1. What is the level of measurement for each of these variables (from the 2002 General Social Survey) (nominal, ordinal, interval/ratio)? Briefly justify your answer. Note: Ignore the response categories of NAP, DK, and NA for purposes of this question.

CHILDS – How many children have you ever had?

0-6. Actual Number; 7. Seven or more; 9. DK or NA

Response TOTAL %
0) None 799 28.9
1) One 469 17
2) Two 657 23.8
3) Three 481 17.4
4) Four 185 6.7
5) Five 73 2.6
6) Six 40 1.4
7) Seven 22 0.8
8) Eight or more 34 1.2
9) No answer 5 0.2
TOTAL 2765 100

Ordinal. At first glance, one might think the levels of measurement with these variables is interval/ratio, since a zero means an absence of children and there are equal distances between each measure. There are no “half-children”, and if you have four children you have twice as many than if you only have two. However, the level of measurement is actually ordinal for this question since there is a grouping of responses at ‘Eight or more’ children. There’s no difference if you have 8, 9 or 15 children for this grouping, but there is a difference between this grouping and any response less than eight, so it would not be considered nominal either.

EDUC – Highest year of school completed

0-20. Actual number of years; 9. DK or NA

Response TOTAL %
0) No formal schooling 5 0.2
1) 1st grade 2 0.1
2) 2nd grade 15 0.5
3) 3rd grade 2 0.1
4) 4th grade 6 0.2
5) 5th grade 8 0.3
6) 6th grade 25 0.9
7) 7th grade 14 0.5
8) 8th grade 66 2.4
9) 9th grade 60 2.2
10) 10th grade 88 3.2
11) 11th grade 137 5
12) 12th grade 818 29.6
13) 1 year of college 265 9.6
14) 2 years 374 13.5
15) 3 years 157 5.7
16) 4 years 377 13.6
17) 5 years 93 3.4
18) 6 years 128 4.6
19) 7 years 43 1.6
20) 8 years 70 2.5
98) Don’t know 5 0.2
99) No answer 7 0.3
TOTAL 2765 100

Levels of measurement here are interval/ratio as a zero level indicates no years of schooling and the intervals equal in length. Four years of schooling is twice as long as 2 years of schooling.

2. A student discovers that his grade on a recent test was in the 72nd percentile. If 90 students took the test, then approximately how many students received a higher grade than he did?

Approximately 25 students

90-(90 * .72)
## [1] 25.2

3. A sample of 7 underweight babies was fed a special diet and the following weight gains (lbs) were observed at the end of three months.

6.7 2.7 2.5 3.6 3.4 4.1 4.8

Find the mean and standard deviation for these 7 babies. Show your calculations (credit will not be given if you simply provide the answers).

First, we find the mean (u), then we substract each variable from the mean (x-u), then square each

x=c(6.7,2.7,2.5,3.6,3.4,4.1,4.8)       # Entering the Dataset into a Vector

u=sum(x)/length(x)                      # Calculate the Mean

u                                       # Print the Mean
## [1] 3.971429
x x-u (x-u)^2
6.7 2.7 7.4
2.7 -1.3 1.6
2.5 -1.5 2.2
3.6 -0.4 0.1
3.4 -0.6 0.3
4.1 0.1 0.0
4.8 0.8 0.7

We then sum those up, divide by number of variables minus one to calculate the variance, and take the square root to get the standard deviation.

Variance = sum((x-u)^2)/(length(x)-1)    # Calculate Variance

sqrt(Variance)                          # Calculate and print Standard Deviation
## [1] 1.437259

4. The Nielsen Company publishes information on the TV-viewing habits of Americans in Nielsen Report on Television. A sample of 20 people yielded the weekly viewing times, in hours, in the table below.

a) Determine the median of these data.

set1=c(25,41,27,32,43,66,35,31,15,5,34,26,32,38,16,30,38,30,20,21) # Dataset into a Vector
n=length(set1)              # Obtain number of variables
u=sum(set1)/length(set1)    # Mean
sorted = sort(set1)         # Sort the data into a new vector
                            # The median is the average of the two middle values, 30 and 31
medianvalues = sorted[(length(set1)/2)] + (sorted[(length(set1)/2)]+1)
medianvalues / 2 
## [1] 30.5

b) The quartiles of these data.

# Q1 = 25th Percentile
quantile(sorted, 0.25)
## 25% 
##  24
# Q3 = 75th Percentile
quantile(sorted, 0.75)
##   75% 
## 35.75
# IQR = Q3 - Q1   <-- The interquartile range
IQR(sorted)
## [1] 11.75

c) Obtain the lower and upper limits.

Lower Limit

24 - 1.5*(IQR(sorted))               # Lower Limit = Q1 – 1.5 × IQR
## [1] 6.375

Upper Limit

35.75 + 1.5*(IQR(sorted))            # Upper Limit = Q3 + 1.5 × IQR
## [1] 53.375

d) Determine potential outliers, if any.

Potential outliers are 5 and 66 since they are outside the range of 6.375 and 53.375. A boxplot can easily visualize this, and shows that 66 is indeed an outlier.

5. Use the Area under the standard normal curve table to obtain the shaded area under the standard normal curves below.

a. Area between z = -1.28 and 1.28

The area under the curve for z = -1.28 is 0.1003. Multiply by two, then substract from 1: 1 - (2 *(0.1003)) = 0.7994 or 79.94% of the area under the normal distribution curve.

b. Area between z = -1.64 and 1.64

The area under the curve for z =-1.64 is 0.0505. Similar to case a: multiply by two, then substract from 1: 1 - (2 *(0.0505)) = 0.899 or 89.90% of the area under the normal distribution curve.

c. Area below z = -1.96 and above 1.96

The area under the curve for z = -1.96 is 0.0250, so the area would be double that: 0.05, or 5% of the total area under the normal distribution curve.

d. Area below z = -2.33 and 2.33

The area under the curve for z = -2.33 is 0.0102, so the area would be double that: 0.0204, or 2.04% of the total area under the normal distribution curve.