Answer: Margin of error = 8.32, and Confidence interval using normal distribution is narrower than confidence interval using t-distribution.
Step-by-step explanation:
Since we have given that
n = 13
Mean repair cost = $85.00
Standard deviation = $15.30
At 95% confidence interval,
z= 1.96
Since it is normally distributed.
Margin of error is given by
[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{15.30}{\sqrt{13}}\\\\=8.32[/tex]
95% confidence interval would be
[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=85\pm 1.96\times \dfrac{15.30}{\sqrt{13}}\\\\=85\pm 8.32\\\\=(85-8.32,85+8.32)\\\\=(76.68,93.32)[/tex]
A 95% confidence interval using the t-distribution was (75.8, 94.2 ).
Confidence interval using normal distribution is narrower than confidence interval using t-distribution.
Final answer:
A 95% confidence interval for the mean repair cost of microwave ovens with a known standard deviation is calculated using the z-score. The margin of error is found to be approximately $8.31, resulting in a confidence interval of ($76.69, $93.31).
Explanation:
To calculate the 95% confidence interval for the population mean when the population standard deviation is known, we can use the z-score associated with the 95% confidence level, which is 1.96. The formula for the margin of error (EBM) is EBM = z * (σ/√n), where σ is the population standard deviation, n is the sample size, and z is the z-score. Given that the sample standard deviation is $15.30, we assume it to be the population standard deviation because the question states that it is known.
With a sample mean (μ) of $85.00, a standard deviation of $15.30, and a sample size of 13, the margin of error is calculated as follows:
EBM = 1.96 * (15.30/√13) = 1.96 * 4.24 ≈ $8.31
The 95% confidence interval is therefore ($85.00 - $8.31, $85.00 + $8.31) = ($76.69, $93.31). The results using the z-distribution are similar to those obtained using the t-distribution, but usually, the t-distribution would be used when the sample size is small and the population standard deviation is unknown, which results in a wider interval due to the extra uncertainty.
A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnel that returns him to his cell after 2 days’ travel. The second leads to a tunnel that returns him to his cell after 4 days’ travel. The third door leads to freedom after 1 day of travel. If it is assumed that the prisoner will always select doors 1, 2, and 3 with respective probabilities .5, .3, and .2, what is the expected number of days until the prisoner reaches freedom?
1. Repeat problem, assuming the prisoner remembers previously chosen doors, and does not re-choose them. Assume the probabilities for the other doors are proportionally larger.
2. Repeat problem but now suppose there is another cell, and that door 1 takes him to the other cell after 2 days of travel. For the other cell, there are two doors, one of which leads to freedom after 3 days of travel, and the other leads back to the prisoner’s original cell after 3 days of travel; each door is equally likely.
Answer:
dont escape lol
Step-by-step explanation:
the answer is 2
A sequential circuit has two flip-flops (A and B), one input (X), and one output (Y). When X = 0, the state of the circuit remains the same. When X = 1, the circuit goes through the state transitions from 00 to 10 to 11 to 01, back to 00, and then repeats. Output Y = 1, when AB = 01. Determine the minimized D FF input equations and output equation.
Answer:
The equation for sequential circuit shown in the attachment.
Step-by-step explanation:
1 point) (Hypothetical.) The average salary of all residents of a city is thought to be about $39,000. A research team surveys a random sample of 200 residents; it happens that the average salary of these 200 is about $40,000 with a SD of $12,000. Make a z-test of the null hypothesis that this difference was just chance (in the sampling).
Answer:
We accept H₀
We don´t have enough evidence to reject H₀
Step-by-step explanation:
Nomal Distribution
population mean μ₀ = 39000
sample size n = 200
sample mean μ = 40000
sample standard deviation s = 12000
Test hypothesis
As we are just interested in look if the difference was just chance, we will do a one tail-test (right)
1.- Hypothesis
H₀ null hypothesis μ₀ = 39000
Hₐ Alternative hypothesis μ₀ > 39000
2.-We considered the confidence interval of 95 % then
α = 0,05 and z(c) = 1.64
3.-Compute z(s)
z(s) = [ ( μ - μ₀ ) ] / 12000/√200 z(s) = [ 40000- 39000)* √200] / 12000
z(s) = 1000*14,14/ 12000 z(s) = 1.1783
4.-Compare
z(s) and z(c)
1.1783 < 1.64
z(s) is inside acceptance region we accep H₀
The test statistic of zequalsnegative 2.40 is obtained when testing the claim that less than 0.32. a. Using a significance level of alphaequals0.10, find the critical value(s). b. Should we reject Upper H 0 or should we fail to reject Upper H 0
Answer:
a) Critical value = -1.285
b) We should reject null hypothesis that the mean equals 0.32
Step-by-step explanation:
Given that the statistic of z equals negative 2.40 is obtained when testing the claim that less than 0.32
i.e. for hypotheses
[tex]H_0: \bar = 0.32\\H_a: \bar x <0.32\\[/tex]
(one tailed test at 10% significance level)
Z critical value for 90% one tailed = -1.285
Since our test statistic is less than -1.285 we reject null hypothesis
a) Critical value = -1.285
b) We should reject null hypothesis that the mean equals 0.32
The following price quotations are for exchange-listed options on Primo Corporation common stock. Company Strike Expiration Call Put Primo 61.12 55 Feb 7.25 0.48 With transaction costs ignored, how much would a buyer have to pay for one call option contract. Assume each contract is for 100
Answer:
$ 725
Step-by-step explanation:
Price of call option = 7.25
buyer have to pay for one call option contract. Assume each contract is for 100 = 100 * 7.25 = $ 725
rewrite the statement in conditional form. lines with the same slope are parallel
Answer:
If lines have the same slope, then they are parallel.
Step-by-step explanation:
The first part of the given statement sets up the condition for the second part to be true. A conditional statement makes that explicit.
(lines described by first part) are (lines described by second part) ⇒
if (lines are described by first part), then (lines are described by second part)
Your conditional statement can be written ...
If lines have the same slope, then they are parallel.
The Hypotenuse of a right triangle is 0.5 units long. The longer leg is 0.1 units longer than the shorter leg. Find the lengths of the sides of the triangle.
Answer:
The lengths of the sides of a right triangle are
Longer leg = 0.4 units.
Shorter leg = 0.3 units.
Step-by-step explanation:
Given:
Hypotenuse = 0.5 units
Let the length of shorter leg of right triangle be x units then
According to the given condition,
length of longer leg will be (0.1 + x) units
Now,we know for a right triangle,by Pythagoras theorem we have
[tex](\textrm{Hypotenuse})^{2} = (\textrm{Longer leg})^{2}+(\textrm{Shorter leg})^{2}[/tex]
substituting the values we get
[tex]0.5^{2}= (x+0.1)^{2}+ x^{2}[/tex]
Applying [tex](a+b)^{2}= a^{2}+2ab+b^{2}[/tex] we get
[tex]0.25= x^{2} +2\times 0.1\times x+ 0.1^{2} + x^{2} \\2x^{2} +0.2x+0.01-0.25=0\\2x^{2} +0.2x-0.24=0\\[/tex]
which is a quadratic equation
dividing the equation throughout by two we get
[tex]x^{2} +0.1x-0.12=0\\\textrm{on factorizing we get}\\x^{2} +0.4x-0.3x-0.12=0\\(x+0.4)(x-0.3)=0[/tex]
[tex]\therefore (x-0.3)= 0\\\therefore x=0.3[/tex]
Since x cannot be negative we take
x = 0.3 units
∴ Longer leg = x + 0.1
= 0.3+0.1
=0.4 units
So, the lengths of the sides of a right triangle are
Longer leg = 0.4 units.
Shorter leg = 0.3 units.
Final answer:
To find the lengths of the sides of the right triangle with a hypotenuse of 0.5 units and one leg being 0.1 units longer than the other, you can use the Pythagorean theorem to set up an equation and solve for the lengths of the legs.
Explanation:
To find the lengths of the sides of the triangle, let's assume the length of the shorter leg is x units. Then, the length of the longer leg would be x + 0.1 units. Applying the Pythagorean theorem, where a and b are the legs and c is the hypotenuse:
a² + b² = c²
(x)² + (x + 0.1)² = (0.5)²
Simplifying the equation and solving for x, we get x ≈ 0.226 units for the shorter leg, and the longer leg would be x + 0.1 ≈ 0.326 units.
A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a volume of 505π ft3, determine the radius and height of the silo that requires the least amount of material to build. Hint: The volume of the silo is πr2h + 2 3 πr3, and the surface area (including the floor) is π(3r2 + 2rh). (Round your answers to one decimal place.)
r = (3V/2π)^(1/3) and h = V/(πr^2), respectively.
Explanation:To determine the dimensions of the silo that requires the least amount of material to build, we need to find the values of the radius and height that minimize the surface area. Let's start by expressing the volume of the silo as a function of one variable. The volume is given by V = πr^2h + (2/3)πr^3, where r is the radius and h is the height. Since we're looking for the minimum surface area, we'll differentiate the surface area expression, equate it to zero, and solve the resulting equation to find the values of r and h that minimize the surface area.
Taking the derivative of the surface area function A(r, h) = π(3r^2 + 2rh) with respect to r and h, we get:
dA/dr = 6πr + 2πh = 0
dA/dh = 2πr = 0
Solving these equations simultaneously, we find that r = 0 and h = 0, which are not meaningful values in this context. Therefore, there are no critical points inside the domain. Instead, we need to examine the endpoints of the domain. Since r and h must be positive, we find that A(r, h) goes to infinity as r or h approaches zero. Therefore, the surface area does not have a minimum in the interior of the domain and we need to consider the endpoints.
First, we'll consider the case where h = 0. In this case, the silo is just a hemispherical dome with no height, so the surface area is equal to the curved surface area of the hemisphere, which is given by 2πr^2. We can rewrite the volume equation as V = πr^2h + (2/3)πr^3 = (2/3)πr^3, since h = 0. Solving this equation for r, we find that r = (3V/2π)^(1/3).Next, we'll consider the case where r = 0. In this case, the silo is just a right circular cylinder with no radius, so the surface area is equal to the curved surface area of the cylinder, which is given by 2πrh. We can rewrite the volume equation as V = πr^2h + (2/3)πr^3 = πr^2h, since r = 0. Solving this equation for h, we find that h = V/(πr^2).Now, we compare the surface areas of the two cases to determine which requires the least amount of material. A(h = 0) = 2πr^2 and A(r = 0) = 2πrh = 2V/r. Substituting the values of r and h from the previous calculations, we find that A(h = 0) = (2π/3)(3V)^(2/3) and A(r = 0) = 6V^(1/3)/(3π)^(2/3). Since the surface area of the dome is smaller than that of the cylinder, the silo with a hemispherical dome on top requires the least amount of material to build. Therefore, the radius and height of the silo are given by r = (3V/2π)^(1/3) and h = V/(πr^2), respectively.Learn more about Optimization here:https://brainly.com/question/33322941
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To determine the dimensions of the silo that requires the least material, set up an equation involving the volume and the surface area. Solve for the values of r and h that minimize the surface area. Use differentiation and the second derivative test to find the minimum point.
Explanation:To determine the radius and height of the silo that requires the least amount of material to build, we need to find the dimensions that minimize the surface area of the silo. The surface area of the silo is given by A = π(3r^2 + 2rh), and the volume of the silo is given by V = πr^2h + (2/3)πr^3.
We are given that the volume of the silo is 505π ft^3. We can use this information to set up an equation involving the volume and the dimensions of the silo. Solving this equation will give us the values of r and h that minimize the surface area.
By differentiating the surface area equation with respect to r and h, and setting the derivatives equal to zero, we can find the critical points. By analyzing the second derivative test, we can determine which point is the minimum.
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A confidence interval (CI) is desired for the true average stray-load loss u (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume that strayload loss is normally distributed with o = 3.0. (1). Construct a 95% CI for u when n = 25 and x = 58.3. (10 points) (2). Construct a 95% CI for u when n = 100 and 1 = 58.3. (10 points) (3). Construct a 99% CI for u when n = 100 and x = 58.3. (10 points) (4). How large must n be if the half-width of the 99% interval for u is to be 0.5? (10 points)
Creating a 95% confidence interval involves capturing 95% of the probability deviation in a normal distribution. The formula for doing so is (x ± (standard deviation/√n)x critical value). By adjusting the sample size or confidence level accordingly, different intervals can be created.
Explanation:The question inquires about the construction of a
confidence interval
for the true average stray-load loss for a certain type of induction motor. When constructing a 95% confidence interval, we are looking to capture the central 95 percent of the probability on a normal distribution, excluding 2.5 percent on each tail of the distribution.
For point one, with n = 25, x = 58.3, and o = 3.0, use the following formula to construct a 95% Confidence Interval: (x ± (standard deviation/√n)x critical value), resulting in (58.3 ± (3/√25)x 1.96). The same method applies to the rest of the points, only changing the sample size, n, or confidence level.
For the final query, the required sample size, n, can be found by rearranging the formula for confidence intervals and solving for n with the half-width of the interval set at 0.5.
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(1). 95% CI (n=25): Standard deviation of 3.0, sample mean of 58.3, yielding a 95% confidence interval of (57.124, 59.476).
(2). 95% CI (n=100): With the same standard deviation and mean, a larger sample size leads to a narrower 95% confidence interval of (57.712, 58.888).
(3). 99% CI (n=100): Expanding the confidence level to 99% increases the margin of error, resulting in a wider interval of (57.5272, 59.0728).
(4). Sample Size (99% CI, ME=0.5): To achieve a half-width of 0.5 for a 99% confidence interval, approximately 956 samples are required.
Given the problem requirements, we will calculate the confidence intervals based on the provided data:
Standard deviation, σ = 3.0Sample mean, x = 58.3Using these values, we’ll calculate the 95% and 99% confidence intervals (CIs) for the population mean, μ:
1. Constructing a 95% CI when n = 25
Calculate the standard error (SE):SE = σ/√n
= 3.0/√25
= 3.0/5
= 0.6
Find the z-value for 95% CI: z = 1.96Margin of error (ME): ME = z * SE= 1.96 * 0.6
= 1.176
Confidence Interval: 58.3 ± 1.176 = (57.124, 59.476)2. Constructing a 95% CI when n = 100
Calculate the standard error (SE):SE = σ/√n
= 3.0/√100
= 3.0/10
= 0.3
Find the z-value for 95% CI: z = 1.96Margin of error (ME): ME = z * SE = 1.96 * 0.3 = 0.588Confidence Interval: 58.3 ± 0.588 = (57.712, 58.888)3. Constructing a 99% CI when n = 100
Calculate the standard error (SE): SE = σ/√n = 3.0/√100 = 3.0/10 = 0.3Find the z-value for 99% CI: z = 2.576Margin of error (ME): ME = z * SE = 2.576 * 0.3 = 0.7728Confidence Interval: 58.3 ± 0.7728 = (57.5272, 59.0728)4. Determining the sample size n for a half-width of 0.5 in the 99% CI
Margin of error (ME): 0.5Find the z-value for 99% CI: z = 2.576Using the formula for margin of error, ME = z * (σ/√n):Solve for n:Therefore, n ≈ 956 to achieve a half-width of 0.5 for a 99% confidence interval.
There are four large groups of people, each with 1000 members. Any two of these groups have 100 members in common. Any three of these groups have 10 members in common. And there is 1 person in all four groups. All together, how many people are in these groups?
Answer:
3,359
Step-by-step explanation:
Let A, B, C and D the four groups of people.
Let us denote with |A| the number of elements in a set A.
Then the number of elements of A∪B∪C∪D is the sum of the elements in each group subtracting the elements that have been counted twice.
That is,
|A∪B∪C∪D |=|A|+|B|+|C|+|D| - |A∩B| - |A∩C| - |A∩D| - |B∩C| - |B∩D|- |C∩D| - |A∩B∩C| - |A∩B∩D| - |A∩C∩D| - |B∩C∩ D| - |A∩ B∩C∩ D| =
1000+1000+1000+1000 - 100 - 100 - 100 - 100 - 100 - 100 - 10- 10- 10- 10 - 1 = 3359
use the functions f(x)=2x and g(x)=x^2+1 to find the value of each expression
1. f(g(3))
2. f(3)+g(4)
3. f(5)-2×g(1)
Answer:
1. 20
2. 23
3. 6
Step-by-step explanation:
We have that:
f(x) = 2x
g(x) = x² + 1
f(g(x)) is the composite function of f and g. So
f(g(x)) = f(x²-1) = 2(x²+1) = 2x² + 2
1. f(g(3))
f(g(x)) = 2x² - 2 = 2(3)² + 2 = 18 + 2 = 20
2. f(3)+g(4)
f(3) = 2(3) = 6
g(4) = 4² + 1 = 17
f(3) + g(4) = 6 + 17 = 23
3. f(5) - 2g(1)
f(5) = 2(5) = 10
g(1) = (1)² + 1 = 2
f(5) - 2g(1) = 10 - 2*2 = 10 - 4 = 6
Find the exact value of csc (–1020)°.
By using reference angle, the exact value of csc (-1020)° is [tex]\frac {-2\sqrt3}{3}[/tex].
How to find the exact value
To find the exact value of csc (-1020)° using the reference angle, first determine the reference angle for -1020°.
To find the reference angle, add or subtract multiples of 360° to bring the angle into the range of 0° to 360°.
-1020° + 360° = -660°
Since -660° is still negative, add another 360°:
-660° + 360° = -300°
Now we have a reference angle of 300°.
The cosecant function (csc) is the reciprocal of the sine function. We know that the sine function is positive in the first and second quadrants, so the cosecant function will be positive in those quadrants.
The exact value of csc(300°) can be found by taking the reciprocal of the sine of 300°:
csc(300°) = 1/sin(300°)
To find the sine of 300°, use the periodicity of the sine function:
sin(300°) = sin(300° - 360°) = sin(-60°)
The sine of -60° is the same as the sine of 60°, but with a negative sign:
sin(-60°) = -sin(60°) = [tex]\frac {-\sqrt3}{2}[/tex]
Now, find the reciprocal:
csc(300°) = 1/(-√3/2) = [tex]\frac {-2}{\sqrt3}[/tex]
To rationalize the denominator, multiply the numerator and denominator by √3:
csc(300°) = [tex]\frac {-2}{\sqrt3}[/tex] * [tex]\frac {\sqrt3}{\sqrt3}[/tex] = [tex]\frac {-2\sqrt3}{3}[/tex]
Therefore, the exact value of csc (-1020)° is [tex]\frac {-2\sqrt3}{3}[/tex].
Complete question
Use reference angle to find the exact value of csc (–1020)°.
In a population of 1000 subjects, 770 possess a certain characteristic. A sample of 40 subjects selected from this population has 24 subjects who possess the same characteristic. What are the values of the population and sample proportions?
Answer:
0.77,0.60
Step-by-step explanation:
Given that in a population of 1000 subjects, 770 possess a certain characteristic.
A sample of 40 subjects selected from this population has 24 subjects who possess the same characteristic
To find out sample proportion:
Sample size n = 40
Favourable x = 24
Sample proportion p = [tex]\frac{24}{40} =0.60[/tex]
To find out population proportion:
Total population N = 1000
Favourable X = 770
population proportion P = [tex]\frac{770}{1000} =0.77[/tex]
Define a bijection between 5-subsets of the set S = {1, 2, 3, 4, 5, 6, 7, 8} and 8-bit strings with exactly five 1's. A subset X of S with five elements maps on to a string x so that j ∈ X if and only if the jth bit of x is 1. What string corresponds to the set {1, 3, 4, 5, 8}?
To define a bijection between 5-subsets of the set S = {1, 2, 3, 4, 5, 6, 7, 8} and 8-bit strings with exactly five 1's, we can assign each element of the subset to a bit position in the string.
Explanation:In mathematics, a bijection is a one-to-one correspondence between two sets, such that each element in one set is paired with a unique element in the other, and vice versa. It implies both injective (no duplicates) and surjective (every element has a match) properties.
To define a bijection between 5-subsets of the set S = {1, 2, 3, 4, 5, 6, 7, 8} and 8-bit strings with exactly five 1's, we can assign each element of the subset to a bit position in the string. For example, the subset {1, 3, 4, 5, 8} corresponds to the 8-bit string 11011001, where the 1st, 3rd, 4th, 5th, and 8th bits are set to 1, representing the elements in the subset.
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An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Suppose that the total weight of the baggage checked by an individual passenger is a random variable x with a mean value of 49 lb and a standard deviation of 18 lb. If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Hint: With n = 100, the total weight exceeds the limit when the average weight x exceeds 6000/100.) (Round your answer to four decimal places.)
Answer:
[tex]P(\bar x>60)=P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]
Is a very improbable event.
Step-by-step explanation:
We want to calculate the probability that the total weight exceeds the limit when the average weight x exceeds 6000/100=60.
If we analyze the situation we this:
If [tex]x_1,x_2,\dots,x_100[/tex] represent the 100 random beggage weights for the n=100 passengers . We assume that for each [tex]i=1,2,3,\dots,100[/tex] for each [tex]x_i[/tex] the distribution assumed is normal with the following parameters [tex]\mu=49, \sigma=18[/tex].
Another important assumption is that the each one of the random variables are independent.
1) First way to solve the problem
The random variable S who represent the sum of the 100 weight is given by:
[tex]S=x_1 +x_2 +\dots +x_100 =\sum_{i=1}^{100} x_i[/tex]
The mean for this random variable is given by:
[tex]E(S)=\sum_{i=1}^{100} E(x_i)=100\mu = 100*49=4900[/tex]
And the variance is given by:
[tex]Var(S)=\sum_{i=1}^{100} Var(x_i)=100(\sigma)^2 = 100*(18)^2[/tex]
And the deviation:
[tex]Sd(S)=\sqrt{100(\sigma)^2} = 10*(18)=180[/tex]
So we have this distribution for S
[tex]S \sim (4900,180)[/tex]
On this case we are working with the total so we can find the probability on this way:
[tex]P(S>6000)=P(z>\frac{6000-4900}{180})=P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]
2) Second way to solve the problem
We know that the sample mean have the following distribution:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}}[/tex]
If we are interested on the probability that the population mean would be higher than 60 we can find this probability like this:
[tex]P(\bar x >60)=P(\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{60-49}{\frac{18}{\sqrt{100}}})[/tex]
[tex]P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]
And with both methods we got the same probability. So it's very improbable that the limit would be exceeded for this case.
A county clerk wants to estimate the proportion of voters who will need special election facilities.
Suppose a sample of 400 voters was taken.
If 150 need special election facilities, what is the upper confidence limit (UCL) for the 90% confidence interval for the population proportion of voters who will need special election facilities?
Round your answer to 3 decimal places.
Answer: The upper confidence limit for the 90% confidence interval would be 0.415.
Step-by-step explanation:
Since we have given that
n = 400
x = 150
So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{150}{400}=0.375[/tex]
At 90% confidence interval, z = 1.645
So, margin of error would be
[tex]z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.645\times \sqrt{\dfrac{0.375\times 0.625}{400}}\\\\=0.0398[/tex]
So, the upper limit would be
[tex]\hat{p}+0.0398\\\\=0.375+0.0398\\\\=0.415[/tex]
Hence, the upper confidence limit for the 90% confidence interval would be 0.415.
USA Today reported that about 47% of the general consumer population in the United States is loyal to the automobile manufacturer of their choice. Suppose Chevrolet did a study of a random sample of 1004 Chevrolet owners and found that 482 said they would buy another Chevrolet. Does this indicate that the population proportion of consumers loyal to Chevrolet is more than 47%? Use α = 0.01.
What is the sample test statistic?
What is the p value of the test statistic?
Answer:
[tex]z=\frac{0.48 -0.47}{\sqrt{\frac{0.47(1-0.47)}{1004}}}=0.635[/tex]
[tex]p_v =P(z>0.635)=1-P(z<0.635)=1-0.737=0.263[/tex]
Step-by-step explanation:
1) Data given and notation n
n=1004 represent the random sample taken
X=482 represent the Chevrolet owners who said they would buy another Chevrolet.
[tex]\hat p=\frac{482}{1004}=0.480[/tex] estimated proportion of Chevrolet owners who said they would buy another Chevrolet.
[tex]p_o=0.47[/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 the proportion of Chevrolet owners who said they would buy another Chevrolet is higher than 47%:
Null hypothesis:[tex]p\leq 0.47[/tex]
Alternative hypothesis:[tex]p > 0.47[/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].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.48 -0.47}{\sqrt{\frac{0.47(1-0.47)}{1004}}}=0.635[/tex]
4) Statistical decision
P value method or p value approach . "This method consists on 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.
If we use the significance level provided, [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.
Since is a one side upper test the p value would be:
[tex]p_v =P(z>0.635)=1-P(z<0.635)=1-0.737=0.263[/tex]
So based on the p value obtained and using the significance level assumed [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of Chevrolet owners who said they would buy another Chevroletis not significantly higher than 0.47 or 47% .
Question 1 (40 pt). You are given designs of 3 caches for a 16-bit address machine: D1: Direct-mapped cache. Each cache line is 1 byte. 10-bit index, 6-bit tag. 1 cycle hit time. D2: 2-way set associative cache. Each cache line is 1 word (4 bytes). 7-bit index, 7-bit tag. 2 cycle hit time. D3: fully associative cache with 256 cache lines. Each cache line is 1 word. 14-bit tag. 5 cycle hit time. Answer the following set of questions: a) What is the size of each cache? b) How much space does each cache need to store tags? c) Which cache design has the most conflict misses? Which has the least? d) The following information is given to you: hit rate for the 3 caches is 50%, 70% and 90% but did not tell you which hit rate corresponds to which cache, which cache would you guess corresponded to which hit rate? Why? e) Assuming the miss time for each is 20 cycles, what is the average service time for each? (Service Time = (hit rate)*(hit time) + (miss rate)*(miss time)). Question 2 (30 pt). Assume we have a computer where the CPI is 1.0 when all memory accesses (including data and instruction accesses) hit in the cache. The cache is a unified (data + instruction) cache of size 256 KB, 4-way set associative, with a block size of 64 bytes. The data accesses (loads and stores) constitute 50% of the instructions. The unified cache has a miss penalty of 25 clock cycles and a miss rate of 2%. Assume 32-bit instruction and data addresses. Now, answer the following questions:
Answer:
Step-by-step explanation:
a) What is the size of each cache?
Direct mapped cache= 2^index * size of cache line= 2^10 * 1B lines = 1KB.
2-way set associative cache= 2^index * size of cache line * 2 ways=2^7 * 4 words *2ways= 128 4B lines * 2 ways = 1KB
Fully associative cache= number of cache lines* size of each line= 256 * 4B lines = 1KB
b) How much space does each cache need to store tags?
Direct mapped cache= 1024 * 6-bit tags = 6Kb
2-way set associative cache= 256 * 7-bit tags = 1792 bits
Fully associative cache= 256 * 14-bit tags = 3584 bits
c) Which cache design has the most conflict misses? Which has the least?
Direct mapped cache has likely the most conflict misses, because it is direct mapped. Fully associative cache has the least since it is fully associative so it can never have conflict misses.
d) The following information is given to you: hit rate for the 3 caches is 50%, 70% and 90% but did not tell you which hit rate corresponds to which cache, which cache would you guess corresponded to which hit rate? Why?
Since the size of all three caches is same size and as we said in the previous answer that direct mapped cache has more conflict misses and fully associative has the least so direct mapped will have 50%, 2-way set associative 70%, and Fully associative will have 90% hit rate.
e) Assuming the miss time for each is 20 cycles, what is the average service time for each? (Service Time = (hit rate)*(hit time) + (miss rate)*(miss time)
We are given hit rates and miss rates. Also miss time=2o cycles for each cache and hit time= 1, 2, 5 for direct mapped, 2-way set associative and fully associative cache respectively.
Direct mapped= 0.5*1 + 0.5*20 = 10.5 cycles
2-way set associative= 0.7*2 + 0.3*20 = 7.4 cycles
Fully associative cache= 0.9*5 + 0.1*20 = 6.5 cycles.
The size of each cache, the space needed to store tags, the cache design with the most and least conflict misses, guessing which cache corresponds to each hit rate, and calculating the average service time for each cache.
Explanation:D1: Direct-mapped cache:Size: Each cache line is 1 byte, so the total cache size is 2^10 * 1 byte = 1024 bytesTag space: Each cache line has a 6-bit tag, so the total tag space required is 2^6 * 1 byte = 64 bytesD2: 2-way set associative cache:D3: Fully associative cache:Size: Each cache line is 1 word, so the total cache size is 256 lines * 2^2 bytes = 1024 bytesTag space: Each cache line has a 14-bit tag, so the total tag space required is 2^14 * 1 byte = 16384 bytesEach cache design has different levels of conflict misses. The direct-mapped cache (D1) has the most conflict misses because multiple memory locations map to the same cache line. The 2-way set associative cache (D2) has fewer conflict misses because each set can hold two cache lines, reducing the chance of multiple memory locations mapping to the same set. The fully associative cache (D3) has the least conflict misses because any memory location can be stored in any cache line, reducing conflicts.
Based on the given hit rates, we can guess which cache corresponds to each hit rate. The cache with the highest hit rate (90%) is likely to be D3 (fully associative cache), as it has the lowest miss rate and therefore the highest hit rate. The cache with the lowest hit rate (50%) is likely to be D1 (direct-mapped cache), as it has the highest miss rate and therefore the lowest hit rate. The cache with the intermediate hit rate (70%) is likely to be D2 (2-way set associative cache).To calculate the average service time for each cache, we use the formula: Average service time = (hit rate) * (hit time) + (miss rate) * (miss time). Given a miss time of 20 cycles for each cache, we can calculate the average service time for each cache using their respective hit rates:D1: Average service time = (0.5) * (1 cycle) + (0.5) * (20 cycles) = 10.5 cyclesD2: Average service time = (0.7) * (2 cycles) + (0.3) * (20 cycles) = 4.6 cyclesD3: Average service time = (0.9) * (5 cycles) + (0.1) * (20 cycles) = 5.5 cyclesThe standard deviation of pulse rates of adult males is less than 12 bpm. For a random sample of 135 adult males, the pulse rates have a standard deviation of 11.5 bpm. Complete parts (a) and (b) below. a. Express the original claim in symbolic form.
Answer:
Step-by-step explanation:
Hello!
The population variance is symbolized σ² and the standard deviation σ (remember that for estimations and statistics test, the parameter of study is always the variance)
The sentence "The standard deviation of pulse rates of adult males is less than 12 bpm." is symbolized σ <12
The sample variance is symbolized S² and the standard deviation is S.
The sample standard deviation is S= 11.5.
I hope it helps!
The symbol σ is used to represent the standard deviation in statistics. So, the standard deviation of pulse rates of adult males being less than 12 bpm in symbolic form is σ < 12. This value tells us that the dispersion of pulse rates among adult males is quite small.
Explanation:The original claim is that the standard deviation of pulse rates of adult males is less than 12 bpm. This can be expressed in symbolic form as follows:
σ < 12
where σ represents the standard deviation.
In the context of this question, the standard deviation is a measure of the dispersion or spread in the pulse rates of adult males. The sample that was taken had a standard deviation of 11.5 bpm, which is less than 12 bpm, thus supporting the original claim.
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Suppose a regional computer center wants to evaluate the performance of its disk memory system.One measure of performance is the average time between failures of its disk drive. To estimate this value, the center recorded the time between failures for a random sample of 45 disk-drive failures. The following sample statistics were computed: s . 215 hours # 1,762 hours Estimate the true mean time between failures with a 90% confidence interval. a) ) If the disk memory system is running properly, the true mean time between failures will exceed 1,700 hours. Based on the interval, in part a, what can you infer about the disk memory system?
Answer:
a) solved in attachment
b) If disk is running properly time between failure will exceed then Upper 90 % would be good indicator.
Step-by-step explanation:
One of the loudest sounds in recent history was that made by the explosion of Krakatoa on August 26-27, 1883. According to barometric measurements, the sound had a decibel level of 180 dB at a distance of 161 km. Assuming the intensity falls off as the inverse of the distance squared, what was the decibel level on Rodriguez Island, 4,800 km away?
Answer:
150.51 dB
Step-by-step explanation:
Data provided in the question:
decibel level of sound at 161 km distance = 180 dB
d₁ = 161 km
d₂ = 4800 km
I₁ = 180 db
The formula for intensity of sound is given as:
I = [tex]10\log(\frac{I_2}{I_1})[/tex]
and the relation between intensity and distance is given as:
I ∝ [tex]\frac{1}{d^2}[/tex]
or
Id² = constant
thus,
I₁d₁² = I₂d₂²
or
[tex]\frac{I_2}{I_1}=\frac{d_1}{d_2}[/tex]
therefore,
I = [tex]10\log(\frac{d_1}{d_2})^2[/tex]
or
I = [tex]10\times2\times\log(\frac{161}{4,800})[/tex]
or
I = 20 × (-1.474)
or
I = -29.49
Therefore,
the decibel level on Rodriguez Island, 4,800 km away
= 180 - 29.49
= 150.51 dB
To find the decibel level on Rodriguez Island 4,800 km away from the explosion of Krakatoa, you can use the inverse square law and the formula dB1 - dB2 = 20log10(d1/d2), where d1 and d2 are the distances from the explosion. Plugging in the values, you can calculate the decibel level on Rodriguez Island to be approximately 201.94 dB.
Explanation:To find the decibel level on Rodriguez Island 4,800 km away from the explosion of Krakatoa, we can use the inverse square law. According to the problem, the decibel level at a distance of 161 km is 180 dB. To find the decibel level at 4,800 km, we can use the formula: dB1 - dB2 = 20log10(d1/d2), where d1 and d2 are the distances from the explosion. Plugging in the values, we get: dB2 = dB1 - 20log10(d1/d2). Substituting dB1 = 180 dB, d1 = 161 km, and d2 = 4800 km, we can calculate the decibel level on Rodriguez Island.
dB2 = 180 - 20log10(161/4800) ≈ 180 - 20(-1.097) ≈ 180 + 21.94 ≈ 201.94 dB
Therefore, the decibel level on Rodriguez Island, 4,800 km away, is approximately 201.94 dB.
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What is the p-value? -- Researcher Jessie is studying how the fear of going to the dentist affects an adult's actual number of visits to the dentist. She asks a random sample of adults whether or not they fear going to the dentist and also how many times they have gone in the past 10 years. She would like to assess if the average number of visits made by adults who fear going to the dentist (Group 1) is higher than the average number of visits for those who don't have that fear (Group 2), that is, test H0: μ1 = μ2 versus Ha: μ1 > μ2, using a 10% significance level. Her random sample of adults resulted in 14 stating they feared going to the dentist and 31 stated they did not fear going to the dentist. The first sample mean was 1.71 pooled standard errors above the second sample mean. Jessie has asked you to provide a complete sketch of the p-value that she can include in her report. You can use the shiny app in R which will give the exact p-value or make a complete sketch by hand and provide the bounds for the p-value using the T table.
2) Jessie also remembers some condition about a normal model required for her two populations of responses. She asks you to check this condition for her. You recall that a QQ plot helps to assess if a population of responses can be considered normally distributed. How many QQ plots do you need to make in this case?
None, since the total sample size of 45 adults is large you can just use the CLT.
One, for the 45 responses (number of visits) reported by the 45 adults that were surveyed.
Two, one for the 14 responses (number of visits) by the adults who fear going to the dentist and one for the 31 responses (number of visits) by the adults who do not fear going to the dentist.
Answer:
Two, one for the 14 responses (number of visits) by the adults who fear going to the dentist and one for the 31 responses (number of visits) by the adults who do not fear going to the dentist.
Step-by-step explanation:
Hello!
1)
You want to test if the average visits to the dentist of people who fear to visit it are greater than the average visits of people that don't fear it.
In this case, the statistic to use is a pooled Student t-test. The reason I've to choose this test is that one of your sample sizes is small (n₁= 14) and the t-test is more accurate for small samples. Even if the second sample is greater than 30, if both variables are normally distributed, the pooled t-test is the one to use.
H₀: μ₁ = μ₂
H₁: μ₁ > μ₂
α: 0.10
t= (X₁[bar]-X₂[bar]) - (μ₁ - μ₂) ~ t[tex]_{n₁+n₂-2}[/tex]
Sₐ√(1/n₁+1/n₂)
Where
X₁[bar] and X₂[bar] are the sample means of both groups
Sₐ is the pooled standard deviation
This is a one-tailed test, you will reject the null hypothesis to big numbers of t. Remember: 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), and in this case, is also one-tailed.
P(t[tex]_{n₁+n₂-2}[/tex] ≥ t[tex]_{H0}[/tex]) = 1 - P(t[tex]_{n₁+n₂-2}[/tex] < t[tex]_{H0}[/tex])
Where t[tex]_{H0}[/tex] is the value of the calculated statistic.
Since you didn't copy the data of both samples, I cannot calculate it.
2)
Well there was one sample taken and separated in two following the criteria "fears the dentist" and "doesn't fear the dentist" making two different samples, so this is a test for two independent samples. To check if both variables are normally distributed you need to make two QQplots.
I hope it helps!
The p-value represents the probability of the observed data under the null hypothesis. Jessie's study regarding dentist visits requires a p-value sketch based on a test statistic of 1.71 standard errors. Two QQ plots are necessary to check the normality condition of the sample data for both groups.
Explanation:The p-value is a statistical measure that indicates the probability of the observed data or something more extreme occurring under the assumption that the null hypothesis is true. In Researcher Jessie's study, the null hypothesis (H0) is that the mean number of dentist visits for adults who fear going to the dentist (Group 1) is equal to the mean number of visits for those who do not fear going to the dentist (Group 2).
The alternative hypothesis (Ha) is that the mean for Group 1 is greater than the mean for Group 2. Jessie has found that the sample mean for Group 1 is 1.71 pooled standard errors above the mean for Group 2. To sketch the p-value, we need to find the area to the right of the test statistic (1.71 standard errors above the mean) on a normal distribution curve. This area represents the p-value, which we could find using statistical software or by consulting a T table with appropriate degrees of freedom.
Regarding the condition about the normal model, we need to check whether the data is normally distributed to correctly apply the t-test. We would create two QQ plots: one for the 14 responses from adults who fear going to the dentist, and one for the 31 responses from adults who do not.
The QQ plots will help determine if the data for each group deviates from a normal distribution, which is important given the sample sizes are less than 30 and normality cannot be assumed based on the Central Limit Theorem (CLT).
The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance in the pounds of the impurities. The sample variance was found to be 6.62. Find a 95% confidence interval for the population variance in the pounds of impurities. State any assumption you need to make to be able to answer this problem.
Answer:
The 95% confidence interval for the variance in the pounds of impurities would be [tex] 3.829 \leq \sigma^2 \leq 14.121[/tex].
Step-by-step explanation:
1) Data given and notation
[tex]s^2 =6.62[/tex] represent the sample variance
s=2.573 represent the sample standard deviation
[tex]\bar x[/tex] represent the sample mean
n=20 the sample size
Confidence=95% or 0.95
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
The Chi square distribution is the distribution of the sum of squared standard normal deviates .
2) Calculating the confidence interval
The confidence interval for the population variance is given by the following formula:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.025,19)" "=CHISQ.INV(0.975,19)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=32.852[/tex]
[tex]\chi^2_{1- \alpha/2}=8.907[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(19)(6.62)}{32.852} \leq \sigma^2 \frac{(19)(6.62)}{8.907}[/tex]
[tex] 3.829 \leq \sigma^2 \leq 14.121[/tex]
So the 95% confidence interval for the variance in the pounds of impurities would be [tex] 3.829 \leq \sigma^2 \leq 14.121[/tex].
The confidence interval for the population variance in the pounds of impurities is calculated using the sample variance, the degrees of freedom, and the critical values from a chi-squared distribution. The values are then plugged into the formulas to give the 95% confidence interval for the population variance.
Explanation:The subject of this question is in the field of statistics and it asks about the calculation of the 95% confidence interval for the population variance. In this case, we are trying to estimate the variance in the pounds of impurities in fertilizer bags based on a sample of twenty 100-pound bags with a variance of 6.62.
To solve this, we make the assumption that the population from which the bags are sampled follows a normal distribution. The confidence interval for the variance of a normal distribution can be found using the chi-squared distribution. The formula for this assumes that our samples are independent and identically distributed.
For a 95% confidence interval and d.f. (degrees of freedom) = n - 1 = 20 - 1 = 19, the upper and lower critical values (in this case for a chi-squared distribution, rounded to three decimal places) are 8.231 and 32.852 respectively. We can then use these values in the following formulas:
Lower Limit = sample variance * (d.f. / upper critical value) = 6.62*(19/32.852)
Upper Limit = sample variance * (d.f. / lower critical value) = 6.62*(19/8.231)
Finally, evaluate these to get our confidence interval.
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Suppose y varies jointly as x and z. Find y when x = –13 and z = 7, if y = 205 when x = –5 and z = –8. Round your answer to the nearest hundredth, if necessary.
Answer: y = - 466.375
Step-by-step explanation:
y varies jointly as x and z.
This means that y varies directly as x and also varies directly as z.
In order to remove the proportionality symbol, we will introduce a constant of proportionality, k. Therefore,
y = kxz
The next step is to determine the value of k
if y = 205 when x = –5 and z = –8.
we will substitute these values into the equation to determine k.
205 = k × -5 × -8
205 = 40k
k = 205/40 = 5.125
Therefore, the equation becomes
y = 5.125xz
We want to determine y when x = - 13 and z = 7
y = 5.125 × - 13 × 7
y = 5.125 × - 91
y = - 466.375
I don't know how to approach it, I've been looking at my notes and I can't grasp it.
Answer:
8
Step-by-step explanation:
You can skip directly to the formula for the sum of an infinite sequence with first term a₁ and common ratio r:
S = a₁/(1-r)
Your values of the variables in this formula are a₁ = 6 and r = 2/8. Putting these into the formula gives ...
S = 6/(1 -2/8) = 6/(6/8) = 8
The sum of the infinite geometric sequence is 8.
_____
The above formula is the degenerate form of the formula for the sum of a finite sequence:
S = a₁((rⁿ -1)/(r -1))
When the common ratio r has a magnitude less than 1, the term rⁿ tends to zero as n gets very large. When that term is zero, the sum of the infinite sequence is ...
S = a₁(-1/(r-1)) = a₁/(1-r)
6) A cylindrical shaped silo has a diameter of 18 feet and a height of 4 feet. How much water can the silo hold, in terms of π?
The amount of water silo can hold in terms of π is 324π cubic feet
Solution:Given that cylindrical shaped silo has a diameter of 18 feet and a height of 4 feet
To find: Amount of water silo can hold in terms of π
This means we are asked to find the volume of cylindrical shaped silo
The volume of cylinder is given as:
[tex]\text {volume of cylinder }=\pi \mathrm{r}^{2} \mathrm{h}[/tex]
Where "r" is the radius of cylinder
"h" is the height of cylinder
π is the constant has a value 3.14
Given diameter = 18 feet
[tex]radius = \frac{diameter}{2} = \frac{18}{2} = 9[/tex]
Substituting the values in formula, we get
[tex]\text {volume of cylinder }=\pi \times 9^{2} \times 4[/tex]
[tex]\text {volume of cylinder }=\pi \times 81 \times 4=324 \pi[/tex]
Thus amount of water silo can hold in terms of π is 324π cubic feet
Answer:324π
Step-by-step explanation:
A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. At a 1% level of significance, an appropriate conclusion is:
Answer: It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.
Step-by-step explanation:
Since we have given that
n = 84
x = 11
So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{11}{84}=0.13[/tex]
p = 0.20
So, hypothesis:
[tex]H_0:p=\hat{p}\\\\H_a:\hat{p}<p[/tex]
so, test statistic value would be
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.13-0.20}{\sqrt{\dfrac{0.2\times 0.8}{84}}}\\\\z=-1.604[/tex]
At 1% level of significance, critical value would be
z= 2.58
Since 2.58>-1.604
So, We will accept the null hypothesis.
Hence, It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.
20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.
To determine whether the means of two populations are equal,
A. a t test must be performed.
B. an analysis of variance must be performed.
C. either a t test or an analysis of variance can be performed.
D. a chi-square test must be performed.
Answer:
The correct option is C. either a t test or an analysis of variance can be performed.
Step-by-step explanation:
Consider the provided information.
The t-test, is used for whether the means of two groups are equal or not. The assumption for the test is that both groups are sampled from normal distributions with equal variances. Analysis of Variance (ANOVA) is a statistical method evaluating variations between two or more methods. ANOVA is used in a study to analyze the gaps between group methods.ANOVA is used not for specific differences between means, but for general testing.The chi-squared test is often used to evaluate whether there was a significant difference in one or more groups between the predicted frequencies and the observed frequencies.Hence, Either a t test or an analysis of variance can be performed to determine whether the means of two population are equal.
Therefore, the correct option is C. either a t test or an analysis of variance can be performed.
To determine the means of two populations are equal or not: C. either a t-test or an analysis of variance can be performed.
What is the t-test and ANOVA used for?The t-test is a statistical test that is used to compare the means of two groups, to determine if there is any significant difference between the two groups or populations.
The ANOVA, like the t-test is also used to compare means, however, it is used when the groups or populations involved are more than two.
Therefore, to determine the means of two populations are equal or not: C. either a t-test or an analysis of variance can be performed.
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A cohort study was conducted to study the association of coffee drinking and anxiety in a population-based sample of adults. Among 10,000 coffee drinkers, 500 developed anxiety. Among the 20,000 non-coffee drinkers, 200 cases of anxiety were observed. What is the relative risk of anxiety associated with coffee use?
The relative risk of anxiety associated with coffee use is 5.
Explanation:The relative risk of anxiety associated with coffee use can be calculated by comparing the incidence rates of anxiety among coffee drinkers and non-coffee drinkers. In this study, among 10,000 coffee drinkers, 500 developed anxiety, while among 20,000 non-coffee drinkers, 200 developed anxiety.
To calculate the relative risk, we can use the formula:
Relative Risk = (Incidence Rate of Anxiety among Coffee Drinkers) / (Incidence Rate of Anxiety among Non-Coffee Drinkers)
In this case, the incidence rate of anxiety among coffee drinkers is 500/10,000 = 0.05, and the incidence rate of anxiety among non-coffee drinkers is 200/20,000 = 0.01. Therefore, the relative risk of anxiety associated with coffee use is 0.05/0.01 = 5.
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Final answer:
To calculate the relative risk of anxiety associated with coffee use, compare the incidence rates of anxiety in coffee drinkers (500/10,000) and non-coffee drinkers (200/20,000) which results in a relative risk of 5.0, indicating that coffee drinkers have a fivefold increased risk of anxiety.
Explanation:
The relative risk of anxiety associated with coffee use can be calculated by comparing the incidence of anxiety among coffee drinkers versus non-coffee drinkers. In the provided scenario, among 10,000 coffee drinkers, 500 developed anxiety, giving us an incidence rate of 500/10,000 or 0.05. Among 20,000 non-coffee drinkers, 200 developed anxiety, with an incidence rate of 200/20,000 or 0.01. Therefore, the relative risk (RR) is calculated as the incidence rate among the exposed (coffee drinkers) divided by the incidence rate among the unexposed (non-coffee drinkers), which is 0.05/0.01 or 5.0. This suggests that the coffee drinkers have a five times higher risk of developing anxiety compared to non-coffee drinkers.
The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 50 and a standard deviation of 11. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 39 and 50?
Answer:
34% of lightbulb replacement requests numbering between 39 and 50.
Step-by-step explanation:
The 68-95-99.7 rule states that, for a normally distributed random variable:
68% are within 1 standard deviation of the mean(34% between one standard deviation below and the mean, 34% between the mean and one standard deviation above the mean).
95% are within 2 standard deviations of the mean.
99.7% are within 3 standard deviations of the mean.
In this problem, we have that:
The distribution of the number of daily requests is bell-shaped and has a mean of 50 and a standard deviation of 11.
Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 39 and 50?
50 is the mean
39 is one standard deviation below the mean.
This means that 34% of lightbulb replacement requests numbering between 39 and 50.