Optional Statistics Practice Sheet Answer and Work Key
1) The following are the ages of a group of students in a college statistics class:
23, 18, 20, 19, 27, 18, 22, 47, 32, 21
Find the following information:
Mean: sum of ages (Σ) ÷ number of people (N)
= 247 ÷ 10 = 24.7
Median: line up all of the ages in ascending order: 18, 18, 19, 20, 21, 22, 23, 27, 32, 47
Find the middle value; in this case it falls between 21 and 22 so the answer is 21.5
Range: subtract the lowest value in the series from the highest value in the series to find the range. =
47 – 18 = 29
Standard deviation:
__
In this case: X = age, X = the mean (24.7), and n = the number of students (10)
Steps 1 & 2: Subtract the mean from each value in the series. Then square each difference.
18 – 24.7 = -6.7
= 44.89
18 – 24.7 = -6.7
= 44.89
19 – 24.7 = -5.7
now
= 32.49
20 – 24.7 = -4.7
square
= 22.09
21 – 24.7 = -3.7
each
= 13.69
22 – 24.7 = - 2.7
of
= 7.29
23 – 24.7 = -1.7
these
= 2.89
27 – 24.7 = 2.3
values
= 5.29
32 – 24.7 = 7.3
= 53.29
47 – 24.7 = 22.3
= 497.29
Step 4: Add up all of the squared differences (Sum of Squares)
44.89 + 44.89 + 32.49 + 22.09 + 13.69 + 7.29 + 2.89 + 5.29 + 53.29 + 497.29 = 724.1
__
So, so far we have found Σ ( X – X )2 which is 724.1
__
Step 5: divide Σ ( X – X )2 by ( n – 1)
724.1 ÷ 9 = 80.455555
Final step:
__
Find the square root of the quotient of Σ ( X – X )2 ÷ ( n – 1)
SQRT of 80.455555 = 8.969702
Standard Deviation = 8.97
2) The ten students earn the following grades on the mid-term exam:
57, 94, 82, 73, 66, 94, 100, 88, 90, 60
Find the following information:
Mean: ____80.4_______
Median: ___85___________
Range: ______43_______
Standard deviation: ___15.39________
57 – 80.4 = -23.4
= 547.56
60 – 80.4 = -20.4
= 416.16
66 – 80.4 = -14.4
now
= 207.36
73 – 80.4 = -7.4
square
= 54.76
82 – 80.4 = 1.6
each
= 2.56
88 – 80.4 = 7.6
of
= 57.76
90 – 80.4 = 9.6
these
= 92.16
94 – 80.4 = 13.6
values
= 184.96
94 – 80.4 = 13.6
= 184.96
100 – 80.4 = 19.6
= 384.16
__
Σ = 2132.4
Σ ( X – X )2 ÷ ( n – 1) = 2132.4 ÷ 9
= 236.933333
SQRT = 15.3926389333776
3) You go on a shopping spree and buy several items of clothing at your favorite store at the
following prices:
$86, $15, $245, $108, $50, $32, $14, $21, $67, $17, $125, $28
Find the following information:
Mean: _____$ 67.30_________
Median: ____$ 41.00___________
Range: _____231____________
Standard deviation: ___$ 67. 59_______
14 – 67.3 = -53.3
= 2840.89
15 – 67.3 = -52.3
= 2735.29
17 – 67.3 = -50.3
now
= 2530.09
21 – 67.3 = -46.3
square
= 2143.69
28 – 67.3 = -39.3
each
= 1544.49
32 – 67.3 = -35.3
of
= 1246.09
50 – 67.3 = -17.3
these
= 299.29
67 – 67.3 = -0.3
values
=
0.09
86 – 67.3 = 18.7
= 349.69
108 – 67.3 = 40.7
= 1656.49
125 – 67.3 = 57.7
= 3329.29
245 – 67.3 = 177.7
= 31577.29
__
Σ = 50252.68
2
Σ ( X – X ) ÷ ( n – 1) = 50252.68 ÷ 11
= 4568.4254545454545
SQRT = 67.590128380891943
4) In a recent survey of men and women on a college campus the following results were obtained:
47 men agree
82 men disagree
92 women agree
70 women disagree
To Solve this problem start by setting up a table with your data:
Men
Women
Row Total
Agree
47
92
139
Disagree
82
70
152
Column Total
129
162
291
Remember that the denominator (bottom) of your fraction will be the number that
represents the phrase that comes directly after “what percent” [see underlined]. Your
numerator (top) will be the number that represents the remaining relevant phrase [see
italicized].
What percent of men disagree? __63.6%____ 82÷129 (column percent)
What percent of women agree? __56.8%____ 92÷162 (column percent)
What percent of respondents agree? __47.8%_ 139÷291 (total percent
What percent of those who agree are women? _66.2%___ 92÷139 (row percent)
What percent of those who agree are men? __33.8%___ 47÷139 (row percent)
What percent of respondents are men who agree? __16.2%___ 47÷291 (total percent)
What percent of those who disagree are men? __54.0%___ 82÷152 (row percent)
5) In a recent survey of men and women on a college campus the following results were obtained:
47 men agree
Men
Women
Row Total
Agree
47
92
139
Disagree
82
70
152
Column Total
129
162
Grand= 291
82 men disagree
92 women agree
70 women disagree
Working off of the “null hypothesis” that there is no difference between men and women:
These are expected frequencies, not percents. The formula for finding an expected
frequency is: (ROW Total x COLUMN Total)/ Grand Total
How many women would we “expect” to agree? ___77.4___ 162 * 139 ÷ 291
How many women would we “expect” to disagree? ___84.6__ 162 * 152 ÷ 291
How many men would we “expect” to agree? ___61.6___ 129 * 139 ÷ 291
How many men would we “expect” to disagree? __67.4__ 129 * 152 ÷ 291
6) What is the margin of error for a sample size of 1,234? __2.9%__
The formula for finding a margin of error for a given sample size is:
m = 1 ÷ √n
m = 1 ÷ √1234
=
1÷ 35.128336
=
0.028467 (change to a percent) = 2.9%
7) What size sample do you need from a population of 200,000 people in order to get a 2% margin
of error? __2500___
The formula for finding a sample size for a given margin of error is:
N = 1 ÷ m2
N = 1 ÷ 0.022 = 1 ÷ 0.02 * 0.02 =
1 ÷ 0.0004 = 2500
8) What size sample do you need for a survey of all Republicans in the United States if you want a
1% margin of error? __10,000__
N = 1 ÷ 0.012 = 1 ÷ 0.01 * 0.01 =
1 ÷ 0.0001 = 10,000
9) You are collaborating on a research project in a city with 500,000 residents. There are 100,000
Caucasians, 150,000 Hispanics, 90,000 African Americans, and 160,000 Asian Americans.
What size sample would you need for a survey if you want a 4% margin of error for each of the
groups? __2500___
You will use the formula for finding a sample size for a given margin of error:
N = 1 ÷ m2
N = 1 ÷ 0.042
= 1 ÷ 0.04 * 0.04 =
1 ÷ 0.0016 = 625
Since you want the margin of error for each of the groups you will need to multiply the
sample size by the number of groups, in this case it is four:
Sample size for a margin of error of 4% = 625 * the number of groups (4)
625 * 4 =
2500
10) What is the margin of error for a sample size of 1,638 people? __2.5%___
11) What is the margin of error for a sample size of 891? __3.4%__
12) If there is a sample of 832 Cherry Hill residents with a mean income of $96,328 with a standard
deviation of $4624, what is the margin of error? __$320.62____
The formula for finding a margin of error for a mean score is:
m = 2 * SD ÷ √n
m = 2 * 4624 ÷ √832
= 9248 ÷ 28.84441
= 320.61671
Answer in dollars = $320.62
13) In a recent survey of men and women who purchased new cars at a Toyota dealership the
following results were obtained:
82 men agree
Men
Women
Row Totals
104 men disagree
Agree
82
97
179
97 women agree
Disagree
104
110
214
Column totals
186
207
393
110 women disagree
What percent of men disagree? __55.9%___
104 ÷ 186
What percent of women agree? __46.9%___
97 ÷ 207
What percent of the respondents agree? __45.6%__
179 ÷ 393
What percent of those who agree are women? __54.2%__
97 ÷ 179
What percent of those who agree are men? __45.8%___
82 ÷ 179
What percent of the respondents are men who agree? __20.9%__
82 ÷ 393
What percent of those who disagree are men? __48.6%__
104 ÷ 214
What percent of the respondents are women? ___52.7%__
207 ÷ 393
What percent of the respondents are women who disagree? __28%__ 110 ÷ 393
14) In a recent survey of men and women who purchased new cars at a Toyota dealership the
following results were obtained:
82 men agree
104 men disagree
97 women agree
110 women disagree
Working off of the “null hypothesis” that there is no difference between men and women:
How many women would we “expect” to agree? __94.3____
How many women would we “expect” to disagree? __112.7___
How many men would we “expect” to agree? __84.7____
How many men would we “expect” to disagree? __101.3____
* Remember: EF = RT * CT ÷ GT
Men
Women
Row Totals
Agree
82
97
179
Disagree
104
110
214
Column totals
186
207
393