Answer:
a. Calculated χ² = 5.576 and critical region χ² ≥ χ² (0.05,3) = 7.82
b. Yes the degree of nausea associated with the brand of motion sickness.
c. n= 4 so we calculated dof= n-1 - 4-1= 3
because of leaving the areas of 1- ∝/2 and ∝/2 respectively to the side.
Step-by-step explanation:
1.Formulate the null and alternate hypothesis as
H0: The degree of nausea is associated with the brand of motion sickness pills.
against
Ha: degree of nausea is not associated with the brand of motion sickness pills
2.The significance level is set at ∝= 0.05
3.The test statistic to use is the Brandt-Snedecor formula i.e
χ² = N²/ AB [∑ ai²/ci - A²/ N]
which has an approximate χ² distribution with n-1 degrees of freedom.
4.The critical region is χ² ≥ χ² (0.05,3) = 7.82
5. Computations:
We calculate the value of χ² as follows
None. Slight Moderate Severe Total
Brand A 18 17 6 4 45= A
Brand B 11 14 14 6 45= B
Total ci 29 31 20 10 90= N
Now χ² = N²/ AB [∑ ai²/ci - A²/ N]
Putting the values
χ² = (90)²/ (45)(45) [ (18)²/29 +(17)²/31 + (6)²/20 + (4)²/10 - (45)²/90]
= 8100/2025 [ 324/29+ 289/31+ 36/20+ 16/10 - 2025/90 ]
= 4[ 11.172+9.322+ 1.8+1.6- 22.5]
= 4 [ 1.394]
χ² = 5.576
5. Conclusion:
Since the calculated value χ² = 5.576 does not fall in the critical region so we are unable to reject H0: We may conclude that the degree of nausea is associated with the brand of motion sickness pills.
c. To construct a two sided confidence interval for σ² we find two values of χ² distribution with (n-1) degrees of freedom .
The χ² distribution with (n-1) degrees of freedom leaves areas of 1- ∝/2 and ∝/2 respectively to the side.
The chi square is positively skewed .The skewness decreases as n increases. The chi- square distribution tends to be normal distribution as n approaches infinity.
