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Statistics Assignment Solutions: Mean, Standard Deviation, and Variance, Schemes and Mind Maps of Statistics

Solutions to problems related to calculating mean, standard deviation, and variance for different samples and populations. It covers scenarios where the mean is not an integer, scores are added or multiplied, and extreme scores are present.

Typology: Schemes and Mind Maps

2021/2022

Uploaded on 09/12/2022

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PSY 216
Assignment 4 Answers
1. Problem 2 from the text
Can SS ever have a value less than zero? Explain your answer.
SS = Σ(X - )2. When you square a real number, the result is always non-negative. The sum of non-negative
numbers must be non-negative.
2. Problem 4 from the text
What does it mean for a sample to have a standard deviation of zero? Describe the scores in such a sample.
If the standard deviation is 0 then the variance is 0 and the mean of the squared deviation scores must be 0.
The sum of the squared deviation scores must be equal to 0 for the mean of the squared deviation scores to be
equal to 0. Since the squared deviation scores must be non-negative because they are squared, all of the
squared deviation scores must be 0; otherwise, the sum would be non-zero. The only way that each squared
deviation score can be equal to 0 is if all of the scores equal the mean. Thus, when the standard deviation
equals 0, all the scores are identical and equal to the mean.
3. Problem 5 from the text
Explain why the formulas for sample variance and population variance are different.
The extreme scores in a population that is approximately normal in shape are less likely to be
included in a sample because of their rarity. Thus, the sample is usually less variable than the
population. By dividing the sum of squares of the sample by a slightly smaller number (n 1 instead
of n), the variance will be slightly larger, correcting for underestimation.
4. Problem 7 from the text
On an exam with a mean of M = 78, you obtain a score of X = 84.
a. Would you prefer a standard deviation of s = 2 or s = 10? (Hint: Sketch each distribution
and find the location of your score.)
One should prefer s = 2 as a score of 84 is an extremely large value in such a distribution.
You likely did much better than everyone else. With s = 10, your score is less extreme and
others likely did better than you.
b. If your score was X = 72, would you prefer s = 2 or s = 10. Explain your answer.
One should prefer s = 10 as your below average performance is not an extreme score. With
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PSY 216

Assignment 4 Answers

  1. Problem 2 from the text Can SS ever have a value less than zero? Explain your answer. SS = Σ(X - )^2. When you square a real number, the result is always non-negative. The sum of non-negative numbers must be non-negative.
  2. Problem 4 from the text What does it mean for a sample to have a standard deviation of zero? Describe the scores in such a sample. If the standard deviation is 0 then the variance is 0 and the mean of the squared deviation scores must be 0. The sum of the squared deviation scores must be equal to 0 for the mean of the squared deviation scores to be equal to 0. Since the squared deviation scores must be non-negative because they are squared, all of the squared deviation scores must be 0; otherwise, the sum would be non-zero. The only way that each squared deviation score can be equal to 0 is if all of the scores equal the mean. Thus, when the standard deviation equals 0, all the scores are identical and equal to the mean.
  3. Problem 5 from the text Explain why the formulas for sample variance and population variance are different. The extreme scores in a population that is approximately normal in shape are less likely to be included in a sample because of their rarity. Thus, the sample is usually less variable than the population. By dividing the sum of squares of the sample by a slightly smaller number ( n – 1 instead of n ), the variance will be slightly larger, correcting for underestimation.
  4. Problem 7 from the text On an exam with a mean of M = 78, you obtain a score of X = 84. a. Would you prefer a standard deviation of s = 2 or s = 10? ( Hint : Sketch each distribution and find the location of your score.) One should prefer s = 2 as a score of 84 is an extremely large value in such a distribution. You likely did much better than everyone else. With s = 10, your score is less extreme and others likely did better than you. b. If your score was X = 72, would you prefer s = 2 or s = 10. Explain your answer. One should prefer s = 10 as your below average performance is not an extreme score. With

s = 2, few, if any, people performed less well than you on the exam.

  1. Problem 8 from the text A population has a mean of μ = 30 and a standard deviation of σ = 5. (Problem 8 from the text) a. If 5 points were added to every score in the population, what would be the new values for the mean and standard deviation? Adding a constant to every score increases the mean by the same constant amount. Thus, μ = 30 + 5 = 35. Adding a constant to every score has no effect on the standard deviation. σ = 5 b. If every score in the population were multiplied by 3 what would be the new values for the mean and standard deviation? Multiplying each score by a constant also multiplies the mean by the same constant. Thus, μ = 30 * 3 = 90. In a population, multiplying each score by a constant also multiplies the standard deviation by the constant. Thus, σ = 5 * 3 = 15
  2. Problem 13 from the text Calculate the mean and SS (sum of squared deviations) for each of the following samples. Based on the value for the mean, you should be able to decide which SS formula is better to use. a. Sample A: 1, 4, 8, 5 M = ΣX / n = (1 + 4 + 8 + 5) / 4 = 18 / 4 = 4. SS = (ΣX^2 – (ΣX)^2 / N ) = ( 12 + 4^2 + 8^2 + 5^2 – (1 + 4 + 8 + 5)^2 / 4) = (1 + 16 + 64 + 25 – 182 / 4) = (106 – 324 / 4) = (106 – 81) = 25 Less preferred method because the mean is not an integer: SS = Σ(X – M )^2 = (1 – 4.5)^2 + (4 – 4.5)^2 + (8 – 4.5)^2 + (5 – 4.5)^2 = - 3.5^2 + - 0.5^2 + 3 .5^2 + 0.5^2 =12.25 + 0.25 + 12.25 + 0. = 25 b. Sample B: 3, 0, 9, 4
  1. Problem 18 from the text Calculate SS , variance, and standard deviation for the following population of N = 7 scores: 8, 1, 4, 3, 5, 3, 4. ( Note: The definitional formula works well with these scores.) M = ΣX / N = (8 + 1 + 4 + 3 + 5 + 3 + 4) / 7 = 28 / 7 = 4 SS = Σ(X – M )^2 = ( 8 – 4 )^2 + ( 1 – 4 )^2 + ( 4 – 4 )^2 + ( 3 – 4 )^2 + ( 5 – 4 )^2 + ( 3 – 4 )^2 + ( 4 – 4 )^2 = 42 + - 32 + 02 + - 12 + 1^2 + - 12 + 0^2 = 16 + 9 + 0 + 1 + 1 + 1 + 0 = 28 σ^2 = SS / N = 28 / 7 = 4 σ = √ σ^2 = √ 4 = 2
  2. Use SPSS to graph a histogram of the number of younger siblings variable in the class data set. For that variable, use SPSS to find the following statistics and report their values in the following table: Statistic Value Mean 0. 95 Median 1. Mode 0 Standard deviation 1. 089 Variance 1.1 86 Interquartile range Q 3 – Q 1 = 75th^ %tile – 25 th^ %tile = 1.25 – 0 = 1. Range 4

Attach your SPSS output to your answers. There a numerous ways of accomplishing this in SPSS. One way is to click on Analyze | Descriptive Statistics | Frequencies. Move the Number of Younger Siblings variable into the Variables list. Click on the Statistics button and select Mean, Median, Mode, Std. deviation, Variance, Range, and Quartiles (for the interquatile range). Click Continue. Click the Charts button. Select Histograms. Click Continue. Click OK. Statistics Number of younger siblings N Valid 38 Missing 0 Mean. Median 1. Mode 0 Std. Deviation 1. Variance 1. Range 4 Percentiles 25. 50 1. 75 1.