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castortr0y [4]
1 year ago
15

Given that the sum of squares for error is 60 and the sum of squares for regression is 140, then the coefficient of determinatio

n is:
Mathematics
1 answer:
Anastasy [175]1 year ago
7 0

Answer:

<h2>0.7</h2>

Step-by-step explanation:

The coefficient of determination which is also known as the R² value is expressed as shown;

R^{2} = \frac{sum\ of \ squares \ of \  regression}{sum\ of \ squares \ of total}

Sum of square of total (SST)= sum of square of error (SSE )+ sum of square of regression (SSR)

Given SSE = 60 and SSR = 140

SST = 60 + 140

SST = 200

Since R² = SSR/SST

R² = 140/200

R² = 0.7

Hence, the coefficient of determination is 0.7. Note that the coefficient of determination always lies between 0 and 1.

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Draw a simple branch diagram to work the probabilities out. You find that the chance of a poisonous mushroom is 0.08 and the chance of a red poisonous is 0.04. So the probability that a poisonous mushroom is red is 1/2 or 0.5.
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A shoe manufacturer compared material A and material B for the soles of shoes. Twelve volunteers each got two shoes. The left wa
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Answer:

a) Are dependent since we are mesuring at the same individuals but on different times and with a different method

b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

c) p_v =P(t_{(12)}>0.969) =0.353

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".

Let put some notation  

x=value for A , y = value for B

A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392

B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411

(a) Are the two samples paired or independent? Explain your answer.

Are dependent since we are mesuring at the same individuals but on different times and with a different method

(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?

The first step is calculate the difference d_i=A_i-B_i and we obtain this:

d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19

In order to do the qqplot we can use the following R code:

d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)

qqnorm(d)

And the graph obtained is attached.

If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem

The system of hypothesis for this case are:

Null hypothesis: \mu_A- \mu_B = 0

Alternative hypothesis: \mu_A -\mu_B \neq 0

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{80}{12}=6.67

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =23.849

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{6.67 -0}{\frac{23.849}{\sqrt{12}}}=0.969

The next step is calculate the degrees of freedom given by:

df=n-1=12-1=11

Now we can calculate the p value, since we have a two tailed test the p value is given by:

p_v =P(t_{(12)}>0.969) =0.353

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

4 0
2 years ago
The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If
frutty [35]
For this case we have the following equation:
 w = 0.9* \sqrt{E}
 Where,
 w: The weight of a spring in pounds
 E: the energy stored by the spring in joules.
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 w = 0.9* \sqrt{12}
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the approximate weight of the spring in pounds is:
 
w=3.12
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Which could be the dimensions of a rectangular prism whose surface area is greater than 140 square feet? Select three options,
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Answer:

These are the rectangular prisms that have a surface area greater than 140 square feet

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Step-by-step explanation:

The formula to find the surface area of a rectangular prism is SA=2(wl+hl+hw)

All you have to do is go through all the options given to you to figure out which retangular prisms have a surface area greater than 140 square feet.

6 feet by 2 feet by 3 feet

SA = 2((2)(6)+(3)(6)+(3)(2))

SA = 2(12 + 18 + 6)

SA = 2(36)

SA = 72

6 feet by 5 feet by 4 feet

SA = 2((6)(5) + (4)(5) + (4)(6))

SA = 2(30 + 20 + 24)

SA = 2(74)

SA = 148

7 feet by 6 feet by 4 feet

SA = 2((7)(6) + (4)(6) + (4)(7))

SA = 2(42 + 24 + 28)

SA = 2(94)

SA = 188

8 feet by 3 feet by 7 feet

SA = 2((8)(3)+(7)(3)+(7)(8))

SA = 2(24 + 21 + 56)

SA = 2(101)

SA = 202

8 feet by 4 feet by 3 feet

SA = 2((8)(4)+(3)(4)+(3)(8))

SA = 2(32 + 12 + 24)

SA = 2(68)

SA = 136

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2 years ago
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Answer: 5 units

Step-by-step explanation:

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