Data Analysis Assignment Help

Data Analysis Assignment Help

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Frequency Tables and Measures of Central Tendencies Below are solutions to questions on frequency tables, measures of central tendencies, and other concepts of data analysis using SPSS. Mean, median, and standard deviation are some of the calculated parameters. A histogram pie chart, dot plot, and frequency polygon are constructed to give a visual presentation of the data.

Creating a Detailed Frequency Table Description : The data set represents the number of minutes a sample of 27 people exercise each week 108 139 120 123 120 132 123 131 131 157 150 124 111 101 135 119 16 117 127 128 139 119 118 114 127 142 130 Statisticsassignmenthelp.com Statisticsassignmenthelp.com

1.- Construct a frequency distribution for the data set. Include class limits, midpoints boundaries, frequencies, and cumulative frequency. Describe each graph.

Solution Midp

Frequenc

Class limits

oint

y

Cumulative frequency

100-112

106

3

3

113-125

119

11

14

126-138

132

8

22

139-151

145

4

26

152-164

158

1

27

Generating a Histogram for the Table 2.- Display the data, using a frequency histogram and a frequency polygon.

Histogram

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Histogram 12

11

10 8

8

6

4

4 3

2

0

1 100-112

113-125

126-138

Frequency Polygon

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139-151

152-164

Frequency Polygon 12 10 8 6 4 2 0

100-112

113-125

126-138

3.- Display the data using a stem and leaf plot. count

steam

leaf

2

10

18

7

11

1467899

8

12

00334778

7

13

0112599

1

14

2

2

15

07 Statisticsassignmenthelp.com Statisticsassignmenthelp.com

139-151

152-164

5.- Make a pie chart graph.

1 Pie Chart 152-164100-112 4% 11%

139-151 15%

100-112 113-125 126-138 139-151 152-164

113-125 41%

126-138 30%

Constructing a Dot Plot

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Dot plot 12 10 8 6 4 2 0 0.5

1

1.5

2

2.5

3

3.5

4

Finding the Measures of Central Tendency

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4.5

5

5.5

Class limits

Midpoint

Frequency

fx

X-M

(X-M)^2

f(X-M)^2

100-112

106

3

318

-20.7037

428.6433

1285.93

-7.7037

59.34705

652.8176

130 113-125

119

11

9 105

126-138

132

8

6

5.296296

28.05075

224.406

139-151

145

4

580

18.2963

334.7545

1339.018

152-164

158

1

158

31.2963

979.4582

979.4582

342 Total

27

1

Mean Median =

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4481.63

Therefore, the median will be in the class limit 113-125 Standard

deviation

=

Finding the coefficient of variation and CI Coefficient of variation = 2.Confidence interval for Mean

The confidence interval for the mean is 122.844 and 131.536 Testing the hypothesis 10.- The coach claims that mean time the 27 people exercise is more than125 minutes each week. Can you support the claim, using a level of significance of 0.05?

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Level of significance =5% = 0.05 Test statistics Test statistic Critical value

1.645

Conclusion: since the test statistic is less than the critical, we fail to reject the null hypothesis because there is no enough evidence to support the claim that the mean time of 27 people exercise is more than 125 minutes each week.

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