Answer:
Null Hypothesis: No first order autocorrelation
Alternative hypothesis: first order correlation exists
The assumptions to run this test are:
1) Errors are normally distributed with mean 0
2) Errors follows an stationary process
3) Independence condition between the erros
The statistic is defined as:
[tex] DW = \frac{\sum_{t=2}^ T (e_t -e_{t-1})^2}{\sum_{t=1}^T e^2_t}[/tex]
And if the value for the DW is near to 0 we can conclude that the assumption of Independence is not satisfied.
Step-by-step explanation:
The Durbin Watson test is a way to check autocorrelation in residuals for a time seeries or a regression.
We need to remember that the autocorrelation is the similarity of the time series in successive intervals. When we conduct this type of test we are checking if the time series can be modeled with and AR(1) process autoregressive.
The system of hypothesis on this case are:
Null Hypothesis: No first order autocorrelation
Alternative hypothesis: First order correlation exists
The assumptions to run this test are:
1) Errors are normally distributed with mean 0
2) Errors follows an stationary process
3) Independence condition between the errors
The statistic is defined as:
[tex] DW = \frac{\sum_{t=2}^ T (e_t -e_{t-1})^2}{\sum_{t=1}^T e^2_t}[/tex]
And if the value for the DW is near to 0 we can conclude that the assumption of Independence is not satisfied.
Which of the following is equivalent to 60 Superscript one-half? StartFraction 60 Over 2 EndFraction StartRoot 60 EndRoot StartFraction 1 Over 60 squared EndFraction StartFraction 1 Over StartRoot 60 EndRoot EndFraction
Answer:
[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]
Step-by-step explanation:
We are given the following expression:
[tex](60)^{\frac{1}{2}}[/tex]
We are given the following options:
[tex]A.~\dfrac{60}{2}\\\\B.~\sqrt{60}\\\\C.~(\dfrac{1}{60})^{2}\\\\D.~\dfrac{1}{\sqrt{60}}[/tex]
Exponent Properties:
[tex]x^{-a} = (\dfrac{1}{x})^{a}\\\\(x^m)^n = x^{mn}\\\\\dfrac{x^m}{x^n} = x^{m-n}[/tex]
Thus, the correct answer is
[tex](60)^{\frac{1}{2}} = \sqrt{60}[/tex]
Answer:
B or [tex]\sqrt{60}[/tex]
Step-by-step explanation:
In an article regarding interracial dating and marriage recently appeared in a newspaper. Of 1719 randomly selected adults, 311 identified themselves as Latinos, 322 identified themselves as blacks, 251 identified themselves as Asians, and 775 identified themselves as whites. Among Asians, 79% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.
NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Construct the 95% confidence intervals for the three Asian responses.
1. Welcome a white person ( , )
2. Welcome a Latino ( , )
3. Welcome a Black person ( , )
Answer:
Step-by-step explanation:
Hello!
The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:
X₁: Number of Asians that would welcome a white person into their families.
X₂: Number of Asians that would welcome a Latino person into their families.
X₃: Number of Asians that would welcome a black person into their families.
Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251
Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.
[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]
1. 95% CI for Asians that would welcome a white person.
If 79% would welcome a white person, then the expected value is:
E(X)= n*p= 251*0.79= 198.29
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409
√V(X)= 6.45
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
198.29±1.965*6.45
[185.62;210.96]
With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.
2. 95% CI for Asians that would welcome a Latino person.
If 71% would welcome a Latino person, then the expected value is:
E(X)= n*p= 251*0.71= 178.21
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809
√V(X)= 7.19
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
178.21±1.965*7.19
[164.08; 192.34]
With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.
3. 95% CI for Asians that would welcome a Black person.
If 66% would welcome a Black person, then the expected value is:
E(X)= n*p= 251*0.66= 165.66
And the Standard deviation is:
V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244
√V(X)= 7.50
You can construct the interval as:
E(X)±Z₁₋α/₂*√V(X)
165.66±1.965*7.50
[150.92; 180.40]
With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.
I hope it helps!
An average light bulb manufactured at The Lightbulb Company lasts and average of 300 days, with a standard deviation of 50 days. Suppose the lifespan of a light bulb from this company is normally distributed. (a) What is the probability that a light bulb from this company lasts less than 210 days? More than 330 days?(b) What is the probability that a light bulb from this company lasts between 280 and 380 days?(c) How would you characterize the lifespan of the light bulbs whose lifespans are among the shortest 2% of all bulbs made by this company?(d) If a pack of 6 light bulbs from this company are purchased, what is the probability that exactly 4 of them last more than 330 days?
Answer:
a). 0.03593, 0.27425
(bl 0.0703
(c) the lifespan of such bulb is less than 210 days
(d) 0.0298.
Step-by-step explanation:
Given that U = 300, sd= 50
Z = X-U/sd
(a) We find the probability
Pr(X<210) = Pr(Z<(210-300)/50)
= Pr(Z<-1.8)
= 0.03593
Pr(X >330) = Pr(Z >(330-300)/50))
= Pr(Z> 0.6)
= 1- PR(Z<0.6) = 1 - 0.72575
Pr(X>330)= 0.27425
(b) Pr(280< X< 330) = Pr( 280 < Z< 380)
= Pr([280-300/50] < Z < [380-300/50])
= Pr(0.4 < Z < 1.6)
= Pr(Z< 1.6) - Pr(Z < 0.4)
=0.72575 - 0.65542
= 0.07033
= 0.0703
(c) the lifespan of such bulbs is less than 210 days
(d) given n = 6, x = 4 ,
Pr(X>330) = 0.27425 = p
q = 1-p = 0.72575
Pr(X=4) = 6C4 × (0.27425)⁴ ×(0.72575)²
Pr(X=4) = 10 × 0.005657 × 0.52671
Pr(X= 4) = 0.029795
Pr(X= 4) = 0.0298
In a certain small dorm, the dorm council will consist of 8 students, ofwhom 3 must be women and 5 must be men, because there are a total of 15 women and 25 men currently living there. What is the total possible number of complete councils that could be selected
Answer:
The total possible number of complete councils that could be selected is 24,174,150.
Step-by-step explanation:
The order of the men and the women is not important. So we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
What is the total possible number of complete councils that could be selected
3 women from a set of 15
5 men from a set of 25
[tex]T = C_{15,3}*C_{25,5} = \frac{15!}{3!12!}*\frac{25!}{5!20!} = 455*53130 = 24174150[/tex]
The total possible number of complete councils that could be selected is 24,174,150.
Define a random variable x = number of cups of coffee consumed on an average day. Let x = 4 represent four or more cups. Round your answers to four decimal places.
Answer:
E (X) = 6.4
Step-by-step explanation:
SOLUTION:
A random variable x = number of cups of coffee consumed on an average day.
∴Let x = 4 represent four or more cups. Round your answers to four decimal places.
X Probability (X)
0 0.1
1 0.15
2 0.3
3 0.75
4 0. 25
5 0.21
∴ E (X) = Ux(Mean)
0x.0.1 + 1 x.15 + 2 x 0.3 + 3 x 0.75 + 4 x 0.25 + 5 x 0.21 = 6.4
Jeremy and Brenda drove their cars in
opposite directions. When they stopped
after some time, they were already
126 miles apart. If Jeremy drove twice
as far as Brenda, how many miles did
Jeremy drive?
Answer: Jeremy drove 84 miles.
Step-by-step explanation:
Let x represent the number of miles that Brenda drove.
If Jeremy drove twice
as far as Brenda, it means that the distance covered by Jeremy would be 2x miles
When they stopped after some time, they were already
126 miles apart. This means that the total distance covered by both of them is 126 miles. Therefore,
x + 2x = 126
3x = 126
x = 126/3
x = 42 miles
The number of miles that Jeremy drove is
42 × 2 = 84 miles
solution to 2/3x = -5/2 + 2
Answer:
x=-0.75
Step-by-step explanation:
Combine Like terms
2/3x= -5/2+2
2/3x= -2.5+2
2/3x= -0.5
Multiply both sides by 3
2x= -1.5
Divide both sides by 2
x= -0.75
Answer:
x = -4/3
Step-by-step explanation:
The LCM for the fractions is 6x.
2/3x = -5/2 + 2
Multiply through by 6x
(6x) 2/3x = (6x) -5/2 + (6x)2
4 = -15x + 12x
4 = -3x
x = -4/3
Enjoy Maths!
The numerical course grades in a statistics course can be approximated by a normal model with a mean of 70 and a standard deviation of 10. The professor must convert the numerical grades to letter grades. She decides that she wants 10% A's, 30% B's, 40% C's, 15% D's, and 5% F's. a. What is the cutoff for an A grade?
Answer:
The cutoff for an A grade is 82.8.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 70, \sigma = 10[/tex]
a. What is the cutoff for an A grade?
The top 10% of the class get an A grade. So the cutoff is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 70}{10}[/tex]
[tex]X - 70 = 1.28*10[/tex]
[tex]X = 82.8[/tex]
The cutoff for an A grade is 82.8.
Montoya (2007) asked 56 men and 82 women to rate 21 different body parts on a scale of 1 (no opinion) to 5 (very desirable). They found that men and women rated the eyes similarly, with an average rating of about 3.77 ± 1.23 (M ± SD). Assuming these data are normally distributed, answer the following questions. (Round your answers to two decimal places.) (a) What percentage of participants rated the eyes at least a 5 (very desirable)? % (b) What percentage rated the eyes at most a 1 (no opinion)? %
Answer:
a. 15.9%
b. 1.2%
Step-by-step explanation:
using the normal distribution we have the following expression
[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{5-M}{SD}) \\[/tex]
Where the first expression in the right hand side is the z-scores, M is the mean of value 3.77 and SD is the standard deviation of value 1.23.
if we simplify and substitute values, we arrive at
[tex]P(X\geq a)=P(\frac{X-u}{\alpha } \geq \frac{a-M}{SD}) \\P(X\geq 5)=P(Z \geq \frac{5-3.77}{1.23})\\P(X\geq 5)=P(Z \geq 1)\\P(X\geq 5)=1- P(Z < 5)\\P(X\geq 5)=1-0.841\\P(X\geq 5)=0.159[/tex]
in percentage, we arrive at 15.9%
b. for the percentage rated the eyes most a 1
[tex]P(X\leq a)=P(\frac{X-u}{\alpha } \leq \frac{a-M}{SD})\\for a=1\\P(X\leq 1)=P(Z \leq \frac{1-3.77}{1.23})\\P(X\leq 1)=P(Z \leq -2.25})\\P(X\leq 1)=0.012\\[/tex]
In percentage we have 1.2%
1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260. Use this sample information to construct a 95% confidence interval for the average daily revenue. The store manager believes that the coffee and pastry strategy would lead to an average daily revenue of $1,200. Is the manager correct based on the 95% confidence interval?
Answer:
No, the manager is not correct based on the 95% confidence interval.
Step-by-step explanation:
We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .
The Pivotal quantity for 95% confidence interval is given by;
[tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, X bar = sample mean = $1080
s = sample standard deviation = $260
n = sample size = 35 {five-week}
So, 95% confidence interval for average daily revenue, [tex]\mu[/tex] is given by;
P(-2.032 < [tex]t_3_4[/tex] < 2.032) = 0.95
P(-2.032 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.032) = 0.95
P(-2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95
P(X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ X bar - 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] , X bar + 2.032 * [tex]{\frac{s}{\sqrt{n} }[/tex] ]
= [ 1080 - 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] , 1080 + 2.032 * [tex]{\frac{260}{\sqrt{35} }[/tex] ]
= [ 990.70 , 1169.30 ]
No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.
Therefore, the store manager believe is not correct.
The 95% confidence interval for the store's average daily revenue is calculated to be approximately ($993.97, $1166.03). Since $1200 is outside this interval, the manager's belief that the coffee and pastry strategy will lead to an average daily revenue of $1200 is not backed by this confidence level.
Explanation:In the field of statistics, a confidence interval (CI) is a type of interval estimate that is used to indicate the reliability of an estimate. The method for calculating a 95% confidence interval for the average daily revenue involves the sample mean, the standard deviation, and the z-score associated with a 95% confidence level, which is approximately 1.96. Let's use the provided data to calculate:
Calculate the standard error by dividing the standard deviation by the square root of the sample size. Here, the standard deviation is $260, and the sample size is 5 weeks * 7 days/week = 35 days. So, the standard error is $260 / sqrt(35) = $43.89.Multiply the standard error by the z-score to get the margin of error. So, $43.89 * 1.96 = $86.03.Calculate the lower and upper bounds of the 95% confidence interval by subtracting and adding the margin of error from/to the sample mean. So, ($1080 - $86.03, $1080 + $86.03) = ($993.97, $1166.03).The range of this 95% confidence interval is from $993.97 to $1166.03. This means we are 95% confident that the true average daily revenue lies within this interval. Since $1200 lies outside this interval, the manager's belief is not supported by this confidence interval.
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Solve using normalcdf
Answer:
0.209
Step-by-step explanation:
Find the sample mean and standard deviation.
μ = 1050
s = 218 / √50 = 30.8
Find the z-score.
z = (1075 − 1050) / 30.8
z = 0.81
Find the probability.
P(Z > 0.81) = 1 − 0.7910 = 0.2090
Is there more wood in a 60-foot-high tree trunk with a radius of 2.2 feet or in a 50-foot-high tree trunk with a radius of 2.6 feet? Assume that the trees can be regarded as right circular cylinders.
Answer:
V₂=1061.85 ft³
Step-by-step explanation:
To determine where more wood is found, just find the volume of each log and see which one has the largest volume
knowing that the volume of a straight cylinder is
V=πR²h, where R=radius and h=height
[tex]V_{1}=\pi (2.2)^{2}60= 912.31ft^{3}\\V_{2}=\pi (2.6)^{2}50= 1061.85ft^{3}[/tex]
As we can observe
V₂>V₁
An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 6 cm, and the height is 12 cm. Find the rate at which the water level is rising when the water level is 5 cm.
Answer:
The volume of a pyramid is 1/3 Sh, where S is the area of the base and h is the height. Since the area of base (Square) is , S = [tex]s^{2}[/tex], where s is the side of the base. So the volume is
V = 1/3 [tex]s^{2}[/tex]h.
Further solution is on paper (Pictures attached)
In a new card game, you start with a well-shuffled full deck and draw 3 cards without replacement. If you draw 3 hearts, you win $50. If you draw 3 black cards, you win $25. For any other draws, you win nothing.
(a) Find the expected winnings for a single game.
(b) Find the standard deviation of the winnings.
(c) If the game costs $5 to play, what would be the expected value of the net profit (or loss)?
(Hint: profit = winnings - cost; X - 5)
(d) If the game costs $5 to play, what would be the standard deviation of the net profit (or loss)?
(e) If the game costs $5 to play, should you play this game?
There are [tex]\binom{52}3=\frac{52!}{3!(52-3)!}[/tex] (or "52 choose 3") ways of drawing any 3 cards from the deck.
There are 13 hearts in the deck, and 26 cards with a black suit. So there are [tex]\binom{13}3[/tex] and [tex]\binom{26}3[/tex] ways of drawing 3 hearts or 3 black cards, respectively. Then the probability of drawing 3 hearts is
[tex]P(\text{3 hearts})=\dfrac{\binom{13}3}{\binom{52}3}=\dfrac{11}{850}[/tex]
and the probability of drawing 3 black cards is
[tex]P(\text{3 black})=\dfrac{\binom{26}3}{\binom{52}3}=\dfrac2{17}[/tex]
All other combinations can be drawn with probability [tex]1-\frac{11}{850}-\frac2{17}=\frac{739}{850}[/tex].
Let [tex]W[/tex] be the random variable for one's potential winnings from playing the game. Then
[tex]P(W=w)=\begin{cases}\frac{11}{850}&\text{for }w=\$50\\\frac2{17}&\text{for }w=\$25\\\frac{739}{850}&\text{otherwise}\end{cases}[/tex]
a. For a single game, one can expect to win
[tex]E[W]=\displaystyle\sum_ww\,P(W=w)=\frac{\$50\cdot11}{850}+\frac{\$20\cdot2}{17}+\frac{\$0\cdot739}{850}=\$3[/tex]
b. For a single game, one's winnings have a variance of
[tex]V[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2[/tex]
where
[tex]E[W^2]=\displaystyle\sum_ww^2\,P(W=w)=\frac{\$50^2\cdot11}{850}+\frac{\$20^2\cdot2}{17}+\frac{\$0^2\cdot739}{850}=\$^2\frac{1350}{17}\approx\$^279.41[/tex]
so that [tex]V[W]=\$^2\frac{1197}{17}\approx\$^270.41[/tex]. (No, that's not a typo, variance is measured in squared units.) Standard deviation is equal to the square root of the variance, so it is approximately $8.39.
c. With a $5 buy-in, the expected value of the game would be
[tex]E[W-\$5]=E[W]-\$5=-\$2[/tex]
i.e. a player can expect to lose $2 by playing the game (on average).
d. With the $5 cost, the variance of the winnings is the same, since the variance of a constant is 0:
[tex]V[W-\$5]=V[W][/tex]
so the standard deviation is the same, roughly $8.39.
e. You shouldn't play this game because of the negative expected winnings. The odds are not in your favor.
The expected winning for a single game defined is : $3.59
The standard deviation of winning is : $10.11
Expected winning if game costs $5 to play is :
- $0.79
The standard deviation of winning if game costs $5 to play is : $11.33
The game should not be played with a game play fee of -$5 as the expected winning value is negative.
Recall : selection is done without replacement :
Number of hearts in a deck = 13
Probability of drawing 3 hearts :
P(drawing 3 Hearts)
First draw × second draw × third draw
13/52 × 12/51 × 11/50 = 1716/132600 = 858/66300
Probability of selecting 3 black cards :
Number of black cards in a deck = 26
P(drawing 3 black cards) :
First draw × second draw × third draw
26/52 × 25/51 × 24/50 = 15600 / 132600 = 7800/66300
Probability of making any other draw :
P(3 hearts) + P(3 blacks) + P(any other draw) = 1
858/66300 + 7800/66300 + P(any other draw) = 1
P(any other draw) = 57642/66300
For a single game :
X _______ $50 ________ $25 _______ $0
P(X)_ 858/66300__ 7800/66300_57642/66300
E(X) = Σ[ X × p(X)]
E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+0]
E(X) = $3.588
The standard deviation = √Var(X)
Var(X) = Σ[ X² × p(X)] - E(X)
Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+0] = 105.88235
Var(X) = 105.88235 - 3.588 = 102.29435
Standard deviation of winning = √102.29435 = $10.114
If the game cost $5 to play :
Net amount won if :
3 hearts are drawn = $50 - $5 = $45
3 blacks are drawn = $25 - $5 = $20
Any other combination are drawn= $0 - $5 = -$5
The distribution becomes :
X _______ $45 ________ $20 _______ -$5
P(X)_ 858/66300__ 7800/66300_57642/66300
E(X) = Σ[ X × p(X)]
E(X) =Σ[(50 × 858/66300)+(25 × 7800/66300)+ (-5 × 57642/66300)]
E(X) = - $0.7588
Standard deviation of winning :
The standard deviation = √Var(X)
Var(X) = Σ[ X² × p(X)] - E(X)
Σ[ X²×p(X)] = Σ[(50² × 858/66300)+(25² × 7800/66300)+ (-5² × 57642/66300)] = 127.61764
Var(X) = 127.61764 - (-0.7588) = 128.37644
Standard deviation of winning :
Std(X) = √Var(X) = √128.37644 = $11.330
With a game cost of - $5 ; the expected winning for a single game gives a negative value, therefore you should not play the game.
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A researcher would like to evaluate the claim that large doses of vitamin C can help prevent the common cold. One group of participants is given a large dose of the vitamin (500 mg per day), and a second group is given a placebo (sugar pill). The researcher records the number of colds each individual experiences during the 3-month winter season. a. Identify the dependent variable for this study.b. Is the dependent variable discrete or continuous?c. What scale of measurement (nominal, ordinal, or interval/ratio) is used to measure the dependent variable?d. What is the independent variable?e. What research method is being used (experimental or correlational?
Answer:
(a) the dependent variable here are the Participants.
(b) the dependent variable is discrete.
(c) The scale measurement of measurement used is interval/ratio to measure the dependent variable.
(d) the independent variable is Vitamin C
(e) The research method being used is experimental.
Step-by-step explanation:
The dependent variable (sometimes known as the responding variable) is what is being studied and measured in the experiment.
Examples of continuous dependent variable may include costs, profits and sales.But some dependent variables are discrete – that is, they take on a relatively small number of integer values.
Interval scale and ratio scale are the two variable measurement scales where they define the attributes of the variables quantitatively.A ratio scale is a measurement scale which has more or less all the properties of an interval scale. Ratio data on this scale has measurable intervals.
Experimental research is a study that strictly adheres to a scientific research design. It includes a hypothesis, a variable that can be manipulated by the researcher, and variables that can be measured, calculated and compared.
Answer: 1. The dependent variable is the common cold.
2. It is a discrete variable.
3. The interval/ratio scale.
4. The independent value is the large doses of Vitamin C administered.
5. The research method used is Experimental.
Step-by-step explanation:
1. The dependent variable is the factor that we are trying to understand. In this case, it is the common cold and how it is affected by large doses of vitamins.
2. A discrete value is computed by counting. So the dependent variable - the common cold was computed by counting the number of occurrences.
3. The Interval/ratio scale is used because both the order of measurement and the differences between them are observed.
4. The Independent variable is the control in the experiment which can be compared to changes in the dependent variable. The large doses of vitamin C affects the common cold.
5. Correlational research observes patterns in variables that occur naturally while Experimental research introduces a change and monitors its effect. The research is Experimental because a change in the form of the placebo is introduced and observed.
(1 point) Find the general solution to the homogeneous differential equation. ????2y????????2−20????y????????+136y=0 Use c1 and c2 in your answer to denote arbitrary constants, and enter them as c1 and c2. y(????)= ?
Answer:
Question is not clear please post question clearly lots of question marks.
Your differential equation is not displayed well. It though looks like this:
2d²y/dx² - 20dy/dx + 136y = 0
If this is not the differential equation, the method of solving this would still be used in solving the correct one.
We first write an auxiliary equation to the differential equation.
The auxiliary equation is:
2m² - 20m + 136 = 0
Dividing by 2, we have
m² - 10m + 68 = 0
Next, we solve the auxiliary equation to obtain the values of m.
Solving using the quadratic formula
m = [-b ± √(b² - 4ac)]/2a
Where a = 1, b = -10, and c = 68
m = [10 ± √(100 - 272)]/2
= 5 ± (1/2)√(-172)
= 5 ± (1/2)i√172
= 5 ± 6.6i
For solutions of the form a ± ib, the complimentary solution is
y = e^(ax)[C1cosbx + C2sinbx]
Therefore, the complimentary solution is
y = e^(5x)[C1cos(6.6x) + C2sin(6.6x)]
2. A canoe requires 8 hours of fabrication. A row boat requires 5 hours of fabrication. The fabrication department has at most 110 hours to labor each week. Write the equation and solve the problems.
Answer:
Let x is the number of the cannon and y is the number of the row boat.Thus the equation will become
8x+5y=110
Answer: 8C+5R≤110
Step-by-step explanation: Let hours of fabrication of canoes = C
Hours of fabrication of rowboat = R
If a canoe requires 8 hours of fabrication. And a row boat requires 5 hours of fabrication. Then
8C + 5R
The fabrication department has at most 110 hours to labor each week.
Therefore
8C+5R≤110
At most means less than or equal to.
Apply the properties of angles to solve for the missing angles. Angle y is what degrees. Angle x is what degrees.
Answer:
y= 65 degrees. and x=30 degrees
Step-by-step explanation:
i haven't done this for a while so i may or may not be correct i tried to help though :)
Answer: y = 52°. x = 64°
Step-by-step explanation:
y = 26 + 26 = 52 degrees ( exterior angle of a triangle is the sum of the two opposite interior angle of the triangle)
x = 180-(52+64)(sum of angles in a triangle is 180°)
x = 180-116
x = 64degrees ( base angles of an isosceles triangle are equal)
Seventy independent messages are sent from an electronic transmission center.Messages are processed sequentially, one after another. Transmission time of each message is Exponential with parameter λ = 5 min−1. Find the probability that all 70 messages are transmitted in less than 12 minutes. Use the Central Limit Theorem.
Answer:
The probability that all 70 messages are transmitted in less than 12 minutes is 0.117.
Step-by-step explanation:
Let X = the transmission time of each message.
The random variable X follows an Exponential distribution with parameter λ = 5 minutes.
The expected value of X is:
[tex]E(X)=\frac{1}{\lambda}=\frac{1}{5}=0.20[/tex]
The variance of X is:
[tex]V(X)=\frac{1}{\lambda^{2}}=\frac{1}{5^{2}}=0.04[/tex]
Now define a random variable T as:
T = X₁ + X₂ + ... + X₇₀
According to a Central limit theorem if a large sample (n > 30) is selected from a population with mean μ and variance σ² then the sum of random variables X follows a Normal distribution with mean, [tex]\mu_{s} = n\mu[/tex] and variance, [tex]\sigma^{2}_{s}=n\sigma^{2}[/tex].
Compute the probability of T < 12 as follows:
[tex]P(T<12)=P(\frac{T-\mu_{T}}{\sqrt{\sigma_{T}^{2}}}<\frac{12-(70\times0.20)}{\sqrt{70\times 0.04}})\\=P(Z<-1.19)\\\=1-P(Z<1.19)\\=1-0.883\\=0.117[/tex]
*Use a z-table for the probability.
Thus, the probability that all 70 messages are transmitted in less than 12 minutes is 0.117.
Following are the solution to the given question:
Let X signify the letter's transmission time [tex]i^{th}[/tex] , and [tex]i =1,2,3,...,70.[/tex] this is assumed that the [tex]X_i \sim Exp (\lambda =5)[/tex].
[tex]\to Mean =\frac{1}{\lambda} =\frac{1}{5}=0.2\\\\ \to Variance =\frac{1}{\lambda^2} =\frac{1}{5^2} =\frac{1}{25}=0.04\\\\[/tex]
Let [tex]T = X_1 + X_2 +...+ X_{70}[/tex], be the total transmission time. According to the Central Limit Theorem, T has a Normal distribution of mean, Variance.mean [tex]= E(T) = 70 \times 0.2=14[/tex]
Variance [tex]=V(T) =70 \times 0.04 = 2.8[/tex]
Therefore
[tex]\to P(T<12) = P (Z < \frac{12-14}{\sqrt{2.8}})[/tex]
[tex]= P(Z<-1.195) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (by sy memetry) \\\\=0.5-P(0<Z<1.19)\\\\=0.5-0.3830 \\\\=0.1170\\\\[/tex]
Therefore, the final answer is "0.1170".
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The accounting records of Blossom Company show the following data. Beginning inventory 3,090 units at $5 Purchases 7,930 units at $8 Sales 9,390 units at $10 Determine cost of goods sold during the period under a periodic inventory system using the FIFO method. (Round answer to 0 decimal places, e.g. 1,250.)
Answer:
The cost of goods sold is 65,850
Step-by-step explanation:
FIFO Perpetual chart is attached.
FIFO Perpetual chart shows purchases, sales and balance of the period.
The cost of goods sold is:
3,090 units x $5=$15,450
6,300 units x $8=$50,400
Total=65,850
5. (20 pts) Consider the array 25, 14, 63, 29, 63, 47, 12, 21. Apply the Split procedure in Quicksort, as described in class, to this array using the first element as the pivot.
Answer:
Step-by-step explanation: see attachment
Consider the following data on x = rainfall volume (m3) and y = runoff volume (m3) for a particular location. x 8 12 14 19 23 30 40 45 55 67 72 79 96 112 127 y 4 10 13 14 15 25 27 44 38 46 53 74 82 99 105 Use the general software output to decide whether there is a useful linear relationship between rainfall and runoff.
There is a useful linear relationship between rainfall and runoff and the relationship is y = 0.85x - 2.65
Deciding whether there is a useful linear relationship between rainfall and runoff
From the question, we have the following parameters that can be used in our computation:
x 8 12 14 19 23 30 40 45 55 67 72 79 96 112 127
y 4 10 13 14 15 25 27 44 38 46 53 74 82 99 105
Using a graphing tool, we have the following summary
Sum of X = 799Sum of Y = 649Mean X = 53.2667Mean Y = 43.2667Sum of squares (SSX) = 20086.9333Sum of products (SP) = 17313.9333The regression equation can be represented as
y = bx + a
Where
b = SP/SSX = 17313.93/20086.93 = 0.86
a = MY - bMX = 43.27 - (0.86*53.27) = -2.65
So, we have
y = 0.85x - 2.65
Hence, there is a useful linear relationship between rainfall and runoff
Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.52 and a standard deviation of 0.38. Using the empirical rule, what percentage of the students have grade point averages that are between 1.76 and 3.28
Using the empirical rule for a bell-shaped distribution, which states that nearly all data for a normal distribution falls within three standard deviations of the mean, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.
Explanation:The student is asking for the percentage of GPA between 1.76 and 3.28 using the Empirical Rule which applies to a Bell-shaped distribution or normal distribution. The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean.
The mean in this case is 2.52 and the standard deviation is 0.38. Therefore the values 1.76 and 3.28 fall within the mean minus two standard deviations and mean plus two standard deviations respectively. Therefore, by the empirical rule, these values represent approximately 95% of the data.
In other words, according to the empirical rule, approximately 95% of undergraduate students at this university have grade point averages between 1.76 and 3.28.
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A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2. Calculate how far the car travels in the time it takes to stop. Round your answer to one decimal place.
Answer:
80 feet
Step-by-step explanation:
Given:
Initial speed of the car ([tex]v_0[/tex]) = 40 ft/sec
Deceleration of the car ([tex]\frac{dv}{dt}[/tex]) = -10 ft/sec²
Final speed of the car ([tex]v_x[/tex]) = 0 ft/sec
Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.
Now, deceleration means rate of decrease of velocity.
So, [tex]\frac{dv}{dt}=-10\ ft/sec^2[/tex]
Negative sign means the velocity is decreasing with time.
Now, [tex]\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})[/tex] using chain rule of differentiation. Therefore,
[tex]\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx[/tex]
Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,
[tex]\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft[/tex]
Therefore, the car travels a distance of 80 feet before stopping.
The least-squares regression model y =−3.4+5.2x and correlation coefficient r=−0.66 were calculated for a set of bivariate data with variables x and y . What is closest to the proportion of the variation in y that cannot be explained by the explanatory variable?
Answer:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For this case the value of r = -0.66
Now we can calculate the determination coeffcient:
[tex] r^2 = (-0.66)^2 = 0.4356[/tex]
And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable
And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable
Step-by-step explanation:
For this case we need to calculate the slope with the following formula:
[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]
[tex]\bar x= \frac{\sum x_i}{n}[/tex]
[tex]\bar y= \frac{\sum y_i}{n}[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x[/tex]
And the model obtained for this case is:
[tex] y = -3.4 +5.2 x[/tex]
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For this case the value of r = -0.66
Now we can calculate the determination coeffcient:
[tex] r^2 = (-0.66)^2 = 0.4356[/tex]
And then we can conclude that 43.56% of the variation in y can be explained by the explanatory variable
And then 100-43.56 = 56.44 % of the variation in y that cannot be explained by the explanatory variable
The Coefficient of determination gives the proportion of variation which can be explained by the regression line.
The proportion of variation that cannot be explained by the given regression line is 56.44%
From the question, the correlation Coefficient, R = - 0.66
The Coefficient of determination, R² can be calculated from the R value given. R² = - 0.66² = 0.4356This means that :
Percentage of variation that can be explained = (0.4356) × 100% = 43.56%The percentage of variation that cannot be explained = 100% - 43.56% = 56.44%Therefore, proportion of the variation in y that cannot be explained is 56.44%
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he purpose of an x-bar chart is to determine whether there has been a: change in the AOQ. change in the number of defects in a sample. change in the percent defective in a sample. change in the dispersion of the process output. change in the central tendency of the process output.
Answer:
E. Change in the central tendency of the process output.
Step-by-step explanation:
The x-bar chart is a control chart for the central tendency of the process output. Therefore it's purpose is to determine whether there has been a change in the central tendency of the process output.
Answer:
The change in the central tendency of the process output.
Step-by-step explanation:
Lets examine each of the listed options
The change in the percent defective in a sample is associated with the change in average outgoing quality (AOQ).
The change in the dispersion of the process output is associated with R-chart since the function of R-chart is to detect changes in the dispersion.
The change in the number of defects in a sample is associated with C-chart since the function of a C-chart is to show the number of flaws per unit in a sample.
The change in the central tendency of the process output is associated with X-bar chart since the function of X-bar chart is to check whether the values are within the appropriate limits (central tendency) of the process output or not.
Therefore, the purpose of an X-bar chart is to determine whether there has been a change in the central tendency of the process output.
A certain college graduate borrows 7864 dollars to buy a car. The lender charges interest at an annual rate of 13%. Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k dollars per year, determine the payment rate that is required to pay off the loan in 3 years. Also determine how much interest is paid during the 3-year period.
Answer:
Therefore rate of payment = $ 3145.72
Therefore the rate of interest = =$1573.17
Step-by-step explanation:
Consider A represent the balance at time t.
A(0)=$ 7864.
r=13 % =0.13
Rate payment = $k
The balance rate increases by interest (product of interest rate and current balance) and payment rate.
[tex]\frac{dB}{dt} = rB-k[/tex]
[tex]\Rightarrow \frac{dB}{dt} - rB=-k[/tex].......(1)
To solve the equation ,we have to find out the integrating factor.
Here p(t)= the coefficient of B =-r
The integrating factor [tex]=e^{\int p(t) dt[/tex]
[tex]=e^{\int (-r)dt[/tex]
[tex]=e^{-rt}[/tex]
Multiplying the integrating factor the both sides of equation (1)
[tex]e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}[/tex]
[tex]\Rightarrow e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt[/tex]
Integrating both sides
[tex]\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt[/tex]
[tex]\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C[/tex] [ where C arbitrary constant]
[tex]\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}[/tex]
Initial condition B=7864 when t =0
[tex]\therefore 7864= \frac{k}{r} - Ce^0[/tex]
[tex]\Rightarrow C= \frac{k}{r} -7864[/tex]
Then the general solution is
[tex]B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}[/tex]
To determine the payment rate, we have to put the value of B(3), r and t in the general solution.
Here B(3)=0, r=0.13 and t=3
[tex]B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}[/tex]
[tex]\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0[/tex]
⇒k≈3145.72
Therefore rate of payment = $ 3145.72
Therefore the rate of interest = ${(3145.72×3)-7864}
=$1573.17
The payment rate that is required to pay off the loan in 3 years is approximately 2949.51 dollars/year. The interest paid during the 3-year period is about 984.53 dollars.
Explanation:To solve this problem, the formula for continuously compounded interest should be used which is A = P * e^(rt), where A is the value of the investment at a future time t, P is the principal amount, r is the annual interest rate, and t is the time the money is invested or borrowed for.
First, set up the equation: 7864 = k * (1/(0.13)) * (e^(0.13 * 3) - 1), then solve for k: k = 7864 * (0.13) / (e^(0.13 * 3) - 1) ≈ 2949.51 dollars/year.
Using the formula A = P * e^(rt), the total amount paid is A = 2949.51 * 3 = 8848.53 dollars. The interest paid during the 3-year period is calculated by subtracting the loan amount from the total amount paid, that is 8848.53 - 7864 = 984.53 dollars.
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A college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section was 460 with a standard deviation of 105, and the mean score for the Quantitative Reasoning was 429 with a standard deviation of 148. Suppose that both distributions are nearly normal. Round calculated answers to 4 decimal places unless directed otherwise.
Answer:
Step-by-step explanation:
Given that a college senior who took the Graduate Record Examination exam scored 530 on the Verbal Reasoning section and 600 on the Quantitative Reasoning section.
Quantitative reasons scores are N (429, 148)
Hence Z score for the college senior = [tex]\frac{530-429}{148} \\=0.6824[/tex]
The mean score for Verbal Reasoning section was 460 with a standard deviation of 105
Verbal reasoning score was N(460, 105)
The score of college senior = 530
Z score for verbal reasoning =[tex]\frac{530-460}{105} \\=0.6667[/tex]
comparing we can say he scored more on quantiative reasoning.
Thus comparison was possible only by converting to Z scores.
Below are the jersey numbers of 11 players randomly selected from a football team. Find the range, variance, and standard deviation for the given sample data. What do the results tell us? 59 19 30 75 42 49 57 81 11 87 91
Answer:
Range = The difference between the highest and lowest which is
The mean is the sum of all values divided by the number of values
= 59+19+30+75+42+49+57+81+11+87+91 divided by n (which is 11)
601÷11
= 54.6
Variance = sum of squared deviations from the mean divided by n-1
=679.65
Standard Deviation = This is the square root of variance which gives us;
26.07
Children inherit many traits from their parents. Some inherited traits are not desirable. For example, if both parents are carriers, there is a 25% chance any child will have sickle-cell disease. If we assume no identical twins in a family, and both individuals in a couple are carriers, what is the probability none of their four children has the disease
The probability that none of their four children has the disease = 0.316
Step-by-step explanation:
Step 1 :
The probability that the child will have the specified disease if both the parents are carriers = 25% = 0.25
So the probability that the child does not have the specified disease = 1 - 0.25 = 0.75
Step 2 :
There are four children. The probability that each child has the disease is independent of the condition of the other children
The probability of more than one independent events happening together can be obtained by multiplying probability of individual events
So, the probability that none of their 4 children has the sickle cell disease = [tex](0.75)^{4}[/tex] = 0.316
Step 3 :
Answer :
The probability no children has the specified cell disease = 0.316