Question

An education researcher would like to test whether 2nd graders retain or lose knowledge during the summer when they are presumably not in school. She asks nine 2nd graders to take a comprehension exam at the end of the school year (May), and then asks those same students to come back after the summer (late August) to retake a different but equivalent exam, to see if their level of comprehension has changed. Using the data below, test this hypothesis using an alpha level of .05.

May August
90 100
65 80
78 92
50 60
89 90
92 98
75 70
90 96
65 87
a) What is the appropriate test? Is it a one or two tailed test?
b) State the null hypothesis (in words and using means).
c) State the alternative hypothesis (in words and using means).
d) Find the critical value.
e) Calculate the obtained value by hand. What does your decision mean?

Answer 1

See step by step explanation

Explanation:

From the problem statement, we understand that we need to investigate if 2nd graders RETAIN OR LOSE knowledge comparing information concerning two different months, therefore we are going to find out if the information gives us enough evidence of a difference in the two groups

a) The test should be a two tail-test

Sample sizes are small ( n₁ = n₂ = 9 ) which means we have to use t-student test

b and c) Test Hypothesis:

For Null Hypothesis H₀, we establish that the two means are equal, which in this case means that the two samples´ means are equal. And as Hypothesis alternative Hₐ that the H₀ is not true, or that the samples mean are different.

Null Hypothesis H₀ x₁ - x₂ = 0 or x₁ = x₂

Alternative Hypothesis Hₐ x₁ - x₂ ≠ 0 or x₁ ≠ x₂

d) Significance level α = 0,05 then α/2 = 0,025

then t(c) with α/2 = 0,025 and degree of freedom

df = n₁ + n₂ - 2 df = 9 + 9 -2 df = 16

From t s-tudent table we find t(c) = 2,12

e) Calculating:

x₁ and σ₁

x₂ and σ₂

σₓ = [ ( n₁ - 1 )* σ₁² + ( n₂ - 1 )*σ₂² ]/ n₁ + n₂ - 2

Using a calculator:

x₁ = 77,11 σ₁ = 13,86

x₂ = 85,89 σ₂ = 12,74

σₓ = [ ( 8*(13,86)² + 8*( 12,74 )² ] / 16

σₓ = (8* 192,1) + 8* ( 162,31) / 16

σₓ = (1536,8 + 1298,48 ) / 16

σₓ = 177,205

t(s) = ( x₁ - x₂ ) /√ σₓ²/ n₁ + σₓ²/ n₂

t(s) = ( 77,11 - 85,89 )/ (177,205)²/n₁ + (177,205)²/n₂

t(s) = - 8,78 / √3489,07 + 3489,07

t(s) = -8,78 / 83,54

t(s) = - 0,11

Comparing t(s) and t(c)

|t(s)| < |t(c)|

0,11 < 2,12

t(s) is in the acceptance region. We accept H₀