A 2-year old mice can solve a simple maze in 30 seconds on average, with a standard deviation of 2 seconds. In a recent experiment, a group of 25 2-year old mice were given the blood of younger mice to see if there is any improvement in terms of cognitive functions. These mice were observed to solve the same maze in 29 seconds. Can we say that the blood transfusion improved the cognitive functions of 2-year old mice (i. e. shorter time to solve the maze)? Required:
a. State the hypotheses.
b. What is the p-value?
c. Using α = 0.05, what is your conclusion?

a)

Null Hypothesis            H₀ : μ = 30

Alternative Hypothesis H₁ : μ < 30

b)

p-value = 0.0099

c)

Since 0.0099 < 0.05

Therefore, there is enough evidence to conclude that blood transfusion improved the cognitive functions of 2-year old mice.  

Step-by-step explanation:

Given the data in the question,  

a) State the hypotheses.

Null Hypothesis            H₀ : μ = 30

Alternative Hypothesis H₁ : μ < 30

b) What is the p-value

Test statistics

t = ( - μ) / s/√n

t = (29 - 30) / (2/√25)

t = -1 / 0.4

t = -2.5

degree of freedom df = n-1 = 25 - 1 = 24

from table

p-value = 0.0099

c) Using α = 0.05, what is your conclusion?

p-value = 0.0099

α = 0.05

0.0099 < 0.05

So, there is enough evidence to conclude that blood transfusion improved the cognitive functions of 2-year old mice.  

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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