Nausea from air sickness affects some travelers. In a comparative study of the effectiveness of two brands of motion sickness pills, brand A pills were given to 45 persons randomly selected from a group of 90 air travelers, while the other 45 persons were given brand B pills. The following results were obtained. None. Slight Moderate Severe
Brand A 18 17 6 4
Brand B 11 14 14 6

Required:
a. Test at a significance level of 0.05 if it matters which brand of pill you take and whether it will help. A normal probability plot and test for equal variance has been provided whether you need it or not.
b. Is the degree of nausea associated with the brand of motion sickness pills? Test at α = 0.05.
c. How do we compute the degrees of freedom (dof) of the test statistic in part (a)? Why?

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.

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.