Answer:
C) The auditor may or may not achieve the desired risk of assessing control risk too low.
Step-by-step explanation:
In a concept of risk sampling, if the sample size is chosen randomly in accordance with random selection procedures, the auditor may or may not achieve the desired risk of assessing risk too low. In other words the auditor may or may not achieve desired precision. This is because a samole chosen randomly may not represent the true population.
This depends largely on the sample size. If the sample size selected is too small, the allowance for sampling risk will be larger than what is required because it will lead to a large standard error of the mean
The area of a circle is 153.86 square meters. What is the diameter of the circle? Use 3.14 for π.
Answer:
Option (2). 14 m
Step-by-step explanation:
Formula to get the area of a circle 'A' = [tex]\pi r^{2}[/tex]
where r = radius of the circle
Given in the question,
Area of the circle = 153.86 square meters
By putting the values in the formula,
153.86 = πr²
r = [tex]\sqrt{\frac{153.86}{\pi } }[/tex]
r = [tex]\sqrt{49}[/tex]
r = 7 meters
Diameter of circle = 2 × (radius of the circle)
= 2 × 7
= 14 meters
Therefore, diameter of the circle is 14 meters.
Option (2) is the answer.
Answer:
14m
Step-by-step explanation:
Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by y 1 3 ex2 /3 , y 0, x 0, and x 3 about the y-axis. Round your answer to three decimal places.
Answer:
Step-by-step explanation:
[tex]y = f(x) =\frac{1}{\sqrt{3 \pi} } e^{-x^{2/3}}[/tex]
y = 0, x = 0 and x = 3
Consider an element of thickness dx at a distance x from the origin. By Cylindirical Shell Method, the volume of the element is given by
[tex]dV=(2\pi rdr)h=(2\pi xdx)f(x) => dV=(2\pi xdx) \frac{1}{\sqrt{3\pi}}e^{-x^{\frac{2}{3}}}[/tex]
[tex]dV=2\sqrt{\frac{\pi}{3}}xe^{-x^{\frac{2}{3}}}dx[/tex]
Integrate the above integral over the limits x=0 to x=3 which implies
[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}xe^{-x^{\frac{2}{3}}}dx[/tex]
Solve by subsititution
[tex]Let,\\ -x^{\frac{2}{3}}=y => \frac{-2}{3}x^{\frac{-1}{3}}dx=dy => x^{\frac{-1}{3}}dx=\frac{-3}{2}dy[/tex]
Also, apply the new limits
[tex]At,\\\\ x=0, y=0 \ and \ At, x=3, y=-\sqrt[3]{9}[/tex]
This implies,
[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}x^{\frac{4}{3}}e^{-x^{\frac{2}{3}}}x^{\frac{-1}{3}}dx=2\sqrt{\frac{\pi}{3}}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}(\frac{-3}{2})dy[/tex]
[tex]V=-\sqrt{3\pi}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]
Let,
[tex]I=\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]
Integrate by parts the above integral
[tex]u=y^2 \ and \ dv=e^ydy => du=2y \ and \ v=e^y[/tex]
Integrate by parts formula
[tex]\int udv=uv-\int vdu => y^2e^y-\int 2ye^ydy[/tex]
Again integrate by parts
[tex]u=y \ and \ dv=e^ydy => du=1 \ and \ v=e^y[/tex]
Integrate by parts formula
[tex]\int udv=uv-\int vdu => y^2e^y-2[ye^y-e^y]=e^y[y^2-2y+2][/tex]
Therefore,
[tex]I=[e^y(y^2-2y+2)]_{0}^{-\sqrt[3]{9}}\\\\=e^{-2.0802}[(2.0802)^2+2(2.0802)+2]-e^{0}[0-0+2]\\\\\frac{(4.3272+4.1604+2)}{8.0061}-2\\\\=\frac{10.4876}{8.0061}-2\\\\=1.3099-2\\\\=-0.6901[/tex]
This implies, the volume is
[tex]V=-\sqrt{3\pi}I\\\\=-\sqrt{3\times 3.142} \times (-0.6901)\\\\=3.0701 \times 0.6901\\\\=2.1186[/tex]
That is, up to three decimal places
[tex]V\approx 2.118[/tex]
Which of the following is the solution to |x-1|=8
Answer:
-7,9
Step-by-step explanation:
x-1=-8
x=-7
x-1=8
x=9
Please answer this correctly
Answer:
1.5 meters
Step-by-step explanation:
The formula for the area of a trapezoid is h * (a+b)/2, where a is the first base and b is the second base. Now, we can work backwards to determine the height of the trapezoid:
3.75=h*(1.7+3.3)/2
3.75=h*2.5
h=3.75/2.5=1.5
Hope this helps!
Answer:
Step-by-step explanation:
use the formula and rearrange for h.
1/2 x h x (a + b) = A
1/2 x (1.7 + 3.3) x h = 3.75
2.5 x h = 3.75
h = 1.5
hope this helps! :)
What expression is equivalent to 6•6•6•6•6
Answer:
6^5
Step-by-step explanation:
6 multiplied with itself 5 times is equal to 6^5
What’s the correct answer for this?
Answer:
The capital B refers to the base of the area
Step-by-step explanation:
Answer:
A
Step-by-step explanation:
The capital B means the area of the base
A teacher figures that final grades in the chemistry department are distributed as: A, 25%; B, 25%;C, 40%;D, 5%; F, 5%. At the end of a randomly selected semester, the following number of grades were recorded. Calculate the chi-square test statistic x^2 to determine if the grade distribution for the department is different than expected. Use α = 0.01.
Grade A B C D F
Number 36 42 60 14 8
a. 6.87
b. 0.6375
c. 5.25
d. 4.82
Answer:
[tex]E_{A} =0.25*160=40[/tex]
[tex]E_{B} =0.25*160=40[/tex]
[tex]E_{C} =0.4*160=64[/tex]
[tex]E_{D} =0.05*160=8[/tex]
[tex]E_{F} =0.05*160=8[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(36-40)^2}{40}+\frac{(42-40)^2}{40}+\frac{(60-64)^2}{64}+\frac{(14-8)^2}{8}+\frac{(8-8)^2}{8} =5.25[/tex]
The answer would be:
c. 5.25
Step-by-step explanation:
The observed values are given by:
A: 36
B: 42
C: 60
D: 14
E: 8
Total =160
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is no difference in the proportions for the final grades
H1: There is a difference in the proportions for the final grades
The level of significance assumed for this case is [tex]\alpha=0.01[/tex]
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
Now we just need to calculate the expected values with the following formula [tex]E_i = \% * total[/tex]
And the calculations are given by:
[tex]E_{A} =0.25*160=40[/tex]
[tex]E_{B} =0.25*160=40[/tex]
[tex]E_{C} =0.4*160=64[/tex]
[tex]E_{D} =0.05*160=8[/tex]
[tex]E_{F} =0.05*160=8[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(36-40)^2}{40}+\frac{(42-40)^2}{40}+\frac{(60-64)^2}{64}+\frac{(14-8)^2}{8}+\frac{(8-8)^2}{8} =5.25[/tex]
The answer would be:
c. 5.25
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(categories-1)=(5-1)=4[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{4} >5.25)=0.263[/tex]
The p value is higher than the significance so we have enough evidence to FAIL to reject the null hypothesis
(DONT REPORT OR ANSWER) pls don’t (what is -2 plus -2)
Answer:
I think it’s -4
Step-by-step explanation:
-2 plus -2 is -4
since 2 plus 2 is 4
Hope this helps ;)
Answer:
It is -4Step-by-step explanation:
It is because 2 plus 2 is 4
so -2 plus -2 is -4
hth!
3
Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.
The product of (3 + 2) and a complex number is (17 + 71).
The complex number is
Answer:
5-i
Step-by-step explanation:
Product=multiplication
Let the complex number=x
(3+2i)*x=17+7i
x=17+7i / 3+2i
x=(17-7i)*(3-2i)/(3+2i)*(3+2i)
=51-34i+21i+14i^2 / 9+6i+6i+4i^2
=51+13i+14i^2 / 9+12i+4i^2
= (51+14 - 13i) / 13
= (65 -13i) / 13
= 65 / 13 - 13 i / 13
= 5 - i.
I need help again♀️,
Answer:
The second choice.
Step-by-step explanation:
Answer:
2nd graph down
Step-by-step explanation:
3a+11 > 5
Subtract 11 from each side
3a+11-11 > 5-11
3a > -6
Divide each side by 3
3a/3 > -6/3
a >-2
Open circle at 02
line going to the right
e of Scores, a publication of the Educational Testing Service, the scores on the verbal portion of the GRE have mean 150 points and standard deviation 8.75 points. Assuming that these scores are (approximately) normally distributed, a. obtain and interpret the quartiles. b. find and interpret the 99th percentile.
Answer:
a) Q1= 144.10
Median = 150
Q3=155.90
b) The 99 percentile would be:[tex]a=150 +2.33*8.75=170.39[/tex]
And represent a value who accumulate 99% of the values below
Step-by-step explanation:
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(150,8.75)[/tex]
Where [tex]\mu=150[/tex] and [tex]\sigma=8.75[/tex]
Part a
Lets begin with the first quartile:
[tex]P(X>a)=0.75[/tex] (a)
[tex]P(X<a)=0.25[/tex] (b)
We can find the quantile in the normal standard distribution and we got z=-0.674.
And we can apply the z score formula and we got:
[tex]z=-0.674<\frac{a-150}{8.75}[/tex]
And if we solve for a we got
[tex]a=150 -0.674*8.75=144.10[/tex]
The median for this case is the mean [tex]Median =150[/tex]
For the third quartile we find the quantile who accumulate 0.75 of the area below and we got z=0.674 and we got:
[tex]a=150 +0.674*8.75=155.90[/tex]
Part b
We can find the quantile in the normal standard distribution who accumulate 0.99 of the area below and we got z=2.33.
And we can apply the z score formula and we got:
[tex]z=2.33<\frac{a-150}{8.75}[/tex]
And if we solve for a we got
[tex]a=150 +2.33*8.75=170.39[/tex]
And represent a value who accumulate 99% of the values below
The data represents the heights of eruptions by a geyser. Use the heights to construct a stemplot. Identify the two values that are closest to the middle when the data are sorted in order from lowest to highest.
Height of eruption
62 33 50 90
80 50 40 70
50 63 74 53
55 64 60 60
78 70 43 82
Required:
Identify the two values that are closest to the middle when the data are sorted in order from lowest to highest. The values closest to the middle are_________inches and_______inches.
Answer:
[tex] Median = \frac{60+60}{2}=60[/tex]
And we see that the closest values to 60 are 62 and 63 and then the answer would be:
The values closest to the middle are 62 inches and 63 inches.
Step-by-step explanation:
We have the following dataser given:
62 33 50 90 80 50 40 70 50 63 74 53 55 64 60 60 78 70 43 82
We can sort the values from the lowest to the highest and we got::
33 40 43 50 50 50 53 55 60 60 62 63 64 70 70 74 78 80 82 90
Now we see that we have n=20 values and the values closest to the middle and we can use the middle as the median and for this case the median can be calculated from position 10 and 11th and we got:
[tex] Median = \frac{60+60}{2}=60[/tex]
And we see that the closest values to 60 are 62 and 63 and then the answer would be:
The values closest to the middle are 62 inches and 63 inches.
The values closest to these middle elements are 60 and 63 inches
The dataset is given as:
62 33 50 90 80 50 40 70 50 63 74 53 55 64 60 60 78 70 43 82
Next, we sort the data elements in ascending order
33 40 43 50 50 50 53 55 60 60 62 63 64 70 70 74 78 80 82 90
The length of the dataset is 20.
So, the elements at the middle are the 10th and the 11 elements.
From the sorted dataset, these elements are: 60 and 62
Hence, the values closest to these middle elements are 60 and 63
Read more about median at:
https://brainly.com/question/14532771
Find the length of both of the unknown sides in the triangle shown here.
Give your answer correct to the nearest metre. [5 marks]
Answer:
[tex] (x+11)^2 = (x+3)^2 +16^2[/tex]
And if we solve this equation for x we got:
[tex] x^2 +22x +121 = x^2 +6x +9 +256[/tex]
We can cancel [tex]x^2[/tex] in both sides and we have this:
[tex] 22x -6x= 256+9-121 =144[/tex]
And then we got:
[tex] 16 x= 144[/tex]
[tex] x =\frac{144}{16}= 9[/tex]
And then the length of the sides are 9+11= 20 m for the hypothenuse, 16 for the adjacent side and 9+3 = 12m for the last side.
Lenght of the smaller unknown side: 12m
Lenght of the larger unknown side: 20m
Step-by-step explanation:
For this case we have a right triangle and we can use the Pythagoras Theorem and using the info given by the triangle we can set up the following equation:
[tex] (x+11)^2 = (x+3)^2 +16^2[/tex]
And if we solve this equation for x we got:
[tex] x^2 +22x +121 = x^2 +6x +9 +256[/tex]
We can cancel [tex]x^2[/tex] in both sides and we have this:
[tex] 22x -6x= 256+9-121 =144[/tex]
And then we got:
[tex] 16 x= 144[/tex]
[tex] x =\frac{144}{16}= 9[/tex]
And then the length of the sides are 9+11= 20 m for the hypothenuse, 16 for the adjacent side and 9+3 = 12m for the last side side.
Lenght of the smaller unknown side: 12m
Lenght of the larger unknown side: 20m
A tank contains 5,000 L of brine with 13 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate.
Required:
a. How much salt is in the tank after t minutes?
b. How much salt is in the tank after 20 minutes?
Answer:
a) [tex]x(t) = 13*e^(^-^\frac{t}{100}^)[/tex]
b) 10.643 kg
Step-by-step explanation:
Solution:-
- We will first denote the amount of salt in the solution as x ( t ) at any time t.
- We are given that the Pure water enters the tank ( contains zero salt ).
- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min
- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.
- The ODE is mathematically expressed as:
[tex]\frac{dx}{dt} =[/tex] ( salt flow in ) - ( salt flow out )
- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0
- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).
- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.
- So any time ( t ) the concentration of salt in the 5,000 L is:
[tex]conc = \frac{x(t)}{1000}\frac{kg}{L}[/tex]
- The amount of salt leaving the tank per unit time can be determined from:
salt flow-out = conc * V( flow-out )
salt flow-out = [tex]\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\[/tex]
salt flow-out = [tex]\frac{x(t)}{100}\frac{kg}{min}[/tex]
- The ODE becomes:
[tex]\frac{dx}{dt} = 0 - \frac{x}{100}[/tex]
- Separate the variables and integrate both sides:
[tex]\int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)[/tex]
- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:
[tex]13 = C*e^0 = C[/tex]
- The solution to the ODE becomes:
[tex]x(t) = 13*e^(^-^\frac{t}{100}^)[/tex]
- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:
[tex]x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg[/tex]
- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg
Can anybody please help me with this one??
Answer:
[tex]the \: answer \: is \: d.(x + 4) {}^{2} = 8(y + 4)[/tex]
3. (05.01)
A pair of linear equations is shown below:
y = -x + 1
y = 2x + 4
Which of the following statements best explains the steps to solve the pair of equations graphically? (4 points)
On a graph, plot the line y = -x + 1, which has y-intercept = -1 and slope = 1, and y = 2x + 4, which has y-intercept = 2 and slope = 4, and write the coordinates of the point of
Intersection of the two lines as the solution.
On a graph, plot the line y = -x + 1, which has y-intercept - 1 and slope = 1, and y = 2x + 4, which has y-intercept = 1 and slope = 4, and write the coordinates of the point of
intersection of the two lines as the solution.
On a graph, plot the line y = -x + 1, which has y-intercept = 1 and slope = -1, and y = 2x + 4, which has y-intercept = -2 and slope = 2, and write the coordinates of the point
of intersection of the two lines as the solution.
On a graph, plot the line y = -x + 1, which has y-intercept = 1 and slope = -1, and y = 2x + 4, which has y-intercept = 4 and slope = 2, and write the coordinates of the point of
intersection of the two lines as the solution.
Answer:
On a graph, plot the line y = -x + 1, which has y-intercept = 1 and slope = -1, and y = 2x + 4, which has y-intercept = 4 and slope = 2, and write the coordinates of the point of intersection of the two lines as the solution.
Step-by-step explanation:
Each equation is in slope-intercept form:
y = mx + b . . . . . where m is the slope, and b is the y-intercept
The first equation is ...
y = -x +1
so the slope is -1, and the y-intercept is +1.
__
The second equation is ...
y = 2x +4
so the slope is 2, and the y-intercept is 4.
__
The slopes and intercepts are properly described in the last selection.
Assume that military aircraft use ejection seats designed for men weighing between 133.8 lb and 208.0 lb. If women’s weights are normally distributed with a mean of 172.6 lb and a standard deviation of 42.4 lb, what percentage of women have weights between the ejection seat’s weight limits (that is, 133.8 to 208.0 lb)? Enter your answer as a percent rounded to one decimal place (do not add a "%"); add a trailing zeros as needed. The percentage of women with weights between 133.8 and 208.0 lb is [EjectPct] percent.
Answer:
61.8
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 172.6, \sigma = 42.4[/tex]
What percentage of women have weights between the ejection seat’s weight limits (that is, 133.8 to 208.0 lb)?
We have to find the pvalue of Z when X = 208 subtracted by the pvalue of Z when X = 133.8 for the proportion. Then we multiply by 100 to find the percentage.
X = 208
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{208 - 172.6}{42.4}[/tex]
[tex]Z = 0.835[/tex]
[tex]Z = 0.835[/tex] has a pvalue of 0.798
X = 133.8
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{133.8 - 172.6}{42.4}[/tex]
[tex]Z = -0.915[/tex]
[tex]Z = -0.915[/tex] has a pvalue of 0.180
0.798 - 0.18 = 0.618
0.618*100 = 61.8%
Without the %, the answer is 61.8.
(5m+100) (2m+10) what’s the value of m
Answer:
m=-30
Step-by-step explanation:
5m+100=2m+10
We want to get the variable on one side of the equation. First we subtract 100 from both sides.
5m=2m-90
Subtract 2m from both sides.
3m=-90
Divide both sides by 3.
m=-30
Which of the following best describes the slope of the line below?
PLSSS HELP
The slope is zero. Slope formula is Y=mx+b and since B is 1.5 and it is a straight line, Y=mx+1.5. What plus 1.5 is 1.5? 0. Hope this helps.
What is the value of n in the equation: 8n+9= -n+5?
Answer:
n = -1
Step-by-step explanation:
So first subtract 9 to both sides
8n = -n - 9
Now you want the n on one side and the constant on the other
so add the single n to the n side
9n = -9
Divide 9 to both sides to solve for n
n = -1
0.580 80 repeating as a simplified fraction
Answer:
979
Step-by-step explanation:
Answer:
115/198
Step-by-step explanation:
khan
I need help with this one
Answer:
2 2/3
Step-by-step explanation:
Which expressions are equivalent to 64^1Check all that apply
The right answers are:
4^38^22^6Hope it helps.
please see the attached picture for full solution
Good luck on your assignment
Graph the image of the figure given the translation. 1. (x, y) → (x +4, y - 1)
Answer:
Y=(-1,0)
G=(0,1)
F=(-1,3)
Step-by-step explanation:
I need HELP PLEASE HELP ME
Answer:
Graph 2
Step-by-step explanation:
You can see that all the shaded numbers are above negative 25 in that graph. Hope this helped!
.
A students received a score of 50 on his history test. The test had a mean of 69 and a standard deviation of 10. Find the z score and assess whether his score is considered unusual.
1.90; unusual
–1.90; not unusual
–1.90; unusual
1.90; not unusual
Answer:
c) The Z-score = - 1.90 unusual
Step-by-step explanation:
Explanation:-
Let 'X' be the random variable in normal distribution
Given student received a score X = 50
Mean of the Population x⁻ = 69
standard deviation of the Population 'σ' = 10
now
[tex]Z = \frac{x^{-}-mean }{S.D}[/tex]
[tex]Z = \frac{x^{-}-mean }{S.D} = \frac{50 -69}{10} = - 1.90[/tex]
The Z-score = - 1.90
Conclusion:-
The Z-score = - 1.90 unusual
Answer: The Z-score = - 1.90 unusual
Step-by-step explanation:
The bottom of a ladder must be placed 3 ft. from a wall. The ladder is 12 feet long. How far above the ground does the ladder touch the wall? Round your answer to the nearest tenth.
Use the Pythagorean theorem to solve.
Height = sqrt(12^2 -3^2)
Height = sqrt(144-9)
Height = sqrt(135)
Height = 11.6189 = 11.6 feet
A mattress store sells only king, queen and twin-size mattresses. Sales records at the store indicate that the number of queen-size mattresses sold is one-fourth the number of king and twin-size mattresses combined. Records also indicate that three times as many king-size mattresses are sold as twin-size mattresses. Calculate the probability that the next mattress sold is either king or queen-size.
Answer:
The probability that the next mattress sold is either king or queen-size is P=0.8.
Step-by-step explanation:
We have 3 types of matress: queen size (Q), king size (K) and twin size (T).
We will treat the probability as the proportion (or relative frequency) of sales of each type of matress.
We know that the number of queen-size mattresses sold is one-fourth the number of king and twin-size mattresses combined. This can be expressed as:
[tex]P_Q=\dfrac{P_K+P_T}{4}\\\\\\4P_Q-P_K-P_T=0[/tex]
We also know that three times as many king-size mattresses are sold as twin-size mattresses. We can express that as:
[tex]P_K=3P_T\\\\P_K-3P_T=0[/tex]
Finally, we know that the sum of probablities has to be 1, or 100%.
[tex]P_Q+P_K+P_T=1[/tex]
We can solve this by sustitution:
[tex]P_K=3P_T\\\\4P_Q=P_K+P_T=3P_T+P_T=4P_T\\\\P_Q=P_T\\\\\\P_Q+P_K+P_T=1\\\\P_T+3P_T+P_T=1\\\\5P_T=1\\\\P_T=0.2\\\\\\P_Q=P_T=0.2\\\\P_K=3P_T=3\cdot0.2=0.6[/tex]
Now we know the probabilities of each of the matress types.
The probability that the next matress sold is either king or queen-size is:
[tex]P_K+P_Q=0.6+0.2=0.8[/tex]
Section Exercise 11-8 Sales of People magazine are compared over a 5-week period at four Borders outlets in Chicago. Weekly Sales Store 1 Store 2 Store 3 Store 4 102 97 89 100 106 77 91 116 105 82 75 87 115 80 106 102 112 101 94 100 Click here for the Excel Data File Fill in the missing data. (Round your p-value to 4 decimal places, mean values to 1 decimal place, and other answers to 2 decimal places.) Treatment Mean n Std. Dev Store 1 Store 2 Store 3 Store 4 Total One-Factor ANOVA Source SS df MS F p-value Treatment Error Total (a) Based on the given hypotheses choose the correct option. H0: μ1 = μ2 = μ3 = μ4 H1: Not all the means are equal α = 0.05 Reject the null hypothesis if F > 3.24 Reject the null hypothesis if F < 3.24 (b) Determine the value of F. (Round your answer to 2 decimal places.) F-value (c) On the basis of the above-determined values, choose the correct decision from below. Fail to reject the null hypothesis. Reject the null hypothesis. (d) Determine the p-value. (Round your answer to 4 decimal places.) p-value Next Visit question mapQuestion 1 of 2 Total1 of 2 Prev
Answer:
Step-by-step explanation:
Hello!
An ANOVA was conducted to analyze the variable
Y: sales of "People" magazine over a 5-week period
This was studied in 4 Borders outlets in Chicago.
So this test has one factor: "Borders outlets" and four treatments: "Store 1, store 2, store 3 and store 4"
For each store you have the data for the weekly sales over a 5-week period so the sample sizes are:
n₁=n₂=n₃=n₄= 5 weeks
Store 1
∑X₁= 540; ∑X₁²= 58434
X[bar]₁= 108
S₁²= 28.50
S₁= 5.34
Store 2
∑X₂= 437; ∑X₂²= 38663
X[bar]₂= 87.40
S₂²= 117.30
S₂= 10.83
Store 3
∑X₃= 455; ∑X₃²= 41899
X[bar]₃= 91
S₃²= 123.50
S₃= 11.11
Store 4
∑X₄=505; ∑X₄²= 51429
X[bar]₄= 101
S₄²= 106
S₄= 10.30
Totals
N= n₁ + n₂ + n₃ + n₄= 4*5= 20
∑Mean= 108+87.40+91+101= 387.4
∑Variance= 28.50+117.30+123.50+106= 375.30
∑Standard deviation= 5.34+10.83+11.11+10.30= 37.58
Hypothesis test:
H₀: μ₁= μ₂= μ₃= μ₄
H₁: At least one population mean is different.
α: 0.05
The statistic for this test is
[tex]F= \frac{MS_{Treatments}}{MS_{Error}} ~~F_{k-1;N-k}[/tex]
k-1= 3 Df of the treatments, k=4 number of treatments
N-k= 16 Df of errors, N=20 total number of observations in all treatments
[tex]F_{H_0}= \frac{441.78}{93.83}= 4.71[/tex]
The critical region and p-value for this test are one-tailed to the right.
Using the critical value approach:
[tex]F_{k-1;N-k;1-\alpha }= F_{3;16;0.95}= 3.25[/tex]
The decision rule is:
If [tex]F_{H_0}[/tex] ≥ 3.25, reject the null hypothesis.
If [tex]F_{H_0}[/tex] < 3.25, do not reject the null hypothesis.
Using the p-value approach:
Little reminder: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
So under the distribution F₃,₁₆ you have to calculate the probability of the calculated [tex]F_{H_0}[/tex]:
P(F₃,₁₆≥4.71)= 1 - P(F₃,₁₆<4.71)= 1 - 0.9847 = 0.0153
p-value= 0.0153
The decision rule for this approach is
If p-value ≤ α, reject the null hypothesis.
If p-value > α, do not reject the null hypothesis.
The p-value is less than the level of significance, so the decision is to reject the null hypothesis.
I hope this helps!
An extremely simple (and surely unreliable) weather prediction model would be one where days are of two types: sunny or rainy. A sunny day is 90% likely to be followed by another sunny day, and a rainy day is 50% likely to be followed by another rainy day. Model this as a Markov chain. If Sunday is sunny, what is the probability that Tuesday (two days later) is also sunny
Answer:
The probability that if Sunday is sunny, then Tuesday is also sunny is 0.86.
Step-by-step explanation:
Let us denote the events as follows:
Event 1: a sunny day
Event 2: a rainy day
From the provided data we know that the transition probability matrix is:
[tex]\left\begin{array}{ccc}1&\ \ \ \ 2\end{array}\right[/tex]
[tex]\text{P}=\left\begin{array}{c}1&2\end{array}\right[/tex] [tex]\left[\begin{array}{cc}0.90&0.10\\0.50&0.50\end{array}\right][/tex]
In this case we need to compute that if Sunday is sunny, what is the probability that Tuesday is also sunny.
This implies that we need to compute the value of P₁₁².
Compute the value of P² as follows:
[tex]P^{2}=P\cdot P[/tex]
[tex]=\left[\begin{array}{cc}0.90&0.10\\0.50&0.50\end{array}\right]\cdot \left[\begin{array}{cc}0.90&0.10\\0.50&0.50\end{array}\right]\\\\=\left[\begin{array}{cc}0.86&0.14\\0.70&0.30\end{array}\right][/tex]
The value of P₁₁² is 0.86.
Thus, the probability that if Sunday is sunny, then Tuesday is also sunny is 0.86.
