Question

Two statistics professors at two rival schools decide to use IQ scores as a measure of how smart the students at their respective schools are. IQ scores are known to be Normally distributed. The two professors will use this knowledge to their advantage. They will randomly select 10 students from their respective schools and determine the students' IQ scores by means of the standard IQ test. The two professors will use the pooled version of the two-sample t test to determine whether the students at the two universities are equally smart. Let m1 and m2 represent the mean IQ scores of the students at the two universities. Let s1 and s2 be the corresponding population standard deviations. The hypotheses they will test are H0: m1 – m2 = 0 versus Ha: m1 –m2 ¹ 0. Based on the two samples of 10 students, the two professors find the following information: = 113, = 122 s1 = 8, and s2 = 12. (Hint: Calculate sp2 first.)

A. In a study comparing four groups with six observations in each group, the MSE = 467 and the MSG = 2345. What is the value of the coefficient of determination? A) 0.484 B) 0.516 C) 0.587 D) 0.939 34.

B. A study compares six groups with five observations in each group. An F statistic of 3.712 is reported. What can we say about the P-value for this F test? A) P-value < 0.001 B) 0.001 < P-value < 0.01 C) P-value > 0.01 D) 0.01 < P-value < 0.05

Answer 1

Since we are testing if the students are equally smart, we use the 2 tailed test.

Given:

= 113, = 8 and n1 = 10

= 122, = 12 and n2 = 10

We find the pooled variance:

Sp = 10.2

Since we are using pooled variance, the degrees of freedom (df) = n1 + n2 - 2 = 10 + 10 - 2 = 18

The Hypothesis:

H0: - = 0: Claim

Ha: - : 0

The Test Statistic:

The p Value: The p value (2 Tail) for t = -1.97, df = 18, is; p value = 0.0644

The Critical Value: The critical value (2 tail) at = 0.05(default), df = 10 , tcritical = +2.1 and -2.1

The Decision Rule: If tobserved is > tcritical or if tobserved is < -tcritical, Then Reject H0.

Also If the P value is < , Then Reject H0

The Decision: Since tobserved (-1.97) lies in between +2.1 and -2.1, We Fail To Reject H0

Also since P value (0.0644) is > (0.05), We Fail to Reject H0.

The Conclusion: There is insufficient evidence at the 95% significance level to conclude that the students of the two rival schools are not equally smart.

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(A) k = no of groups = 4, n = number of observations/group = 6

Therefore N = Total observation = 4 * 6 = 24

Degrees of freedom, (between) = k - 1 = 4 - 1 = 3

Degrees of Freedom, (error) = N - k = 24 - 4 = 20

MSG = SSG/df between. Therefore SSG = MSG * df between = 2345 * 3 = 7035

MSE = SSE/df error. Therefore SSE = MSE * df error = 467 * 20 = 9340

Therefore SST = SSG + SSE = 7035 + 9340 = 16375

The Co - efficient of determination R² = SSG/SST = 7035/16375 = 0.4296 0.423

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(B) k = 6, and n = 5. Therefore Total observations = 5 * 6 = 30

dfbetween = k - 1 = 6 - 1 = 5 and

dferror = N - k = 30 - 6 = 24

Given F observed = 3.712.

Therefore the p value for dfbetween = 5, dferror = 24 for F = 3.712 is ; p value = 0.0125

Therefore the correct range is Option D: 0.01 < p value < 0.05

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