Question

1a)

Here is a bivariate data set.

x	y
58.1	46.5
57.2	50.7
41	53.9
47	50.6
65.7	50.9

Find the correlation coefficient and report it accurate to three decimal places.
r =

What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place.
r² =

1b)

Test the claim that the mean GPA of Orange Coast students is larger than the mean GPA of Coastline students at the 0.05 significance level.

The null and alternative hypothesis would be:

H0:μO≤μCH0:μO≤μC
H1:μO>μCH1:μO>μC

H0:μO≥μCH0:μO≥μC
H1:μO<μCH1:μO<μC

H0:pO≤pCH0:pO≤pC
H1:pO>pCH1:pO>pC

H0:pO=pCH0:pO=pC
H1:pO≠pCH1:pO≠pC

H0:μO=μCH0:μO=μC
H1:μO≠μCH1:μO≠μC

H0:pO≥pCH0:pO≥pC
H1:pO<pCH1:pO<pC

The test is:

two-tailed

left-tailed

right-tailed

The sample consisted of 65 Orange Coast students, with a sample mean GPA of 3.3 and a standard deviation of 0.06, and 65 Coastline students, with a sample mean GPA of 3.27 and a standard deviation of 0.08.

The test statistic is:  (to 2 decimals)

The p-value is:  (to 2 decimals)

Based on this we:

• Reject the null hypothesis
• Fail to reject the null hypothesis

Answer 1

1a)

Find the correlation coefficient:

We calculate correlation by using excel data analysis tool.

Given,

x	y
58.1	46.5
57.2	50.7
41	53.9
47	50.6
65.7	50.9

The output of excel is,

	x	y
x	1
y	-0.544	1

Therefore, the correlation coefficient between x and y is

r = -0.544

What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place.
r² = (-0.544) * (-0.544)

r² = 0.3.

1b)

Test the claim that the mean GPA of Orange Coast students is larger than the mean GPA of Coastline students at the 0.05 significance level.

The null and alternative hypothesis would be:

H0:μO≤μC
H1:μO>μC

The test is:

Notice the inequality points to the right
Therefore,

Test is right-tailed.

Given information,

Orange coast students: n1 = 65, x1_bar = 3.3, sd1 = 0.06

Coastline students: n2 = 65, x2_bar = 3.27, sd2 = 0.08.

The test statistic is:

we are calculating two sample t-test statistic,

We calculate first Sp2,

Therefore,

P-value:

We calculate p-value with (n1+n2-2 = 65+65-2 = 128) df and alpha 0.05 by using excel function =TDIST(2.42,128,1)

P-value = 0.01

Conclusion: The p-value (0.01) is less than 0.05 then reject null hypothesis.

That is, test the claim that the mean GPA of Orange Coast students is larger than the mean GPA of Coastline students.