# Learning Activity: Statistics Exercises Student Template

You MUST show your work to be eligible for partial credit.

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1. (20 Pts, 1 pt each). Calculate the mean, median, mode, standard deviation, and range for the following sets of measurements (fill out the table):
1. 20, 18, 17, 17, 19
2. 15, 10, 7, 6, 4
3. 28, 28, 28, 28, 28
4. 10, 10, 7, 6, 4, 79

Mean – add all numbers up then divide by the amount of numbers there are.

1. 91 divided by 5 =18.2
2. 42 divided by 5 = 8.4
3. 140 divided by 5 = 28
4. 116 divided by 6 = 19.33333

Median – arrange numbers from lowest to highest, if you have an odd amount of numbers you will find median, if you have even set of numbers, add the two middle number than divide by 2.

1. 17, 17, 18, 19, 20 = 18
2. 4, 6, 7, 10, 15 = 7
3. 28, 28, 28, 28, 28 = 28
4. 4, 6, 7, 10, 10, 79 = 7 + 10 = 17 divided by 2 = 8.5

Mode – What number appears the most? Some data won’t have a mode and some data may have two modes.

1. 20, 18, 17, 17, 19 = 17
2. 15, 10, 7, 6, 4 = no mode
3. 28, 28, 28, 28, 28 = 28
4. 10, 10, 7, 6, 4, 79 = 10

SD – First subtract each number in the data set by the mean. Then square answers. Then add all square answers together to get a total number for each data set. Then divide that answer by 1 less number in the data set. So if there are 5 numbers divide the answer by 4 etc. Once you get that total find the square root of it to get your standard deviation.

1. 20 – 18.2 = 1.8 squared = 3.24
• 18 – 18.2 = – 0.2 squared = 0.04
• 17 – 18.2 = – 1.2 squared = 1.44
• 17 – 18.2 = – 1.2 squared =1.44
• 19 – 18.2 = 0.8 squared = 0.64

Total squared answers added together = 6.8 divided by 5 = 1.36 square root = 1.17

1. 15 – 8.4 = 6.6 squared = 43.56
• 10 – 8.4 = 1.6 squared = 2.56
• 7 – 8.4 = – 1.4 squared = 1.96
• 6 – 8.4 = – 2.4 squared = 5.76
• 4 – 8.4 = – 4.4 squared = 19.36

Total squared answers added together = 25.12 divided by 5 = 5.24 square root = 2.29

1. (5 TIMES) 28 – 28 = 0 squared = 0

Total squared answers added together = 0 divided by 5 = 0 square root = 0

1. 10 – 19.33333 = – 9.33333 squared = 87.11
• 10 = 19.33333 = – 9.33333 squared = 87.11
• 7 – 19.33333 = – 12.33333 squared = 152.11
• 6 – 19.33333 = – 13.33333 squared = 177.78
• 4 – 19.33333 = – 15.33333 squared = 235.11
• 79 – 19.33333 = 59.66667 squared = 3,560.11

Total squared answers added together = 4,299.33 divided by 6 = 716.56 square root = 26.77

Range – Subtract the highest value from the lowest.

1. 20 – 17 = 3
2. 15 – 4 = 11
3. 28 – 28 = 0
4. 79 – 4 = 75

 DISTRIB MEAN MEDIAN MODE SD RANGE a. 18.2 18 17 1.17 3 b. 8.4 7 No Mode 2.29 11 c. 28 28 28 0 0 d. 19.33333 8.5 10 26.77 75

1. (20 Pts, 5 pts each) Answer the following questions.

1. Why is the SD in (d) so large compared to the SD in (b)?
• Data values that vary more from the mean have a larger standard deviation and the data is more spread out.

1. Why is the mean so much higher in (d) than in (b)?
• The mean is higher because of the values and not only is the sum larger ; it also has a value of 79.

1. Why is the median relatively unaffected?
• Due to its reliance on the number in the middle of any set of data, the median remains unaffected by number distributions.

1. Which measure of central tendency best represents the set of scores in (d)? Why?

• As a measure of central tendency in (d) the mean best represents the set scores because the mean uses a set of values in order to compute an average value.

1. (4 pts) Determine the semi-interquartile range for the following set of scores.

92        95        89        65        99        100      85        67        72        99        85        100

• 65, 67, 72, 85, 85, 89, 92, 95, 99, 99, 100, 100 = 12 numbers total
• Median = 89 + 92 divided by 2 = 90.5
• Q1 65, 67, 72, 85, 85, 89 = 5
• Q3 92, 95, 99, 99, 100, 100 = 99
• Q3 – Q1 DIVIDED BY 2 = 25 = the semi – interquartile range

1. (24 pts, 2 pts each) Fill in the blanks on the table with the appropriate raw scores, z-scores, T-scores, and approximate percentile ranks. You may refer to the distribution curve below.

Note:  the Mean = 50, SD = 5.

 RAW z T Percentile 40 2.5 35 84.13

1. (6 pts, 3 pts each) The following are the means and standard deviations of some well-known standardized tests, referred to as Test A, Test B, and Test C. All three yield normal distributions.
 Test Mean Standard Deviation Test A 300 75 Test B 250 4 Test C 40 12

1. (3 pts) A score of 275 on Test A corresponds to what score on Test B? ____
2. (3 pts) A score of 400 on Test A corresponds to what score on Test C? ____
3. (12 pts, 2 pts each) The Graduate Record Exam (GRE) has a combined verbal and quantitative mean of 1000 and a standard deviation of 200. Scores range from 200 to 1600 and are approximately normally distributed. For each of the following problems, indicate the percentage or score called for by the problem and select the appropriate distribution curve (from below) that relates to the problem.

1. (2 pts) What percentage of the persons who take the test score below 600? ___

1. (2 pts) What percentage of the persons who take the test score below 1200? ___

1. (2 pts) Above what score do the top 2.27% of the test-takers score? ___

1. (14 pts, varied) Refer to the following data and scatterplots to respond to questions 7a-e.
 Individual Years of School Body Mass Index A 21 18 B 18 20 C 17 33 D 17 29 E 14 31 F 11 32 G 22 19 H 23 21 I 16 33 J 22 36 K 17 30 L 15 28 M 17 20 N 12 28 O 14 33 P 13 29

Figure A represents a scatterplot constructed from the data; Figure B represents a regression line drawn through the scatterplot that “fits” the data points reasonably well; Figure C represents an ellipse drawn around the data points.

1. (2 pts.) What is the overall direction of the correlation? ___

1. (2 pts.) Estimate the strength of the correlation coefficient: ___

Consider Figure D (below).

1. (2 pts.) Using only the data points associated with the years of school above 16; what effect does this have on the direction and strength of the correlation coefficient?

1. (4 pts.) Explain why this is the case.

1. (4 pts.) Identify how likely it is that a causal relationship has been indicated.