Datasets:
bnolton
/
AP_Stat_Inference_Questions

Dataset card Data Studio Files
xet
Community
1
Question
stringlengths
35
1.85k
Answer
stringlengths
1
1.44k
⌀
A hotel chain is interested in evaluating reservation processes. Guests can reserve a room by using either a telephone system or an online system that is accessed through the hotel’s web site. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. Based on these data, is it reasonable to conclude that the proportion who are satisfied is greater for those who reserve a room online? Test the appropriate hypotheses using a significance level of 0.05.
H0: pr - po = 0, Ha: pr – po < 0, z = -1.667, P-value = 0.048, reject H0. There is convincing evidence that the proportion who are satisfied is higher for those who reserve a room online.
“Smartest People Often Dumbest About Sunburns” is the headline of an article that appeared in the San Luis Obispo Tribune (July 19, 2006). The article states that “those with a college degree reported a higher incidence of sunburn than those without a high school degree—43% versus 25%.” Suppose that these percentages were based on random samples of size 200 from each of the two groups of interest (college graduates and those without a high school degree). Is there convincing evidence that the proportion experiencing a sunburn is greater for college graduates than it is for those without a high school degree? Test the appropriate hypotheses using a significance level of 0.01.
H0: pc - ph = 0, Ha: pc – ph < 0, z = 3.800, P-value ≈ 0, reject H0. There is convincing evidence that the proportion experiencing sunburn is higher for college graduates than for those without a high school degree.
The report "Young People Living on the Edge" (Greenberg Quinlan Rosner Research, 2008) summarizes a survey of people in two independent random samples. One sample consisted of 600 young adults (ages 19 to 35), and the other sample consisted of 300 parents of young adults ages 19 to 35. The young adults were presented with a variety of situations (such as getting married or buying a house) and were asked if they thought that their parents were likely to provide financial support in that situation. The parents of young adults were presented with the same situations and asked if they would be likely to provide financial support in that situation. When asked about getting married, 41% of the young adults said they thought parents would provide financial support and 43% of the parents said they would provide support. Carry out a hypothesis test to determine if there is convincing evidence that the proportion of young adults who think their parents would provide financial support and the proportion of parents who say they would provide support are different.
H0: p1 – p2 = 0, Ha: p1 – p2 ≠ 0, z = -0.574, P-value = 0.566. Fail to reject H0. There is not convincing evidence of a difference between the proportion of young adults who think that their parents would provide financial support for marriage and the proportion of parents who say they would provide financial support for marriage.
In December 2001, the Department of Veterans' Affairs announced that it would begin paying benefits to soldiers suffering from Lou Gehrig's disease who had served in the Gulf War (The New York Times, December 11, 2001). This decision was based on an analysis in which the Lou Gehrig's disease incidence rate (the proportion developing the disease) for the approximately 700,000 soldiers sent to the Persian Gulf between August 1990 and July 1991 was compared to the incidence rate for the approximately 1.8 million other soldiers who were not in the Gulf during this time period. Based on these data, explain why it is not appropriate to perform a hypothesis test in this situation and yet it is still reasonable to conclude that the incidence rate is higher for Gulf War veterans than for those who did not serve in the Gulf War.
Since the values given are population characteristics, an inference procedure is not applicable. It is known that the rate of Lou Gehrig’s disease among soldiers sent to the war is higher than for those not sent to the war.
The news release referenced in the previous exercise also included data from independent samples of teenage drivers and parents of teenage drivers. In response to a question asking if they approved of laws banning the use of cell phones and texting while driving, 74% of the teens surveyed and 95% of the parents surveyed said they approved. The sample sizes were not given in the news release, but suppose that 600 teens and 400 parents of teens were surveyed and that these samples are representative of the two populations. Do the data provide convincing evidence that the proportion of teens who approve of banning cell phone and texting while driving is less than the proportion of parents of teens who approve? Test the relevant hypotheses using a significance level of 0.05.
H0: p1 – p2 = 0, Ha: p1 – p2 < 0, z = -8.543, P-value ≈ 0, reject H0. There is convincing evidence that the proportion of teens who approve of the proposed laws is less than the proportion of parents of teens who approve.
The report “Audience Insights: Communicating to Teens (Aged 12–17)” (www.cdc.gov, 2009) described teens’ attitudes about traditional media, such as TV, movies, and newspapers. In a representative sample of American teenage girls, 41% said newspapers were boring. In a representative sample of American teenage boys, 44% said newspapers were boring. Sample sizes were not given in the report. Suppose that the percentages reported were based on samples of 58 girls and 41 boys. Is there convincing evidence that the proportion who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using α = 0.05.
H0: pG – pB = 0, Ha: pG – pB < 0, z = -0.298, P-value = 0.766, reject H0, fail to reject H0. There is not convincing evidence that the proportion who think that newspapers are boring is different for teenage girls and boys.
The report “Audience Insights: Communicating to Teens (Aged 12–17)” (www.cdc.gov, 2009) described teens’ attitudes about traditional media, such as TV, movies, and newspapers. In a representative sample of American teenage girls, 41% said newspapers were boring. In a representative sample of American teenage boys, 44% said newspapers were boring. Sample sizes were not given in the report. Suppose that the percentages reported were based on samples of 2,000 girls and 2,500 boys. Is there convincing evidence that the proportion who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using α = 0.05.
H0: pG – pB = 0, Ha: pG – pB ≠ 0, z = -2.022, P-value = 0.043, reject H0. There is convincing evidence that the proportion who think that newspapers are boring is different for teenage girls and boys.
A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 9.
100, 3.333
A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 15.
100, 2.582
A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 36.
100, 1.667
A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 50.
100, 1.414
A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 100.
100, 1.000
A random sample is selected from a population with mean µ = 100 and standard deviation σ = 10. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 400.
100, 0.500
The article “How Business Students Spend Their Time—Do They Really Know?” (Research in Higher Education Journal [May 2009]: 1–10) describes a study of 212 business students at a large public university. Each student kept a log of time spent on various activities during a 1-week period. For these 212 students, the mean time spent studying was 9.66 hours and the sample standard deviation was 6.62 hours. Suppose that this sample was a random sample from the population of all business majors at this university and that you are interested in learning about the value of µ, the mean time spent studying for this population. How do you know the sampling distribution of x-bar is centered at the actual (but unknown) value of the population mean.
The mean of the sampling distribution of x-bar is equal to the population mean µ.
The article “How Business Students Spend Their Time—Do They Really Know?” (Research in Higher Education Journal [May 2009]: 1–10) describes a study of 212 business students at a large public university. Each student kept a log of time spent on various activities during a 1-week period. For these 212 students, the mean time spent studying was 9.66 hours and the sample standard deviation was 6.62 hours. Suppose that this sample was a random sample from the population of all business majors at this university and that you are interested in learning about the value of µ, the mean time spent studying for this population. How do you know an estimate of the standard deviation of x-bar, which describes how the x-bar values spread out around the population mean µ, is 0.455.
σx = σ / sqrt(n) = 6.62 / sqrt(212)
The article “How Business Students Spend Their Time—Do They Really Know?” (Research in Higher Education Journal [May 2009]: 1–10) describes a study of 212 business students at a large public university. Each student kept a log of time spent on various activities during a 1-week period. For these 212 students, the mean time spent studying was 9.66 hours and the sample standard deviation was 6.62 hours. Suppose that this sample was a random sample from the population of all business majors at this university and that you are interested in learning about the value of µ, the mean time spent studying for this population. How do you know the sampling distribution of x-bar is approximately normal.
The sampling distribution of x-bar is approximately normal because the sample size (n = 212) is greater than 30.
Explain the difference between µ and µx.
The quantity µ is the population mean, while µx is the mean of the x distribution. It is the mean value of x-bar for all possible random samples of size n.
The time that people have to wait for an elevator in an office building has a uniform distribution over the interval from 0 to 1 minute. For this distribution, µ = 0.5 and σ = 0.289. Let x-bar be the average waiting time for a random sample of 16 waiting times. What are the mean and standard deviation of the sampling distribution of x-bar?
µx = µ = 0.5, σx = σ / sqrt(n) = 0.289 / sqrt(16) = 0.07225
The time that people have to wait for an elevator in an office building has a uniform distribution over the interval from 0 to 1 minute. For this distribution, µ = 0.5 and σ = 0.289. Let x-bar be the average waiting time for a random sample of 50 waiting times. What are the mean and standard deviation of the sampling distribution of x-bar?
When  n = 50, µx = µ = 0.5, and σx = σ / sqrt(n) = 0.289 / sqrt(50) = 0.07225. Since n ≥ 30, the distribution of x-bar is approximately normal
A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 12.
200, 4.330
A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 20.
200, 3.354
A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 25.
200, 3.000
A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 40.
200, 2.371
A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 90.
200, 1.581
A random sample is selected from a population with mean µ = 200 and standard deviation σ = 15. Determine the mean and standard deviation of the x-bar sampling distribution for a sample size of n = 300.
200, 0.866
Explain the difference between x-bar and µx.
x-bar is the mean of a single sample, while µ is the mean of the x-bar distribution. It is the mean value of x-bar for all possible random samples of size n.
Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm. If the distribution of interpupillary distance is normal and a random sample of n = 25 adult males is to be selected, what is the probability that the sample mean distance x-bar for these 25 will be between 64 and 67 mm? At least 68 mm?
0.8185, 0.0013
Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm. Suppose that a random sample of 100 adult males is to be selected. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and 67 mm? At least 68 mm?
0.9772, 0.0000
A manufacturing process is designed to produce bolts with a diameter of 0.5 inches. Once each day, a random sample of 36 bolts is selected and the bolt diameters are recorded. If the resulting sample mean is less than 0.49 inches or greater than 0.51 inches, the process is shut down for adjustment. The standard deviation of bolt diameters is 0.02 inches. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an x-bar in the shutdown range when the actual process mean is 0.5 inches.)
P(0.49 < x-bar < 0.51 = 0.9974: the probability that the manufacturing line will be shut down unnecessarily is 1 – 0.9974 = 0.0026.
What percentage of the time will a variable that has a t distribution with 10 degrees of freedom fall between −1.81 and 1.81
0.9
What percentage of the time will a variable that has a t distribution with 24 degrees of freedom fall between −2.06 and 2.06
0.95
What percentage of the time will a variable that has a t distribution with 24 degrees of freedom fall outside the interval from −2.80 to 2.80
0.01
What percentage of the time will a variable that has a t distribution with 10 degrees of freedom fall to the left of −1.81
0.05
What is the appropriate t critical value a 95% confidence, with a sample size of n = 17? 
0.02
What is the appropriate t critical value a 99% confidence, with a sample size of n = 24? 
0.03
What is the appropriate t critical value a 90% confidence, with a sample size of n = 13?
0.02
The two intervals (114.4, 115.6) and (114.1, 115.9) are confidence intervals for µ = mean resonance frequency (in hertz) for all tennis rackets of a certain type. The two intervals were computed using the same sample data. What is the value of the sample mean resonance frequency?
1.15
The two intervals (114.4, 115.6) and (114.1, 115.9) are confidence intervals for µ = mean resonance frequency (in hertz) for all tennis rackets of a certain type. The two intervals were computed using the same sample data. The confidence level for one of these intervals is 90%, and for the other it is 99%. Which is which, and how can you tell?
The 99% confidence interval is wider than the 90% confidence interval. The 90% interval is (114.4, 115.6), and the 99% interval is (114.1, 115.9).
In June, 2009, Harris Interactive conducted its Great Schools Survey. In this survey, the sample consisted of 1,086 adults who were parents of school-aged children. The sample was selected to be representative of the population of parents of school-aged children. One question on the survey asked respondents how much time per month (in hours) they spent volunteering at their children’s school during the previous school year. The following summary statistics for time volunteered per month were given: n = 1,086, x-bar = 5.6, median = 1. What does the fact that the mean is so much larger than the median tell you about the distribution of time spent volunteering at school per month?
That the mean is much greater than the median suggests that the distribution of times spent volunteering in the sample was positively skewed.
In June, 2009, Harris Interactive conducted its Great Schools Survey. In this survey, the sample consisted of 1,086 adults who were parents of school-aged children. The sample was selected to be representative of the population of parents of school-aged children. One question on the survey asked respondents how much time per month (in hours) they spent volunteering at their children’s school during the previous school year. The following summary statistics for time volunteered per month were given: n = 1,086, x-bar = 5.6, median = 1. Explain why it is not reasonable to assume that the population distribution of time spent volunteering is approximately normal.
With the sample mean much greater than the sample median, and with the sample regarded as representative of the population, it seems very likely that the population is strongly positively skewed and, therefore, not normally distributed.
In June, 2009, Harris Interactive conducted its Great Schools Survey. In this survey, the sample consisted of 1,086 adults who were parents of school-aged children. The sample was selected to be representative of the population of parents of school-aged children. One question on the survey asked respondents how much time per month (in hours) they spent volunteering at their children’s school during the previous school year. The following summary statistics for time volunteered per month were given: n = 1,086, x-bar = 5.6, median = 1. Explain why it is appropriate to use the one-sample t confidence interval to estimate the mean time spent volunteering for the population of parents of school-aged children even though the population distribution is not approximately normal.
Since n = 1086 ≥ 30, the sample size is large enough for the one-sample t confidence interval to be appropriate, even though the population distribution is not approximately normal.
In June, 2009, Harris Interactive conducted its Great Schools Survey. In this survey, the sample consisted of 1,086 adults who were parents of school-aged children. The sample was selected to be representative of the population of parents of school-aged children. One question on the survey asked respondents how much time per month (in hours) they spent volunteering at their children’s school during the previous school year. The following summary statistics for time volunteered per month were given: n = 1,086, x-bar = 5.6, median = 1. Suppose that the sample standard deviation was s = 5.2 . Use the five-step process for estimation problems (EMC3) to compute and interpret a 98% confidence interval for µ, the mean time spent volunteering for the population of parents of school-aged children.
(5.232, 5.968). You can be 98% confident that the mean time spent volunteering for the population of parents of school-age children is between 5.232 and 5.968 hours.
In a study of academic procrastination, the authors of the paper “Correlates and Consequences of Behavioral Procrastination” (Procrastination, Current Issues and New Directions [2000]) reported that for a sample of 411 undergraduate students at a mid-size public university, the mean time spent studying for the final exam in an introductory psychology course was 7.74 hours and the standard deviation of study times was 3.40 hours. Assume that this sample is representative of students taking introductory psychology at this university. Construct a 95% confidence interval estimate of µ, the mean time spent studying for the introductory psychology final exam.
(7.411, 8.069). You can be 95% confident that the mean time spent studying for the final exam is between 7.411 and 8.069 hours.
The paper referenced in the previous exercise also gave the following sample statistics for the percentage of study time that occurred in the 24 hours prior to the final exam: n = 411, x-bar = 43.18, s = 21.46 Construct and interpret a 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the final exam.
(41.439, 44.921). You can be 90% confident that the mean percentage of study time that occurs in the 24 hours prior to the final exam is between 41.439% and 44.921%.
The Bureau of Alcohol, Tobacco, and Firearms (BATF) has been concerned about lead levels in California wines. In a previous testing of wine specimens, lead levels ranging from 50 to 700 parts per billion were recorded. How many wine specimens should be tested if the BATF wishes to estimate the mean lead level for California wines with a margin of error of 10 parts per billion?
Using (sample range)/4 = 162.5 as an estimate of the population standard deviation, a sample size of 1,015 is needed
The article “Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes” (Ipsos-Reid, October 24, 2005) summarized a survey of 1,000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween. The resulting sample mean and standard deviation were $46.65 and $83.70, respectively. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense.
This could happen if the sample distribution was very positively skewed.
The article “Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes” (Ipsos-Reid, October 24, 2005) summarized a survey of 1,000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween. The resulting sample mean and standard deviation were $46.65 and $83.70, respectively. Is it reasonable to think that the distribution of anticipated Halloween expense is approximately normal? Explain why or why not.
No. Since the sample distribution is very skewed to the right, it is very unlikely that the variable anticipated Halloween expense is approximately normally distributed.
The article “Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes” (Ipsos-Reid, October 24, 2005) summarized a survey of 1,000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween. The resulting sample mean and standard deviation were $46.65 and $83.70, respectively. Is it appropriate to use the one-sample t confidence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not.
Yes, since the sample is large.
The article “Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes” (Ipsos-Reid, October 24, 2005) summarized a survey of 1,000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween. The resulting sample mean and standard deviation were $46.65 and $83.70, respectively. If appropriate, construct and interpret a 99% confidence interval for the mean anticipated Halloween expense for Canadian residents.
(39.821, 53.479). You can be 99% confident that the mean anticipated Halloween expense for the population of Canadian residents is between 39.821 and 53.479 dollars.
What is the appropriate t critical value for a 95% confidence and a sample size of n = 15. 
2.15
What is the appropriate t critical value for a 99% confidence and a sample size of n = 20.  
2.86
What is the appropriate t critical value for a 90% confidence and a sample size of n = 26.
1.71
Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were: 6 17 11 22 29 Assuming that these five students can be considered as representative of all students participating in the free checkup program, construct a 95% confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program.
(5.776, 28.224). You can be 95% confident that the mean number of months elapsed since the last visit for the population of students participating in the program is between 5.776 and 28.224.
Use the information given in the previous exercise to construct a 95% confidence interval for the mean number of partners on a mating flight for queen honeybees. For purposes of this exercise, assume these 30 queen honeybees are representative of the population of queen honeybees.
(3.301, 5.899). You can be 95% confident that the mean number of partners on a mating flight is between 3.301 and 5.899.
Give as much information as you can about the P-value of a t test with the following information: Upper-tailed test, df = 8, t = 2.0
0.04
Give as much information as you can about the P-value of a t test with the following information: Lower-tailed test, df = 10, t = -2.4
0.019
Give as much information as you can about the P-value of a t test with the following information: Lower-tailed test, n = 22, t = -4.2
0
Give as much information as you can about the P-value of a t test with the following information: Two-tailed test, df = 15, t = -1.6
0.13
The paper “Playing Active Video Games Increases Energy Expenditure in Children” (Pediatrics [2009]: 534–539) describes a study of the possible cardiovascular benefits of active video games. Mean heart rate for healthy boys ages 10 to 13 after walking on a treadmill at 2.6 km/hour for 6 minutes is known to be 98 beats per minute (bpm). For each of 14 boys, heart rate was measured after 15 minutes of playing Wii Bowling. The resulting sample mean and standard deviation were 101 bpm and 15 bpm, respectively. Assume that the sample of boys is representative of boys ages 10 to 13 and that the distribution of heart rates after 15 minutes of Wii Bowling is approximately normal. Does the sample provide convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is different from the known mean heart rate after 6 minutes walking on the treadmill? Carry out a hypothesis test using α = 0.05.
H0: µ = 98, Ha: µ > 98, t = 0.748, P-value = 0.468, fail to reject H0. There is not convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is different from the mean after 6 minutes of walking on a treadmill.
The paper “Playing Active Video Games Increases Energy Expenditure in Children” (Pediatrics [2009]: 534–539) describes a study of the possible cardiovascular benefits of active video games. Mean heart rate for healthy boys ages 10 to 13 after walking on a treadmill at 2.6 km/hour for 6 minutes is known to be 98 beats per minute (bpm). For each of 14 boys, heart rate was measured after 15 minutes of playing Wii Bowling. The resulting sample mean and standard deviation were 101 bpm and 15 bpm, respectively. Assume that the sample of boys is representative of boys ages 10 to 13 and that the distribution of heart rates after 15 minutes of Wii Bowling is approximately normal. The paper also states that the known resting mean heart rate for boys in this age group is 66 bpm. Is there convincing evidence that the mean heart rate after Wii Bowling for 15 minutes is higher than the known mean resting heart rate for boys of this age? Use α = 0.01.
H0: µ = 66, Ha: µ > 66, t = 8.731, P-value ≈ 0, reject H0. There is convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is higher than the mean resting heart rate.
The Economist collects data each year on the price of a Big Mac in various countries around the world. A sample of McDonald’s restaurants in Europe in May 2009 resulted in the following Big Mac prices (after conversion to U.S. dollars): 3.80 5.89 4.92 3.88 2.65 5.57 6.39 3.24 The mean price of a Big Mac in the U.S. in May 2009 was $3.57. For purposes of this exercise, you can assume it is reasonable to regard the sample as representative of European McDonald’s restaurants. Does the sample provide convincing evidence that the mean May 2009 price of a Big Mac in Europe is greater than the reported U.S. price? Test the relevant hypotheses using α=0.05.
H0: µ = 3.57, Ha: µ > 3.57, t = 2.042, P-value = 0.040, reject H0. There is convincing evidence that the mean price of a Big Mac in Europe is higher than $3.57.
In a study of computer use, 1,000 randomly selected Canadian Internet users were asked how much time they spend online in a typical week (Ipsos Reid, August 9, 2005). The sample mean was 12.7 hours. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent online by Canadians in a typical week is greater than 12.5 hours.
H0: µ = 12.5, Ha: µ > 12.5, t = 1.26, P-value = 0.103, fail to reject H0. There is not convincing evidence that the mean time spent online is greater than 12.5 hours.
In a study of computer use, 1,000 randomly selected Canadian Internet users were asked how much time they spend online in a typical week (Ipsos Reid, August 9, 2005). The sample mean was 12.7 hours. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of 0.05 to decide if there is convincing evidence that the mean time spent online by Canadians in a typical week is greater than 12.5 hours.
H0: µ = 12.5, Ha: µ > 12.5, t = 3.16, P-value = 0.0008, reject H0. There is convincing evidence that the mean time spent online is greater than 12.5 hours.
The paper titled "Music for Pain Relief" (The Cochrane Database of Systematic Reviews, April 19, 2006) concluded, based on a review of 51 studies of the effect of music on pain intensity, that "Listening to music reduces pain intensity levels ... However, the magnitude of these positive effects is small, and the clinical relevance of music for pain relief in clinical practice is unclear." Are the authors of this paper claiming that the pain reduction attributable to listening to music is not statistically significant, not practically significant, or neither statistically nor practically significant? Explain.
By saying that listening to music reduces pain levels, the authors are telling you that the study resulted in convincing evidence that pain levels are reduced when music is being listened to. (In other words, the results of the study were statistically significant.) By saying, however, that the magnitude of the positive effects was small, the authors are telling you that the effect was not practically significant.
The eating habits of 12 vampire bats were examined in the article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica [1991]: 63–67). These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was x-bar = 21.9 minutes. Suppose that the standard deviation was s = 7.7  minutes. Is there convincing evidence that the mean supper time of a vampire bat whose meal consists of a frog is greater than 20 minutes? What assumptions must be reasonable for the one-sample t test to be appropriate?
H0: µ = 20, Ha: µ > 20, t = 0.85, P-value = 0.205, fail to reject H0. There is not convincing evidence that the mean suppertime is greater than 2 minutes.
An article titled "Teen Boys Forget Whatever It Was" appeared in the Australian newspaper The Mercury (April 21, 1997). It described a study of academic performance and attention span and reported that the mean time to distraction for teenage boys working on an independent task was 4 minutes. Although the sample size was not given in the article, suppose that this mean was based on a random sample of 50 teenage boys and that the sample standard deviation was 1.4 minutes. Is there convincing evidence that the average attention span for teenage boys is less than 5 minutes? Test the relevant hypotheses using α = 0.01.
H0: µ = 5, Ha: µ < 5, t = -5.051, P-value = 0.000, reject H0. There is convincing evidence that the mean attention span for teenage boys is less than 5 minutes.
An automobile manufacturer decides to carry out a fuel efficiency test to determine if it can advertise that one of its models achieves 30 mpg (miles per gallon). Six people each drive a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: 27.2 29.3 31.2 28.4 30.3 29.6 Assuming that fuel efficiency is normally distributed, do these data provide evidence against the claim that actual mean fuel efficiency for this model is (at least) 30 mpg?
H0: µ = 30, Ha: µ < 30, t = -1.160, P-value = 0.149, fail to reject H0. There is not convincing evidence that the mean fuel efficiency under these circumstances is less than 30 miles per gallon.
A hot tub manufacturer advertises that a water temperature of 100°F can be achieved in 15 minutes or less. A random sample of 25 tubs is selected, and the time necessary to achieve a 100°F temperature is determined for each tub. The sample mean time and sample standard deviation are 17.5 minutes and 2.2 minutes, respectively. Does this information cast doubt on the company's claim? Carry out a test of hypotheses using significance level 0.05.
H0: µ = 15, Ha: µ > 15, t = 5.682, P-value = 0.000, reject H0. There is convincing evidence that the mean time to 100°F is greater than 15 minutes.
Let x denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of the x distribution are µ = 2  minutes and σ = 0.8 minutes, respectively. If x-bar is the sample mean time for a random sample of n = 9 students, where is the x-bar distribution centered, and what is the standard deviation of the x-bar distribution?
µx = 2, σx = 0.267
Let x denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of the x distribution are µ = 2  minutes and σ = 0.8 minutes, respectively. If x-bar is the sample mean time for a random sample of size of n = 20, where is the x-bar distribution centered, and what is the standard deviation of the x-bar distribution? How about for a sample of size n = 100. How do the centers and spreads of the three x-bar distributions (include n = 9) compare to one another? Which sample size would be most likely to result in an x-bar value close to µ, and why?
In each case µi = 2. When n = 20, σx = 0.179, and when n = 100, σx = 0.08 . All three centers are the same, and the larger the sample size, the smaller the standard deviation of x-bar. Since the distribution of x-bar when n = 100 is the one with the smallest standard deviation of the three, this sample size is most likely to result in a value of x-bar close to µ.
Suppose that a random sample of 50 bottles of a particular brand of cough medicine is selected, and the amount of cough medicine in each bottle is determined. Let µ denote the mean amount of cough medicine for the population of all bottles of this brand. Suppose that this sample of 50 results in a 95% confidence interval for µ of (7.8, 9.4). Would a 90% confidence interval have been narrower or wider than the given interval? Explain your answer.
Narrower
Suppose that a random sample of 50 bottles of a particular brand of cough medicine is selected, and the amount of cough medicine in each bottle is determined. Let µ denote the mean amount of cough medicine for the population of all bottles of this brand. Suppose that this sample of 50 results in a 95% confidence interval for µ of (7.8, 9.4). Consider the following statement: There is a 95% chance that µ is between 7.8 and 9.4. Is this statement correct? Why or why not?
The statement is not correct. The population mean, µ, is a constant, and it is not appropriate to talk about the probability that it falls within a certain interval.
Suppose that a random sample of 50 bottles of a particular brand of cough medicine is selected, and the amount of cough medicine in each bottle is determined. Let µ denote the mean amount of cough medicine for the population of all bottles of this brand. Suppose that this sample of 50 results in a 95% confidence interval for µ of (7.8, 9.4). Consider the following statement: If the process of selecting a random sample of size 50 and then computing the corresponding 95% confidence interval is repeated 100 times, exactly 95 of the resulting intervals will include µ. Is this statement correct? Why or why not?
The statement is not correct. You can say that in the long run about 95 out of every 100 samples will result in confidence intervals that will contain µ, but we cannot say that in 100 such samples, exactly 95 will result in confidence intervals that contain µ.
How much money do people spend on graduation gifts? In 2007, the National Retail Federation (www.nrf.com) surveyed 2,815 consumers who reported that they bought one or more graduation gifts that year. The sample was selected to be representative of adult Americans who purchased graduation gifts in 2007. For this sample, the mean amount spent per gift was $55.05. Suppose that the sample standard deviation was $20. Construct and interpret a 98% confidence interval for the mean amount of money spent per graduation gift in 2007.
(54.172, 55.928). You can be 98% confident that the mean amount of money spent per graduation gift in 2007 was between $54.172 and $55.928.
Suppose that the researchers who carried out the study described in the previous exercise wanted to estimate the mean reaction time with a margin of error of 5 msec. Using the sample standard deviation as a preliminary estimate of the population standard deviation, compute the required sample size.
753
Medical research has shown that repeated wrist extension beyond 20 degrees increases the risk of wrist and hand injuries. Each of 24 students at Cornell University used a proposed new computer mouse design, and while using the mouse, each student's wrist extension was recorded. Data consistent with summary values given in the paper "Comparative Study of Two Computer Mouse Designs" (Cornell Human Factors Laboratory Technical Report RP7992) are given. Use these data to test the hypothesis that the mean wrist extension for people using this new mouse design is greater than 20 degrees. Are any assumptions required in order for it to be appropriate to generalize the results of your test to the population of all Cornell students? To the population of all university students?
H0: µ = 20, Ha: µ > 20, t = 14.836, P-value = 0.000, reject H0. There is convincing evidence that the mean wrist extension for all people using the new mouse design is greater than 20 degrees. To generalize the result to the population of Cornell students, you need to assume that the 24 students used in the study are representative of all students at the university. To generalize the result to the population of all university students, you need to assume that the 24 students used in the study are representative of all university students.
In the following study, the two populations of interest are the students at a particular university who live on campus and the students who live off campus. Which of these studies have samples that are independently selected? Study 1: To determine if there is evidence that the mean amount of money spent on food each month differs for the two populations, a random sample of 45 students who live on campus and a random sample of 50 students who live off campus are selected.
Independently selected
In the following study, the two populations of interest are the students at a particular university who live on campus and the students who live off campus. Which of these studies have samples that are independently selected? Study 2: To determine if the mean number of hours spent studying differs for the two populations, a random sample students who live on campus is selected. Each student in this sample is asked how many hours he or she spends working each week. For each of these students who live on campus, a student who lives off campus and who works the same number of hours per week is identified and included in the sample of students who live off campus.
Not independently selected
In the following study, the two populations of interest are the students at a particular university who live on campus and the students who live off campus. Which of these studies have samples that are independently selected? Study 3: To determine if the mean number of hours worked per week differs for the two populations, a random sample of students who live on campus and who have a brother or sister who also attends the university but who lives off campus is selected. The sibling who lives on campus is included in the on campus sample, and the sibling who lives off campus is included in the off-campus sample.
Not independently selected
In the following study, the two populations of interest are the students at a particular university who live on campus and the students who live off campus. Which of these studies have samples that are independently selected? Study 4: To determine if the mean amount spent on textbooks differs for the two populations, a random sample of students who live on campus is selected. A separate random sample of the same size is selected from the population of students who live off campus.
Independently selected
For the following hypothesis testing scenario, indicate whether or not the appropriate hypothesis test would be about a difference in population means. If not, explain why not. Scenario 1: The international polling organization Ipsos reported data from a survey of 2,000 randomly selected Canadians who carry debit cards (Canadian Account Habits Survey, July 24, 2006). Participants in this survey were asked what they considered the minimum purchase amount for which it would be acceptable to use a debit card. You would like to determine if there is convincing evidence that the mean minimum purchase amount for which Canadians consider the use of a debit card to be acceptable is less than $10.
The appropriate test is not about a difference in population means. There is only one sample.
For the following hypothesis testing scenario, indicate whether or not the appropriate hypothesis test would be about a difference in population means. If not, explain why not. Scenario 2: Each person in a random sample of 247 male working adults and a random sample of 253 female working adults living in Calgary, Canada, was asked how long, in minutes, his or her typical daily commute was ("Calgary Herald Traffic Study," Ipsos, September 17, 2005). You would like to determine if there is convincing evidence that the mean commute times differ for male workers and female workers.
The appropriate test is about a difference in population means.
For the following hypothesis testing scenario, indicate whether or not the appropriate hypothesis test would be about a difference in population means. If not, explain why not. Scenario 3: A hotel chain is interested in evaluating reservation processes. Guests can reserve a room using either a telephone system or an online system. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. You would like to determine if it reasonable to conclude that the proportion who are satisfied is higher for those who reserve a room online.
The appropriate test is not about a difference in population means. There are two samples, but the variable is categorical rather than numerical
Do male college students spend more time using a computer than female college students? This was one of the questions investigated by the authors of the paper "An Ecological Momentary Assessment of the Physical Activity and Sedentary Behaviour Patterns of University Students" (Health Education Journal [2010]: 116–125). Each student in a random sample of 46 male students at a university in England and each student in a random sample of 38 female students from the same university kept a diary of how he or she spent time over a three-week period. For the sample of males, the mean time spent using a computer per day was 45.8 minutes and the standard deviation was 63.3 minutes. For the sample of females, the mean time spent using a computer was 39.4 minutes and the standard deviation was 57.3 minutes. Is there convincing evidence that the mean time male students at this university spend using a computer is greater than the mean time for female students? Test the appropriate hypotheses using α = 0.05.
H0: µM - µF = 0, Ha: µM - µF > 0, t = 0.49, P-value = 0.314, fail to reject H0. There is not convincing evidence that the mean time is greater for males than for females.
A study compared students who use Facebook and students who do not use Facebook ("Facebook and Academic Performance," Computers in Human Behavior [2010]: 1237–1245). In addition to asking the students in the samples about GPA, each student was also asked how many hours he or she spent studying each day. The two samples (141 students who were Facebook users and 68 students who were not Facebook users) were independently selected from students at a large, public Midwestern university. Although the samples were not selected at random, they were selected to be representative of the two populations. For the sample of Facebook users, the mean number of hours studied per day was 1.47 hours and the standard deviation was 0.83 hours. For the sample of students who do not use Facebook, the mean was 2.76 hours and the standard deviation was 0.99 hours. Do these sample data provide convincing evidence that the mean time spent studying for Facebook users is less than the mean time spent studying for students who do not use Facebook? Use a significance level of 0.01.
H0: µNF - µSF = 0, Ha: µNF - µSF < 0, t = -9.29, P-value ≈ 0.000, reject H0. There is convincing evidence that the mean time spent studying is less for Facebook users than for those who do not use Facebook.
The paper "Ladies First?" A Field Study of Discrimination in Coffee Shops" (Applied Economics [2008]: 1-19) describes a study in which researchers observed wait times in coffee shops in Boston. Both wait time and gender of the customer were observed. The mean wait time for a sample of 145 male customers was 85.2 seconds. The mean wait time for a sample of 141 female customers was 113.7 seconds. The sample standard deviations (estimated from graphs in the paper) were 50 seconds for the sample of males and 75 seconds for the sample of females. Suppose that these two samples are representative of the populations of wait times for female coffee shop customers and for male coffee shop customers. Is there convincing evidence that the mean wait time differs for males and females? Test the relevant hypotheses using a significance level of 0.05.
H0: µM - µF = 0, Ha: µM - µF ≠ 0, t = -3.77, P-value ≈ 0.000, reject H0. There is convincing evidence that the mean wait time differs for males and females.
Some people believe that talking on a cell phone while driving slows reaction time, increasing the risk of accidents. The study described in the paper "A Comparison of the Cell Phone Driver and the Drunk Driver" (Human Factors [2006]: 381–391) investigated the braking reaction time of people driving in a driving simulator. Drivers followed a pace car in the simulator, and when the pace car's brake lights came on, the drivers were supposed to step on the brake. The time between the pace car brake lights coming on and the driver stepping on the brake was measured. Two samples of 40 drivers participated in the study. The 40 people in one sample used a cell phone while driving. The 40 people in the second sample drank a mixture of orange juice and alcohol in an amount calculated to achieve a blood alcohol level of 0.08% (a value considered legally drunk in most states). For the cell phone sample, the mean braking reaction time was 779 milliseconds and the standard deviation was 209 milliseconds. For the alcohol sample, the mean breaking reaction time was 849 milliseconds and the standard deviation was 228. Is there convincing evidence that the mean braking reaction time is different for the population of drivers talking on a cell phone and the population of drivers who have a blood alcohol level of 0.08%? For purposes of this exercise, you can assume that the two samples are representative of the two populations of interest.
H0: µC - µA = 0, Ha: µC - µA ≠ 0, t = -1.43, P-value = 0.156 , fail to reject H0. There is not convincing evidence that the mean reaction time differs for those talking on a cell phone and those who have a blood alcohol level of 0.08%.
Research has shown that, for baseball players, good hip range of motion results in improved performance and decreased body stress. The article “Functional Hip Characteristics of Baseball Pitchers and Position Players” (The American Journal of Sports Medicine, 2010: 383–388) reported on a study of independent samples of 40 professional pitchers and 40 professional position players. For the pitchers, the sample mean hip range of motion was 75.6 degrees and the sample standard deviation was 5.9 degrees, whereas the sample mean and sample standard deviation for position players were 79.6 degrees and 7.6 degrees, respectively. Assuming that the two samples are representative of professional baseball pitchers and position players, test hypotheses appropriate for determining if mean range of motion for pitchers is less than the mean for position players.
H0: µpf - µpp = 0, Ha: µpf - µpp < 0, t = -2.63, P-value = 0.005, for α = 0.05 or 0.01 reject H0. There is convincing evidence that the mean range of motion is less for pitchers than for position players.
The article “Plugged In, but Tuned Out” (USA Today, January 20, 2010) summarizes data from two surveys of kids ages 8 to 18. One survey was conducted in 1999 and the other was conducted in 2009. Data on number of hours per day spent using electronic media, consistent with summary quantities in the article, are given below (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, you can assume that the two samples are representative of kids ages 8 to 18 in each of the 2 years when the surveys were conducted. 2009: 5 9 5 8 7 6 7 9 7 9 6 9 10 9 8 1999: 4 5 7 7 5 7 5 6 5 6 7 8 5 6 6 Because the given sample sizes are small, what assumption must be made about the distributions of electronic media use times for the two-sample t test to be appropriate? Use the given data to construct graphical displays that would be useful in determining whether this assumption is reasonable. Do you think it is reasonable to use these data to carry out a two-sample t test?
Since boxplots are roughly symmetrical and since there are no outliers in either sample, the assumption of normality is plausible, and it is reasonable to carry out a two-sample t test.
The article “Plugged In, but Tuned Out” (USA Today, January 20, 2010) summarizes data from two surveys of kids ages 8 to 18. One survey was conducted in 1999 and the other was conducted in 2009. Data on number of hours per day spent using electronic media, consistent with summary quantities in the article, are given below (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, you can assume that the two samples are representative of kids ages 8 to 18 in each of the 2 years when the surveys were conducted. 2009: 5 9 5 8 7 6 7 9 7 9 6 9 10 9 8 1999: 4 5 7 7 5 7 5 6 5 6 7 8 5 6 6 Do the given data provide convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999? Test the relevant hypotheses using a significance level of 0.01.
H0: µ2009 - µ1999 = 0, Ha: µ2009 - µ1999 > 0, t = 3.32, P-value = 0.001, reject H0. There is convincing evidence that the mean time spent using electronic media was greater in 2009 than in 1999.
In a study of malpractice claims where a settlement had been reached, two random samples were selected: a random sample of 515 closed malpractice claims that were found not to involve medical errors and a random sample of 889 claims that were found to involve errors (New England Journal of Medicine [2006]: 2024–2033). The following statement appeared in the paper: “When claims not involving errors were compensated, payments were significantly lower on average than were payments for claims involving errors ($313,205 vs. $521,560, P = 0.004).” What hypotheses did the researchers test to reach the stated conclusion?
µ1 = mean payment for claims not involving errors; µ2 = mean payment for claims involving errors; H0: µ1 - µ2 = 0, Ha: µ1 - µ2 < 0.
In a study of malpractice claims where a settlement had been reached, two random samples were selected: a random sample of 515 closed malpractice claims that were found not to involve medical errors and a random sample of 889 claims that were found to involve errors (New England Journal of Medicine [2006]: 2024–2033). The following statement appeared in the paper: “When claims not involving errors were compensated, payments were significantly lower on average than were payments for claims involving errors ($313,205 vs. $521,560, P = 0.004).” Which of the following could have been the value of the test statistic for the hypothesis test? Explain your reasoning. i. t = 5.00 iii. t = 2.33 ii. t = 2.65 iv. t = 1.47
Since the samples are large, a t distribution with a large number of degrees of freedom is used, which can be approximated with the standard normal distribution. P(z > 2.65 = 0.004 , which is the P-value given. None of the other possible values of t gives the correct P-value.
The paper “Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children in a National Household Survey” (Pediatrics [2004]: 112–118) investigated the effect of fast-food consumption on other dietary variables. For a representative sample of 663 teens who reported that they did not eat fast food during a typical day, the mean daily calorie intake was 2,258 and the sample standard deviation was 1,519. For a representative sample of 413 teens who reported that they did eat fast food on a typical day, the mean calorie intake was 2,637 and the standard deviation was 1,138. Use the given information and a 95% confidence interval to estimate the difference in mean daily calorie intake for teens who do eat fast food on a typical day and those who do not.
(−538.606, −219.394). You can be 95% confident that the mean daily calorie intake for teens who do not eat fast food on a typical day minus the mean daily calorie intake for teens who do eat fast food on a typical day is between −538.606 and −219.394.
In the study described in the paper “Exposure to Diesel Exhaust Induces Changes in EEG in Human Volunteers” (Particle and Fibre Toxicology [2007]), 10 healthy men were exposed to diesel exhaust for 1 hour. A measure of brain activity (called median power frequency, or MPF) was recorded at two different locations in the brain both before and after the diesel exhaust exposure. The resulting data are given in the accompanying table. For purposes of this exercise, assume that the sample of 10 men is representative of healthy adult males. Construct and interpret a 90% confidence interval estimate for the difference in mean MPF at brain location 2 before and after exposure to diesel exhaust.
(−3.069, −0.791). You can be 90% confident that the mean difference in MPF at brain location 1 is between −3.069 and −0.791.
In the study described in the paper “Exposure to Diesel Exhaust Induces Changes in EEG in Human Volunteers” (Particle and Fibre Toxicology [2007]), 10 healthy men were exposed to diesel exhaust for 1 hour. A measure of brain activity (called median power frequency, or MPF) was recorded at two different locations in the brain both before and after the diesel exhaust exposure. The resulting data are given in the accompanying table. For purposes of this exercise, assume that the sample of 10 men is representative of healthy adult males. Construct and interpret a 90% confidence interval estimate for the difference in mean MPF at brain location 2 before and after exposure to diesel exhaust.
(−2.228, −0.852). You can be 90% confident that the mean difference in MPF at brain location 1 is between −2.228 and −0.852.
Do girls think they don’t need to take as many science classes as boys? The article “Intentions of Young Students to Enroll in Science Courses in the Future: An Examination of Gender Differences” (Science Education [1999]: 55–76) describes a survey of randomly selected children in grades 4, 5, and 6. The 224 girls participating in the survey each indicated the number of science courses they intended to take in the future, and they also indicated the number of science courses they thought boys their age should take in the future. For each girl, the authors calculated the difference between the number of science classes she intends to take and the number she thinks boys should take. Explain why these data are paired.
There is a single sample of girls, with two data values for each girl.
Do girls think they don’t need to take as many science classes as boys? The article “Intentions of Young Students to Enroll in Science Courses in the Future: An Examination of Gender Differences” (Science Education [1999]: 55–76) describes a survey of randomly selected children in grades 4, 5, and 6. The 224 girls participating in the survey each indicated the number of science courses they intended to take in the future, and they also indicated the number of science courses they thought boys their age should take in the future. For each girl, the authors calculated the difference between the number of science classes she intends to take and the number she thinks boys should take. The mean of the differences was -0.83 (indicating girls intended, on average, to take fewer science classes than they thought boys should take), and the standard deviation was 1.51. Construct and interpret a 95% confidence interval for the mean difference.
(−1.029, −0.631). You can be 95% confident that the mean difference between the number of science classes a girl intends to take and the number she thinks boys should take is between −1.029 and −0.631.
An individual can take either a scenic route to work or a nonscenic route. She decides that use of the nonscenic route can be justified only if it reduces the mean travel time by more than 10 minutes. If µ1 refers to the mean travel time for scenic route and µ2 to the mean travel time for nonscenic route, what hypotheses should be tested?
H0: µ1 - µ2 = 10 versus Ha: µ1 - µ2 > 10
An individual can take either a scenic route to work or a nonscenic route. She decides that use of the nonscenic route can be justified only if it reduces the mean travel time by more than 10 minutes. If µ1 refers to the mean travel time for nonscenic route and µ2 to the mean travel time for scenic route, what hypotheses should be tested?
H0: µ1 - µ2 = -10 versus Ha: µ1 - µ2 < -10
Indicate whether or not the appropriate hypothesis test would be for a difference in population means for the following scenario. If not, explain why not. Scenario 1: The authors of the paper “Adolescents and MP3 Players: Too Many Risks, Too Few Precautions” (Pediatrics [2009]: e953–e958) studied independent random samples of 764 Dutch boys and 748 Dutch girls ages 12 to 19. Of the boys, 397 reported that they almost always listen to music at a high volume setting. Of the girls, 331 reported listening to music at a high volume setting. You would like to determine if there is convincing evidence that the proportion of Dutch boys who listen to music at high volume is greater than this proportion for Dutch girls.
no