Answer:
Correct Answer: CA larger sample size results in a smaller margin error, i.e. with a plant of 100 plots instead of 25, the margin error will be smaller.
Compute a 99% confidence interval rather than a 90% confidence interval, because a higher confidence level will result in a smaller margin of error
What is Statistics?Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data.
An agricultural researcher plants 25 plots with a new variety of yellow corn.
Assume that the yield per acre for the new variety of yellow corn follows a Normal distribution with unknown mean and standard deviation of 10.
We need to find a confidence interval with a smaller margin of error than the 90% confidence interval
n=25, x=150,s=10,a=0.95
Unknown mean u means we use t table.
[tex]150 ± t_{0.975} \frac{10}{\sqrt{25}}[/tex]
150±[tex]t_{0.975}[/tex]×2
Compute a 99% confidence interval rather than a 90% confidence interval, because a higher confidence level will result in a smaller margin of error
Hence, option C is correct. Compute a 99% confidence interval rather than a 90% confidence interval, because a higher confidence level will result in a smaller margin of error
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Fine the value of d. 2d-5= 17
Answer:
d = 11
Step-by-step explanation:
What is the g?
-15 - g/3 = -5
Answer:
g = - 30
Step-by-step explanation:
[tex] - 15 - \frac{g}{3} = - 5 \\ - \frac{g}{3} = - 5 + 15 \\ - \frac{g}{3} = 10 \\ - g = 10 \times 3 \\ - g = 30 \\ \huge \red{ \boxed{g = - 30}}[/tex]
For a healthy human, a body temperature follows a normal distribution with Mean of 98.2 degrees Fahrenheit and Standard Deviation of 0.26 degrees Fahrenheit. For an individual suffering with common cold, the average body temperature is 100.6 degrees Fahrenheit with Standard deviation of 0.54 degrees Fahrenheit. Simulate 10000 healthy and 10000 unhealthy individuals and answer questions 14 to 16. 14. If person A is healthy and person B has a cold, which of the events are the least likely
Complete Question:
For a healthy human, a body temperature follows a normal distribution with Mean of 98.2 degrees Fahrenheit and Standard Deviation of 0.26 degrees Fahrenheit. For an individual suffering with common cold, the average body temperature is 100.6 degrees Fahrenheit with Standard deviation of 0.54 degrees Fahrenheit. Simulate 10000 healthy and 10000 unhealthy individuals and answer questions 14 to 16.
14. If person A is healthy and person B has a cold, which of the events are the most likely? Pick the closest answer.
a. Person B will have higher temperature than 101 degrees.
b. Person A will have temperature higher than 99 degrees
c. Person B will have temperature lower than 100 degrees
d. Person A will have temperature lower than 97.5 degrees
15. What would be a range [A to B], which would contain 68% of healthy individuals? Pick the closest answer.
a. Between 97.9 and 98.46
b. Between 99.5 and 101.6
c. Between 100.06 and 101.14
d. Between 100.1 and 102.2
16. What is the approximate probability that a randomly picked, unhealthy individual (one with the cold) would have body temperature above 101 degrees Fahrenheit? Pick the closest answer.
a. About 10%
b. About 15%
c. About 22%
c. About 34%
Answer:
14. option a
15. option a
16. option a
explanation:
14 option (a)
person B have higher temperature than 101 degrees,
(b) cannot be true if person a has higher temperature than 99 then he has cold most likely and as mentioned by the standard dev and mean of healthy people
(c) AND (d) also cannot be true as they also do not satisfy the given standard dev and mean
15 option (a) because of the A to B range and not B to A range ....... as the healthy person's standard dev is 0.26. Hence the 68% data will lie in 97.94 - 98.46
but, if it was B to A, then the (c) option will be true
16 option (a) because most of the unhealthy individuals above the 0.54 standard dev will come somewhat near 90%
Find the average rate of change of g from x=-5 to x=5
The average rate of change can be calculated using the formula: (g(b) - g(a)) / (b - a), substituting -5 for a and 5 for b in our case.
Explanation:Average Rate of Change in MathematicsIn order to find the average rate of change of a function, in this case, function g, from x = -5 to x = 5, we use the formula: (g(b) - g(a)) / (b - a), where b is the ending value and a is the starting value. This gives us the slope of the tangent line that connects these two points on the graph, which reflects the average rate of change of the function over the interval. In this case, to find the average rate of change of function g from x = -5 to x = 5, we would substitute -5 for a and 5 for b in the formula to get: (g(5) - g(-5)) / (5 - (-5)). From here we can solve based on the known values of g(5) and g(-5).
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The null and alternative hypotheses for a test are given, as well as some information about the actual sample and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. Hypotheses: H0:p=0.5 vs Ha:p>0.5 Sample: p^=0.7 , n=30 Randomization statistic =p^ Enter your answer in accordance to the question statement
Answer:
The randomization will be centered around 0.5
Step-by-step explanation:
n = 30
Null hypothesis, H₀ : p = 0.5
Alternative hypothesis, : p > 0.5
The randomization will be centered around 0.5
The indication of where the randomization distribution will be centered shows that the randomization distribution is always centered at the value of the parameter given in the null hypothesis, which is p = 0.5.
Where is the randomization distribution centered?The randomization distribution is centered on the value in the null hypothesis. Since the null hypothesis is given as H0:p=0.5 and the alternate hypothesis is Ha:p>0.5 with sample values of p^=0.7 and n=30, the value in the null hypothesis is used to center the randomization distribution.What is the randomization distribution?The randomization distribution is the graphical representation of data or all the values for the statistic in the form of a histogram based on the random assignment of the sample. The variability in a sample statistic results from the induced variability of the random sample.Thus, the randomization distribution is always centered at the value of the parameter given in the null hypothesis, which is p = 0.5.
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A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 420 gram setting. It is believed that the machine is underfilling of overfilling the bags. A 61 bag sample had a mean of 424 grams with a standard deviation of 26. Assume the population is normally distributed. A level of significance of 0.01 will be used. Specify the type of hypothesis test.
Answer:
[tex]t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202[/tex]
[tex]p_v =2*P(t_{(60)}>1.202)=0.234[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from 420. So the specification is satisfied.
Step-by-step explanation:
Data given and notation
[tex]\bar X=424[/tex] represent the sample mean
[tex]s=26[/tex] represent the sample standard deviation
[tex]n=61[/tex] sample size
[tex]\mu_o =420[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] 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 check if the true mean is different from 420 or not, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 420[/tex]
Alternative hypothesis:[tex]\mu \neq 420[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=61-1=60[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{(60)}>1.202)=0.234[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from 420. So the specification is satisfied.
A number between 0 and 1, expressed as one number over another, is called a _ _ _ _ _ _ _ _.
Answer:
Fraction
Step-by-step explanation:
Answer:
fraction
Step-by-step explanation:
pls mark brainliest
You are given the polar curve r=eθ. (a) List all of the points (r,θ) where the tangent line is horizontal. In entering your answer, list the points starting with the smallest value of r and limit yourself to 1≤r≤1000 ( note the restriction on r!) and 0≤θ<2π. If two or more points share the same value of r, list those starting with the smallest value of θ. If any blanks are unused,
Answer:
θ [tex]0.75\pi[/tex] [tex]1.75\pi[/tex]
r 10.551 244.151
Step-by-step explanation:
The maximum value for [tex]\theta[/tex] is:
[tex]\theta_{max} = \ln r[/tex]
[tex]\theta_{max} = 2.199\pi\,rad[/tex]
The formula for the slope of the tangent line in polar coordinates is:
[tex]m = \frac{r'\cdot \sin \theta + r \cdot \cos \theta}{r' \cdot \cos \theta - r \cdot \cos \theta}[/tex]
Horizontal tangent lines have a slope of zero. So, the following relation must be satisfied:
[tex]r'\cdot \sin \theta + r \cdot \cos \theta = 0[/tex]
[tex]r'\cdot \sin \theta = - r \cdot \cos \theta[/tex]
[tex]\tan \theta = - \frac{r}{r'}[/tex]
[tex]\tan \theta = -\frac{e^{\theta}}{e^{\theta}}[/tex]
[tex]\tan \theta = -1[/tex]
[tex]\theta = \tan^{-1}(-1)[/tex]
[tex]\theta = \frac{3}{4}\pi + i\cdot \pi[/tex], for all [tex]i \in \mathbb{N}_{O}[/tex].
The maximum value of i is:
[tex]i = \frac{\theta_{max}-\frac{3}{4}\pi }{\pi}[/tex]
[tex]i = \frac{2.199-0.75}{1}[/tex]
[tex]i = 1.449[/tex] ([tex]i_{max} = 1[/tex]).
Then, values are listed below:
θ [tex]0.75\pi[/tex] [tex]1.75\pi[/tex]
r 10.551 244.151
Answer:
{ ( 2.193 , π / 4) , ( 10.551 , 3π / 4) , ( 50.754 , 5π / 4) , ( 244.151 , 7π / 4) }
Step-by-step explanation:
Given:-
- The polar curve has the equation:
r = e^θ
- list the points starting with the smallest value of r such that:
1 ≤ r ≤ 1000 , 0 ≤ θ < 2π.
Find:-
List all of the points (r,θ) where the tangent line is horizontal
Solution:-
- We will first transform the polar curve to cartesian coordinate system using the parametric relations:
x = r*cos (θ)
y = r*sin (θ)
- The tangent line is horizontal when the " dy / dθ " = 0 and " dx / dθ " = 0, so:
x = e^θ*cos (θ) , y = e^θ*sin (θ)
dx / dθ = e^θ*cos (θ) - e^θ*sin(θ)
= e^θ*[cos (θ) - sin(θ)]
dx / dθ = e^θ*[cos (θ) - sin(θ)] = 0,
e^θ = 0 , [cos (θ) - sin(θ)] = 0
e^θ ≠ 0 for the given interval 0 ≤ θ< 2π
cos (θ) - sin(θ) = 0 , tan ( θ ) = 1 - (1st quad and 3rd quad)
θ = { π / 4 , 5π / 4 } , 0 ≤ θ< 2π
- Similarly, evaluate dy/dθ = 0;
dy/dθ = e^θ*cos (θ) + e^θ*sin(θ)
= e^θ*[cos (θ) + sin(θ)]
dy / dθ = e^θ*[cos (θ) + sin(θ)] = 0,
e^θ = 0 , [cos (θ) + sin(θ)] = 0
e^θ ≠ 0 for the given interval 0 ≤ θ< 2π
cos (θ) + sin(θ) = 0 , tan ( θ ) = -1 , (2nd quad and 4th quad)
θ = { 3π / 4 , 7π / 4 } , 0 ≤ θ< 2π
- All possibilities of " θ " over the interval satisfying the a horizontal tangent line to the given polar curve:
θ = { π / 4, 3π / 4 , 5π / 4 , 7π / 4 } , 0 ≤ θ < 2π
- We will plug the evaluated list of values of "θ " in the given polar curve and determine the corresponding values of "r":
r = e^θ
θ = π / 4 , r = e^(π / 4) = 2.193
1: ( r , θ ) = ( 2.193 , π / 4)
θ = 3π / 4 , r = e^(3π / 4) = 10.55072
2: ( r , θ ) = ( 10.551 , 3π / 4)
θ = 5π / 4 , r = e^(5π / 4) = 50.754
3: ( r , θ ) = ( 50.754 , 5π / 4)
θ = 7π / 4 , r = e^(7π / 4) = 244.151
4: ( r , θ ) = ( 244.151 , 7π / 4)
Mrs.Mroc has 35 students in her science class.If her class increases to 42 students next
year, what will be the percent of change?
Answer:
20%
Step-by-step explanation:
Find the increase (subtract original number with final):
42-35= 7
Find the percent change (divide the increase by original number, then multiply by 100 to find the percentage):
[tex]\frac{7}{35}[/tex]*100=20
The percent change from 35 students to 42 students is 20%. This is calculated by using the formula for percent change: ((New Amount - Original Amount) / Original Amount) * 100.
Explanation:In this case, we are trying to calculate the percent change in the number of students in Mrs. Mroc's class. The formula for percent change is:
((New Amount - Original Amount) / Original Amount) * 100
The New Amount is 42 (the number of students next year), and the Original Amount is 35 (the current number of students). So the percent change would be (((42-35)/35)*100), which equals 20%.
So, if Mrs. Mroc's class increases from 35 students to 42 students next year, the percent change will be 20%.
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For the data set: 12, 8, 6, 6, 9, 8, 7, 11, 6, “6” is the value for which measure?
In the given data set, 12, 8, 6, 6, 9, 8, 7, 11, 6, the value 6 represents the mode.
We are given the following dataset: 12, 8, 6, 6, 9, 8, 7, 11, 6.
Count the frequency of each number:
12 appears 1 time.
8 appears 2 times.
6 appears 3 times.
9 appears 1 time.
7 appears 1 time.
11 appears 1 time.
The number 6 appears 3 times, more frequently than any other number.
By the definition of mode, it is the value that appears most frequently in a data set.
Since 6 occurs more times than any other number in the list, then it is the mode
At a family reunion, there were only blood relatives, consisting of children, parents, and grandparents, in attendance. There were 400 people total. There were twice as many parents as grandparents, and 50 more children than parents. How many children, parents, and grandparents were in attendance
The number of people in attendance are:
Children=70
Parents=140
Grandparents=190
Let
C represent children
P represent parents
G represent grandparents
Hence:
children+parents+ grandparent=400
P=2G
C=P+50 = 2G+50
2G+50 + 2G + G =400
5G=400-50
5G = 350
G=350/5
G =70 grandparents
P=70×2
P =140 parents
C=140+50
C =190 children
Inconclusion The number of people in attendance are:
Children=70
Parents=140
Grandparents=190
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The question is a math problem requiring a system of linear equations to solve. There were 70 grandparents, 140 parents, and 190 children at the family reunion. This was determined by assigning variables to each group and using the given conditions to form equations, which were then solved sequentially.
Explanation:This problem can be solved using a system of linear equations. We can represent each group at the family reunion (children, parents, and grandparents) with a variable. To create equations from the information provided:
Let G represent the number of grandparents.Let P represent the number of parents, which is twice the number of grandparents (P = 2G).Let C represent the number of children, which is 50 more than the number of parents (C = P + 50).The total number of all groups equals 400 (C + P + G = 400).First, replace P and C in the total equation with the equivalent as per the other equations so it reads (2G + 50 + 2G + G = 400). Simplify this equation and you get 5G + 50 = 400. Solving for G, we find that there were 70 grandparents.
Plug the value of G (70) back in the equation for P (2*70), we find that there were 140 parents.
Similarly plug P (140) in the equation for C, we find that there were 190 children.
So, in conclusion, there were 70 grandparents, 140 parents, and 190 children at the family reunion.
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A random sample of 340 people in Chicago showed that 66 listened to WJKT-1450, a radio station in South Chicago Heights. Based on this sample information, what is the point estimate for the proportion of people in Chicago that listen to WJKT-1450? Question 29 options: 1450 340 About 0.194 66
Answer:
Point of estimate of the true Proportion 'p' = 0.1941
Step-by-step explanation:
Explanation:-
Given data a random sample of 340 people in Chicago showed that 66 listened to WJKT-1450, a radio station in South Chicago Heights.
Given sample size 'n'= 340
Point of estimate of the true Proportion P = [tex]\frac{x}{n}[/tex]
Properties of Estimation
An estimator is not expected to estimate the Population parameter without error.
An estimator should be close to the true value of un-known parameter.
Point of estimate of the true Proportion = [tex]p = \frac{66}{340}= 0.1941[/tex]
Conclusion:-
Point of estimate of the true Proportion = 0.1941
Stefan’s neighborhood has a community garden. The garden has 15 equally sized rectangular plots for people to grow fruits and vegetables. Each rectangular plot measures 8 feet by 10 2 5 feet. A rectangle has a base of 8 feet and height of 10 and two-fifths feet. What is the area of each plot? Of the garden? The area of each plot is ft2. The total area of the garden is ft2.
Answer:
top 83 1/5
bottom 1,248
Step-by-step explanation:
i got it right
Answer:
Top 83 1/5
Bottom 1,248
Step-by-step explanation:
Just took it
A charity organization had to sell 18 tickets to their fundraiser just to cover necessary production costs.
They sold each ticket for $45.
Let y represent the net profit (in dollars) when they have sold 2 tickets.
Which of the following information about the graph of the relationship is given?
Answer: slope and x-intercept
y-intercept and a point that is not an intercept is the information about the graph of the relationship is given. So, the correct answer is E. y-intercept and a point that is not an intercept
Let's analyze the information given in the problem:
1. The charity organization sold 18 tickets to cover necessary production costs.
2. They sold each ticket for $45.
3. They want to determine the net profit when they have sold 2 tickets.
We are given a specific point on the graph: (x = 2, y = net profit when 2 tickets are sold).
Now, let's calculate the net profit when 2 tickets are sold:
Net profit when 2 tickets are sold = Total revenue - Total production costs
Total revenue = 2 tickets * $45 per ticket = $90
Total production costs = $18 (to cover necessary production costs)
Net profit = $90 - $18 = $72
Now, let's look at the options:
A. Slope and x-intercept: We do not have the slope or the x-intercept of the graph.
B. Slope and y-intercept: We do not have the slope or the y-intercept of the graph.
C. Slope and a point that is not an intercept: We have a point on the graph, but we do not have the slope.
D. x-intercept and y-intercept: We do not have the x-intercept or the y-intercept of the graph.
E. y-intercept and a point that is not an intercept: We have the y-intercept (net profit when 0 tickets are sold, which is $18) and a point on the graph (net profit when 2 tickets are sold, which is $72).
F. Two points that are not intercepts: We have one point on the graph (net profit when 2 tickets are sold, which is $72), but we do not have another point.
So, Based on the information given, the correct option is:
E. y-intercept and a point that is not an intercept
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Correct Question is:
A charity organization had to sell 18 tickets to their fundraiser just to cover necessary production costs.
They sold each ticket for $45.
Let y represent the net profit (in dollars) when they have sold 2 tickets.
Which of the following information about the graph of the relationship is given?
A. Slope and x- intercept
B. Slope and y-intercept
C. Slope and a point that is not an intercept
D. x-intercept and y intercept
E. y-intercept and a point that is not an intercept
F. Two points that are not intercepts
In the growth equation y = 3(1.2)^5 what is the rate of increase?
The rate of increase is 1.2 as it represents b in the equations A+b power of x+c then +D.
Answer:
y=7.46496
Step-by-step explanation:
A plant grows 12 mm every four months how many centimeters will it grow in one year?
Step-by-step explanation:
One year is 12 months. 12 mm is 1.2 cm.
Write and solve a proportion
1.2 cm / 4 months = x / 12 months
x = 3.6 cm
Use the surface integral in Stokes' Theorem to calculate the circulation of the field Bold Upper F equals x squared Bold i plus 4 x Bold j plus z squared Bold kF=x2i+4xj+z2k around the curve C: the ellipse 25 x squared plus 16 y squared equals 525x2+16y2=5 in the xy-plane, counterclockwise when viewed from above.
Stokes' theorem says the integral of the curl of [tex]\vec F[/tex] over a surface [tex]S[/tex] with boundary [tex]C[/tex] is equal to the integral of [tex]\vec F[/tex] along the boundary. In other words, the flux of the curl of the vector field is equal to the circulation of the field, such that
[tex]\displaystyle\iint_S\nabla\times\vec F\cdot\mathrm d\vec S=\int_C\vec F\cdot\mathrm d\vec r[/tex]
We have
[tex]\vec F(x,y,z)=x^2\,\vec\imath+4x\,\vec\jmath+z^2\,\vec k[/tex]
[tex]\implies\nabla\times\vec F(x,y,z)=4\,\vec k[/tex]
Parameterize the ellipse [tex]S[/tex] by
[tex]\vec s(u,v)=\dfrac{u\cos v}{\sqrt5}\,\vec\imath+\dfrac{u\sqrt5\sin v}4\,\vec\jmath[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le2\pi[/tex].
Take the normal vector to [tex]S[/tex] to be
[tex]\dfrac{\partial\vec s}{\partial\vec u}\times\dfrac{\partial\vec s}{\partial\vec v}=\dfrac u4\,\vec k[/tex]
Then the flux of the curl is
[tex]\displaystyle\iint_S4\,\vec k\cdot\dfrac u4\,\vec k\,\mathrm dA=\int_0^{2\pi}\int_0^1u\,\mathrm du\,\mathrm dv=\boxed{\pi}[/tex]
Before sending track and field athletes to the Olympics, the U.S. holds a qualifying meet.
The upper box plot shows the top 12men's long jumpers at the U.S. qualifying meet. The lower box plot shows the distances (in meters) achieved in the men's long jump at the2012 Olympic games.
Which pieces of information can be gathered from these box plots?
Choose all answers that apply:
Choose all answers that apply:
(Choice A)
A
The Olympic jumps were farther on average than the U.S. qualifier jumps.
(Choice B)
B
All of the Olympic jumps were farther than all of the U.S. qualifier jumps.
(Choice C)
C
The Olympic jumps vary noticeably more than the U.S. qualifier jumps.
(Choice D)
D
None of the above
2 horizontal boxplots titled U.S. Qualifier and Olympics are graphed on the same horizontal axis, labeled Distance, in meters. The boxplot titled U.S. Qualifier has a left whisker which extends from 7.68 to 7.7. The box extends from 7.7 to 7.89 and is divided into 2 parts by a vertical line segment at 7.74. The right whisker extends from 7.9 to 7.99. The boxplot titled Olympics has a left whisker which extends from 7.7 to 7.83. The box extends from 7.83 to 8.12 and is divided into 2 parts by a vertical line segment at 8.04. The right whisker extends from 8.12 to 8.31. All values estimated.
Answer:
a and c
Step-by-step explanation:
Looking at the pieces of information, the pieces of information that can be gathered from these box plots are:
The Olympic jumps were farther on average than the U.S. qualifier jumps.The Olympic jumps vary noticeably more than the U.S. qualifier jumps.What is Olympics?Olympics actually refers to the major international event which usually features different sports and normally holds once in every four years. It is also known as the Games of the Olympiad.
From the pieces of information given, we can actually conclude that the above options can be gathered from the information.
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Which line of music shows a reflection?
solve for x
5x^2-x-4=0
Answer:
x = -5/4 x=1
Step-by-step explanation:
5x^2 -x -4 =0
Factor
(5x+4) (x-1)=0
Using the zero product property
5x+4 =0 x-1 =0
5x=-4 x=1
x = -5/4 x=1
Answer:
x = -0.8, 1
Step-by-step explanation:
5x² - x - 4 = 0
5x² - 5x + 4x - 4 = 0
5x(x - 1) + 4(x - 1) = 0
(5x + 4)(x - 1) = 0
x = -⅘, 1
A triangle is translated by using the rule (XY)
(X-4.Y+1). Which describes how the figure is moved?
Answer:
4 units to the right & 1 unit up.
Answer: it’s B)
The answer is the second one
The desired percentage of SiO2 in a certain type of aluminous cement is 5.5. To test whether the true average
percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. Suppose that
the percentage of SiO2 in a sample is normally distributed with (sigma) = 0.3 and that x-bar = 5.25.
a) Does this indicate conclusively that the true average percentage differs from 5.5?
b) If the true average percentage is (mu) = 5.6 and a level a = 0.01 test based on n = 16 is used,
what is the probability of detecting this departure from H0?
Answer:
A. There is sufficient evidence to prove and conclude that the true average percentage differs from desired percentage.
B. P(Z<-4.93333) = 0.0000
C. n = 5945 samples
Step-by-step explanation:
A)
The rightnull and alternative hypotheses are given as below:
Null hypothesis H0: µ = 5.5
Alternative hypothesis Ha: µ ≠ 5.5
Conducting the one sample z test for population mean.
Test statistic formula is given as below:
Z = (Xbar - µ)/[σ/√(n)]
Given,
Xbar = 5.23
µ = 5.5
σ = 0.30
n = 16
Z = (5.23 – 5.5) / [ 0.30 /√(16)]
Z = -3.6000
P-value = 0.0003
α value = 0.05
P-value < α value
So, we reject the null hypothesis. There is sufficient evidence to prove and conclude that the true average percentage differs from desired percentage.
B
From the information given
Xbar = 5.23
µ = 5.6
σ = 0.30
n = 16
α value = 0.01
Z = (Xbar - µ)/[σ/√(n)]
Z = (5.23 - 5.6)/(0.30/√(16))
Z = -4.93333
P(Z<-4.93333) = 0.0000
Required probability = 0.0000
C
We are given α =0.01
Therefore, critical Z value = 2.57
E = 0.01
σ = 0.30
n = (Z*σ/E)^2
n = (2.57*0.30/0.01)^2 = 5944.41
n = 5945 samples
Answer true or false. If false, explain briefly. a) Some of the residuals from a least squares linear model will be positive and some will be negative. b) Least squares means that some of the squares of the residuals are minimized. c) We write ModifyingAbove y with caret to denote the predicted values and y to denote the observed values.
a) True. Some residuals will be positive and some will be negative. b) True. Least squares minimizes the squares of the residuals. c) False. y with a caret denotes predicted values.
Explanation:a) True. Some of the residuals from a least squares linear model will be positive and some will be negative. Residuals measure the difference between the observed values and the predicted values, and they can be positive if the observed value is higher than the predicted value, or negative if the observed value is lower.
b) True. Least squares regression aims to minimize the sum of squared residuals. By minimizing the squares of the residuals, we ensure that the line of best fit is as close as possible to the data points.
c) False. We write y with a caret (^) to denote the predicted values and y without a caret to denote the observed values.
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The diameter of the sun is 1.391 million kilometers. Represent this number in scientific notation.
Answer:
1391000000
Step-by-step explanation:
Claim: Students should be required to take at least one online class.
Counterclaim: Some students can focus better when they are in a traditional classroom setting.
Which statement is the best rebuttal to the counterclaim given?
(A). When working on an online class, students have more freedom to choose an environment that is best for them.
(B). A traditional classroom is the best environment for learning, but an online class can be good for students too.
(C). Students should only be taught in a traditional classroom environment, because it works best for everyone.
Answer:
A. when working on an online class students have more freedom to choose an environment in which they are most comfortable
Step-by-step explanation:
some students do better at home some do better in a traditional setting but being online lets you change your setting to anywhere with online access
In the right triangle shown, m\angle A=45\degreem∠A=45°m, angle, A, equals, 45, degree and AB = 12AB=12A, B, equals, 12. How long is BC
Answer:
[tex]BC=6\sqrt{2}\ units[/tex]
Step-by-step explanation:
In this problem i will assume that the side AB is the hypotenuse
The picture in the attached figure
we know that
In a right triangle, if one angle is 45 degrees, then the other angle complementary is equal to 45 degrees too
so
[tex]m\angle B=45^o[/tex]
In the right triangle ABC
[tex]cos(B)=\frac{BC}{AB}[/tex] ----> by CAH (adjacent side divided by the hypotenuse)
substitute the given values
we have
[tex]cos(45^o)=\frac{\sqrt{2}}{2}[/tex]
[tex]AB=12\ units[/tex]
substitute
[tex]BC=cos(B)(AB)[/tex]
[tex]BC=\frac{\sqrt{2}}{2}(12)=6\sqrt{2}\ units[/tex]
Answer: 6 square root 2
Step-by-step explanation:
6 of 25
Camden spent a total of $599.38 for 63 meals. How much did he average on each meal?
dollars
A college entrance exam company determined that a score of 23 on the mathematics portion of the exam suggests mat a student is ready for college-level mathematics. To achieve this goal, the company recommends that students take a core curriculum of math courses in high school. Suppose a random sample of 200 students who completed this core set of courses results n a mean math score of 23.6 on the college entrance exam with a standard deviation of 3.2. Do these results suggest mat students who complete the core curriculum are ready for college-level mathematics? That is, are they scoring above 23 on the math portion of the exam? Complete parts a through d below. a. State the appropriate null and alternative hypotheses. Choose the correct answer below i. H_0: mu = 23.6 versus H_1 = mu notequalto 23.6 ii.H_0: mu = 23.6 versus H_1 = mu > 23.6 iii.H_0: mu = 23.6 versus H_1 = mu < 23.6 iv. H_0: mu = 23.6 versus H_1 = mu > 23 v. H_0: mu = 23.6 versus H_1 = mu < 23 b. Verify that the requirements to perform the test using the t-distribution are satisfied. Is the sample obtained using simple random sampling or from a randomized experiment? i. Yes, because only high school students were sampled. ii. No, because not all students complete the courses.iii. No, because only high school students were sampled.iv. Yes, because the students were randomly sampled. Is the population from which the sample is drawn normally distributed or is the sample size, n, large (n Greaterthanorequalto 30)? i. No, neither of these conditions are true ii. Yes, the sample size is larger man 30 iii. Yes, the population is normally distributed It is impossible to determine using the given information c. Are the sampled values independent of each other? i. Yes, because each student's test score does not affect other students' test scores ii. No. because students from the same class will affect each other's performance iii. Yes, because the students each take their own tests iv. No, because every student takes the same test d. Use the P-value approach at the alpha = 0 10 level of significance to test the hypotheses. (Round to three decimal places as needed) State the conclusion for the test Choose the correct answer below i. Do not reject the null hypothesis because the P-value is greater than the alpha = 0 10 level of significance ii. Reject the null hypothesis because the P-value a less than the alpha = 0 10 level of significance iii. Do not reject (he null hypothesis because the P value is less than the alpha = 0 10 level of Significance. iv. Reject the nun hypothesis because the P-value greater than the alpha = 010 level of significance e. Write a conclusion based on the results. Choose the correct answer below. i. There is sufficient evidence to conclude that the population mean is greater than 23. ii. There is sufficient evidence to conclude that the population mean is less than 23 iii.There is not sufficient evidence to conclude that the population mean is greater than 23 iv. There is not sufficient evidence to conclude that the population mean is less man 23.
Answer:
a) The null hypothesis is represented as
H₀: μ ≤ 23
The alternative hypothesis is given as
Hₐ: μ > 23
b) Check the Explanation
The conditions for a t-test to be performed are satisfied or not?
- Yes, because the students were randomly sampled.
- Yes, the sample size is larger man 30.
And the central limit theorem allows us to approximate that the random sample obtained from the population is a normal distribution.
c) Are the sampled values independent of each other?
Yes, because each student's test score does not affect other students' test scores.
d) p-value obtained = 0.004
Reject the null hypothesis because the P-value a less than the alpha = 0.10 level of significance
e) There is sufficient evidence to conclude that the population mean is greater than 23.
Step-by-step explanation:
For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
For this question, we want to check if results suggest that students who complete the core curriculum are ready for college-level mathematics.
The only condition to be ready for college is scoring above 23.
So, the null hypothesis would be that the mean of test scores of students that complete core curriculum is less than or equal to 23. That is, there isn't significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics.
And the alternative hypothesis would be that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.
Mathematically
The null hypothesis is represented as
H₀: μ ≤ 23
The alternative hypothesis is given as
Hₐ: μ > 23
b) The conditions required before performing t-test.
- The sample should be a random sample
- The dependent variable should be approximately normally distributed.
- The observations are independent of one another.
- The dependent variable should not contain any outliers
All of these conditions are satisfied for our distribution.
c) Are the sampled values independent of each other?
Yes, because each student's test score does not affect other students' test scores.
d) To do this test, we will use the t-distribution because no information on the population standard deviation is known
So, we compute the t-test statistic
t = (x - μ₀)/σₓ
x = sample mean = 23.6
μ₀ = 23
σₓ = standard error = (σ/√n)
where n = Sample size = 200
σ = Sample standard deviation = 3.2
σₓ = (3.2/√200) = 0.226
t = (23.6 - 23) ÷ 0.226 = 2.65
checking the tables for the p-value of this t-statistic
- Degree of freedom = df = n - 1 = 200 - 1 = 199
- Significance level = 0.10
- The hypothesis test uses a one-tailed condition because we're testing only in one direction.
p-value (for t = 2.65, at 0.10 significance level, df = 199, with a one tailed condition) = 0.004348 = 0.004 to 3 d.p.
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.10
p-value = 0.004
0.004 < 0.10
Hence,
p-value < significance level
This means that we reject the null hypothesis, accept the alternative hypothesis and say that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.
e) The result of the p-value obtained is that there is significant evidence to conclude that the results suggest that students who complete the core curriculum are ready for college-level mathematics. That is, the mean score of those that complete the core curriculum is above 23 and are ready for college-level mathematics.
Hope this Helps!!!
For a community service project, of the students in a class bring paper towels 3/4 2/5 of the students bring towels and soap What fraction of the students who bring paper towels also bring soap ? es
Step-by-step explanation:
For a community service project, the no of the students in a class bring paper towels = [tex]\frac{3}{4}[/tex]
The students bring towels and soap = [tex]\frac{2}{5}[/tex]
The students who bring paper towels = [tex]\frac{3}{4}[/tex]
The fraction of students who bring paper towels also bring soap = [tex]\frac{2}{5}[/tex]
What is the word form for 355.3 kilograms
Answer:
Three hundred fifty-five and three tenths.
Step-by-step explanation:
The word form for 355.3 kilograms is "Three hundred fifty-five point three kilograms." This representation breaks down the number into its hundreds, tens, ones, tenths, and specifies the unit of measurement, kilograms.
The word form for 355.3 kilograms is "Three hundred fifty-five point three kilograms."In word form, numbers are expressed using words, making it easier to understand and communicate quantities. Here's a breakdown of the conversion:"Three hundred" represents the hundreds place value, which is 300."Fifty" represents the tens place value, which is 50."Five" represents the ones place value, which is 5."Point" indicates the decimal point."Three" represents the tenths place value, which is 3."Kilograms" specifies the unit of measurement.So, when you combine these components, you get "Three hundred fifty-five point three kilograms." This accurately describes the quantity of 355.3 kilograms in word form, emphasizing the hundreds, tens, ones, tenths, and the unit of measurement. This notation is especially important in contexts where precise measurement and clear communication of quantities are necessary, such as in scientific, industrial, or trade-related scenarios.For more questions on kilograms -
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