Psych Stat Chapter 4 Practice Problems

A population has a mean of µ = 50 and a standard deviation of σ = 10. If 3 points were added to every score in the population, what would be the new values for the mean and standard deviation? -The mean is still μ = 50, and the new standard deviation is σ = 13. -The new mean is μ = 53, and the standard deviation is still σ = 10. -The new mean is μ = 47, and the new standard deviation is σ = 13. -The new mean is μ = 47, and the standard deviation is still σ = 10.

-The new mean is μ = 53, and the standard deviation is still σ = 10.

A population has a mean of µ = 50 and a standard deviation of σ = 20. a. Would a score of X = 70 be considered an extreme value (out in the tail) in this population? No Yes b. If the standard deviation were σ = 5, would a score of X = 70 be considered an extreme value? No Yes

a. No. X = 70 is 20 points away from the mean, only 1.0 standard deviation. b. Yes. With σ = 5, 20 points is equal to a distance of 4.0 standard deviations.

Calculate SS, variance, and standard deviation for the following sample of n = 8 scores. (Round your answers to two decimal places where needed.) 0, 4, 1, 3, 2, 1, 1, 0

SS: 14 Variance: 2 Standard deviation: 1.41 The sum of these scores is ∑X = 0 + 4 + 1 + 3 + 2 + 1 + 1 + 0 = 12, and the mean is ∑X/n = 12/8 = 1.5. Therefore, the computational method for finding SS works best with these scores. SS = ∑X^2−(∑X)^2/n = 0+16+1+9+4+1+1+0−(12)280+16+1+9+4+1+1+0−1228 = 14 Then s2 = SS/n − 1 = 14/7 = 2 s = √s2 = √2 = 1.41

Calculate SS, variance, and standard deviation for the following population of N = 5 scores: 2, 13, 4, 10, 6 (Note: The definitional formula for SS works well with these scores.)

SS:80 Variance:16 Standard deviation:4 The mean is (2 + 4 + 6 + 10 + 13)/5 = 7. Therefore, SS = (2 - 7)² + (13 - 7)² + (4 - 7)² + (10 - 7)² + (6 - 7)² = 25 + 36 + 9 + 9 + 1 = 80. The population variance is SS/N = 80/5 = 16, and the standard deviation is √16 = 4.

What does it mean for a sample to have a standard deviation of zero? Describe the scores in such a sample. -When standard deviation is zero, you cannot measure variability. -When standard deviation is zero, no two scores are exactly the same. -When standard deviation is zero, the scores have no variability. -When standard deviation is zero, there are no data points in the sample.

-When standard deviation is zero, the scores have no variability A standard deviation of zero indicates there is no variability. In this case, all of the scores in the sample have exactly the same value.

Is it possible to obtain a negative value for the variance or the standard deviation? -No -Yes

-No Variance and standard deviation are always greater than or equal to zero. They are measures of distance that are based on squared deviations, which are always positive.

If every score in the population were multiplied by 2, what would be the new values for the mean and standard deviation? -The new mean is μ = 100, and the new standard deviation is σ = 20. -The new mean is µ = 52, and the new standard deviation is σ = 20. -The mean is still μ = 50, and the new standard deviation is σ = 20. -The new mean is µ = 100, and the new standard deviation is still σ = 10.

-The new mean is µ = 52, and the new standard deviation is σ = 20.

There are two different formulas or methods that can be used to calculate SS (sum of squared deviations). Under what circumstances is the definitional formula easy to use? Check all that apply. -When the mean is a whole number and there are relatively few scores -When the mean is a whole number and the range is small -When the mean is a whole number and there are many scores -When there are relatively few scores and many of them are zero

-When the mean is a whole number and there are relatively few scores The definitional formula is easy to use when the mean is a whole number and there are relatively few scores.

Under what circumstances is the computational formula preferred? Check all that apply. -When the mean is not a whole number -When the mean is negative -When the range is large -When there are many scores

-When the mean is not a whole number -When there are many scores The computational formula is preferred when the mean is not a whole number or when there are many scores.

For the following sample of n = 6 scores: 0, 11, 5, 10, 5, 5 On the basis of your plot, find the sample mean. Make an estimate of the standard deviation from the frequency distribution histogram. (Note: An estimate of the standard deviation can be made by taking the average of the smallest and the largest deviations (from the mean) of the distribution) Compute SS, variance, and standard deviation for the sample. How well does your estimate compare with the actual value of s? -The calculated standard deviation is in excellent agreement with the estimate. -The calculated standard deviation differs greatly from the estimate. -The calculated standard deviation differs slightly from the estimate.

Mean = M =6 Mean = M = ∑X/n = (0 + 5 + 5 + 5 + 10 + 11)/6 = 36/6 = 6 For this distribution, the smallest distance from the mean is 1 point and the largest distance is 6 points . The estimated standard deviation = 3.5 . SS =80 Variance =16 Standard deviation =4 SS = ∑(X - M)² = (0 - 6)² + (11 - 6)² + (5 - 6)² + (10 - 6)² + (5 - 6)² + (5 - 6)² = 36 + 25 + 1 + 16 + 1 + 1 = 80 Variance = s² = SS/n - 1 = 80/6 - 1 = 16 Standard deviation = s = √s² = √16 = 4 -The calculated standard deviation differs slightly from the estimate. Your estimated value for s should have been 3.5. The calculated standard deviation differs slightly from this estimate.

Explain why the formulas for sample variance and population variance are different. -Variance is defined as the mean squared deviation, and, for a population, is computed as the sum of squared deviations divided by N. The sample variance will be biased and will consistently underestimate the corresponding population value. -Variance is defined as the mean deviation, and, for a population, is computed as the sum of deviations divided by N. The sample variance will be biased and will consistently underestimate the corresponding population value. -Variance is defined as the mean squared deviation, and, for a population, is computed as the sum of deviations divided by N. The sample variance will be biased and will consistently underestimate the corresponding population value.

-Variance is defined as the mean squared deviation, and, for a population, is computed as the sum of squared deviations divided by N. The sample variance will be biased and will consistently underestimate the corresponding population value. Variance is defined as the mean squared deviation, and, for a population, is computed as the sum of squared deviations divided by N. However, if this same definition is used for a sample, the sample variance will be biased and will consistently underestimate the corresponding population value. Therefore, the formula for sample variance includes an adjustment to correct for the bias. The adjustment involves dividing by df = n - 1 rather than n.

Calculate SS, variance, and standard deviation for the following sample of n = 5 scores: 2, 9, 5, 5, 9

SS:36 Variance:9 Standard deviation:3 The sum of these scores is ∑X = 2 + 5 + 5 + 9 + 9 = 30, and the mean is ∑X/n = 30/5 = 6. Therefore, the definitional method for finding SS can be used with these scores. SS = (2 - 6)² + (9 - 6)² + (5 - 6)² + (5 - 6)² + (9 - 6)² = 16 + 9 + 1 + 1 + 9 = 36 Then s2 = SS/n − 1 = 36/4 = 9 s = √s^2 = √9=3

For the following population of N = 6 scores: 2, 9, 6, 8, 9, 8 a. Calculate the range and the standard deviation. (Use either definition for the range.) -The range is either 8 or 9, and the standard deviation is σ = 6. -The range is either 7 or 8, and the standard deviation is σ = 2.45. -The range is either 6 or 7, and the standard deviation is σ = 36. b. Add 2 points to each score, and compute the range and standard deviation again. -The range is either 6 or 7, and the standard deviation is σ = 36. -The range is either 8 or 9, and the standard deviation is σ = 6. -The range is either 7 or 8, and the standard deviation is σ = 2.45. c. Describe how adding a constant to each score influences measures of variability. -Adding a constant to every score affects measures of variability. -Adding a constant to every score does not affect measures of variability.

a. -The range is either 7 or 8, and the standard deviation is σ = 2.45. The range is either 9 - 2 = 7 or 9.5 - 1.5 = 8. b. -The range is either 7 or 8, and the standard deviation is σ = 2.45. The scores are now 4 + 11 + 8 + 10 + 11 + 10. The range is either 11 - 4 = 7 or 11.5 - 3.5 = 8 as for the original scores. c. -Adding a constant to every score does not affect measures of variability. Since the range and the standard deviation remain the same after adding 2 points to each score, adding a constant to every score does not affect measures of variability.