Coaching companies claim that their courses can raise the SAT scores of high school students. But students who retake the SAT without paying for coaching also usually raise their scores. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were uncoached. Starting with their verbal scores on the first and second tries, we have these summary statistics: Try 1 Try 2 Gain n \overline x s \overline x s \overline x sCoached 427 500 92 529 97 29 59Uncoached 2733 506 101 527 101 21 52Use Table C to estimate a 90% confidence interval for the mean gain of all students who are coached. toat 90% confidence. Now test the hypothesis that the score gain for coached students is greater than the score gain for uncoached students. Let \mu_1 be the score gain for all coached students. Let \mu_2 be the score gain for uncoached students.(a) Give the alternative hypothesis: \mu_1 - \mu_20.(b) Give the t test statistic:(c) Give the appropriate critical value for \alpha =5%: .

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Coaching companies claim that their courses can raise the SAT scores of high school students. But students who retake the SAT without paying for coaching also usually raise their scores. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were uncoached. Starting with their verbal scores on the first and second tries, we have these summary statistics:

Try 1 Try 2 Gain

n x s xsx s

Coached4275009252997 29 59

Uncoached 273350610152710121 52

Use Table C to estimate a 90% confidence interval for the mean gain of all students who are coached.

at 90% confidence.

Now test the hypothesis that the score gain for coached students is greater than the score gain for uncoached students. Let μ1 be the score gain for all coached students. Let μ2 be the score gain for uncoached students.

(a) Give the alternative hypothesis:

μ1 - μ2.

(b) Give the t test statistic:

(c) Give the appropriate critical value for \alpha =5%:

Solution:

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = sample mean score gain for all coached students

x2 = sample mean score gain for all uncoached students

s1 = sample standard deviation score gain of coached students

s2 = sample standard deviation score gain of uncoached students

For a 90% confidence interval, the z score is 1.645

From the information given,

x1 = 29

s1 = 59

n1 = 427

x2 = 21

s2 = 52

n2 = 2733

x1 - x2 = 29 - 21 = 8

z√(s1²/n1 + s2²/n2) = 1.645√(59²/427 + 52²/2733) = 4.97

90% Confidence interval = 8 ± 4.97

a) The population standard deviations are not known. it is a two-tailed test. The random variable is μ1 - μ2 = difference in the score gain for coached and uncoached students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 > μ2 H1 : μ1 - μ2 > 0

This is a right tailed test.

b) Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (29 - 21)/√(59²/427 + 52²/2733)

t = 2.65

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [59²/427 + 52²/2733]²/[(1/427 - 1)(59²/427)² + (1/2733 - 1)(52²/2733)²] = 9.14/0.1564

df = 58

c) from the t distribution table, the critical value is 1.67

In order to reject the null hypothesis, the test statistic must be smaller than - 1.67 or greater than 1.67

Since - 2.65 < - 1.67 and 2.65 < 1.67, we would reject the null hypothesis.

Therefore, at 5% significance level, we can conclude that the score gain for coached students is greater than the score gain for uncoached students.

21/58.

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