1. Categorizing Inference Questions

2. 1. Infants who cry easily may be more easily stimulated than others
1. Infants who cry easily may be more easily stimulated than others. This may be a sign of higher IQ. Child development researchers explored the relationship between the crying of infants 4 to 10 days old and their later IQ test scores. A snap of a rubber band on the sole of the foot caused the infants to cry. The researchers recorded the crying and measured its intensity by the number of peaks in the most active 20 seconds. They later measured the children’s IQ at age 3 years. The table below contains data from a random sample of 38 infants. Do these data provide convincing evidence that there is a positive linear relationship between the cry counts and IQ in the population of infants?

3. A) bivariate – positive linear relationship
The researchers recorded the crying intensity by the number of peaks in the most active 20 seconds. They later measured the children’s IQ at age 3 years. The table below contains data from a random sample of 38 infants. Do these data provide convincing evidence that there is a positive linear relationship between the cry counts and IQ in the population of infants?
A) bivariate – positive linear relationship
B) the true slope of the relationship between cry counts and IQ in the population of infants
C) 1 sample of 38 infants
D) T-test for slope of a regression line t = 𝑏1 −𝛽 𝑠𝑏1
E) Assumptions:
SRS, relationship is linear, responses vary normally about the LSRL, standard deviation of the residuals is constant

4. C) one sample of 10 wells - two measurements at each well
2. Trace metals found in wells affect the taste of drinking water, and high concentrations can pose a health risk. Researchers measured the concentration of zinc (in milligrams per liter) near the top and the bottom of 10 randomly selected wells in a large region. The data are provided in the table below. Construct and interpret a 95% confidence interval for the mean difference in the zinc concentrations from these two locations in the wells.
A) means
B) the mean difference in the zinc concentrations from these two locations in the wells
C) one sample of 10 wells - two measurements at each well

5. 𝑥 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 ± 𝑡 ∗ 𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑛 E) Assumptions
Researchers measured the concentration of zinc (in milligrams per liter) near the top and the bottom of 10 randomly selected wells in a large region. The data are provided in the table below. Construct and interpret a 95% confidence interval for the mean difference in the zinc concentrations from these two locations in the wells.
D) This will be a one-sample t-interval for matched pairs. There are two measurements for each well. Subtract the matching measurements and use the difference data to do the t-interval
𝑥 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 ± 𝑡 ∗ 𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑛
E) Assumptions
σ unknown, use t
random sample of wells - given
population of wells > 100
will have to check graph of difference data to see if a normal approximation is valid since sample size is too small.

6. 3. Researchers designed a survey to compare the proportions of children who come to school without eating breakfast in two low-income elementary schools. An SRS of 80 students from School 1 found 19 had not eaten breakfast. At School 2, an SRS of 150 students included 26 who had not had breakfast. More than 1500 students attend each school. Do these data give convincing evidence of a difference in the population proportions?
A) proportions
B) the true difference in the population proportions of children who come to school without eating breakfast in two low income elementary schools
C) two samples

7. Researchers designed a survey to compare the proportions of children who come to school without eating breakfast in two low-income elementary schools. An SRS of 80 students from School 1 found 19 had not eaten breakfast. At School 2, an SRS of 150 students included 26 who had not had breakfast. More than 1500 students attend each school. Do these data give convincing evidence of a difference in the population proportions?
D) this will be a two proportion z test for difference in proportions 𝑧= 𝑝 1 − 𝑝 2 −(𝑝1−𝑝2) 𝑝(1−𝑝) 𝑛1 + 1 𝑛2
Remember the plain “p” is the pooled p, putting the two samples together
E) Assumptions
SRS of 80 students from one school and independent SRS of 150 students in second school- given
n1p1=19 n1(1-p1) = 61; n2p2 = 26 n2(1-p2) = 124 since all are greater than 10 we can use a normal approximation
population of students in school 1 > 800 and school 2 >1500 – question states both populations >1500

8. 4. Bottles of a popular cola are supposed to contain 300 milliliters of cola. There is some variation from bottle to bottle because the filling machinery is not perfectly precise. From experience, the distribution of the contents of the bottles is approximately normal. An inspector measures the contents of six randomly selected bottles from a single day’s production. Do these data provide convincing evidence that the mean amount of cola in all the bottles filled that day differs from the target value of 300 ml?
A) means
B) the mean amount of cola in all the bottles filled that day
C) one sample

9. D) This will be a one-sample t-test for means t = 𝑥 − 𝜇 𝑠 𝑛
From experience, the distribution of the contents of the bottles is approximately normal. An inspector measures the contents of six randomly selected bottles from a single day’s production. Do these data provide convincing evidence that the mean amount of cola in all the bottles filled that day differs from the target value of 300 ml?
D) This will be a one-sample t-test for means t = 𝑥 − 𝜇 𝑠 𝑛
E) Assumptions:
σ unknown, use t
Random sample of bottles from day’s production - given
Contents of bottles given to be approximately normally distributed - given
Population of bottles in one day’s production > 60

10. 5. Some doctors have begun to use medical magnets to treat patients with chronic pain. Scientists wondered whether this type of therapy really worked, so they designed an experiment to find out. Fifty patients with chronic pain were recruited for the study. A doctor identified a painful site on each patient and asked him or her to rate the pain on a scale from 0 to 10. Then, the doctor selected a sealed envelope containing a magnet from a box that contained both active and inactive magnets. The chosen magnet was applied to the site of the pain for 45 minutes. After treatment, each patient was again asked to rate the level of pain from 0 to 10. In all, 29 patients were given active magnets and 21 patients received inactive magnets. All but one of the patients rated their initial pain as 8, 9, or 10, so scientists decided to focus on the patients’ final pain ratings. Do these data show statistical evidence to suggest that the active magnets help reduce pain?

11. In all, 29 patients were given active magnets and 21 patients received inactive magnets. All but one of the patients rated their initial pain as 8, 9, or 10, so scientists decided to focus on the patients’ final pain ratings. Do these data show statistical evidence to suggest that the active magnets help reduce pain?
A) This question doesn’t explicitly say whether it’s proportions or means, but there are no percentages, and we could find the mean pain rating, so it is means.
B) The true reduction(difference) in mean pain rating for patients using magnets as opposed to a placebo
C) Two samples – even though there were 50 patients originally, they get split into two separate treatments. Since whether they get the magnet or not is random, we can say the two samples are independent

12. In all, 29 patients were given active magnets and 21 patients received inactive magnets. All but one of the patients rated their initial pain as 8, 9, or 10, so scientists decided to focus on the patients’ final pain ratings. Do these data show statistical evidence to suggest that the active magnets help reduce pain?
D) Two-sample t-test for difference of means t = 𝑥 1 − 𝑥 2 − 𝜇1 − 𝜇 𝑠1 2 𝑛1 + 𝑠2 2 𝑛2
E) Assumptions:
σ unknown, use t
Random sample of 29 patients given magnets and independent sample of 21 patients with inactive magnets – given
Reasonable to assume true population of patients with chronic pain is greater than 210 and 290
Since each sample size is less than 30, we will need to look at graphs of the pain ratings for each group to determine if we can use a normal approximation

13. C) One sample of students
6. Tonya wants to estimate what proportion of the seniors in her school plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom.
A) proportions
B) The true proportion of seniors at Tonya’s school who plan to go to the prom
C) One sample of students
D) One-proportion z-interval since we are estimating the parameter 𝑝 ± 𝑧 ∗ 𝑝 (1− 𝑝 ) 𝑛
E) Assumptions:
SRS of seniors - given
np = 36, n(1-p) = 14. Since both are >10 we can use normal approx.
Population of seniors = 750 which is greater than 500 (10n)

14. 7. A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of 200 of the over 2500 students at the school. In all, 83 students say that they approve of the new parking arrangement. Is this evidence that the change was effective?
A) Proportions – you have a % and fraction 83/200
B) The true difference in proportion of students satisfied with the parking situation before and after the change.
C) Two samples of students

15. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of 200 of the over 2500 students at the school. In all, 83 students say that they approve of the new parking arrangement. Is this evidence that the change was effective?
D) two-proportion z-test for difference of proportions z = 𝑝 1 − 𝑝 2 − 𝑝1 −𝑝2 𝑝(1−𝑝) 𝑛 𝑛2 2
E) Assumptions
Random sample of students before change (not given) and SRS of students after change - given
n1p1 = (2500)(.37) = 925 n1(1-p1) = 1575
n2p2 = 83 n2(1-p2) = 117
Since all these values > 10 we can use a normal approximation
Population of students is 2500 which is greater than 2000 (10n)

16. 8. Here are data on the time (in minutes) Professor Moore takes to swim 2000 yards and his pulse rate (beats per minute) after swimming on a random sample of 23 days. Is there statistically significant evidence of a linear relationship between Professor Moore’s swim time and his pulse rate in the population of days on which he swims 2000 yards?
A) bivariate data
B) the true slope of the relationship between Mr. Moore’s swim time and his pulse rate
C) one sample of 23 days
D) t-test for slope of LSRL t = 𝑏1 −𝛽 𝑠𝑏1
E) Same assumptions as question #1

17. B) the true mean healing rate of skin wounds C) one sample
9. Biologists studying the healing of skin wounds measured the rate at which new cells closed a cut made in the skin of an anesthetized newt. Here are data from a random sample of 18 newts, measured in micrometers per hour. We want to estimate the mean healing rate with 95% confidence.
A) means
B) the true mean healing rate of skin wounds
C) one sample
D) one-sample t-interval for means 𝑥 ± 𝑡 ∗ 𝑠 𝑛
E) Assumptions:
σ unknown, use t
Random sample of newts – given
Population of newts > 180
Since the sample size is <30 we would need to look at a graph of the data to determine if we can use a normal approximation

18. 10. Breast-feeding mothers secrete calcium into their milk
10. Breast-feeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers compared a random sample of 47 breast-feeding women with a random sample of 32 women of similar age who were neither pregnant nor lactating. They measured the percent change in the bone mineral content of the women’s spines over three months. Comparative data is given below. Is the mean change in bone mineral content significantly lower for the mothers who are breast-feeding?
A) means
B) the true difference in the mean change of bone mineral content between breast feeding women and those not breastfeeding

19. Researchers compared a random sample of 47 breast-feeding women with a random sample of 32 women of similar age who were neither pregnant nor lactating. They measured the percent change in the bone mineral content of the women’s spines over three months. Comparative data is given below. Is the mean change in bone mineral content significantly lower for the mothers who are breast-feeding?
C) Two samples
D) Two-sample t-test for difference of means t = 𝑥 1 − 𝑥 2 − 𝜇1 − 𝜇 𝑠1 2 𝑛1 + 𝑠2 2 𝑛2
E) Assumptions
Neither σ is known, so use t
Random sample of 47 breast-feeding women and random sample of 32 women not breast-feeding – given
Both sample sizes are greater than 30, so we can use a normal approximation
Population of breast-feeding women > 470 and population of women not breast-feeding > 320

20. 11. Some doctors argue that “normal” human body temperature is not really 98.6oF. One researcher took the oral temperature reading for each of 130 randomly chosen, healthy 18- to 40-year olds. The mean temperature was 98.25oF, with a standard deviation of 0.73oF. Do these data provide convincing evidence that normal body temperature is not 98.6oF?
A) means
B) the true mean temperature of healthy 18- to 40-year-olds
C) one sample
D) one-sample t-test for means t = 𝑥 −𝜇 𝑠 𝑛

21. E) Assumptions σ unknown, use t
One researcher took the oral temperature reading for each of 130 randomly chosen, healthy 18- to 40-year olds. The mean temperature was 98.25oF, with a standard deviation of 0.73oF. Do these data provide convincing evidence that normal body temperature is not 98.6oF?
E) Assumptions
σ unknown, use t
Random sample of healthy 18- to 40-year-olds – given
Since the sample size is greater than 30, we can use a normal approximation
Population of 18- to 40-year-olds > 1300

22. 12. A study followed a random sample of 8474 people with normal blood pressure for about four years. All the individuals were free of heart disease at the beginning of the study. Each person took a test which measures how prone a person is to sudden anger. Researchers also recorded whether each individual developed coronary heart disease. Do the data provide convincing evidence of an association between anger level and heart disease in the population of interest?
A) categorical data
B) whether there is an association between anger level and heart disease in the population of adults
C) one sample – two variables

23. D) Chi-square test of independence E) Assumptions
A study followed a random sample of 8474 people with normal blood pressure for about four years. All the individuals were free of heart disease at the beginning of the study. Each person took a test which measures how prone a person is to sudden anger. Researchers also recorded whether each individual developed coronary heart disease. Do the data provide convincing evidence of an association between anger level and heart disease in the population of interest?
D) Chi-square test of independence
𝑋 2 = 𝑜𝑏𝑠−𝑒𝑥𝑝 2 𝑒𝑥𝑝
E) Assumptions
random sample of people – given
Data are counts
All expected counts are greater than 5 – we would need to actually check the individual counts, but considering the sample is so large, this condition will probably be met

24. 13. A drug manufacturer claims that less than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300 out of 5000 Alzheimer’s patients whose families have given informed consent for the patients to participate in the study. In all, 25 of the subjects experience nausea.
A) proportions
B) the true proportion of Alzheimer’s patients experiencing nausea when taking a drug
C) one sample of patients
D) one sample z-test of proportions z = 𝑝 −𝑝 𝑝(1−𝑝) 𝑛

25. E) Assumptions Random sample of 300 Alzheimer’s patients - given
A drug manufacturer claims that less than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300 out of 5000 Alzheimer’s patients whose families have given informed consent for the patients to participate in the study. In all, 25 of the subjects experience nausea.
E) Assumptions
Random sample of 300 Alzheimer’s patients - given
n1p1 = n1(1-p1) Since both of these values are >10 we can use a normal approximation
Population of Alzheimer’s patients > 3000

26. C) one sample of students
14, Glenn wonders what proportion of students at his school think that tuition is too high. He interviews an SRS of 50 of the 240 students at his college. Thirty-eight of those interviewed think tuition is too high.
A) proportions
B) the true proportion of students at Glenn’s school who think that tuition is too high
C) one sample of students
D) Since Glenn is trying to determine the proportion, or estimate the proportion, this will be a confidence interval – z-interval for one sample proportion 𝑝 ± 𝑧 ∗ 𝑝 (1− 𝑝 ) 𝑛
E) Assumptions
SRS of students - given
np = n(1-p) = 12 Since both values > 10 we can use a normal approx.
Population of students is 240, which is not 10n!!! We can’t assume independence in our sample. We can proceed, but our results may not be reliable.

27. 15. Market researchers suspect that background music may affect the mood and buying behavior of customers. One study in a supermarket compared three randomly assigned treatments: no music, French accordion music, and Italian string music. Under each condition, the researchers recorded the numbers of bottles of French, Italian, and other wine purchased. Are the distributions of wine purchases under the three music treatments similar or different?
A) categorical data
B) if the population of customers has the same buying habits with different music treatments
C) 3 samples (treatment groups)

28. D) Chi-square test of homogeneity E) Assumptions
One study in a supermarket compared three randomly assigned treatments: no music, French accordion music, and Italian string music. Under each condition, the researchers recorded the numbers of bottles of French, Italian, and other wine purchased. Are the distributions of wine purchases under the three music treatments similar or different?
D) Chi-square test of homogeneity
𝑋 2 = 𝑜𝑏𝑠−𝑒𝑥𝑝 2 𝑒𝑥𝑝
E) Assumptions
Random sample of supermarket customers - given
Data are counts
All expected counts > 5 – would need the data to check this

29. 16. A surprising number of young adults (ages 19 to 25) still live in their parents’ homes. The National Institutes of Health planned to estimate the difference in proportions of women and men in this age group who live at home. The random sample included 2253 men and 2629 women in this age group. The survey found that 986 of the men and 923 of the women lived with their parents.
A) proportions
B) the true difference in proportions of men and women ages 19 – 25 who still live with their parents
C) two samples – men and women

30. The National Institutes of Health planned to estimate the difference in proportions of women and men in this age group who live at home. The random sample included 2253 men and 2629 women in this age group. The survey found that 986 of the men and 923 of the women lived with their parents.
D) two-sample z-interval for difference of proportions 𝑝 1 − 𝑝 2 ± 𝑧 ∗ 𝑝 1 (1− 𝑝 1 ) 𝑛1 + 𝑝 2 (1− 𝑝 2 ) 𝑛2
E) Assumptions
Random sample of year old men and random sample of year old women - given
n1p1 = n1(1-p1) = 1267
n2p2 = n2(1 – p2) = 1706
Since all these values are >10, I can use a normal approximation
Population of year old men > and population of year old women > 2629

31. 17. The Wade Tract Preserve in Georgia is an old-growth forest of long-leaf pines that has survived in a relatively undisturbed state for hundreds of years. One question of interest to foresters who study the area is “How do the sizes of long-leaf pine trees in the northern and southern halves of the forest compare?” To find out, researchers took random samples of 30 trees from each half of the forest and measured the trees’ diameter in centimeters. What is the difference in mean diameters of long-leaf pines in the northern and southern halves?
A) mean – it’s not proportions, and we can find the mean diameter
B) the true difference of the mean diameters of trees in the northern and southern halves of the Wade Tract Preserve

32. To find out, researchers took random samples of 30 trees from each half of the forest and measured the trees’ diameter in centimeters. What is the difference in mean diameters of long-leaf pines in the northern and southern halves?
C) two samples of 30 trees
D) We want to find/estimate the difference, so we use a two – sample t-interval for the difference of means 𝑥 1 − 𝑥 2 ± 𝑡 ∗ 𝑠1 2 𝑛1 + 𝑠2 2 𝑛2
E) Assumptions:
Random sample of trees from northern part and independent random sample of trees from southern part of the forest - given
Since both sample sizes are 30, we can use a normal approximation
Population of the trees in each the northern and southern parts of the forest are > 300
σ unknown, use t

33. 18. Environmentalists, government officials, and vehicle manufacturers are all interested in studying the auto exhaust emissions produced by motor vehicles. The major pollutants in auto exhaust from gasoline engines are hydrocarbons, carbon monoxide, and nitrogen oxides (NOX). Researchers collected data on the NOX levels (in grams per mile) for a random sample of 40 light-duty engines of the same type. The mean NOX reading was and the standard deviation was
A) means
B) The true mean NOX level of the population of light-duty engines
C) one sample

34. Researchers collected data on the NOX levels (in grams per mile) for a random sample of 40 light-duty engines of the same type. The mean NOX reading was and the standard deviation was
D) We’re not testing a claim, so we are going to estimate the true mean using a one-sample t-interval for means 𝑥 ± 𝑡 ∗ 𝑠 𝑛
E) Assumptions
Random sample of light-duty engines – given
Sample size 40 > 30 so we can use a normal approximation
Population of light-duty engines > 400
σ unknown, use t

35. 19. Do experienced computer game players earn higher scores when they play with someone present to cheer them on or when they play alone? Fifty teenagers who are experienced at playing a particular computer game have volunteered for a study. We randomly assign 25 of them to play the game alone and the other 25 to play the game with a supporter present. Each player’s score is recorded.
A) means (mean score)
B) the true difference in the mean scores of players with and without a supporter present
C) two samples (or randomly assigned groups)

36. D) two – sample t-test of difference of means
Fifty teenagers who are experienced at playing a particular computer game have volunteered for a study. We randomly assign 25 of them to play the game alone and the other 25 to play the game with a supporter present. Each player’s score is recorded.
D) two – sample t-test of difference of means
t = 𝑥 1 − 𝑥 2 − 𝜇1 − 𝜇 𝑠1 2 𝑛1 + 𝑠2 2 𝑛2
E) Assumptions:
Randomly assigned players in each group – with and without a supporter present - given
Both sample sizes, 25, are less than 30 so we will need to look at the data to determine if we can use a normal approximation
Population of computer game players >
σ unknown, use t

37. 20. As part of the Pew Internet and American Life Project, researchers conducted two surveys in late The first survey asked a random sample of 800 U.S. teens about their use of social media and the Internet. A second survey posed the same questions to a random sample of 2253 U.S. adults. In these two studies, 73% of teens and 47% of adults said that they use social-networking sites. Construct and interpret a 95% confidence interval for the difference in the proportion of all US teens and adults who use social-networking sites.
A) proportions
B) the true difference in the proportion of US teens and adults who use social-networking sites

38. The first survey asked a random sample of 800 U. S
The first survey asked a random sample of 800 U.S. teens about their use of social media and the Internet. A second survey posed the same questions to a random sample of 2253 U.S. adults. In these two studies, 73% of teens and 47% of adults said that they use social-networking sites. Construct and interpret a 95% confidence interval for the difference in the proportion of all US teens and adults who use social-networking sites.
C) two samples
D) two-sample z-interval for difference of proportions 𝑝 1 − 𝑝 2 ± 𝑧 ∗ 𝑝 1 (1− 𝑝 1 ) 𝑛1 + 𝑝 2 (1− 𝑝 2 ) 𝑛2
E) Assumptions
Random sample of 800 US teens and independent random sample of 2253 US adults – given
n1p1 = (800)(.73)=584 n1(1-p1)=216
n2p2 = (2253)(.47)=1059 n2(1-p2)=1194
Since all these values > 10, we can use a normal approximation
Population of US teens > 8000 and population of US adults > 22530