Answer:
a) Q = 122 units/order
b) Number of orders = 2.05 orders/year
c) Average inventory = 61 units
d) Ordering costs = 125 $/order
Step-by-step explanation:
The economic quantity order (EOQ) formula allow us to minimize the ordering cost, in function of the demand, ordering cost and holding cost.
The EOQ formula is:
[tex]EOQ=\sqrt{\frac{2DS}{H} }[/tex]
where:
D: demand in units/year
S: Order costs, per order
H: holding or carrying cost, per unit a year
a) In this case:
D: 250 u/year
S: 30 $/order
H: 1 $/year-unit
[tex]EOQ=\sqrt{\frac{2DS}{H} }=\sqrt{\frac{2*250*30}{1} }=\sqrt{15000}=122.47\approx122[/tex]
b) If we have a demand of 250 units/year and we place orders of 122 units, the amount of orders/year is:
[tex]\#orders=\frac{D}{EOQ}=\frac{250\,units/year}{122\,units/order}=2.05\, \frac{orders}{year}[/tex]
c) We assume that there is no safety stock, so everytime the stock hits 0 units, a new order enter the inventory.
In this case, the average inventory can be estimated as the average between the inventory when a new order enters the inventory (122 u.) and the inventory right before a order enters (0 u.)
[tex]\#av.inventory=\frac{122+0}{2}=61[/tex]
The average inventory is 61 units.
d) If 250 units is the optimal quantity for an order, it means it is equal to the EOQ. We can calculate the new ordering costs as:
[tex]EOQ=\sqrt{\frac{2DS}{H} }=\sqrt{\frac{2*250*S}{1} }=250\\\\2*250*S=250^2\\\\S=250/2=125\,\$/order[/tex]
A) To minimize cost, the number of units that should be ordered each time an order is placed is; 122 units
B) The number of orders per year needed with the optimal policy is; 2 orders per year.
C) The average inventory if costs are minimized is; 61 units
D) For the order policy of Q = 250 to be optimal, the ordering cost would have to be; $125 per order
The formula for economic quantity order (EOQ) is given as;
EOQ = √(2DS/H)
Where;
D is demand rate S is set up costs H is holding costWe are given;
Annual demand; D = 250 units/year
Holding Cost; H = $1 per unit per year
Set up costs; S = $30 per order
A) EOQ here is;EOQ = √(2 × 250 × 30/1)
EOQ = 122.47
But EOQ has to be a whole number and so we approximate to the nearest whole number to get;
EOQ = 122
B) With the optimal policy, the number of orders per year is gotten from the formula;n = D/EOQ
Plugging in the relevant values gives;
n = 250/122
n = 2.049
But number of orders has to be a whole number. Thus, we approximate to the nearest whole number to get;
n = 2 orders per year
C) If costs are minimized, the average inventory is defined as the average between of the inventory when a new order enters and the inventory just before a new order enters.Before a new order enters the inventory is 0 if we assume that there is no safe stock. Thus;
average inventory = (122 + 0)/2 = 61 units
D) We are told that ordering cost is not $30 but Optimal order EOQ is 250 and so;EOQ = √(2DS/H)
⇒ 250 = √(2 × 250 × S/1)
Square both sides to get;
250² = 500S
S = 250²/500
S = $125 per order
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In the Country A legal system, a defendant is presumed innocent until proven guilty. Consider a null hypothesis, Upper H 0, that the defendant is innocent, and an alternative hypothesis, Upper H 1, that the defendant is guilty. A jury has two possible decisions: Convict the defendant (i.e., reject the null hypothesis) or do not convict the defendant (i.e., do not reject the null hypothesis). Explain the meaning of the risks of committing either a Type I or Type II error in this example.
(A) A Type I error would be incorrectly convicting the defendant when he is guilty. A Type II error would be incorrectly failing to convict the defendant when he is innocent.
(B) A Type I error would be incorrectly convicting the defendant when he is innocent. A Type II error would be incorrectly failing to convict the defendant when he is guilty.
(C) A Type I error would be incorrectly failing to convict the defendant when he is guilty. A Type II error would be incorrectly convicting the defendant when he is innocent.
(D) A Type I error would be incorrectly failing to convict the defendant when he is innocent. A Type II error would be incorrectly convicting the defendant when he is guilty.
Answer:
(C) A Type I error would be incorrectly failing to convict the defendant when he is guilty. A Type II error would be incorrectly convicting the defendant when he is innocent.
Step-by-step explanation:
Type I error is rejecting the true null hypothesis and type II error is not rejecting the false null hypothesis. Hence in this scenario, it will be:
A Type I error would be incorrectly convicting the defendant when he is innocent. A Type II error would be incorrectly failing to convict the defendant when he is guilty.
Option C is correct.
Euro Coin. Statistics students at the Akademia Podlaka conducted an experiment to test the hypothesis that the one-Euro coin is biased (i.e., not equally likely to land heads up or tails up). Belgian-minted one-Euro coins were spun on a smooth surface, and 140 out of 250 coins landed heads up. Does this result support the claim that one-Euro coins are biased
Answer:
[tex]z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{250}}}=1.897[/tex]
[tex]p_v =2*P(z>1.897)=0.0578[/tex]
If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of heads in the Euro coins is not significantly different from 0.5.
Step-by-step explanation:
1) Data given and notation
n=250 represent the random sample taken
X=140 represent the number of heads obtained
[tex]\hat p=\frac{140}{250}=0.56[/tex] estimated proportion of heads
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that that one-Euro coins are biased, so the correct system of hypothesis are:
Null hypothesis:[tex]p=0.5[/tex]
Alternative hypothesis:[tex]p \neq 0.5[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Check for the assumptions that he sample must satisfy in order to apply the test
a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.
b) The sample needs to be large enough
[tex]np_o =250*0.5=125>10[/tex]
[tex]n(1-p_o)=250*(1-0.5)=125>10[/tex]
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{250}}}=1.897[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z>1.897)=0.0578[/tex]
If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of heads in the Euro coins is not significantly different from 0.5.
According to the Pew report, 14.6% of newly married couples in 2008 reported that their spouse was of another race or ethnicity (CNNLiving, June 7, 2010). In a similar survey in 1980, only 6.8% of newlywed couples reported marrying outside their race or ethnicity. Suppose both of these surveys were conducted on 120 newly married couples. Use Table 1.
a.
Specify the competing hypotheses to test the claim that there is an increase in the proportion of people who marry outside their race or ethnicity.
H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0
H0: p1 − p2 = 0; HA: p1 − p2 ≠ 0
H0: p1 − p2 ≥ 0; HA: p1 − p2 < 0
H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0 is the competing hypotheses to test the claim that there is an increase in the proportion of people who marry outside their race or ethnicity. The correct option is option A.
A hypothesis is a put forth explanation of a phenomenon (plural: hypotheses). A hypothesis must be testable according to the scientific method for it to be considered a scientific hypothesis. Scientific hypotheses are typically based on prior observations that cannot be adequately explained by the current body of knowledge.
H0:
p1 − p2 = 0;
HA: p1 −
p2 ≠ 0
H0:
p1 − p2 ≥ 0;
HA: p1 −
p2 < 0
H0:
p1 − p2 ≤ 0;
HA: p1 −
p2 > 0
H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0
Therefore, the correct option is option A.
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In ΔDEF, the measure of ∠F=90°, FE = 39, ED = 89, and DF = 80. What ratio represents the sine of ∠D?
Answer:Sin of ∠D = 39/89
Step-by-step explanation:
The diagram of the right angle DEF is shown in the attached photo.
To determine the ratio representing the sine of ∠D, we would apply the sine trigonometric ratio. It is expressed as
Sin# = opposite side/hypotenuse
Looking at the triangle
Opposite side = 39
Hypotenuse = 89
# = ∠D
Therefore
Sin of ∠D = 39/89
The sine of ∠D in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. So, in ΔDEF, sine of ∠D is FE/ED = 39/89.
Explanation:In trigonometry, the sine of an angle in a right triangle is defined as the ratio of the length of the side that is opposite that angle, to the length of the hypotenuse. Considering this, the ratio representing the sine of ∠D in ΔDEF would be the length of side FE (opposite ∠D) divided by the length of side ED (the hypotenuse in ΔDEF because ∠F is a right angle and ED lies opposite to it).
Therefore, the ratio representing the sine of ∠D is FE/ED = 39/89.
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Nielsen Media Research wants to estimate the mean amount of time, in minutes, that full-time college students spend texting each weekday.Find the sample size necessary to estimate that mean with a 15 minute margin of error. Assume that a 96% confidence level is desired and that the standard deviation is estimated to be 112.2 minutes.
Answer:
n=237
Step-by-step explanation:
Previous concepts
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma=112.2)[/tex]
We know that the margin of error for a confidence interval is given by:
[tex]Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.96=0.04[/tex] and [tex]\alpha/2=0.02[/tex]
Using the normal standard table, excel or a calculator we see that:
[tex]z_{\alpha/2}=2.054[/tex]
If we solve for n from formula (1) we got:
[tex]\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}[/tex]
[tex]n=(\frac{z_{\alpha/2} \sigma}{Me})^2[/tex]
And we have everything to replace into the formula:
[tex]n=(\frac{2.054(112.2)}{15})^2 =236.05[/tex]
And if we round up the answer we see that the value of n to ensure the margin of error required [tex]\pm=15 min[/tex] is n=237.
Find an equation of the slant asymptote. Do not sketch the curve. y = 5x4 + x2 + x x3 − x2 + 5
Final answer:
The equation of the slant asymptote is y = 5x + 4. The equation of the slant asymptote for the given function is found by performing polynomial long division and taking the quotient without the remainder.
Explanation:
To find the equation of the slant asymptote for the function y = (5x4 + x2 + x) / (x3 - x2 + 5), we need to divide the numerator by the denominator using polynomial long division or synthetic division. The quotient, without the remainder, will give us the equation of the slant asymptote because as x approaches infinity, the remainder becomes insignificant compared to the terms in the quotient.
Step 1: Perform the division. (This will be specific to the given function.)
Step 2: The quotient (without the remainder) is the equation of the slant asymptote.
To find the equation of the slant asymptote, we need to divide the numerator (5x^4 + x^2 + x) by the denominator (x^3 - x^2 + 5) using long division.
The quotient is 5x + 4, so the equation of the slant asymptote is y = 5x + 4. It is important to note that the slant asymptote represents the behavior of the function as x approaches positive or negative infinity.
Depreciation is the decrease or loss in value of an item due to age, wear, or market conditions. We usually consider depreciation on expensive items like cars. Businesses use depreciation as a loss when calculating their income and taxes. One company buys a new bulldozer for $138000. The company depreciates the bulldozer linearly over its useful life of 23 years. Its salvage value at the end of 23 years is $14950.
Express the value of the bulldozer, V, as a function of how many years old it is, t. Preview The value of the bulldozer after 3 years is $
Answer:The value of the bulldozer after 3 years is $121950
Step-by-step explanation:
We would apply the straight line depreciation method. In this method, the value of the asset(bulldozer) is reduced linearly over its useful life until it reaches its salvage value. The formula is expressed as
Annual depreciation expense =
(Cost of the asset - salvage value)/useful life of the asset.
From the given information,
Useful life = 23 years
Salvage value of the bulldozer = $14950
Cost of the new bulldozer is $138000
Therefore
Annual depreciation = (138000 - 14950)/ 23 = $5350
The value of the bulldozer at any point would be V. Therefore
5350 = (138000 - V)/ t
5350t = 138000 - V
V = 138000 - 5350t
The value of the bulldozer after 3 years would be
V = 138000 - 5350×3 = $121950
(a) Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x)=x/(10 x**2+1) sum_(n=0)^infinity (-1)**n 10**n x**(2n+1) sum_(n=0)^infinity (-1)**n (x**(2n+1))/(10**n) sum_(n=0)^infinity (-1)**n 10**n x**(2n) 10 sum_(n=0)^infinity (-1)**n x**(2n+1) sum_(n=0)^infinity (-1)**n 10**(n+1) x**(n+1) Correct: Your answer is correct. (b) Determine the interval of convergence.
One way to do this is to recall that for [tex]|x|<1[/tex], we have
[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]
so that
[tex]\displaystyle\frac x{10x^2+1}=\frac x{1-(-10x^2)}=x\sum_{n=0}^\infty(-10x^2)^n=\sum_{n=0}^\infty(-10)^nx^{2n+1}[/tex]
(which seems to match the first option) so long as [tex]|-10x^2|=10x^2<1[/tex], or [tex]-\frac1{\sqrt{10}}<x<\frac1{\sqrt{10}}[/tex], which is the interval of convergence.
the Earth has an average distance from the Sun of 1.5* 10 to the 8th power kilometers and the Earth has an approximate distance from the Moon of 4.0 x 10 to the 5th km how many times greater is Earth's average distance from the Sun than is from the Moon
Answer:
Earth's average distance from the Sun is [tex]375[/tex] times more than is from the Moon.
Step-by-step explanation:
Given the average distance of Earth from the Sun is [tex]1.5\times10^{8}[/tex] kilometers.
Also, the approximate distance of Earth from the Moon is [tex]4\times10^{5}[/tex] kilometers.
For this, we will divide the average distance of Earth from the Sun by approximate distance of Earth from the Moon.
So,
[tex]\frac{1.5\times10^{8}}{4\times10^{5}}=375\ km[/tex]
An analyst is forecasting net income for Excellence Corporation for the next fiscal year. Her low-end estimate of net income is $250,000, and her high-end estimate is $350,000. Prior research allows her to assume that net income follows a continuous uniform distribution. The probability that net income will be greater than or equal to $337,500 is
Answer:
[tex]P(X \geq 337,500) = 0.125 = 12.5%[/tex]
Step-by-step explanation:
probability distribution fucntion is given as
[tex] F_x = P(X \leq x)[/tex]
[tex] = \frac{x -a}{b -a} a<x< b[/tex]
where a indicate lower end estimate = $250,000
b indicate high end estimate = $350,000
probability greater than $337500
[tex] P(X \geq 337,500) = 1- P(X < 337500)[/tex]
[tex] = 1 - \frac{x -a}{b -a}[/tex]
[tex] = 1 - \frac{337500 - 250000}{350000 - 250000}[/tex]
[tex]P(X \geq 337,500) = 0.125 = 12.5%[/tex]
Calculate the probability of net income being greater than or equal to $337,500 for Excellence Corporation following a continuous uniform distribution. Probability will be equal to 12.5%.
The probability that net income will be greater than or equal to $337,500 is 0.25. To calculate this probability for a continuous uniform distribution, we need to find the proportion of the area under the distribution curve that corresponds to values greater than $337,500.
Step-by-step calculation:
Calculate the total range of net income: $350,000 - $250,000 = $100,000
Calculate the proportion of the area for values greater than $337,500: ($350,000 - $337,500) / $100,000 = 0.125
Since the distribution is continuous and uniform, the probability is equal to the proportion calculated: 0.125 = 12.5%
Jill and Kim each improved their yards by planting hostas and geraniums. They bought their supplies from the same store. Jill spent $101 on 11 hostas and 2 geraniums. Kim spent $80 on 8 hostas and 2 geraniums. What is the cost of one hosta and one geramium?
Answer: the cost of one hosta is $7
the cost of one geramium is $12
Step-by-step explanation:
Let x represent the cost of one hosta.
Let y represent the cost of one geramium.
They bought their supplies from the same store. Jill spent $101 on 11 hostas and 2 geraniums. This means that
11x + 2y = 101 - - - - - - - -1
Kim spent $80 on 8 hostas and 2 geraniums. This means that
8x + 2y = 80 - - - - - - - - -2
We will eliminate y by subtracting equation 2 from equation 1, it becomes
3x = 21
x = 21/3 = 7
Substituting x = 7 into equation 2, it becomes
8 × 7 + 2y = 80
56 + 2y = 80
2y = 80 - 56 = 24
y = 24/2 = 12
Of the 50 students in an undergraduate statistics class, 60% send email and/or text messages during any given lecture. They have a 50/50 chance of being caught by their professor, who is not amused by such unprofessional conduct. The eagle-eyed professor never charges an innocent student. What is the probability a student sends an email and/or text message during lecture and gets caught?
Answer:
0.3 or 30%
Step-by-step explanation:
Since no innocent student will ever be caught, the probability that a student sends an email and/or text message during a lecture AND gets caught is given by the product of the probability of a student sending a message (60%) by the probability of the professor catching them (50%) :
[tex]P = 0.5*0.6 = 0.3[/tex]
The probability is 0.3 or 30%.
An electronics company produces transistors, resistors, and computer chips. Each transistor requires 1 unit of copper, 2 units of zinc, and 3 units of glass. Each resistor requires 2 units of copper and 3 units of zinc. Each computer chip requires 1 unit of zinc, and 5 units of glass. How many of each product can be made knowing that the company has 104 units of copper, 169 units of zinc, and 74 units of glass?
Answer:
The capacity is 18 transistors, 43 resistors and 4 computer chips.
Step-by-step explanation:
This problem can be solved by a system of equations.
I am going to say that:
x is the number of transistors that are made.
y is the number of resistors that are made.
z is the number of computer chips that are made.
Building the system:
A transistor requires 1 unit of copper. A resistor requires 2 units of copper. There are 104 units of copper. So [tex]x + 2y = 104[/tex]
A transistor requires 2 units of zinc. A resistor requires 3 units of zinc. A chip requires 1 unit of zing. There are 169 units of zinc. So [tex]2x + 3y + z = 169[/tex].
A transistor requires 3 units of glass. A chip requires 5 units of glass. There are 74 units of glass. So [tex]3x + 5z = 74[/tex]
We have the following system:
[tex]x + 2y = 104[/tex]
[tex]2x + 3y + z = 169[/tex]
[tex]3x + 5z = 74[/tex]
The solution is:
[tex]x = 18, y = 43, z = 4[/tex]
The capacity is 18 transistors, 43 resistors and 4 computer chips.
The endpoints of the side of a square lie on the circle and the opposite side of the square tangent to the circle at its midpoint. Find the ratio of the side and the radius.
Answer:
[tex]\displaystyle \frac{L}{r}=\frac{8}{5}[/tex]
Step-by-step explanation:
Circle and Square
We have a geometric construction as shown in the image below. We can see that
[tex]\displaystyle r+h=L[/tex]
Or, equivalently
[tex]\displaystyle h=L-r[/tex]
The triangle formed by r,h and L/2 is right, because the opposite side of the square is tangent to the circle at its midpoint. This means we can use Pythagoras's theorem:
[tex]\displaystyle r^2=h^2+\left(\frac{L}{2}\right)^2[/tex]
Replacing h
[tex]\displaystyle r^2=(L-r)^2+\left(\frac{L}{2}\right)^2[/tex]
Expanding squares
[tex]\displaystyle r^2=L^2-2Lr+r^2+\frac{L^2}{4}[/tex]
Simplifying
[tex]\displaystyle 2Lr=L^2+\frac{L^2}{4}[/tex]
Multiplying by 4
[tex]\displaystyle 8r=4L+L[/tex]
Joining terms
[tex]\displaystyle 8r=5L[/tex]
Solving for the ratio L/R as required
[tex]\displaystyle \frac{L}{r}=\frac{8}{5}[/tex]
The ratio of the side and the radius is 8:5
calculation;Since
r + h = L
So we can write h = L - r
Now we applied the Pythagoras theorem
[tex]r^2 = h^2 +( \frac{L}{2}) ^2\\\\r^2 = (L-r)^2 + ( \frac{L^}{2}) ^2\\\\r^2 = L^2- 2Lr+ r^2 + \frac{L^2}{4}\\\\ 2Lr = L^2 \frac{L^2}{4}\\\\[/tex]
Now
8r = 4L + L
8r = 5L
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Suppose we want a 90% confidence interval for the average amount of time (in minutes) spent per week on homework by the students in a large introductory statistics course at a large university. The interval is to have a margin of error of 3 minutes, and the amount spent has a Normal distribution with a standard deviation σ = 40 minutes. The number of observations required is closest to:
1180.
683.
482.
22.
Answer: 482
Step-by-step explanation:
Formula to find the sample size is given by :-
[tex]n= (\dfrac{z^*\times \sigma}{E})^2[/tex] (1)
, where z* = critical z-value (two tailed).
[tex]\sigma[/tex] = Population standard deviation and E = Margin of error.
As per given , we have
Margin of error : E= 3
[tex]\sigma=40[/tex]
Confidence level = 90%
Significance level =[tex]\alpha=1-0.90=0.10[/tex]
Using z-table , the critical value for 90% confidence=[tex]z^*=z_{\alpha/2}=z_{0.05}=1.645[/tex]
Required minimum sample size = [tex]n= (\dfrac{(1.645)\times (40)}{3})^2[/tex] [Substitute the values in formula (1)]
[tex]n=(21.9333333333)^2[/tex]
[tex]n=481.07111111\approx482[/tex] [ Round to the next integer]
Hence, the number of observations required is closest to 482.
Suppose an ecologist collects data on the weights and basal trunk diameters of a randomly selected sample of felled trees. The ecologist fits a simple linear regression model, predicting tree weight from the trunk's cross-sectional area. After ensuring that all assumptions of the linear model are met, the ecologist computes a 95% confidence interval for the predicted mean weight of trees with a trunk cross-sectional area of 452 in?.
Select the correct interpretation of this confidence interval.
a. The weight of another tree with a trunk cross-sectional area of 452 in- sampled at random from the same population will fall in the ecologist's interval with 95% probability.
b.The true mean weight for trees in the population with a trunk cross-sectional area of 452 in- will fall within the ecologist's interval with 95% probability.
c.The weight of 95% of trees in the population with a trunk cross-sectional area of 452 inwill fall within the ecologist's interval.
d. The ecologist's interval has a 95% chance of containing the true sample mean weight of trees with a trunk cross- sectional area of 452 in².
e. There is a 95% chance that the ecologist's interval contains the true mean weight of trees in the population with a trunk cross-sectional area of 452 in?.
Answer:
e) There is a 95% chance that the ecologist's interval contains the true mean weight of trees in the population with a trunk cross-sectional area of 452 in
Step-by-step explanation:
Given that an ecologist collects data on the weights and basal trunk diameters of a randomly selected sample of felled trees. The ecologist fits a simple linear regression model, predicting tree weight from the trunk's cross-sectional area. After ensuring that all assumptions of the linear model are met, the ecologist computes a 95% confidence interval for the predicted mean weight of trees with a trunk cross-sectional area of 452 inches
We can interpret the confidence interval as
we are 95% confident that for samples of trees largely drawn with 452 inches the mean of these trees would fall within this interval.
e) There is a 95% chance that the ecologist's interval contains the true mean weight of trees in the population with a trunk cross-sectional area of 452 in
Let F=(2x,2y,2x+2z)F=(2x,2y,2x+2z). Use Stokes' theorem to evaluate the integral of FF around the curve consisting of the straight lines joining the points (1,0,1), (0,1,0) and (0,0,1). In particular, compute the unit normal vector and the curl of FF as well as the value of the integral
Stokes' theorem equates the line integral of [tex]\vec F[/tex] along the curve to the surface integral of the curl of [tex]\vec F[/tex] over any surface with the given curve as its boundary. The simplest such surface is the triangle with vertices (1,0,1), (0,1,0), and (0,0,1).
Parameterize this triangle (call it [tex]T[/tex]) by
[tex]\vec s(u,v)=(1-v)((1-u)(1,0,1)+u(0,1,0))+v(0,0,1)[/tex]
[tex]\vec s(u,v)=((1-u)(1-v),u(1-v),1-u+uv)[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]T[/tex] to be
[tex]\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,1-v,1-v)[/tex]
Divide this vector by its norm to get the unit normal vector. Note that this assumes a "positive" orientation, so that the boundary of [tex]T[/tex] is traversed in the counterclockwise direction when viewed from above.
Compute the curl of [tex]\vec F[/tex]:
[tex]\vec F=(2x,2y,2x+2z)\implies\mathrm{curl}\vec F=(0,-2,0)[/tex]
Then by Stokes' theorem,
[tex]\displaystyle\int_{\partial T}\vec F\cdot\mathrm d\vec r=\iint_T\mathrm{curl}\vec F\cdot\mathrm d\vec S[/tex]
where
[tex]\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\,\mathrm dS[/tex]
[tex]\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\mathrm d\vec S=\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv[/tex]
The integral thus reduces to
[tex]\displaystyle\int_0^1\int_0^1(0,-2,0)\cdot(0,1-v,1-v)\,\mathrm du\,\mathrm dv=\int_0^12(v-1)\,\mathrm dv=\boxed{-1}[/tex]
The question requires the use of Stokes' theorem to compute the integral of a given vector field around a defined curve. This involves calculating the curl of the vector field and a surface integral. However, specific computations were not provided.
Explanation:The question is asking for the use of the Stokes' theorem to compute the integral of the vector field F=(2x,2y,2x+2z)F=(2x,2y,2x+2z) around the curve consisting of the straight lines joining the points (1,0,1), (0,1,0) and (0,0,1). In Stokes theorem, we're interested in the curl of the vector field and the surface integral of that curl over some surface S, with orientation determined by a unit normal vector. The problem also requires the computation of the curl of F which is a vector field whose components are the partial derivatives of the components of F. However, with this given information, we're unable to proceed with computations as detailed integral computations weren't provided. Therefore, a complete answer can't be provided at this time.
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Ask Your Teacher An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 183 of these passed the probe. Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe. (Round your answers to three decimal places.)
Answer:
The 95% (two-sided) confidence interval for the proportion of all dies that pass the probe is (0.462, 0.566).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
356 dies were examined by an inspection probe and 183 of these passed the probe. This neabs that [tex]n = 356, \pi = \frac{183}{356} = 0.514[/tex]
95% confidence interval
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.514 - 1.96\sqrt{\frac{0.514*0.486}{356}} = 0.462[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.514 + 1.96\sqrt{\frac{0.514*0.486}{356}}{119}} = 0.566[/tex]
The 95% (two-sided) confidence interval for the proportion of all dies that pass the probe is (0.462, 0.566).
The regression equation NetIncome = 2,159 + .0312 Revenue was estimated from a sample of 100 leading world companies (variables are in millions of dollars).
(a-1) Calculate the residual for the x, y pair ($45,533, $2,697). (A negative value should be indicated by a minus sign. Round your answer to 4 decimal places.) Residual
(a-2) Did the regression equation underestimate or overestimate the net income? The regression equation overestimated the net income. The regression equation underestimated the net income.
(b-1) Calculate the residual for the x, y pair ($60,417, $3,497). (A negative value should be indicated by a minus sign. Round your answer to 4 decimal places.) Residual
(b-2) Did the regression equation underestimate or overestimate the net income? The regression equation overestimated the net income. The regression equation underestimated the net income.
Answer:
a) -882.6296
b) The regression equation overestimated the net income.
c) -547.0104
d) The regression equation overestimated the net income.
Step-by-step explanation:
We are given the following information in the question:
The regression equation for net income is given by:
[tex]\text{Net Income} = y = 2159 + .0312(\text{Revenue})[/tex]
a) residual for the (x, y) pair ($45,533, $2,697)
Calculated net income =
[tex]2159 + .0312(45533) = 3579.6296[/tex]
Residual = Observed net income - Calculated net income
[tex]\text{Residual} = 2697-3579.6296 = -882.6296[/tex]
b) The regression equation overestimated the net income.
c) residual for the (x, y) pair ($60,417, $3,497)
Calculated net income =
[tex]2159 + .0312(60417) = 4044.0104[/tex]
Residual = Observed net income - Calculated net income
[tex]\text{Residual} = 3497-4044.0104 = -547.0104[/tex]
d) The regression equation overestimated the net income.
The answers are : (a-1) The residual for the [tex](x,y)[/tex] pair [tex](\$45,533, $2,697)[/tex] is [tex]\[\text{Residual} = -881.6136\][/tex]. (a-2) The regression equation overestimated the net income. (b-1) The residual for the x, y pair ([tex]\$60,417, $3,497[/tex]) is [tex]\[\text{Residual} = -546.9704\][/tex]. (b-2) The regression equation overestimated the net income.
1. Calculate the predicted value [tex](\(\hat{y}\))[/tex] using the regression equation:
[tex]\[ \hat{y} = 2{,}159 + 0.0312 \times x \][/tex]
2. Compute the residual by subtracting the predicted value from the actual value:
[tex]\[ \text{Residual} = y - \hat{y} \][/tex]
Part (a-1)
For the [tex]\( x, y \) pair \((45,533, 2,697)\)[/tex]
1. Calculate the predicted net income
[tex]\[ \hat{y} = 2{,}159 + 0.0312 \times 45{,}533 \][/tex]
[tex]\[ \hat{y} = 2{,}159 + 1{,}419.6136 \][/tex]
[tex]\[ \hat{y} = 3{,}578.6136 \][/tex]
2. Compute the residual
[tex]\[ \text{Residual} = y - \hat{y} \][/tex]
[tex]\[ \text{Residual} = 2{,}697 - 3{,}578.6136 \][/tex]
[tex]\[ \text{Residual} = -881.6136 \][/tex]
Part (a-2)
The residual is negative, which means the actual net income is less than the predicted net income. Therefore, the regression equation overestimated the net income.
Part (b-1)
For the [tex]\( x, y \) pair \((60,417, 3,497)\)[/tex]
1. Calculate the predicted net income
[tex]\[ \hat{y} = 2{,}159 + 0.0312 \times 60{,}417 \][/tex]
[tex]\[ \hat{y} = 2{,}159 + 1{,}884.9704 \][/tex]
[tex]\[ \hat{y} = 4{,}043.9704 \][/tex]
2. Compute the residual
[tex]\[ \text{Residual} = y - \hat{y} \][/tex]
[tex]\[ \text{Residual} = 3{,}497 - 4{,}043.9704 \][/tex]
[tex]\[ \text{Residual} = -546.9704 \][/tex]
Part (b-2)
The residual is negative, which means the actual net income is less than the predicted net income. Therefore, the regression equation overestimated the net income.
A coupon for $5 off any lunch price states that a 15% tip will be added to the price before the $5 is subtracted. So, C(x) = x - 5 represents the price after the coupon reduction. T(x) = 1.15 x represents the price after the tip is applied. Write the simplified composite functions C(T(x)) and T(C(x)). Which composite function represents the conditions of the coupon?
Answer:
[tex]C(T(x))=1.15x-5[/tex] and [tex]T(C(x))=1.15x-5.75[/tex]
C(T(x)) represents the conditions of the coupon.
Step-by-step explanation:
The price after the coupon reduction is represented by the function
[tex]C(x)=x-5[/tex]
The price after the tip is applied is represented by the function
[tex]T(x)=1.15x[/tex]
We need to find the composite functions C(T(x)) and T(C(x)).
[tex]C(T(x))=C(1.15x)[/tex] [tex][\because T(x)=1.15x][/tex]
[tex]C(T(x))=1.15x-5[/tex] [tex][\because C(x)=x-5][/tex]
This function represents that 15% tip will be added first after that $5 is subtracted.
Similarly,
[tex]T(C(x))=T(x-5)[/tex] [tex][\because C(x)=x-5][/tex]
[tex]T(C(x))=1.15(x-5)[/tex] [tex][\because T(x)=1.15x][/tex]
[tex]T(C(x))=1.15x-5.75[/tex]
This function represents that $5 is subtracted first after that 15% tip will be added.
It is given that a coupon for $5 off any lunch price states that a 15% tip will be added to the price before the $5 is subtracted.
It means 15% tip will be added first after that $5 is subtracted. So, C(T(x)) represents the conditions of the coupon.
The functions C(T(x)) and T(C(x)) represent the application of a tip and a coupon to a lunch price, respectively, in different order. The function that correctly represents the specific conditions given by the coupon in the problem statement is T(C(x)).
Explanation:The composite function C(T(x)) is found by substituting T(x) into the function C(x). So, C(T(x)) = T(x) - 5 = 1.15x - 5.
The composite function T(C(x)) is calculated by substituting C(x) into the function T(x). So, T(C(x)) = 1.15 * (x - 5) = 1.15x - 5.75.
In the context of the coupon conditions, the right composite function is T(C(x)). This composite function first applies the $5 coupon reduction (C(x)) and then the 15% tip (T(x)), which exactly follows the procedure described by the coupon.
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Cyril, who is 68 years of age, received Social Security benefits of $12,000, wages of $5,000, interest and dividends of $4,000, unemployment compensation of $3,000 and municipal bond interest of $1,500. Calculate Cyril's adjusted gross income.
Answer:
Cyril's adjusted gross income is $12000.
Step-by-step explanation:
First, we have that the adjusted gross income is given by the sum of the income that is not taxable.
For our case, we have the following data:
Social security: 12000 (This is not taxable)
Wages: 5000 (This is taxable)
Interest and dividends: 4000 (This is taxable)
Unemployment compensation: 3000 (This is taxable)
municipal bond interest: $1,500 (This is not taxable)
Then, adding up the income that's deductible, we have that Cyril's adjusted gross income is:
AGI=5000+4000+3000=$12000
A researcher wishes to estimate the proportion of adults who have high-speed internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 95% confidence ifa) she uses a previous estimate of 0.36?b) she does not use any prior estimates?
Answer: a) 984 b) 1068
Step-by-step explanation:
When the prior estimate of the population proportion(p) is available .
Then the formula to find the sample size :-
[tex]n=p(1-p)(\dfrac{z^*}{E})^2[/tex]
, where E = margin of error
and z* = Critical z-value .
a) p= 0.36
E= 0.03
Critical value for 95% confidence level = z*= 1.96
Required sample size=[tex]n= 0.36(1-0.36)(\dfrac{1.960}{0.03})^2[/tex]
[tex]n= 0.36(0.64)(65.3333333333)^2[/tex]
[tex]n=(0.2304)(4268.44444444)=983.4496\approx984[/tex]
Hence, the required sample size is 984.
b) When the prior estimate of the population proportion is unavailable .
Then we use formula to find the sample size :-
[tex]n= 0.25(\dfrac{z^*}{E})^2[/tex]
, where E = margin of error
and z* = Critical z-value
Put E= 0.03 and z*= 1.960
Required sample size =[tex]n= 0.25(\dfrac{1.960}{0.03})^2[/tex]
[tex]n= 0.25(65.3333333333)^2[/tex]
[tex]n= 0.25(4268.44444444)=1067.11111111\approx1068[/tex]
Hence, the required sample size is 1068.
Which is a reasonable first step that can be used to solve the equation 2(x+6)=3(x-4)+5? Combine like terms, 3 and 5, on the right side of the equation. Distribute 2 to (x + 6) and 3 to (x – 4). Distribute 2 to x on the left side of the equation and 3 to x on the right side of the equation. Apply the subtraction property of equality to move the 6 from the left side of the equation to the right side.
Answer:
Distribute 2 to (x + 6) and 3 to (x – 4)
Step-by-step explanation:
A good first step for an equation such as this is to eliminate parentheses. That is accomplished by using the distributive property on the left and right. That property requires each term inside parentheses be multiplied by the factor outside parentheses. The process can be described as ...
Distribute 2 to (x + 6) and 3 to (x – 4)
_____
Comment on further steps
The result of the first step above would be ...
2x + 12 = 3x - 12 + 5
19 = x . . . . . . . . . . . next step: add 7-2x to both sides and collect terms
_____
Comment on answer wording
The last answer choice uses the wording "move the 6 from the left side of the equation to the right side". This sort of description is often invoked when a constant needs to be removed from one side of the equation or the other.
There is no property of equality that supports this action.
If you want to "move" a constant, you can invoke the addition property of equality that lets you add the same number to both sides of the equation. The number you would choose to add is the additive inverse of the one you want to "move."
In the above solution we have the sum -12+5 on the right after the first step. We want to eliminate that sum, so in the next step we show adding the additive inverse of the value of that sum. The sum of -12+5 is -7, and we add 7 to make that sum into the additive identity element, 0. (Likewise, we add -2x to turn the term 2x into the additive identity element, 0.) After these additions, we have ...
2x -2x +12 +7 = 3x -2x -12 +5 +7
0 + 19 = x + 0
19 = x
The relevant property of equality is the addition property of equality, which says I can add the same amount to both sides of the equation. I can't "move" anything without violating the equal sign, but I can use the addition and multiplication properties of equality to do the same thing to both sides of the equation.
Answer:
The correct answer is Distribute 2 to (x + 6) and 3 to (x – 4).
Step-by-step explanation:
In how many ways can 9 hearts be chosen if 12 cards are chosen from a well-shuffled deck of 52 playing cards? a) 715 b) 220 c) 108 d) 6,534,385 e) 117 f) None of the above.
Answer: d) 6,534,385
Therefore, the number of ways of selecting 9 hearts is 6,534,385
Step-by-step explanation:
Given;
Number of cards to be chosen = 12
Number of hearts to be chosen = 9
Number of non-hearts to be selected = 3
Total number of cards = 52
Number of hearts total = 13
Number of non-hearts total= 39
The number of ways of selecting hearts can be given by the illustration below.
Number of ways of selecting hearts × number of ways of selecting non-hearts
N = 13C9 × 39C3 (it's combination since order is not important)
N = 13!/(9! ×4!) × 39!/(36! × 3!)
N = 6,534,385
Therefore, the number of ways of selecting 9 hearts is 6,534,385
The number of ways to choose 9 hearts from 12 cards is 220, option (b) is correct.
To solve this problem, we'll use the combination formula, which is used when selecting objects from a larger set without regard to the order. The formula for combinations is:
[tex]\[ \text{C}(n, k) = \frac{n!}{k!(n - k)!} \][/tex]
where [tex]\( n \)[/tex] is the total number of items, [tex]\( k \)[/tex] is the number of items to choose, and (!) denotes factorial.
Given that we need to choose 12 cards from a deck of 52, and we want to know the number of ways to choose 9 hearts, which is [tex]\( k = 9 \)[/tex] hearts out of a total of [tex]\( n = 12 \)[/tex] cards.
Step 1:
Calculate the number of ways to choose 9 hearts from the 12 chosen cards:
[tex]\[ \text{C}(12, 9) = \frac{12!}{9!(12 - 9)!} \][/tex]
Step 2:
Simplify the expression:
[tex]\[ \text{C}(12, 9) = \frac{12!}{9!3!} \][/tex]
[tex]\[ \text{C}(12, 9) = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \][/tex]
[tex]\[ \text{C}(12, 9) = \frac{1320}{6} \][/tex]
[tex]\[ \text{C}(12, 9) = 220 \][/tex]
Therefore, there are 220 ways to choose 9 hearts from 12 cards. So, the correct answer is option (b).
find 72 degrees and 40 degrees from the picture
Answer:a) 5 sides
b) 9 sides
Step-by-step explanation:
a) Since the polygon is a regular polygon, all the interior angles are equal. If one exterior angle is 72 degrees, the interior angle would be
180 - 72 = 108 degrees(sum of the angles on a straight line is 180 degrees)
The sum of the interior angles of a polygon is expressed as
180(n - 2)
Where n represents the number of sides that the polygon has. Since the interior angles are equal, then
108n = 180(n - 2)
108n = 180n - 360
180n - 108n = 360
72n = 360
n = 360/72 = 5
b) If one exterior angle is 40 degrees, the interior angle would be
180 - 40 = 140 degrees
Since the interior angles are equal, then
140n = 180(n - 2)
140n = 180n - 360
180n - 140n = 360
40n = 360
n = 360/40 = 9
To play the lottery in a certain state, a person has to correctly select 5 out of 45 numbers, paying $1 for each five-number selection. If the five numbers picked are the same as the ones drawn by the lottery, an enormous sum of money is bestowed. What is the probability that a person will one combination of five numbers will win? What is the probability of winning if 100 different lottery tickets are purchased?
Answer:
Step-by-step explanation:
Given
There are 45 numbers out of which 5 numbers are required to win an enormous sum of money
No of ways in which 5 numbers can be selected out of 45 numbers.
[tex]^nC_r=\frac{n!}{(n-r)!r!}[/tex]
here n=45, r=5
[tex]^{45}C_5=\frac{45!}{40!5!}[/tex]
[tex]^{45}C_5=1221759[/tex]
out of which there is only combination foe contest winning
[tex]P=\frac{1}{1221759}[/tex]
When 100 different tickets are bought then
Probability of winning[tex]=\frac{100}{1221759}[/tex]
The probability of winning with one combination of five numbers is [tex]\( \frac{1}{1,221,759} \)[/tex] .
The probability of winning if 100 different lottery tickets are purchased is [tex]\( 1 - \left(1 - \frac{1}{1,221,759}\right)^{100} \)[/tex], approximately [tex]\( 0.0000182 \)[/tex].
Step 1
The probability of winning the lottery with one combination of five numbers can be calculated using the formula for combinations:
[tex]\[ P(\text{win with one ticket}) = \frac{1}{\binom{45}{5}} \][/tex] Where [tex]\( \binom{45}{5} \)[/tex]represents the number of ways to choose 5 numbers from 45 without regard to the order. Calculating [tex]\( \binom{45}{5} \)[/tex] :
[tex]\[ \binom{45}{5} = \frac{45 \times 44 \times 43 \times 42 \times 41}{5 \times 4 \times 3 \times 2 \times 1} = 1,221,759 \][/tex]
So, the probability [tex]\( P(\text{win with one ticket}) \)[/tex] is:
[tex]\[ P(\text{win with one ticket}) = \frac{1}{1,221,759} \approx 8.19 \times 10^{-7} \][/tex]
If 100 different lottery tickets are purchased, the probability of winning at least once is calculated using the complement rule:
[tex]\[ P(\text{win with 100 tickets}) = 1 - \left( 1 - \frac{1}{\binom{45}{5}} \right)^{100} \][/tex]
Step 2
Substituting [tex]\( \binom{45}{5} \)[/tex] :
[tex]\[ P(\text{win with 100 tickets}) = 1 - \left( 1 - \frac{1}{1,221,759} \right)^{100} \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 1 - (1 - 8.19 \times 10^{-7})^{100} \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 1 - (0.999999181)^{100} \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 1 - 0.9999818 \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 0.0000182 \][/tex]
The probability of winning the lottery with one ticket is approximately [tex]\( 8.19 \times 10^{-7} \)[/tex], while the probability of winning if 100 different tickets are purchased increases to approximately 0.0000182. Purchasing more tickets improves the chances of winning, but it remains a very low probability event due to the large number of possible combinations.
You manage a call center that processes orders. One measure of efficiency in your department is the time spent on each call. You need to report to your superiors on the time-per-call in your call center each month. You’ve identified some of your most efficient workers, and you know that efficiency tends to decrease, in general on Fridays.
1. Identify the ethical issues involved in developing a sampling protocol:
a. There are none. It's a simple mathematical operation
b. In order to get a random and representative sample, you need to consider how efficiency fluctuates over the month, and sample call times on a frequency that includes every day of the week.
c. The analysis and reporting should consider the impact of extreme values in the data.
d. Both "b." and "c." above.
Answer:
The answer is option A.
Step-by-step explanation:
If the report is made once a month, the daily variations (including the tendency to decrease efficiency on Friday) will be masked within the monthly result.
It would only generate an ethical problem in the case that Fridays fall in different samples, but in the case of the monthly report there are usually 4 or 5 Fridays included in each sample.
Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body bone mineral content (TBBMC). A highly skilled operator is required to take the measurements. Recently, a new DXA machine was purchased by a research lab and two operators were trained to take the measurements. TBBMC for eight subjects was measured by both operators. The units are grams (g). A comparison of the means for the two operators provides a check on the training they received and allows us to determine if one of the operators is producing measurements that are consistently higher than the other. Here are the data:
Subject
Operator 1 2 3 4 5 6 7 8
1 1.326 1.337 1.079 1.229 0.936 1.009 1.179 1.289
2 1.323 1.322 1.073 1.233 0.934 1.019 1.184 1.304
Take the difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2. (Operator 1 minus Operator 2. Round your answers to four decimal places.)
X-bar=
S=
Answer:
[tex]\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095[/tex]
-The sample is too small to make judgments about skewness or symmetry.
H0:[tex]\mu_{1}=\mu_{2}[/tex]
H1:[tex]\mu_{1} \neq \mu_{2}[/tex]
[tex]t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013[/tex]
[tex]p_v =2*P(t_{(14)}<-0.0133)=0.990[/tex]
So the p value is a very high value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.
Step-by-step explanation:
First we need to find the difference defined as:
(Operator 1 minus Operator 2)
d1=1.326-1.323=0.003 d2=1.337-1.322=0.015
d3=1.079-1.073=0.006 d4=1.229-1.233=-0.004
d5=0.936-0.934=0.002 d6=1.009-1.019=-0.01
d7=1.179-1.184=-0.005 d8=1.289-1.304=-0.015
Now we can calculate the mean of differences given by:
[tex]\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001[/tex]
And for the sample deviation we can use the following formula:
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095[/tex]
Describe the distribution of these differences using words. (which one is correct)
We can plot the distribution of the differences with the folowing code in R
differences<-c(0.003,0.015,0.006,-0.004,0.002,-0.01,-0.005,-0.015)
hist(differences)
And we got the image attached. And we can see that the distribution is right skewed but we don't have anough info to provide a conclusion with just 8 differnences.
-The sample is too small to make judgments about skewness or symmetry.
Use a significance test to examine the null hypothesis that the two operators have the same mean. Give the test statistic. (Round your answer to three decimal places.)
[tex]\bar X_{1}=1.173[/tex] represent the mean for the operator 1
[tex]\bar X_{2}=1.174[/tex] represent the mean for the operator 2
[tex]s_{1}=0.1506[/tex] represent the sample standard deviation for the operator 1
[tex]s_{2}=0.1495[/tex] represent the sample standard deviation for the operator 2
[tex]n_{1}=8[/tex] sample size for the operator 1
[tex]n_{2}=8[/tex] sample size for the operator 2
t would represent the statistic (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:
H0:[tex]\mu_{1}=\mu_{2}[/tex]
H1:[tex]\mu_{1} \neq \mu_{2}[/tex]
If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
We can replace in formula (1) like this:
[tex]t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013[/tex]
Statistical decision
For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:
[tex]df=n_{1}+n_{2}-2=8+8-2=14[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(t_{(14)}<-0.0133)=0.990[/tex]
So the p value is a very high value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.
At a nationwide travel agency, newly hired employees are classified as beginners (B). Every six months the performance of each agent is reviewed. Past records indicate that transitions through the ranks to intermediate (I) and qualified (Q) are according to the following Markov chain, where F indicates workers that were fired BIQ F B 45.4 0.15 Q 0 0 1 0 F 0 0 0 1 (a) What fraction eventually become qualified? (b) What is the expected time until a beginner is fired or becomes qualified?
Answer
The answer and procedures of the exercise are attached in the following archives.
The information complete of the exercise is attached in the sheet.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Ben swims 50,000 yards per week in his practices. Given this amount of training, he will swim the 100-yard butterfly in 51.5 seconds and place 10th in a big upcoming meet. Ben's coach calculates that if Ben increases his practice to 60,000 yards per week, his time will decrease to 46.4 seconds and he will place 7th in the meet. If Ben practices 70,000 yards per week, his time will be 44.6 seconds and he will win the meet. In terms of Ben's time in the big meet, what is his marginal productivity of the number of yards he practices?
Answer:
MArginal productivity: [tex]\frac{dt}{dL}=-0.0002[/tex]
We can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.
Step-by-step explanation:
The marginal productivy is the instant rate of change in the result for an increase in one unit of a factor.
In this case, the productivity is the time he last in the 100-yard. The factor is the amount of yards he train per week.
The marginal productivity can be expressed as:
[tex]\frac{dt}{dL}[/tex]
where dt is the variation in time and dL is the variation in training yards.
We can not derive the function because it is not defined, but we can approximate with the last two points given:
[tex]\frac{dt}{dL}\approx\frac{\Delta t}{\Delta L} =\frac{t_2-t_1}{L_2-L_1}=\frac{44.6-46.4}{70,000-60,000}=\frac{-2.0}{10,000}=-0.0002[/tex]
Then we can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.
This is an approximation that is valid in the interval of 60,000 to 70,000 yards of training.
The marginal productivity of the number of yards Ben practices is 0.00051 seconds per yard.
Explanation:The marginal productivity of the number of yards Ben practices can be calculated by dividing the change in his performance (measured in seconds) by the change in the number of yards he practices per week.
To find the marginal productivity between 50,000 and 60,000 yards per week, subtract Ben's initial time of 51.5 seconds from his new time of 46.4 seconds. This gives us a change in time of 5.1 seconds. Then divide the change in time by the change in yards, which is 10,000 yards (60,000 - 50,000).
Marginal productivity = Change in time / Change in yards = 5.1 seconds / 10,000 yards = 0.00051 seconds per yard.
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