Answer:
Kindly check explanation
Step-by-step explanation:
Given the data :
________Hours of sleep______
Age ___<6__6-6.9___ 7-7.9 __≥8 ___total
≤ 39___38___60_____77____65____240
≥ 40___36___57_____75____92____260
A.) test of independence for row and column variables :
Null: row and column variables are independent
Alternative : row and column variables are not independent
Expected frequencies :
[(row total) * (column total] / sample size
E1: ((38+60+77+65)*(38+36))/500 = 35.52
E2: ((38+60+77+65)*(60+57))/500 = 56.16
E3: ((38+60+77+65)*(77+75))/500 = 72.96
E4: ((38+60+77+65)*(65+92))/500 = 75.36
E5: ((36+57+75+92)*(38+36))/500 = 38.48
E6: ((36+57+75+92)*(60+57))/500 = 60.84
E7: ((36+57+75+92)*(77+75))/500 = 79.04
E8: ((36+57+75+92)*(65+92))/500 = 81.64
Then performing the Chisquare independence test X²
X² = Σ[(f1-e1)²/e1 + (f2-e2)²/e2... (fn - en)²/en]
F1 to f8 = 38,60,77,65,36,57,75,92
X² = ((38-49.92)^2)/35.52 + ((60-56.16)^2)/56.16 + ((77-72.96)^2)/72.96 + ((65-75.36)^2)/75.36 + ((36-38.48)^2)/38.48 + ((57-60.84)^2)/60.84 + ((75-79.04)^2)/79.04 + ((92-81.64)^2)/81.64
= 4.0070162
X² = 4.00
α = 0.05
Reject Null if p < α
df = (row - 1)*( column - 1)
df = (2 - 1)*(4-1) = 1*3 = 3
To obtain the p value, we can use an online calculator to compute the P-value from the X² square score at α = 0.05 and 3 degree of freedom
p = 0.261
Since ; p > α ; 0.261 > 0.05 ; The result is not significant at α 0.05 ; Hence we fail to reject the Null.
