2.1 Homework

7a) China had the most internet users in 2010

7b) The U.K had approx. 50 million users

7c) China had approx. 350 million more users than Germany

7d) Since China’s population is so big, it will look like they had so many more users. Using the relative frequency will show internet users relative to each population

9a) ~69% believed divorce is morally acceptable

9b) 55,200,000 believed it was morally wrong (assuming there were 240 mill american adults in 2010)

9c) inferential statement

11a) .43 (43%) of 18-34 year olds were more likely to by if American made – .61 (61%) of 35-44 year olds were more likely to buy if American made

11b) 55+ age group has the largest proportion that is more likely to buy if American made

11c) the apparent association is that the older the age group, the more likely it is that they will want to buy something if it is American made

13a) Response Frequency Relative Frequency Never 125 0.0262
Rarely 324 0.0678 Sometimes 552 0.1156 Most times 1257 0.2632 Always 2518 0.5272 4776 (total)

13b) 52.7% of respondents answered “always”

13c) A total of 9.4% of respondents answered “never” or “rarely”

13d-f)

my_data <- c(125, 324, 552, 1257, 2518)

groups <- c("never", "rarely", "sometimes", "most times", "always")

barplot(my_data, main = "How often do you wear your seatbelt in someone else's car?", names.arg = groups)

barplot(my_data, main = "How often do you wear your seatbelt in someone else's car?", names.arg = groups, col = c("red", "blue", "green", "yellow", "orange"))

rel_freq <- my_data / sum(my_data)

barplot(rel_freq, main = "How often do you wear your seatbelt in someone else's car?", names.arg = groups, col = c("red", "blue", "green", "yellow", "orange"))

pie(my_data, labels = groups, main = "How often do you wear your seatbelt in someone else's car?")

13g) Inferential statement

15a) Response Frequency Relative Frequency More than 1 hr a day 377 0.3678 Up to 1 hr a day 192 0.1873 Few times a week 132 0.1288 Few times a month or less 81 0.0790 Never 243 0.2370 1025 (total)

15b) 23.7% of people surveyed never use the internet

15c-e)

my_data <- c(377, 192, 132, 81, 243)

groups <- c("More than 1 hour a day", "Up to 1 hour a day", "A few times a week", "A few times a month or less", "Never")

barplot(my_data, main = "Use the Internet?", names.arg = groups)

barplot(my_data, main = "Use the Internet?", names.arg = groups, col = c("red", "blue", "green", "yellow", "orange"))

rel_freq <- my_data / sum(my_data)

barplot(rel_freq, main = "Use the Internet?", names.arg = groups, col = c("red", "blue", "green", "yellow", "orange"))

pie(my_data, labels = groups, main = "Use the Internet?")

15f) This statement doesn’t have a confidence level

2.2 Homework

9a) most frequent outcome was a value of 8

9b) least frequent outcome was a value of 2

9c) a value of 7 was observed 15 times

9d) 5’s were observed four more times than 4’s

9e) 15% of the time 7’s were observed

9f) left skewed

10a) 4 is the most frequent number of cars sold in a week

10b) 9 weeks occured where 2 cars were sold

10c) 17.3% of the time 2 cars were sold

10d) right skewed

11a) 200 students sampled

11b) class wedth = 10

11c classes and frequencies: 60-69, 2; 70-79, 3; 80-89, 13; 90-99, 42; 100-109, 58; 110-119, 40; 120-129, 31; 130-139, 8; 140-149, 2; 150-159, 1

11d) IQ of 100-109 had the highest frequency

11e) 150-159 had the lowest frequency

11f) 5% of studets had an IQ of at least 130

11g) No one had an IQ of 165

12a) class width = 200

12b) classes and frequencies: 0-199, 26; 200-399, 14; 400-599, 7; 600-799, 0; 800-999, 0; 1000-1199, 2; 1200-1399, 0; 1400-1599, 1

12c) class of 0-199 has the highest frequency

12d) skewed right

12e) the right thing to do would be to compare the number of alcohol-related traffic deaths (15 in VT and 1463 in TX) to the total number of licensed drivers to give a more accurate proportion

13a) right skewed - because with income there are going to be less people making more money so the tail descending on the right side of the graph shows that as the income increase, the people making that kind of money decreases.

13b) bell curve - with SAT scores, you will have a small amount of people do extremely will, a very larg majority of people in in the middle who did average, and a small amount of people who did very badly. So the graph will hit peak frequency in the middle

13c) right skewed - with number of people in a household/family, the national average is somewhere around 4. The frequency will peak there, with a small amount of people who have less than 4 or up to 4, then less and less people will have families of 5, 6, 7, 8, 9 and so on people.

13d) left skewed - with ages of people diagnosed with Alzheimer’s disease, the number of people diagnosed at a young to middle age is very small then steadily increases as people get much older. The frequency will peak on the far right part of the graph at the oldest ages

14a) right skewed - with the number of drinks people consume in a week, the frequency of people who consume a low number of drinks a week will be higher than those who drink heavily. The frequency will be highest on the left side of the graph and the tail will descend to the right

14b) uniform - in general, the ages of kids in a public school distric per grade is about the same. There are roughly the same number of kids entering each grade each year

14c) left skewed - with ages of people who need hearing-aids, the frequency will peak with the older age groups. The tail will ascend up increasing frequency as you more further to the right side of the graph

14d) bell - the height of full grown men is a bell curve because the average height will be the highest frequency. Some men will be very short or very tall, but mostly men will fall in the middle

hist(iris$Sepal.Length)