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Sagot :
Answer:
a. Therefore the expected values are 8,8,8,8,8,8,8
b. The df = n-1= 7-1= 6
The critical value at ∝= 0.05 with 6 d.f is χ²≥χ²(d.f) 0.05= 12.59
c. The null hypothesis is the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week
against the claim the nights for the highest number of students doing the majority of their homework do not occur with equal frequencies during a week.
d. χ² = 4.25
e.Results : accept the null hypothesis
ei = 8,8,8,8,8,8,8
df = 6
H0: σ²1 ≤ σ²2 against the claim Ha: σ²1> σ²2
The critical region is χ²≥χ²(0.05) 6= 12.59
The calculated value of χ² = 4.25 < χ²(0.05) 6= 12.59 therefor accept the null hypothesis.
Step-by-step explanation:
Sunday Monday Tuesday Wednesday
No of Students 11 8 10 7
Thursday Friday Saturday
10 5 5
a. The expected values are = Total number /no of days = 56/7= 8
Therefore the expected values are 8,8,8,8,8,8,8
b. The df = n-1= 7-1= 6
The critical value at ∝= 0.05 with 6 d.f is χ²≥χ²(d.f) 0.05= 12.59
c. The null hypothesis is the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week
against the claim the nights for the highest number of students doing the majority of their homework do not occur with equal frequencies during a week
H0: σ²1 ≤ σ²2 against the claim Ha: σ²1> σ²2 as the goodness of fit test is always right tailed
d.
Days No.of Students
Observed Expected oi-ei (oi-ei)² (oi-ei)²/ei
Sunday 11 8 3 9 1.125
Monday 8 8 0 0 0
Tuesday 10 8 2 4 0.5
Wednesday 7 8 -1 1 0.125
Thursday 10 8 2 2 0.25
Friday 5 8 -3 9 1.125
Saturday 5 8 -3 9 1.125
∑ 4.25
e. since the calculated value of χ² = 4.25 is less than the critical value of
χ²= 12.59 we fail to reject H0 and conclude that the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week.
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