Question

Scores for men on the verbal portion of the SAT-I test are normally distributed with a mean of 509 and a standard deviation of 112. Randomly selected men are given the Columbia Review Course before taking the SAT test. Assume that the course has no effect.

a. If 1 of the men is randomly selected, find the probability his score is at least 590.

b. If 16 of the men are randomly selected, find the probability that their mean score is at least 590.

c. In finding the probability for (b), why can the Central Limit Theorem be used even though the sample size does not exceed 30?

d. If the random sample of 16 men does result in a mean score of 590, is there strong evidence to support the claim that the course is actually effective? Why or why not?

Answer 1

Given Mean, = 509

Standard Deviation, = 112

a.)

If 1 men is randomly selected, n = 1

Score is atleast 590 which means, x = 590

Using formula for z test:

z = (x - ) / ( / )

z = (590 - 509) / (112 / 1)

z = 81 / 112 = 0.723

We need to find P(Z 0.723)

P(Z 0.723) = 1 - P(Z < 0.723)

P(Z 0.723) = 1 - 0.7642 = 0.2358

So, p = 0.2358

b.)

Here n = 16

z = (x - ) / ( / )

So,

z = (590 - 509) / (112 / )

z = 81 / (112 / 4)

z = 81 / 28

z = 2.89

We need to find P(Z 2.89)

P(Z 2.89) = 1 - P(Z < 2.89)

P(Z 2.89) = 1 - 0.9981 = 0.0019

So, p = 0.0019

c.)

We can use Central limit theorem because it was given that Scores are normally distributed

The condition to use Central limit theorem is that data should be either normally distributed or sample must have size of at least 30.

Since our normal distribution condition is satisfied we can say that our data is normally distributed.

d.)

Yes there is strong evidence to support the claim that the course is actually effective.

We need to check if our course was effective for the sample of men taking SAT test. Their mean score of 590 is greater than population mean by chance or because of course.

For this we will compare means of population and sample and perform t test.

So, x = 590

The Hypotheses are:

Ho: x ( Population mean is more than sample mean, course has no impact)

Ha: < x (Population mean is less than sample mean, Course has impact on scores)

Formula for t statistic,

t = (x - ) / ( / )

t = (590 - 509) / (112 / )

t = 81 / (112 / 4)

t = 81 / 28

t = 2.89

Now. degrees of freedom, df = 16 - 1 = 15

We will assume the significance level of 0.01(99% Confidence)

Critical value of t for 15 degrees of freedom at 0.01 significance level is, t_cri = 2.60

Since t >  t_cri we can reject the NULL Hypotheses.

So we can say with 99% confidence that course is actually effective for SAT test and sample who took Course has mean Score greater than Population mean Score.