Answer:
-4
-4×-4=16
Step-by-step explanation:
The answer is the first one; it is an arithmetic sequence with a common difference of -40.
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
Yes the researcher can conclude that the supplement has a significant effect on cognitive skill
b

c
The result of this hypothesis test shows that there is sufficient evidence to that the supplement had significant effect.The measure of effect size is large due to the large value of Cohen's d (0.5778 > 0.30 )
Step-by-step explanation:
From the question we are told that
The sample size is n=16
The sample mean is 
The standard deviation is 
The population mean is 
The level of significance is 
The null hypothesis is 
The alternative hypothesis is
Generally the test statistics is mathematically represented as

=> 
=> 
Generally the p-value is mathematically represented as


From the z-table

=> 
=> 
From the obtained values we see that 
Decision Rule
Reject the null hypothesis
Conclusion
There is sufficient evidence to conclude that the supplement has a significant effect on the cognitive skill of elderly adults
Generally the Cohen's d for this study is mathematically represented as

=> 
=> 
Answer:
a) 
And replacing we got:

b) ![E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148](https://tex.z-dn.net/?f=%20E%2880Y%5E2%29%20%3D80%5B%200%5E2%2A0.45%20%2B1%5E2%2A0.2%20%2B2%5E2%2A0.3%20%2B3%5E2%2A0.05%5D%3D%20148)
Step-by-step explanation:
Previous concepts
In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".
The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).
And the standard deviation of a random variable X is just the square root of the variance.
Solution to the problem
Part a
We have the following distribution function:
Y 0 1 2 3
P(Y) 0.45 0.2 0.3 0.05
And we can calculate the expected value with the following formula:

And replacing we got:

Part b
For this case the new expected value would be given by:

And replacing we got
![E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148](https://tex.z-dn.net/?f=%20E%2880Y%5E2%29%20%3D80%5B%200%5E2%2A0.45%20%2B1%5E2%2A0.2%20%2B2%5E2%2A0.3%20%2B3%5E2%2A0.05%5D%3D%20148)