5 Problems with Solutions - Statistical Analysis - Assignment 1 | STAT 200, Assignments of Statistics

Material Type: Assignment; Class: Statistical Analysis; Subject: Statistics; University: University of Illinois - Urbana-Champaign; Term: Unknown 1989;

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Pre 2010

Uploaded on 03/11/2009

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Stat 200
Homework 1 Solutions
1.4 Use R as you would a calculator to find numeric answers to the following:
1. 1 + 2(3 + 4)
2. 4
3
+ 3
2+1
3.
(4 +3)(2 +1)
4.
1+2
3+4
2
Solutions
1. > 1 + 2 * (3 + 4)
[1] 15
2. > 4^3 + 3^(2+1)
[1] 91
3. > sqrt((4+3)*(2+1))
[1] 4.582576
4. > ((1+2) / (3+4))^2
[1] 0.1836735
1.9
The asking price of unused MINI Coopers varies from seller to seller. An online
classifieds listing has these values in thousands:
15.9 21.4 19.9 21.9 20.0 16.5 17.9 17.5
1. What is the smallest amount? The largest?
2. Find the average amount.
3. Find the differences of the largest and smallest amounts from the mean.
Enter in the data and apply one of R’s functions to find answers to the above questions.
Solutions
> cooper <- c(15.9,21.4,19.9,21.9,20.0,16.5,17.9,17.5)
1. > range(cooper)
[1] 15.9 21.9
2. > mean(cooper)
[1] 18.875
3. > range(cooper) - mean(cooper)
[1] -2.975 3.025
1.12
Create the following sequences:
1. “a”, “a”, “a”, “a”, “a”
2. 1, 3, …, 99 (the odd numbers in [1,100])
pf3

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Stat 200

Homework 1 Solutions

1.4 Use R as you would a calculator to find numeric answers to the following:

  1. 1 + 2(3 + 4)

  2. 43 + 32+

  3. (4 + 3)(2 + 1)

2

Solutions

  1. 1 + 2 * (3 + 4) [1] 15

  2. 4^3 + 3^(2+1) [1] 91

  3. sqrt((4+3)*(2+1)) [1] 4.

  4. ((1+2) / (3+4))^ [1] 0.

1.9 The asking price of unused MINI Coopers varies from seller to seller. An online classifieds listing has these values in thousands: 15.9 21.4 19.9 21.9 20.0 16.5 17.9 17.

  1. What is the smallest amount? The largest?
  2. Find the average amount.
  3. Find the differences of the largest and smallest amounts from the mean. Enter in the data and apply one of R’s functions to find answers to the above questions.

Solutions

cooper <- c(15.9,21.4,19.9,21.9,20.0,16.5,17.9,17.5)

  1. range(cooper) [1] 15.9 21.

  2. mean(cooper) [1] 18.

  3. range(cooper) - mean(cooper) [1] -2.975 3.

1.12 Create the following sequences:

  1. “a”, “a”, “a”, “a”, “a”
  2. 1, 3, …, 99 (the odd numbers in [1,100])

using : seq(), or rep() as appropriate.

Solutions

  1. rep("a", 5) [1] "a" "a" "a" "a" "a"

  2. seq(1, 100, by=2) [1] 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 [24] 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 [47] 93 95 97 99

  3. rep(c(1,2,3), c(3,3,3)) [1] 1 1 1 2 2 2 3 3 3

  4. rep(c(1,2,3), c(3,2,1)) [1] 1 1 1 2 2 3

  5. c(seq(1,5), seq(4,1)) [1] 1 2 3 4 5 4 3 2 1

1.23 For the data set treering, which contains tree-ring widths in dimensionless units, use an R function to answer the following:

  1. How many observations are there?
  2. Find the smallest observation.
  3. Find the largest observation.
  4. How many are bigger than 1.5?

Solutions

  1. length(treering) [1] 7980

  2. min(treering) [1] 0 which(treering == 0) [1] 1395

  3. max(treering) [1] 1. which(treering == 1.908) [1] 2185

  4. sum(treering > 1.5) [1] 219