than in stores where they aren't. Let group 1 be the group of stores where customers are approached by salespeople and let group 2 be the group of stores where customers are not approached by salespeople.
null: m1=m2 alternative: m1>m2
Are children with higher exposure to pesticides more likely to develop ADHD (attention-deficit/hyperactivity disorder)? In a study, authors measured levels of urinary dialkyl phosphate (DAP, a common pesticide) concentrations and ascertained ADHD diagnostic status (Yes/No) for 1139 children who were representative of the general US population.1 The subjects were divided into two groups based on high or low pesticide concentrations, and we compare the proportion with ADHD in each group. (a) Define the relevant parameter(s) and state the null and alternative hypotheses
null: p1=p2 Alternative: p1>p2
State the null and alternative hypotheses for the statistical test described below Testing to see if there is evidence that the correlation between two variables is negative.
null: p= 0 Alternative: p<0
A reporter on cnn.com stated in July 2010 that 95% of all court cases that go to trial result in a guilty verdict. To test the accuracy of this claim, we collect a random sample of 2000 court cases that went to trial and record the proportion that resulted in a guilty verdict. State the null and alternative hypotheses.
null: p=.95 Alternative: p does not equal .95
State the null and alternative hypotheses for the statistical test described below Testing to see if there is evidence that a correlation between height and salary is significant (that is, different than zero.)
null: p=0 alternative: p does not equal 0
Is the following an experiment or an observational study? To examine whether eating brown rice affects metabolism, we ask a random sample of people whether they eat brown rice and we also measure their metabolism rate.
observational, since nothing is being manipulated only observed and then recorded
Do the following data come from an experiment or an observational study? A gene called ACTN3 encodes a protein which functions in fast-twitch muscles. Some people have a variant of this gene that cannot yield this protein. (So we might call the gene variant a possible non-sprinting gene.) To address the question of whether this gene is associated with sprinting ability, geneticists tested people from three different groups: world-class sprinters, world-class marathon runners, and a control group of non-athletes. In the samples tested, 6% of the sprinters had the gene variant, compared with 18% of the non-athletes and 24% of the marathon runners. This study1 suggests that sprinters are less likely than non-sprinters to have the gene variant.
observational, since they can manipulate the gene
matched pairs experiment
two groups, each get both treatments nut in a random order
bootstrap
we sample with replacement from the original sample to get a new sample of the same size
