1.
Aud before The correlation value is 0.992 which is higher than our critical value (0.987). We can infer that the data is normally distributed.
Aud afterThe correlation value is 0.995 which is higher than our critical value (0.987). We can infer that the data is normally distributed.
Vis before The correlation value is 0.994 which is higher than our critical value (0.987). We can infer that the data is normally distributed.
Vua after The correlation value is 0.997 which is higher than our critical value (0.987). We can infer that the data is normally distributed.
According to our QQ plot, all of the data is normally distributed.
2. I used a hypothesis test to determine if H0; U=600. HA;> 600.
Our given alpha value is 0.05. After testing our hypothesis, i found the P value to be 0.0016. P value is less than alpha, therefor we reject the null hypothesis. There is not sufficent evidence to conclude that HA; U>600.
3.At this time, there is not sufficent evidence to conclude that the medication has helped. We will reject the hypothesis because the P value is less than the alpha value.
4.
There's not a large variance between auditory before and after in the given data. This can be interpreted as, the responce has not decreased. Meaning, the effect of the trial did not play a significant role in the "after auditory".
5.
There is enough evidence to conclude that this is unusually low. Considering out P value was less than our alpha value.
6.
There is not enough information collected from the before medication data to conclude that the mean time is greater than 440. In this case we would reject our hypothesis because p value is less than alpha.
7. Finally, we are able to conclude that there is enough evidence to support the statement that the medication is improving visual response times. We would reject the hypothesis because our p value is now 0.0017 which is smaller than our alpha value (0.05).
One sample T hypothesis test:
μ : Mean of variable H_{0} : μ = 600 H_{A} : μ > 600 Hypothesis test results:

Two sample T hypothesis test:
μ_{1} : Mean of AudBefore μ_{2} : Mean of AudAfter μ_{1}  μ_{2} : Difference between two means H_{0} : μ_{1}  μ_{2} = 0 H_{A} : μ_{1}  μ_{2} > 0 (without pooled variances) Hypothesis test results:

Two sample variance hypothesis test:
σ_{1}^{2} : Variance of AudBefore σ_{2}^{2} : Variance of AudAfter σ_{1}^{2}/σ_{2}^{2} : Ratio of two variances H_{0} : σ_{1}^{2}/σ_{2}^{2} = 0 H_{A} : σ_{1}^{2}/σ_{2}^{2} > 0 Hypothesis test results:

One sample proportion hypothesis test:
Outcomes in : AudBefore Success : 121 p : Proportion of successes H_{0} : p = 0.5 H_{A} : p < 0.5 Hypothesis test results:

One sample T hypothesis test:
μ : Mean of variable H_{0} : μ = 440 H_{A} : μ > 440 Hypothesis test results:

Two sample T hypothesis test:
μ_{1} : Mean of VisBefore μ_{2} : Mean of VisAfter μ_{1}  μ_{2} : Difference between two means H_{0} : μ_{1}  μ_{2} = 0 H_{A} : μ_{1}  μ_{2} > 0 (without pooled variances) Hypothesis test results:

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