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evablogger [386]
3 years ago
15

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let µ denote the true average re

flectometer reading for a new type of paint under consideration. A test of H0: µ = 20 versus Ha: µ > 20 will be based on a random sample of size n from a normal population distribution. What conclusion is appropriate in each of the following situations? (Round your P-values to three decimal places.)(a) n = 16, t = 3.3, a = 0.05P-value =(b) n = 8, t = 1.8, a = 0.01P-value =(c) n = 26,t = -0.6P-value =
Mathematics
1 answer:
Zolol [24]3 years ago
3 0

Answer:

a) df=n-1=16-1=15

The statistic calculated is given by t=3.3  

Since is a one-side upper test the p value would be:      

p_v =P(t_{15}>3.3)=0.0024  

So since the p value is lower than the significance level pv we reject the null hypothesis.

b) df=n-1=8-1=7

The statistic calculated is given by t=1.8  

Since is a one-side upper test the p value would be:      

p_v =P(t_{7}>1.8)=0.057  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

c) df=n-1=26-1=25

The statistic calculated is given by t=-0.6  

Since is a one-side upper test the p value would be:      

p_v =P(t_{25}>-0.6)=0.723  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation      

\bar X represent the sample mean

s represent the standard deviation for the sample

n sample size      

\mu_o =20 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)      

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the true mean is higher than 20, the system of hypothesis would be:      

Null hypothesis:\mu \leq 20      

Alternative hypothesis:\mu > 20      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

(a) n = 16, t = 3.3, a = 0.05, P-value =

First we need to calculate the degrees of freedom given by:  

df=n-1=16-1=15

The statistic calculated is given by t=3.3  

Since is a one-side upper test the p value would be:      

p_v =P(t_{15}>3.3)=0.0024  

So since the p value is lower than the significance level pv we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

First we need to calculate the degrees of freedom given by:  

df=n-1=8-1=7

The statistic calculated is given by t=1.8  

Since is a one-side upper test the p value would be:      

p_v =P(t_{7}>1.8)=0.057  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

 First we need to calculate the degrees of freedom given by:  

df=n-1=26-1=25

The statistic calculated is given by t=-0.6  

Since is a one-side upper test the p value would be:      

p_v =P(t_{25}>-0.6)=0.723  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

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Read 2 more answers
1.) What are the zeros of the polynomial? f(x)=x^4-x^3-16x^2+4x+48.
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Answer:

3.) \displaystyle [x - 2][x^2 + 2][x + 4]

2.) \displaystyle 2\:complex\:solutions → x^2 + 3x + 6 >> -\frac{3 - i\sqrt{15}}{2}, -\frac{3 + i\sqrt{15}}{2}

1.) \displaystyle 4, -3, 2, and\:-2

Step-by-step explanation:

3.) By the Rational Root Theorem, we would take the Least Common Divisor [LCD] between the leading coefficient of 1, and the initial value of −16, which is 1, but we will take 2 since it is the <em>fourth root</em> of 16; so this automatically makes our first factor of \displaystyle x - 2.Next, since the factor\divisor is in the form of \displaystyle x - c, use what is called Synthetic Division. Remember, in this formula, −c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

2| 1 2 −6 4 −16

↓ 2 8 4 16

__________________

1 4 2 8 0 → \displaystyle x^3 + 4x^2 + 2x + 8

You start by placing the <em>c</em> in the top left corner, then list all the coefficients of your dividend [x⁴ + 2x³ - 6x² + 4x - 16]. You bring down the original term closest to <em>c</em> then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an x³, the 4x² follows right behind it, bringing 2x right up against it, and bringing up the rear, 8, giving you the quotient of \displaystyle x^3 + 4x^2 + 2x + 8.

However, we are not finished yet. This is our first quotient. The next step, while still using the Rational Root Theorem with our first quotient, is to take the Least Common Divisor [LCD] of the leading coefficient of 1, and the initial value of 8, which is −4, so this makes our next factor of \displaystyle x + 4.Then again, we use Synthetic Division because \displaystyle x + 4is in the form of \displaystyle x - c:

−4| 1 4 2 8

↓ −4 0 −8

_____________

1 0 2 0 → \displaystyle x^2 + 2

So altogether, we have our four factors of \displaystyle [x^2 + 2][x + 4][x - 2].

__________________________________________________________

2.) By the Rational Root Theorem again, this time, we will take −1, since the leading coefficient & variable\degree and the initial value do not share any common divisors other than the <em>special</em><em> </em><em>number</em> of 1, and it does not matter which integer of 1 you take first. This gives a factor of \displaystyle x + 1.Then start up Synthetic Division again:

−1| 1 3 5 −3 −6

↓ −1 −2 −3 6

__________________

1 2 3 −6 0 → \displaystyle x^3 + 2x^2 + 3x - 6

Now we take the other integer of 1 to get the other factor of \displaystyle x - 1,then repeat the process of Synthetic Division:

1| 1 2 3 −6

↓ 1 3 6

_____________

1 3 6 0 → \displaystyle x^2 + 3x + 6

So altogether, we have our three factors of \displaystyle [x - 1][x^2 + 3x + 6][x + 1].

Hold it now! Notice that \displaystyle x^2 + 3x + 6is unfactorable. Therefore, we have to apply the Quadratic Formula to get our two complex solutions, \displaystyle a + bi[or zeros in this matter]:

\displaystyle -b ± \frac{\sqrt{b^2 - 4ac}}{2a} = x \\ \\ -3 ± \frac{\sqrt{3^2 - 4[1][6]}}{2[1]} = x \\ \\ -3 ± \frac{\sqrt{9 - 24}}{2} = x \\ \\ -3 ± \frac{\sqrt{-15}}{2} = x \\ \\ -3 ± i\frac{\sqrt{15}}{2} = x \\ \\ -\frac{3 - i\sqrt{15}}{2}, -\frac{3 + i\sqrt{15}}{2} = x

__________________________________________________________

1.) By the Rational Root Theorem one more time, this time, we will take 4 since the initial value is 48 and that 4 is the root of the polynomial. This gives our automatic factor of \displaystyle x - 4.Then start up Synthetic Division again:

4| 1 −1 −16 4 48

↓ 4 12 −16 −48

___________________

1 3 −4 −12 0 → \displaystyle x^3 + 3x^2 - 4x - 12

We can then take −3 since it is a root of this polynomial, giving us the factor of \displaystyle x + 3:

−3| 1 3 −4 −12

↓ −3 0 12

_______________

1 0 −4 0 → \displaystyle x^2 - 4 >> [x - 2][x + 2]

So altogether, we have our four factors of \displaystyle [x - 2][x + 3][x + 2][x - 4],and when set to equal zero, you will get \displaystyle 4, -3, 2, and\:-2.

I am delighted to assist you anytime.

3 0
3 years ago
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