Answer:
0.00205
Step-by-step explanation:
This is binomial distribution problem.
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = 7
x = Number of successes required = 2
p = probability of success = 0.830
q = probability of failure = 1 - 0.83 = 0.17
P(X =2) = ⁷C₂ 0.83² 0.17⁷⁻²
P(X =2) = ⁷C₂ 0.83² 0.17⁵ = 0.00205
Fewer young people are driving. In 1983, of -year-olds had a driver's license. Twenty-five years later that percentage had dropped to (University of Michigan Transportation Research Institute website, April 7, 2012). Suppose these results are based on a random sample of -year-olds in 1983 and again in 2008. a. At confidence, what is the margin of error and the interval estimate of the number of -year-old drivers in 1983
Answer:
Hence the values for MOE and are interval estimate of 19 yr old age drivers in 1983 are 1%(0.01) and 19±0.0196
Step-by-step explanation:
Given: Refer the paper university of Michigan transportation research Institute website ,April,2012.
It say that 19 year old group age people ,about 87% has driving license .
And of 1200 has a random sample space ,in 1983.
To Find : MOE(margin of error ) and Interval Estimate in 1983.
Solution:
Given that ,there is sample space of 1200 19 yr old people and of which 87% has driving license.
hence , we get that
87% of 1200 = sample size.
sample size =1044 members has driving license out of 1200.
Consider 95 % of confidence level,
for that Z-score is required.
calculating the Alpha for that=1-confidence level.
=1-0.95=0.05
therefore Z-alpha=Z(0.05)=1.96
1)MOE=margin of error is given by ,
=[tex]Z-alpha*\sqrt{\frac{p(1-p)}{N} }[/tex]
here p=0.87 and N=total size =1200.
MOE=1.96*[tex]\sqrt{\frac{0.87(1-0.87)}{1200} }[/tex] =1 %.
2) Interval estimate is given by ,
For that we should know mean,standard deviation and sample size,
we are calculating for 19 year old age of entire age of population,
Hence mean will be 19 yr-old age.
Mean=19.
Sample size=1044.
standard deviation given by,
=[tex]p*(\sqrt{(1-p)}[/tex]
=0.87*[tex]\sqrt{0.13}[/tex]=0.3136.
Hence now calculating the interval estimate ,with 95% confidence level,
μ=M±Z(Standard error )
standard error=standard deviation/sqrt(sample size)
=0.3136/32.31=0.01.
μ=19±1.96*0.01
μ=19±0.0196.
A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times. We'll assume we can make "perfect folds," where each fold makes the folded paper exactly twice as thick as before - and we can make as many folds as we want. Write a function g g that determines the thickness of the folded paper (in mm) in terms of the number folds made, n n. (Notice that g ( 0 )
Answer:
g(n) = 0.05·2^n
Step-by-step explanation:
The paper with no folds is 0.05 mm thick. Each fold multiplies the thickness by 2, so the function is ...
g(n) = 0.05·2^n
_____
Comment on paper folding
In practice, where the paper must bend around the fold, it is impossible to fold an ordinary piece of paper 9 times. You may be able to fold a very large, very thin piece of paper that many times.
The thickness of a standard piece of paper, after it has been folded n times, can be expressed as g(n) = 0.05 * 2^n. This formula takes into account that each fold doubles the thickness of the paper.
Explanation:In order to calculate the thickness of a standard piece of paper after it has been folded a certain number of times, we may use the concept of exponential growth. In this case, we use a power of 2 corresponding to the number of folds made, because with each fold, the thickness doubles. This gives us the function g(n) = 0.05 * 2^n, where n corresponds to the number of folds.
For example, if we fold the paper once (n=1), we get a thickness of 0.05 * 2^1 = 0.10 mm. If we fold the paper twice (n=2), we get a thickness of 0.05 * 2^2 = 0.20 mm, and so on, with the thickness doubling each time we make a new fold.
This mathematical model assumes perfect folding and does not account for physical limitations we would encounter when attempting to fold paper in reality.
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Find the probability that the proportion of individuals in the sample of 225 who hold multiple jobs is between 0.14 and 0.18. Round the answer to at least four decimal places. The probability that the proportion of individuals in the sample of 225 hold multiple jobs is between 0.14 and 0.18 i
Answer:
Step-by-step explanation:
Hello!
The variable of interest is X: number of individuals that hold multiple jobs in a sample of 225.
And you need to calculate the probability of the proportion of individuals being between 0.14 and 0.18.
Considering that the parameter of interest is the proportion and the sample is large enough you can use the standard normal approximation to calculate this interval.
[tex]Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]
Unfortunately, there is no given value for the population proportion of individuals that have multiple jobs. Let's say, for example, that his proportion is 10%.
You can symbolize the probabilities as:
P(0.14≤X≤0.18)= P(X≤0.18)-P(X≤0.14)
Using the approximation of the standard normal you can standardize these proportion values:
P(Z≤(0.18-0.1)/√[(0.1*0.9)/225])-P(Z≤(0.14-0.1)/√[(0.1*0.9)/225])
P(Z≤4)-P(Z≤2)
Now you have to look for the corresponding values in the table of the Z distribution (right table, positive numbers of Z)
P(Z≤4)-P(Z≤2)= 1 - 0.97725= 0.02275
I hope it helps!
Answer:
0.1276
Step-by-step explanation:
Let sample space = S
∴ n(S) = 225
Probability of multiple jobs A, that is, P(A) = 0.14
Probability of multiple jobs B, that is, P(B) = 0.18
Hence, P(A) = n(A)/n(S)
∴ 0.14 = n(A)/225
n(A) = 0.14 X 225 = 31.5%
Also, P(B) = n(B)/n(S)
∴ 0.18 = n(B)/225
n(B) = 0.18 X 225 = 40.5%
The probability that the multiple jobs, P( A2 ∩ B2) = P(A2) X P(B2)
P(A2) = 0.315 and P(B2) = 0.405
∴ P(A2 ∩ B2) = P(A2) X P(B2) = 0.315 X 0.405 = 0.1276
1.) Evaluate the indicated function, where f(x) = x2 − 8x + 3and g(x) = 7x − 5.
(f + g)(5)=
2.) Evaluate the indicated function, where f(x) = x2 − 7x + 4 and g(x) = 6x − 7.
(f + g)(1/2)=
3.) Evaluate the indicated function, where f(x) = x2 − 3x + 4 and g(x) = 4x − 2.
(f − g)(−1)=
4.) Evaluate the indicated function, where f(x) = x2 − 4x + 3and g(x) = 3x − 2.
(fg)(7)=
5.) Evaluate the indicated function, where f(x) = x2 − 3x + 2and g(x) = 4x − 8.
(f/g)(−2)=
6.) Find (g ○ f)(x) and (f ○ g)(x) for the given functions f and g. f(x) = 2x − 8, g(x) = 3x + 1
7.) Find (g ○ f)(x) and (f ○ g)(x) for the given functions f and g.
f(x) = 3/x+5, g(x) = 3x − 6
8.) Evaluate the composite function, where f(x) = 2x + 3,g(x) = x2 − 5x, and h(x) = 4 − 3x2.
(f ○ g)(5)=
Answer:
1) (f + g) (5) = 18, 2) (f + g) (1/2) = - 13/4, 3) (f - g) (-1) = 14, 4) (f * g) (7) = 456, 5) (f/g) (- 2) = - 3/4, 6) (g ○ f)(x) = 6 x - 23, (f ○ g)(x) = 6 x - 6, 7) [tex](g \circ f) (x) = \frac{9}{x} - 24[/tex], [tex](f \circ g) (x) = \frac{1}{x - 2} + 5[/tex], 8) (f ○ g)(5) = 3
Step-by-step explanation:
1) [tex](f + g) (x) = x^{2}-8\cdot x + 3 + 7 \cdot x - 5\\(f + g) (x) = x^{2} - x - 2\\(f + g) (5) = 18[/tex]
2) [tex](f + g) (x) = x^{2} - 7 \cdot x + 4 + 6 \cdot x - 7\\(f + g) (x) = x^{2} - x - 3\\(f + g) (\frac{1}{2} ) = - \frac{13}{4}[/tex]
3) [tex](f - g) (x) = x^{2} - 3 \cdot x + 4 - 4 \cdot x + 2\\(f - g) (x) = x^{2} - 7 \cdot x + 6\\(f - g) (-1) = 14[/tex]
4) [tex](f \cdot g) (x) = (x^{2}-4\cdot x + 3) \cdot (3\cdot x - 2)\\(f \cdot g) (7) = 456[/tex]
5) [tex](f / g) (x) = \frac{x^{2}-3\cdot x + 2}{4 \cdot x - 8} \\(f / g) (-2) = - \frac{3}{4}[/tex]
6) [tex](g \circ f) (x) = 3 \cdot (2 \cdot x - 8) + 1\\(g \circ f) (x) = 6 \cdot x - 23\\(f \circ g) (x) = 2 \cdot (3 \cdot x + 1) - 8\\(f \circ g) (x) = 6 \cdot x - 6[/tex]
7) [tex](g \circ f) (x) = 3 \cdot (\frac{3}{x} - 6) - 6\\(g \circ f) (x) = \frac{9}{x} - 24\\(f \circ g) (x) = \frac{3}{3 \cdot x - 6} + 5\\(f \circ g) (x) = \frac{1}{x - 2} + 5[/tex]
8) [tex](f \circ g) (x) = 2 \cdot (x^{2} - 5 \cdot x) + 3\\(f \circ g) (x) = 2 \cdot x ^{2} - 10 \cdot x + 3\\(f \circ g) (5) = 3[/tex]
1. (f + g)(5) = 18.
2. (f + g)(1/2) = -13/4.
3. (f - g)(-1) = 14.
4. (fg)(7) = 456.
5. (f/g)(-2) = -3/4.
6. (g ○ f)(x) = 6x - 23, (f ○ g)(x) = 6x - 6.
7. (g ○ f)(x) = [tex]\(\frac{9}{x+5} - 6\), (f ○ g)(x) = \(\frac{3}{3x-1}\).[/tex]
8. (f ○ g)(5) = 3.
Let's tackle each problem step by step:
1.Evaluate (f + g)(5):
Step 1: Find f(5) and g(5).
[tex]\(f(5) = (5)^2 - 8(5) + 3 = 25 - 40 + 3 = -12\)[/tex]
[tex]\(g(5) = 7(5) - 5 = 35 - 5 = 30\)[/tex]
Step 2: Add f(5) and g(5).
[tex]\((f + g)(5) = f(5) + g(5) = -12 + 30 = 18\)[/tex]
2. Evaluate (f + g)(1/2):
Step 1: Find f(1/2) and g(1/2).
[tex]\(f(1/2) = (1/2)^2 - 7(1/2) + 4 = 1/4 - 7/2 + 4 = 1/4 - 14/4 + 16/4 = 3/4\)[/tex]
[tex]\(g(1/2) = 6(1/2) - 7 = 3 - 7 = -4\)[/tex]
Step 2: Add f(1/2) and g(1/2).
[tex]\((f + g)(1/2) = f(1/2) + g(1/2) = 3/4 - 4 = -13/4\)[/tex]
3.Evaluate (f − g)(−1):
Step 1: Find f(−1) and g(−1).
[tex]\(f(-1) = (-1)^2 - 3(-1) + 4 = 1 + 3 + 4 = 8\)[/tex]
[tex]\(g(-1) = 4(-1) - 2 = -4 - 2 = -6\)[/tex]
Step 2: Subtract g(-1) from f(-1).
[tex]\((f - g)(-1) = f(-1) - g(-1) = 8 - (-6) = 8 + 6 = 14\)[/tex]
4.Evaluate (fg)(7):
Step 1: Find f(7) and g(7).
[tex]\(f(7) = (7)^2 - 4(7) + 3 = 49 - 28 + 3 = 24\)[/tex]
[tex]\(g(7) = 3(7) - 2 = 21 - 2 = 19\)[/tex]
Step 2: Multiply f(7) by g(7).
[tex]\((fg)(7) = f(7) \times g(7) = 24 \times 19 = 456\)[/tex]
5. Evaluate (f/g)(−2):
Step 1: Find f(-2) and g(-2).
[tex]\(f(-2) = (-2)^2 - 3(-2) + 2 = 4 + 6 + 2 = 12\)[/tex]
[tex]\(g(-2) = 4(-2) - 8 = -8 - 8 = -16\)[/tex]
Step 2: Divide f(-2) by g(-2).
[tex]\((f/g)(-2) = \frac{f(-2)}{g(-2)} = \frac{12}{-16} = -\frac{3}{4}\)[/tex]
6. Find (g ○ f)(x) and (f ○ g)(x) for f(x) = 2x − 8, g(x) = 3x + 1:
[tex]\((g ○ f)(x) = g(f(x)) = g(2x - 8) = 3(2x - 8) + 1 = 6x - 24 + 1 = 6x - 23\)[/tex]
[tex]\((f ○ g)(x) = f(g(x)) = f(3x + 1) = 2(3x + 1) - 8 = 6x + 2 - 8 = 6x - 6\)[/tex]
7. Find (g ○ f)(x) and (f ○ g)(x) for f(x) = 3/x+5, g(x) = 3x − 6:
[tex]\((g ○ f)(x) = g(f(x)) = g(\frac{3}{x+5}) = 3(\frac{3}{x+5}) - 6 = \frac{9}{x+5} - 6\)[/tex]
[tex]\((f ○ g)(x) = f(g(x)) = f(3x - 6) = \frac{3}{3x-6+5} = \frac{3}{3x-1}\)[/tex]
8. Evaluate (f ○ g)(5):
Step 1: Find g(5).
[tex]\(g(5) = 5^2 - 5 \times 5 = 25 - 25 = 0\)[/tex]
Step 2: Find f(g(5)).
[tex]\(f(g(5)) = f(0) = 2 \times 0 + 3 = 3\)[/tex]
In the beginning of the study, a randomly selected group of 121 students scored an average of 252 words per minute on the reading speed test. Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M
Answer:
Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M will be approximately normal.
Step-by-step explanation:
We use the central limit theorem to solve this question.
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a sample size larger than 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
So
Since the sample size is larger than 30, the cognitive psychologist can assume that the sampling distribution of M will be approximately normal.
The area of a rectangle is represented by the function x3 − 2x2 − 40x − 64. The width of the rectangle is x + 4. Find the expression representing the length of the rectangle.
Answer: Length = x² - 6x - 16
Step-by-step explanation:
The formula for determining the area of a rectangle is expressed as. Area = length × width
Length = Area/width
The area of a rectangle is represented by the function
x³ - 2x² - 40x - 64. The width of the rectangle is x + 4. Therefore,
Length = (x³ - 2x² - 40x - 64)/(x + 4)
We would apply the method of long division. The steps are shown in the attached photo. From the photo,
Length = x² - 6x - 16
Hey love! <3
Answer:
⋆ ☄.
·˚ * I see your answer up in the stars! It's x^2 − 6x − 16 or C on your FLVS quiz!
Step-by-step explanation:
If area = length * width and our width is given as x + 4, then the length is found by dividing the area by the width; You could do that using long division, but it's easier using synthetic division. That's what we'll do here for example.
-4 | 1 -2 -40 -64
This is only the start! The -4 inside the box comes from the factor you are dividing by. If x + 4 = 0, then x = -4.. The numbers after are the coefficients from each descending power of x. Multiply the -4 by the 1 and put that product up under the -2 and add to get:
-4 | 1 -2 -40 -64
After you finish the multiplication and simplification process you receive your given answer = x^2 − 6x − 16
Hope this helped you baby! Be sure to drop me a brainliest (no pressure!) (*・∀・*)人(*・∀・*) Sincerely, Kelsey from Brainly.
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Chase consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Chase's body decreases exponentially. The 10-hour decay factor for the number of mg of caffeine in Chase's body is 0.2542. a. What is the 5-hour growth/decay factor for the number of mg of caffeine in Chase's body? b. What is the 1-hour growth/decay factor for the number of mg of caffeine in Chase's body?c. If there were 171 mg of caffeine in Chase's body 1.39 hours after consuming the energy drink, how many mg of caffeine is in Chase's body 2.39 hours after consuming the energy drink?
Answer:
(a) The 5-hour decay factor is 0.5042.
(b) The 1-hour decay factor is 0.8720.
(c) The amount of caffeine in Chase's body 2.39 hours after consuming the drink is 149.112 mg.
Step-by-step explanation:
The amount of caffeine in Chase's body decreases exponentially.
The 10-hour decay factor for the number of mg of caffeine is 0.2542.
The 1-hour decay factor is:
[tex]1-hour\ decay\ factor=(0.2542)^{1/10}=0.8720[/tex]
(a)
Compute the 5-hour decay factor as follows:
[tex]5-hour\ decay\ factor=(0.8720)^{5}\\=0.504176\\\approx0.5042[/tex]
Thus, the 5-hour decay factor is 0.5042.
(b)
The 1-hour decay factor is:
[tex]1-hour\ decay\ factor=(0.2542)^{1/10}=0.8720[/tex]
Thus, the 1-hour decay factor is 0.8720.
(c)
The equation to compute the amount of caffeine in Chase's body is:
A = Initial amount × (0.8720)ⁿ
It is provided that initially Chase had 171 mg of caffeine, 1.39 hours after consuming the drink.
Compute the amount of caffeine in Chase's body 2.39 hours after consuming the drink as follows:
[tex]A = Initial\ amount \times (0.8720)^{2.39} \\=[Initial\ amount \times (0.8720)^{1.39}] \times(0.8720)\\=171\times 0.8720\\=149.112[/tex]
Thus, the amount of caffeine in Chase's body 2.39 hours after consuming the drink is 149.112 mg.
The 5-hour growth/decay factor is 0.561, the 1-hour growth/decay factor is 0.869, and there are 154 mg of caffeine in Chase's body 2.39 hours after consuming the energy drink.
Explanation:a. To find the 5-hour growth/decay factor, we raise the 10-hour decay factor to the power of (5/10). So, the 5-hour growth/decay factor is 0.2542^(5/10) = 0.561.
b. To find the 1-hour growth/decay factor, we raise the 10-hour decay factor to the power of (1/10). So, the 1-hour growth/decay factor is 0.2542^(1/10) = 0.869.
c. To find the number of mg of caffeine in Chase's body 2.39 hours after consuming the energy drink, we multiply the initial amount of caffeine by the 2.39-hour decay factor. So, the number of mg is 171 * 0.2542^(2.39/10) = 154.
Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478]. Use this to answer all parts. What was the point estimate used to estimate the mean height of all adult males in Idaho?
Answer:
The point estimate used to estimate the mean height of all adult males in Idaho was 69.505 inches.
Step-by-step explanation:
The point estimate is the halfway point of the confidence interval, that is, the lower bound added to the upper bound, and then this sum is divided by 2. So
Lower bound: 62.535
Upper bound: 76.478
Point estimate:
[tex]P_{e} = \frac{62.535 + 76.478}{2} = 69.505[/tex]
The point estimate used to estimate the mean height of all adult males in Idaho was 69.505 inches.
The point estimate used to estimate the mean height of all adult males in Idaho was 69.505 inches.
Calculation of the estimation of the pointSince Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478].
So, the estimation of the point is
[tex]= (62.532 + 76.478) \div 2[/tex]
= 69.505 inches
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Suppose that we have conducted a Simple Linear Regression for Exam 1 score by Homework 1 score and found the predicted line equation to be y_hat = 58.52 + 2.19x, where x represents Homework 1 score and y represents Exam 1 score. What Exam 1 score can a student who did not submit the homework expect to receive based on this predicted line equation? Group of answer choices y_hat 117.04 x 2.19 58.52
Answer:
58.52
Step-by-step explanation:
The predicted regression equation for predicting Exam 1 score is
y_hat=58.52+2.19x.
We have to find the predicted exam score 1 for student who did not submit homework. If the student did not submit homework 1 score then the homework 1 score will be zero. So,
y_hat=58.52+2.19(0)
y_hat=58.52.
Thus, the predicted exam score 1 for student who did not submit homework 1 score is 58.52.
. Find sets of parametric equations and symmetric equations of the line through the point parallel to the given vector or line (if possible).
Point (-4,0,2)
Parallel to v=2i + 8j - 7k
(a) parametric equations (Enter your answers as a comma-separated list.)
(b) symmetric equations
A. 2x= y/8 = 7z
B. (x+4)/2 = y/8 = (2-z)/7
C x/2 = y = z/7
D. (x-4)/2 = y = z/7
Answer:
a) L(x,y,z) = (-4,0,2)+(2,8,-7)*t
b) (2-z)/7= y/8=(x+4)/2 (option B)
Step-by-step explanation:
the parametric equation of the line passing through the point P₀= (-4,0,2) and parallel to the vector v=2i + 8j - 7k is
L(x,y,z)=P₀+v*t
therefore
L(x,y,z) = (-4,0,2)+(2,8,-7)*t
or
x=x₀+vx*t = -4 + 2*t
y=y₀+vy*t = 8*t
z=z₀+vz*t = 2 -7*t
solving for t in the 3 equations we get the symmetric equation of the line:
(2-z)/7= y/8=(x+4)/2
thus the option B is correct
Answer:
a) Parametric equations
x = -4 + 2t
y = 8t
z = 2-7t
b) symmetric equations
[tex]\frac{x+4}{2}= \frac{y}{8} = \frac{z+2}{-7}[/tex] The answer is the option B
Step-by-step explanation:
For writing the vectorial equation of a line, we need a point in the line and its director vector, thus:
[tex]L: (x_{0},y_{0},z_{0} ) + t(a,b,c)[/tex]
Where [tex](x_{0},y_{0},z_{0})[/tex] is a point in the line
(a,b,c) is the director vector
Then
[tex]L: (-4,0,2) + t(2,8,-7)[/tex]
a) Parametric equations
Since the vectorial equation, we can obtain the parametric equations writing the equation for each component
x = -4 + 2t
y = 8t
z = 2-7t
b)Symmetric equations
Since the parametric equations, we isolate the parameter t
[tex]\frac{x+4}{2}= \frac{y}{8} = \frac{z+2}{-7}[/tex]
The Toylot company makes an electric train with a motor that it claims will draw an average of only 0.8 ampere (A) under a normal load. A sample of eleven motors was tested, and it was found that the mean current was x = 1.20 A, with a sample standard deviation of s = 0.42 A. Do the data indicate that the Toylot claim of 0.8 A is too low? (Use a 1% level of significance.)
What is the value of the sample test statistic?
Answer:
Test statistic = 3.1587
Step-by-step explanation:
We are given that the Toy-lot company makes an electric train with a motor that it claims will draw an average of only 0.8 ampere (A).
Also, a sample of eleven motors was tested, and it was found that the mean current was x = 1.20 A, with a sample standard deviation of s = 0.42 A.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 0.8 { claim of 0.8 A is not low}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 0.8 { claim of 0.8 A is too low}
Now, the test statistics used here will be;
T.S. = [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, X bar = sample mean = 1.20 A
s = sample standard deviation = 0.42 A
n = sample size = 11 motors
So, Test statistics = [tex]\frac{1.20 - 0.8}{\frac{0.42}{\sqrt{11} } }[/tex] ~ [tex]t_1_0[/tex]
= 3.1587
At 1% level of significance, t table gives a critical value of 2.764 at 10 degree of freedom. Since our test statistics is higher than the critical value so we have sufficient evidence to reject null hypothesis .
Therefore, we conclude that Toy-lot claim of 0.8 A is too low.
To determine if the Toylot company claim of 0.8 A is too low, a one-sample t-test can be used. The sample test statistic is calculated by subtracting the hypothesized mean from the sample mean, divided by the sample standard deviation divided by the square root of the sample size. This statistic can then be compared to the critical value from the t-distribution at the specified level of significance to determine if the claim is too low.
Explanation:To determine if the Toylot company claim of 0.8 A is too low, we can conduct a hypothesis test using the sample data. We will use a one-sample t-test with a null hypothesis that the mean current is equal to 0.8 A. The alternative hypothesis will be that the mean current is greater than 0.8 A.
The sample test statistic is calculated by subtracting the hypothesized mean from the sample mean, and dividing by the sample standard deviation divided by the square root of the sample size. In this case, the sample test statistic is (1.20 - 0.8) / (0.42 / sqrt(11)) = 5.11.
To determine if this test statistic is statistically significant, we compare it to the critical value from the t-distribution with 10 degrees of freedom at a 1% level of significance. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the Toylot claim of 0.8 A is too low.
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A pair of fair dice is rolled. (a) What is the probability that both dice show the same number? (b) What is the probability that both dice show different numbers? (c) What is the probability that the second die lands on a lower value than does the first?
Answer:
a) 16.7% probability that both dice show the same number
b) 83.3% probability that both dice show different numbers
c) 41.67% probability that the second die lands on a lower value than does the first.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this problem, we have these possible outcomes:
Format(Dice A, Dice B)
(1,1), (1,2), (1,3), (1,4), (1,5),(1,6)
(2,1), (2,2), (2,3), (2,4), (2,5),(2,6)
(3,1), (3,2), (3,3), (3,4), (3,5),(3,6)
(4,1), (4,2), (4,3), (4,4), (4,5),(4,6)
(5,1), (5,2), (5,3), (5,4), (5,5),(5,6)
(6,1), (6,2), (6,3), (6,4), (6,5),(6,6)
There are 36 possible outcomes.
(a) What is the probability that both dice show the same number?
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
6 outcomes in which both dice show the same number.
6/36 = 0.167
16.7% probability that both dice show the same number
(b) What is the probability that both dice show different numbers?
The other 30 outcomes
30/36 = 0.833
83.3% probability that both dice show different numbers
(c) What is the probability that the second die lands on a lower value than does the first?
(2,1)
(3,1), (3,2)
(4,1), (4,2), (4,3)
(5,1), (5,2), (5,3), (5,4)
(6,1), (6,2), (6,3), (6,4), (6,5)
15 outcomes in which the second die lands on a lower value than does the first.
15/36 = 0.4167
41.67% probability that the second die lands on a lower value than does the first.
When you generate a random number using your calculator, the random number you get is uniformly distributed over the interval (0, 1). Suppose 50 people in Stat 322 class generate one random number each (independently). X
Answer:
uniformly distributed random numbers here the MATLAB code to find for any number of people and interval
X = rand
X = rand(n)
X = rand(sz1,...,szN)
X = rand(sz)
X = rand(___,typename)
X = rand(___,'like',p)
where
X = rand returns a single uniformly distributed random number in the interval (0,1).
X = rand(n) returns an n-by-n matrix of random numbers.
X = rand(sz1,...,szN) returns an sz1-by-...-by-szN array of random numbers where sz1,...,szN indicate the size of each dimension. For example, rand(3,4) returns a 3-by-4 matrix.
X = rand(sz) returns an array of random numbers where size vector sz specifies size(X). For example, rand([3 4]) returns a 3-by-4 matrix.
X = rand(___,typename) returns an array of random numbers of data type typename. The typename input can be either 'single' or 'double'. You can use any of the input arguments in the previous syntaxes.
X = rand(___,'like',p) returns an array of random numbers like p; that is, of the same object type as p. You can specify either typename or 'like', but not both.
help, ill mark brainliest
Answer: FIRST OPTION.
Step-by-step explanation:
First, it is important to remember that the Slope-Intercept form of the equation of a line is the shown below:
[tex]y=mx+b[/tex]
Where "m" is the slope of the line and "b" is the y-intercept.
By definiton, given a System of Linear equations, if they are exactly the same line, then the System of equations have Infinely many solutions.
In this case you have the following System of Linear equations given in the exercise:
[tex]\left \{ {{y=-2x+5} \atop {y=ax+b}} \right.[/tex]
So, since you need the system has Infinite solutions, you know that the slope and the y-intercept of both lines must be equal.
Therefore, you can identify that the value of "a" and "b" must be the following:
[tex]a=-2\\\\b=5[/tex]
So the Linear System would be the shown below:
[tex]\left \{ {{y=-2x+5} \atop {y=-2x+5}} \right.[/tex]
Researchers are interested in studying how to maintain weight loss. Based on a survey of almost 3000 adults, researchers Wyatt et al. (Obesity Research, 2002) reported that those who ate breakfast regularly tended to be more successful at maintaining their weight loss. Based on this study, could we conclude "eating breakfast regularly" and "maintaining weight loss" are in the "cause-and-effect relationship? Give reasons to support your answer
Answer:
No , the cause and effect can be finished up through analyses as it were.
What we have in the inquiry is only an observational investigation where we basically study 3000 grown-ups and attempt to outline the outcomes with no trial proof.
Imagine a scenario in which individuals who had breakfast normally were inclined to maintain their weight reduction, for sure if individuals who keep up weight reduction will in general have breakfast routinely.
henceforth the circumstances and logical results relationship cannot be built up
So as to do so , one must lead measurable trials, for example, autonomous example t test or ANOVA examination
In an area of the Great Plains, records were kept on the relationship between the rainfall (in inches) and the yield of wheat (bushels per acre). Find the equation of the regression line for the given data.
The student's question pertains to calculating the regression line equation that represents the relationship between rainfall and wheat yield. This can be done using a calculator's statistical functions to provide the least-squares regression line equation, which allows for the prediction of one variable based on another.
Explanation:The question asks about finding the equation for a linear regression line based on the relationship between rainfall and wheat yield. To accomplish this, one must enter the given data into a calculator, generate a scatter plot, and then use the calculator's regression function to find the least-squares regression line's equation. This regression equation typically takes the form î = a + bx, where 'a' is the y-intercept and 'b' is the slope of the line.
In terms of finding the regression line manually, a commonly used method is the median-median line approach, but for this question, we are instructed to use a calculator. The process involves statistical analysis to determine the line that minimally deviates from all data points in the scatter plot. Once the equation is calculated, one can predict values such as wheat yield for a given amount of rainfall.
An instructor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the problems, what is the probability that he or she will answer correctly (a) all 5 problems? (b) at least 4 of the problems?
Answer:
P
Step-by-step explanation:
(all 5 correctly) = (5 chosen among 7) / (5 chosen among 10)
Assessment: To help assess student learning in her developmental math courses, a mathematics professor at a community college implemented pre- and posttests for her students. A knowledge-gained score was obtained by taking the difference of the two test scores.
(a) What type of experimental design is this?
The experimental design used in this assessment is a pretest-posttest design where the difference between pre- and posttest scores is calculated.
Explanation:The experimental design used in this assessment is a pretest-posttest design. A pretest is administered before the students receive the intervention (in this case, the math course), and a posttest is administered after the intervention. The difference between the pre- and posttest scores is then calculated to determine the knowledge-gained score.
This design allows for the comparison of students' scores before and after the math course, giving insight into the effectiveness of the course in improving their knowledge. By comparing the two test scores, the professor can assess how much knowledge students have gained as a result of the course.
Example:
Before the math course, a student's pretest score is 50. After completing the course, the student's posttest score is 75. The knowledge-gained score is calculated as 75 - 50 = 25. This indicates that the student gained 25 points of knowledge between the pre- and posttests.
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Please help!!!
If Jimmy invests $250 twice a year at 4% compounded semi-annually, how much will his investment be worth after 3 years?
Please give a step by step!
Answer: his investment will be worth $1606.5 after 3 years
Step-by-step explanation:
We would apply the formula for determining future value involving deposits at constant intervals. It is expressed as
S = R[{(1 + r)^n - 1)}/r][1 + r]
Where
S represents the future value of the investment.
R represents the regular payments made(could be weekly, monthly)
r = represents interest rate/number of interval payments.
n represents the total number of payments made.
From the information given,
R = $250
r = 0.04/2 = 0.002
n = 2 × 3 = 6
Therefore,
S = 250[{(1 + 0.02)^6 - 1)}/0.02][1 + 0.02]
S = 250[{(1.02)^6 - 1)}/0.02][1.02]
S = 250[{(1.126 - 1)}/0.02][1.02]
S = 250[{0.126}/0.02][1.02]
S = 250[6.3][1.02]
S = 250 × 6.426
S = $1606.5
A quality control inspector has drawn a sample of 1616 light bulbs from a recent production lot. Suppose 30%30% of the bulbs in the lot are defective. What is the probability that exactly 44 bulbs from the sample are defective? Round your answer to four decimal places.
Answer:
0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.
Step-by-step explanation:
For each bulb, there are only two possible outcomes. Either it is defective, or it is not. The probability of a bulb being defective is independent from other bulbs, so we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[\tex]
In which [tex]C_{n,x}[\tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[\tex]
And p is the probability of X happening.
Suppose 30% of the bulbs in the lot are defective.
This means that [tex]p = 0.3[/tex]
A quality control inspector has drawn a sample of 16 light bulbs from a recent production lot.
This means that [tex]n = 16[/tex]
What is the probability that exactly 4 bulbs from the sample are defective?
This is P(X = 4).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[\tex]
[tex]P(X = 4) = C_{16,4}.(0.3)^{4}.(0.7)^{12} = 0.2040[\tex]
0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.
(a) Consider the following system of equations for the growth of rabbits and foxes from year to year: R' = 1.5R-.2F + 100 write this system in matrix form, where p R, F] and p' (b) Write a matrix equation for p", the vector of rabbits and foxes after V (c) Write a matrix equation for pa, the vector of rabbits and foxes (d) Using summation notation (2), write a matrix equation for pon, the 2 years after 3 years. er n years.
Answer:
Step-by-step explanation:
The detailed steps and analysis is as shown in the attached file.
To write the system of equations in matrix form, use p = [R, F] and p' = [1.5R - 0.2F + 100]. To find the vector of rabbits and foxes after time V, integrate p' twice. To find the vector of rabbits and foxes after time a, integrate p' once. To find the vector of rabbits and foxes after 2 years, substitute n=2 and use summation notation.
Explanation:(a) To write the system of equations in matrix form, we have:
R' = 1.5R - 0.2F + 100
We can represent the variables R and F as a vector p, so that p = [R, F]. The derivative of p with respect to time (p') is then given by:
p' = [1.5R - 0.2F + 100]
(b) To find the vector of rabbits and foxes after time V, we can integrate the equation given in part (a) with respect to time twice:
p'' = ∫[1.5R - 0.2F + 100]dt
(c) To find the vector of rabbits and foxes after time a, we can integrate the equation given in part (a) with respect to time:
p(a) = ∫[1.5R - 0.2F + 100]dt
(d) To find the vector of rabbits and foxes after 2 years, we can substitute n=2 into the equation given in part (c) and use summation notation:
p(2) = ∫[1.5R - 0.2F + 100]dt + ∑[∫[1.5R - 0.2F + 100]dt]^n
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The weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 4004 grams and a variance of 103,684. If a newborn baby boy born at the local hospital is randomly selected, find the probability that the weight will be less than 4293 grams. Round your answer to four decimal places.
Answer:
0.8151 is the probability that the weight will be less than 4293 grams.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4004 grams
Variance = 103,684
[tex]\sigma = \sqrt{103684} = 322[/tex]
We are given that the distribution of weight of newborn baby is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(weight will be less than 4293 grams)
P(x < 4293)
[tex]P( x < 4293) \\\\= P( z < \displaystyle\frac{4293 - 4004}{322}) \\\\= P(z < 0.8975)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 4293) =0.8151 = 81.51\%[/tex]
0.8151 is the probability that the weight will be less than 4293 grams.
Answer:
Probability that the weight will be less than 4293 grams is 0.8133.
Step-by-step explanation:
We are given that weights of newborn baby boys born at a local hospital are believed to have a normal distribution with a mean weight of 4004 grams and a variance of 103,684.
Let X = weight of newborn baby boys
So, X ~ N([tex]\mu =4004,\sigma^{2}=322^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean
[tex]\sigma[/tex] = population standard deviation
(a) Probability that weight will be less than 4293 grams is given by = P(X < 4293 grams)
P(X < 4293) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{4293-4004}{322 }[/tex] ) = P(Z < 0.89) = 0.8133
Therefore, if a newborn baby boy born at the local hospital is randomly selected, probability that the weight will be less than 4293 grams is 0.8133.
If the probability of a student taking a calculus class is 0.10, the probability of taking a statistics class is 0.90, and the probability of taking a calculus class and a statistics class is 0.07, what is the probability of a student taking a calculus class or a statistics class
The probability of a student taking a calculus or a statistics class is found using the Addition Rule of Probability. For non-mutually exclusive events like these, the formula is P(A or B) = P(A) + P(B) - P(A and B). Substituting in the given values, we find the probability is 93%.
Explanation:In probability theory, there's a rule called the Addition Rule of Probability. The rule states: the probability of the occurrence of either of two mutually exclusive events A and B is given by the sum of the probabilities of A and B.
However, if the two events aren't mutually exclusive (they can occur together), like our case here with the calculus and statistics classes, we need to adjust the formula. We subtract the probability of both of them happening. Hence, the formula becomes: P(A or B) = P(A) + P(B) - P(A and B).
If we plug in the given values: P(calculus or statistics) = P(calculus) + P(statistics) - P(calculus and statistics) = 0.10 + 0.90 - 0.07 = 0.93 or 93%.
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The probability that a student is taking either a calculus class or a statistics class is 0.93 or 93%. This uses the principle of inclusion-exclusion in probability.
To determine the probability of a student taking a calculus class or a statistics class, we use the principle of inclusion-exclusion for probabilities. This principle states:
P(A or B) = P(A) + P(B) - P(A and B)
Here, we are given the following probabilities:
Probability of taking calculus class, P(C) = 0.10Probability of taking statistics class, P(S) = 0.90Probability of taking both calculus and statistics classes, P(C and S) = 0.07Using the inclusion-exclusion principle, we get:
P(C or S) = P(C) + P(S) - P(C and S)
= 0.10 + 0.90 - 0.07
= 0.93
Thus, the probability that a student is taking either a calculus class or a statistics class is 0.93 or 93%.
A publisher reports that 79% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually more than the reported percentage. A random sample of 100 found that 89% of the readers owned a personal computer. Is there sufficient evidence at the 0.02 level to support the executive's claim
Answer:
Yes, we have sufficient evidence at the 0.02 level to support the executive's claim.
Step-by-step explanation:
We are given that a publisher reports that 79% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually more than the reported percentage. A random sample of 100 found that 89% of the readers owned a personal computer.
Let Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 0.79 {means that the percentage is actually less than or equal to the reported percentage}
Alternate Hypothesis, [tex]H_1[/tex] : p > 0.79 {means that the percentage is actually more than the reported percentage}
The test statics that will be used here is One-sample proportions test;
T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = % of the readers who owned a personal computer in a sample of 100 = 89%
n = sample size = 100
So, test statistics = [tex]\frac{0.89 -0.79}{\sqrt{\frac{0.89(1-0.89)}{100} } }[/tex]
= 3.196
Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.
Therefore, we conclude that percentage is actually more than the reported percentage which means we have sufficient evidence at the 0.02 level to support the executive's claim.
The accompanying data is on cube compressive strength (MPa) of concrete specimens.
112.1 97.0 92.6 86.0 102.0 99.2 95.8 103.5 89.0 86.9
(a) Is it plausible that the compressive strength for this type of concrete is normally distributed?
A. The normal probability plot is acceptably linear, suggesting that a normal population distribution is not plausible.
B. The normal probability plot is not acceptably linear, suggesting that a normal population distribution is plausible.
C. The normal probability plot is not acceptably linear, suggesting that a normal population distribution is not plausible.
D. The normal probability plot is acceptably linear, suggesting that a normal population distribution is plausible.
Answer:
B
Step-by-step explanation:
The data doesn't follow any linearity. Infact, the given data set is in a range of values. It is distributed around some mean value. So it is plausible that it is normally distributed
The normal probability plot is not acceptably linear, suggesting that a normal population distribution is not plausible.
Explanation:The question asks whether it is plausible that the compressive strength for this type of concrete is normally distributed. In order to answer this question, we can examine the normal probability plot. If the plot is acceptably linear, it suggests that a normal population distribution is plausible. On the other hand, if the plot is not acceptably linear, it suggests that a normal population distribution is not plausible.
According to the options provided, the correct answer would be C. The normal probability plot is not acceptably linear, suggesting that a normal population distribution is not plausible. This implies that the compressive strength for this type of concrete is not normally distributed.
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What is the equation of the line
Answer:
y=1/2x+2
Step-by-step explanation:
Answer: y = 1/2x + 2
Step-by-step explanation:
The equation of a straight line can be represented in the slope-intercept form, y = mx + c
Where c = intercept
Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis represent
change in the value of y = y2 - y1
Change in value of x = x2 -x1
y2 = final value of y
y 1 = initial value of y
x2 = final value of x
x1 = initial value of x
From the graph,
y2 = 4
y1 = 2
x2 = 4
x1 = 0
Slope,m = (4 - 2)/(4 - 0) = 2/4 = 1/2
To determine the y intercept, we would substitute x = 4, y = 4 and
m = 1/2 into y = mx + c. It becomes
4 = 1/2 × 4 + c
4 = 2 + c
c = 4 - 2 = 2
The equation becomes
y = x/2 + 2
MARSVECTORCALC6 3.4.020. My Notes A rectangular box with no top is to have a surface area of 64 m2. Find the dimensions (in m) that maximize its volume.
Answer:
We would have
[tex]l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}[/tex]
where " l " is length, " w" is width and "h" is height.
Step-by-step explanation:
Step 1
Remember that
Surface area for a box with no top = [tex]lw+2lh+2wh = 64[/tex]
where " l " is length, " w" is width and "h" is height.
Step 2.
Remember as well that
Volume of the box = [tex]l*w*h[/tex]
Step 3
We can now use lagrange multipliers. Lets say,
[tex]F(l,w,h) = lwh[/tex]
and
[tex]g(l,w,h) = lw+2lh+2wh = 64[/tex]
By the lagrange multipliers method we know that
[tex]\nabla F = \lambda \nabla g[/tex]
Step 4
Remember that
[tex]\nabla F = (wh,lh,lw)[/tex]
and
[tex]\nabla g = (w+2h,l+2h , 2w+2l)[/tex]
So basically you will have the system of equations
[tex]wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)[/tex]
Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get
[tex]lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)[/tex]
Then you would get
[tex]l\lambda (w+2h) = w\lambda (l+2h) = h\lambda (2w+2l)[/tex]
You can get rid of [tex]\lambda[/tex] from these equations and you would get
[tex]lw+2lh = lw+2wh = 2wh+2lh[/tex]
And from those equations you would get
[tex]l = w =2h[/tex]
Now remember the original equation
[tex]lw+2lh+2wh = 64[/tex]
If we plug in what we just got, we would have
[tex]l^{2} + l^{2} + l^2 = 64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}[/tex]
To maximize the volume of an open-topped rectangular box with a fixed surface area, express the volume in terms of a single variable using the provided surface area equations. Solve these using calculus methods to find the maximum volume.
Explanation:This question involves the field of mathematics, specifically calculus and optimization. As the task is about maximizing the volume of an open-topped rectangular box with a fixed surface area, we can look into the relation between length, width, height, surface area, and volume of the box.
Let's denote the length of the box as x, width as y, and height as z. The volume (V) of a the box is found by multiplying length, width, and height: V = xyz.
Since there is no top, the surface area (A) is calculated by adding together the areas of the bottom and the four sides, which gives us: A = xy + 2xz + 2yz = 64 m².
To maximize the volume, we need to express the volume in terms of a single variable. From the surface area equation, we can solve for z to get: z = (64 - xy) / (2x + 2y). Now substitute z into the volume equation to get V = xy * [(64 - xy) / (2x + 2y)]. Now we can take derivative of V with respect to x and y, set that equal to zero and find when the volume will be maximum. After solving, we will find the dimensions that maximize the volume of the box under given condition.
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Suppose that L1 : V → W and L2 : W → Z arelinear transformations and E, F, and G are orderedbases for V, W, and Z, respectively. Show that, if Arepresents L1 relative to E and F and B representsL2 relative to F and G, then the matrix C = BA representsL2 ◦ L1: V → Z relative to E and G. Hint:Show that BA[v]E = [(L2 ◦ L1)(v)]G for all v ∈ V.
Answer:
a) v ∈ ker(L) if only if [tex][V]_{E}[/tex] ∈ N(A)
b) w ∈ L(v) if and only if [tex][W]_{F}[/tex] is in the column space of A
See attached
Step-by-step explanation:
See attached the proof Considering the vector spaces V and W with other bases E and F respectively.
Let L be the Linear transformation form V and W and A is the matrix representing L relative to E and F
In a study of feeding behavior, zoologists recorded the number of grunts of a warthog feeding by a lake in a 15 minute time period following the addition of food. The data showing the weekly number of grunts and the age of the warthog (in days) are listed below. Compute the sum of the squared residuals of the least squared line for the given data Week 1 Number of Grunts 90 age(days) 125 Week 2 Number of Grunts 68 age(days) 141 Week 3 Number of Grunts 39 age(days)155 Week 4 Number of Grunts 44 age(days)160 Week 5 Number of Grunts 63 age(days) 167 Week 6 Number of Grunts 40 Age(days) 174 Week 7 Number of Grunts 62 Age(days) 183 Week 8 Number of Grunts 17 Age(days) 189 Week 9 Number of Grunts 20 Age(days) 195 Can you please show me how to do this by hand I am not allowed to use a scientific calculator?
The sum of the squared residuals of the least squares line for the given data is 8691.90.
To compute the sum of the squared residuals for the given data, we will perform the steps of calculating the least squares line manually. The least squares line represents the linear regression line that minimizes the sum of the squared differences between the observed data points and the predicted values.
Week 1: Number of Grunts = 90, Age (days) = 125
Week 2: Number of Grunts = 68, Age (days) = 141
Week 3: Number of Grunts = 39, Age (days) = 155
Week 4: Number of Grunts = 44, Age (days) = 160
Week 5: Number of Grunts = 63, Age (days) = 167
Week 6: Number of Grunts = 40, Age (days) = 174
Week 7: Number of Grunts = 62, Age (days) = 183
Week 8: Number of Grunts = 17, Age (days) = 189
Week 9: Number of Grunts = 20, Age (days) = 195
Calculate the means of the Number of Grunts (Y) and Age (X) variables:
Mean of Y (Number of Grunts) = (90 + 68 + 39 + 44 + 63 + 40 + 62 + 17 + 20) / 9 = 48.78
Mean of X (Age) = (125 + 141 + 155 + 160 + 167 + 174 + 183 + 189 + 195) / 9 = 165
Calculate the deviations from the means for each data point:
For each data point, subtract the mean of X (Age) from the specific X value and the mean of Y (Number of Grunts) from the specific Y value.
Deviation from mean for each X value:
125 - 165.67 = -40.67
141 - 165.67 = -24.67
155 - 165.67 = -10.67
160 - 165.67 = -5.67
167 - 165.67 = 1.33
174 - 165.67 = 8.33
183 - 165.67 = 17.33
189 - 165.67 = 23.33
195 - 165.67 = 29
Deviation from mean for each Y value:
90 - 48.78 = 41.22
68 - 48.78 = 19.22
39 - 48.78 = -9.78
44 - 48.78 = -4.78
63 - 48.78 = 14.22
40 - 48.78 = -8.78
62 - 48.78 = 13.22
17 - 48.78 = -31.78
20 - 48.78 = -28.78
Calculate the sum of the products of the deviations:
Sum of (Deviation from mean for X * Deviation from mean for Y)
= (-40.67 * 41.22) + (-24.67 * 19.22) + (-10.67 * -9.78) + (-5.67 * -4.78) + (1.33 * 14.22) + (8.33 * -8.78) + (17.33 * 13.22) + (23.33 * -31.78) + (29.33 * -28.78)
= -5.02 + -0.90 + 1.04 + 0.27 + 18.95 + -73.14 + 228.44 + -739.97 + -845.44
= -1407.77
Calculate the sum of the squared deviations for X:
Sum of (Deviation from mean for X)^2
= (-40.67)^2 + (-24.67)^2 + (-10.67)^2 + (-5.67)^2 + (1.33)^2 + (8.33)^2 + (17.33)^2 + (23.33)^2 + (29.33)^2
= 16572.86
Calculate the sum of the squared residuals:
Sum of squared residuals = Sum of (Deviation from mean for Y)^2
= (41.22)^2 + (19.22)^2 + (-9.78)^2 + (-4.78)^2 + (14.22)^2 + (-8.78)^2 + (13.22)^2 + (-31.78)^2 + (-28.78)^2
= 8691.90
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The sum of the squared residuals of the least squared line for the given data is 8399.
To compute the sum of the squared residuals of the least squared line by hand, you will need to follow these steps:
1. Start by organizing the given data into two columns: one for the number of grunts and one for the age of the warthog.
Week 1: Number of Grunts = 90, Age (days) = 125
Week 2: Number of Grunts = 68, Age (days) = 141
Week 3: Number of Grunts = 39, Age (days) = 155
Week 4: Number of Grunts = 44, Age (days) = 160
Week 5: Number of Grunts = 63, Age (days) = 167
Week 6: Number of Grunts = 40, Age (days) = 174
Week 7: Number of Grunts = 62, Age (days) = 183
Week 8: Number of Grunts = 17, Age (days) = 189
Week 9: Number of Grunts = 20, Age (days) = 195
2. Calculate the mean (average) of the age and the number of grunts. To do this, add up all the values in each column and divide by the total number of data points.
Mean of the age (days):
(125 + 141 + 155 + 160 + 167 + 174 + 183 + 189 + 195) / 9 = 165
Mean of the number of grunts:
(90 + 68 + 39 + 44 + 63 + 40 + 62 + 17 + 20) / 9 = 52
3. Subtract the mean of the age from each age value to get the deviation of each data point from the mean. Similarly, subtract the mean of the number of grunts from each number of grunts value.
Deviation of the age:
125 - 165 = -40
141 - 165 = -24
155 - 165 = -10
160 - 165 = -5
167 - 165 = 2
174 - 165 = 9
183 - 165 = 18
189 - 165 = 24
195 - 165 = 30
Deviation of the number of grunts:
90 - 52 = 38
68 - 52 = 16
39 - 52 = -13
44 - 52 = -8
63 - 52 = 11
40 - 52 = -12
62 - 52 = 10
17 - 52 = -35
20 - 52 = -32
4. Square each deviation value obtained in step 3.
Squared deviation of the age:
[tex](-40)^2 = 1600[/tex]
[tex](-24)^2 = 576[/tex]
[tex](-10)^2 = 100[/tex]
[tex](-5)^2 = 25[/tex]
[tex]2^2 = 4[/tex]
[tex]9^2 = 81[/tex]
[tex]18^2 = 324[/tex]
[tex]24^2 = 576[/tex]
[tex]30^2 = 900[/tex]
Squared deviation of the number of grunts:
[tex]38^2 = 1444[/tex]
[tex]16^2 = 256[/tex]
[tex](-13)^2 = 169[/tex]
[tex](-8)^2 = 64[/tex]
[tex]11^2 = 121[/tex]
[tex](-12)^2 = 144[/tex]
[tex]10^2 = 100[/tex]
[tex](-35)^2 = 1225[/tex]
[tex](-32)^2 = 1024[/tex]
5. Sum up all the squared deviation values obtained in step 4.
Sum of squared residuals:
1600 + 576 + 100 + 25 + 4 + 81 + 324 + 576 + 900 + 1444 + 256 + 169 + 64 + 121 + 144 + 100 + 1225 + 1024 = 8399
Therefore, the sum of the squared residuals of the least squared line for the given data is 8399.
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1. Identifying related samples Aa Aa E For each of the following research scenarios, decide whether the design uses a related sample. If the design uses a related sample, identify whether it uses matched subjects or repeated measures. (Note: Researchers can match subjects by matching particular characteristics, or, in some cases, matched subjects are naturally paired, such as siblings or married couples.) John Cacioppo was interested in possible mechanisms by which loneliness may have deleterious effects on health. He compared the sleep quality of a random sample of lonely people to the sleep quality of a random sample of nonlonely people. The design described You are interested in whether husbands or wives care more about how clean their cars are. You collect a random sample of 100 married couples and collect ratings from each partner, indicating the importance each places on car cleanliness. You want to know if the husbands ratings tend to be different than the wives ratings The design described
John Cacioppo's study uses independent samples while the car cleanliness study uses matched subjects because the samples are dependent as couples are paired.
Explanation:In two given scenarios, the research designs differ in whether they use related samples or not. Related samples refer to a study design where the same subjects are measured more than once, or their measures are compared to those of individuals with whom they share a distinct relationship (e.g., married couples).
In John Cacioppo's study, the research design uses independent samples. Participants were randomly assigned to either the group of lonely people or nonlonely people and the similarities or differences in sleep quality were observed. As there is no mention of matching based on particular characteristics, and individuals from one group are not directly linked to those in the other, these samples are not related.
In contrast, the car cleanliness study deals with matched subjects, as the samples are paired together based on marriage. The husbands' ratings are compared directly with their respective wives' ratings, giving rise to related, paired measurements within each couple.
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The correct answer for the first scenario is that the design uses a related sample with matched subjects. For the second scenario, the design uses a related sample with repeated measures.
Explanation for the first scenario:
In the first scenario, John Cacioppo is comparing the sleep quality of lonely people to nonlonely people. Since the researcher is comparing two different groups of people based on their loneliness status, this is an example of a matched subjects design. The subjects are matched on the characteristic of loneliness, meaning that each lonely individual is paired with a nonlonely individual who is similar in other relevant characteristics that could affect sleep quality, such as age, gender, or health status. This allows the researcher to control for these extraneous variables and focus on the effect of loneliness on sleep quality.
Explanation for the second scenario:
In the second scenario, the researcher is interested in the difference in car cleanliness importance ratings between husbands and wives. By collecting data from both partners within each of the 100 married couples, the researcher is using a repeated measures design. Each couple serves as their own control, with the husband and wife providing two sets of measurements (ratings) on the same variable (importance of car cleanliness). This design allows for the examination of within-subject differences, controlling for the influence of other variables that might differ between different families, such as socioeconomic status or cultural background.
In summary, the first scenario is an example of a related sample with matched subjects, while the second scenario is an example of a related sample with repeated measures.