Problem 1 Please consider the following matrix of scores hnicity 34 H6 56 OK ASI

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Problem 1 Please consider the following matrix of scores hnicity 34 H6 56 OK ASIAN AF AMER 1/4 MS SHKWHTE 1/4 i. For each variable, calculate the appropriate measure of central Provide graphical representation of each of the variables (2 In no more than 150 words, provide a statistical summary tendency (2 points) points) (report) of the data for 10 clients measured on the 6 variables (5 points) Problem 2 i. In words explain what is measured by each of the following (3 points): Sum of Squares 2. Variance 3. Standard deviation Is it possible to obtain a negative standard deviation? Explairn your answer (1 point) What does it mean for a sample to have a standard deviation of zero? Give an example of scores in such a sample (1 point) Problem 3 i. A sample of n= 16 scores has a mean of M-20. After one score was removed from the sample, the new mean is found to be M=19. What is the value of the score that has been removed from the sample? (1 point) i. A sample of n-7 has a mean of M-5. One score was added to the sample. The new mean is found to be M-6. What is the value of the score that is added to the sample? (1 point) iA sample of scores as a mean M-9. One score was chang from X= 19 to X-5. What is the value of the ne samp mean? (1 point)

Explanation / Answer

Solution:

Problem 2
i) 1. Sum of Squares :
It measures the variability.

2. Variance:
The variance measures how far each number in the set is from the mean.

3. Standard Deviation :
A measure of the dispersion of a set of data from its mean.

ii) Standard deviation is defined as positive square root of variance, hence standard deviation cannot be negative.

iii) When standard deviation is zero, the scores have no variability.

Standard deviation measure the variability in scores and zero score means no variability in data

Problem 3
i. Sample size = 16
mean = 20
total sum of sample = 16*20 =320
New mean = 19
new sample size =15
total sum of new sample = 15*19 = 285
removed size = 320-285 = 35

ii. Sample size = 7
mean = 5
total sum of sample = 7*5 =35
New mean = 6
new sample size =8
total sum of new sample = 8*6 = 48
removed size = 48-35 = 13

iii. We need sample size to calculate new mean.

Problem 4
i) The mean would increase by 5 as well, but the standard deviation remains.
Hence,
New mean = a + 5
New stadard deviation = b

ii)Both the mean and standard deviation will be multiplied by 3 in that case.
Hence,
New mean = 3a
New stadard deviation = 3b

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