A survey was conducted to determine whether hours of sleep per night are independent of age. A sample of individuals was asked to indicate the number of hours of sleep per night with categorical options: fewer than 6 hours, 6 to 6.9 hours, 7 to 7.9 hours, and 8 hours or more. Later in the survey, the individuals were asked to indicate their age with categorical options: age 39 or younger and age 40 or older. Sample data follow. Hours Of Sleep 39 or Younger 40 or Older
Fewer than 6 38 36
6 to 6.9 60 57
7 to 7.9 77 75
8 or More 65 92

Required:
a. Conduct a test of independence to determine whether hours of sleep are independent of age. Using α= 0.05 level of significance
b. Compute the value of the 2 test statistic.
c. The p-value is:
d. What is your conclusion?

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___607765240

≥ 40___36___577592260

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.

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step-by-step explanation:

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