Question

Question Set 2: Chapter 6

A. Scores on the Wechsler Intelligence Scale for Children (WISC) are standardized to be normally distributed with a mean of 100 and standard deviation of 15.

i. Compute the z score for a WISC score of 110.

ii. Compute the z score for a WISC score of 90.

iii. What is the WISC score of a child who scored 2 standard deviations above the mean?

iv. What is the WISC score of a child who scored half a standard deviation below the mean?

v. What is the WISC score for a child whose z score was 0?

B. SAT-Math scores have a mean of 500 and standard deviation of 100. ACT-Math scores have a mean of 18 and standard deviation of 3. A student scored 640 on the SAT-Math and 22 on the ACT-Math. Compute the student’s z score for each test. On which test did they score more impressively? Explain why.

C. In your own words, explain the Central Limit Theorem. What is it? What is its purpose?

D. In the population, the salaries of NFL players are positively skewed with a mean of $2.2 million and a standard deviation of $3.1 million. Answer the following questions given that a researcher is repeatedly pulling random samples of n=100 to construct a distribution of sample means.

i. What would be the shape of the distribution of sample means? Explain why.

ii. What would be the mean of the distribution of sample means? Explain why.

iii. What would be the standard deviation of the distribution of sample means (i.e., the standard error)?

E. Scores on the Wechsler Adult Intelligence Scale- Third Edition (WAIS-III) are standardized to be normally distributed with a mean of 100 and standard deviation of 15. A psychologist has a dataset containing the WAIS-III scores from a random sample of 50 adults who are members of a specific organization. They want to know if there is evidence that the mean WAIS-III score in the population of all members of this organization is greater than 100.

i. Compute the standard deviation of the distribution of sample means (i.e., the standard error).

ii. Explain why the standard error you computed in part i is smaller than the standard deviation of the raw WAIS-III scores.

iii. In this scenario, what is the hypothesized value of the population mean?

iv. In the sample of 50 adults, the observed sample mean was 105 points. Using the standard error you computed in part i and the hypothesized population mean you identified in part iii, compute the z test statistic.

Answer 1

(A) Let us represent the WISC score by X, and X~N(100,15)

(i) Simply put, a z-score (also called a standard score) gives you an idea of how far from      the mean a data point is. But more technically it’s a measure of how many standard deviations below or above the population mean a raw score is.

Formula Used:

z = (x – μ) / σ

Here x=110

     μ=100

     σ=15

Hence, using above formula we get:

z = (110-100)/15 = 0.667

I.e 0.67 std deviations above the mean.

(ii) Similarly for x= 90

                 μ=100

                 σ=15

Using the formula given in part (I)

z = (90- 100)/15 = -0.667

i.e. 0.67 std dev below the mean

(iii) Now to find the WISC score of the child I.e X when z=2

=> (X-100)/15 =2

=> (X-100) = 30

=> X = 130

(iv) For z=-0.5

=> (x-100)/15= -0.5

=> X = 92.5

(v) If the z-score of the child is zero than :

=> (X-100)/15=0

=> X= 100(mean)

(B) Let us represent SAT math score by X where X~ ( μ=500, σ =100),

                 and ACT math score by Y where Y~ ( μ=18, σ =3)

So we can’t just look at the absolute score and decide which one is better as they are on a different scale. What we can do is assume its a normal dist or relatively close to the Normal distribution and hence we can calculate the z-score for each test (I.e. how many std dev above the mean).

For SAT score: z = (640-500)/100

                 = 140/100

                 = 1.4

ie 1.4 std deviations above the mean

For ACT score: z = (22-18)/3

                 = 4/3

                 = 1.33

ie 1.33 std deviations above the mean

So now if we compare the student did relatively better in the SAT test as in SAT he did 1.4 std deviations and in ACT he did 1.33 std deviations, so the first zvalue is slightly more far away from the mean value.

(C)

Central Limit Theorem:

CLT is a statistical theory stating that given a sufficiently large sample size from a population with a finite level of variance, the mean of all samples from the same population will be approximately equal to the mean of the population.Also, all the samples will follow an approximate normal distribution pattern, with all variances being approximately equal to the variance of the population, divided by each sample's size.

Mathematically if X1,X2,…….,Xn is a random sample of size n taken from a population with mean μ and std deviation  σ and if is the sample mean then

Z=(-μ)/ (σ/) as n-> , is the std normal distribution

Purpose:

A simple example of the central limit theorem is rolling many identical, unbiased dice. The distribution of the sum (or average) of the rolled numbers will be well approximated by a normal distribution. Since real-world quantities are often the balanced sum of many unobserved random events, the central limit theorem also provides a partial explanation for the prevalence of the normal probability distribution. It also justifies the approximation of large-sample statistics to the normal distribution in controlled experiments.

(D)

(i) A theorem states that if the sample size is large (generally n ≥ 30), and the standard deviation of the population is finite, then the distribution of sample means will be approximately normal

Hence the shape of sample means will be bell shaped.

(ii) Since n=100 hence by using central limit theorem we can say that the mean of the sample means will be approximately equal to the population mean that is 2.2 million dollar as the variance is finite for the population

(iii) Now as the number of samples increases ie n then two thing happens:

I. The distribution becomes more normal.

II. The standard deviation decreases

Hence,

If we know the std deviation of the original population (σ) and the number of samples (n) then the variance of the sample means ()

Hence = σ^2/n

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