Question

A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and perform well on their test. During finals week, she randomly samples 45 students and asks them at the door what they ate for breakfast. She categorizes 25 students into Group 1 as those who ate a healthy breakfast that morning and 20 students into Group 2 as those who did not. After grading the final, she finds that 48% of the students in Group 1 earned an 80% or higher on the test, and 40% of the students in Group 2 earned an 80% or higher. Can it be concluded that eating a healthy breakfast improves test scores? Use a 0.05 level of significance.

H₀: P₁ = P₂

H_1: P₁ > P₂

Enter the test statistic - round to 4 decimal places.

Enter the p-value: round to 4 decimal places.

Answer 1

To Test :-

H₀: P₁ = P₂

H_1: P₁ > P₂

p̂1 = 12 / 25 = 0.48
p̂2 = 8 / 20 = 0.4

Test Statistic :-
Z = ( p̂1 - p̂2 ) / √( p̂ * q̂ * (1/n1 + 1/n2) ))
p̂ is the pooled estimate of the proportion P
p̂ = ( x1 + x2) / ( n1 + n2)
p̂ = ( 12 + 8 ) / ( 25 + 20 )
p̂ = 0.4444
q̂ = 1 - p̂ = 0.5556
Z = ( 0.48 - 0.4) / √( 0.4444 * 0.5556 * (1/25 + 1/20) )
Z = 0.5367

Test Criteria :-
Reject null hypothesis if Z > Z(α)
Z(α) = Z(0.05) = 1.645
Z < Z(α) = 0.5367 < 1.645, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Decision based on P value
P value = P ( Z > 0.5367 )
P value = 0.2958
Reject null hypothesis if P value < α = 0.05
Since P value = 0.2958 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Test Statistic :-  Z = 0.5367

P value =  0.2958

There is insufficient evidence to support the claim that eating a healthy breakfast improves test scores.