Final answer:
To test if there has been a significant decrease in the average price of homes sold in the U.S., we will conduct a hypothesis test using the p-value approach. We will define the null and alternative hypotheses, calculate the Z-test statistic, compare the p-value to the significance level, and make a conclusion based on the results.
Explanation:
To test if there has been a significant decrease in the average price of homes sold in the U.S., we will conduct a hypothesis test using the p-value approach. Let's define our hypotheses:
Null hypothesis (H0): The average price of homes sold in the U.S. is not significantly different from $220,000.
Alternative hypothesis (Ha): The average price of homes sold in the U.S. is significantly less than $220,000.
Next, we need to calculate the test statistic. Since the population standard deviation is known, we can use a Z-test. The formula for the Z-test statistic is:
Z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
In this case, the sample mean is $210,000, the population mean is $220,000, the population standard deviation is $36,000, and the sample size is 81. Plugging these values into the formula:
Z = (210,000 - 220,000) / (36,000 / sqrt(81))
Calculating this gives us a Z-statistic of -1.6875.
The final step is to compare the p-value of the test statistic to the significance level. Since the significance level is 1%, the critical value is 0.01. We can use a Z-table or calculator to find the p-value corresponding to a Z-statistic of -1.6875. The p-value is approximately 0.0466.
Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. This means that there is sufficient evidence to conclude that there has been a significant decrease in the average price of homes sold in the U.S.
The position vector for particle A is cos(t)i, and the position vector for particle B is sin(t)j. What is the difference in acceleration (i.e. the relative acceleration) between particle A and B at any time t? The acceleration vector of a particle moving in space is the second derivative of the position vector
Answer:
sin(t)j - cos(t)i
Step-by-step explanation:
Let's start with A:
Position vector = cos(t)i
Velocity vector = -sin(t)i (differentiating the position vector)
acceleration vector = -cos(t)i (differentiating the velocity vector)
Then we go to B:
Position vector = sin(t)j
Velocity vector = cos(t)j
acceleration vector = -sin(t)j
Relative acceleration = -cos(t)i - (-sin(t)j) = sin(t)j - cos(t)i
What can you say about a solution of the equation y' = - y2 just by looking at the differential equation? The function y must be decreasing (or equal to 0) on any interval on which it is defined. The function y must be increasing (or equal to 0) on any interval on which it is defined.
Answer:
The function y must be decreasing (or equal to 0) on any interval on which it is defined.
Step-by-step explanation:
The derivative of a function gives us the rate at which that function is changing. In this case, -y^2, yields a negative value for every possible value of y, thus, the rate of change is always negative and the function y is decreasing (or equal to 0) on any interval on which it is defined.
The differential equation y' = -[tex]y^2[/tex] implies that y is either decreasing or constant wherever it is defined, because the derivative y' is non-positive.
Explanation:By examining the differential equation y' = -[tex]y^2[/tex], we can infer some characteristics about the solutions without solving it. If y is a solution to this equation, then y' represents the derivative of y with respect to x. This derivative tells us about the rate of change of the function y.
Since the right side of the equation is -[tex]y^2[/tex], and a square of a real number is always non-negative, multiplying by -1 makes it non-positive. This implies that the derivative y' is either less than or equal to zero. Therefore, wherever the function y is defined, it must be either decreasing or constant (equal to zero). If y is positive, y will decrease because of the negative sign in front of the square. If y is negative, squaring it results in a positive number, but the negative sign still ensures that the rate of change is non-positive.
Conclusion: the function y is decreasing or remains constant on any interval it is defined; it cannot be increasing.
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Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodology for Probabilistic Life Prediction of Multiple-Anomaly Materials"� proposes a Poisson distribution for X. Suppose that ? = 4. (Round your answers to three decimal places.) (a) Compute both P(X ? 4) and P(X < 4). (b) Compute P(4 ? X ? 5). (c) Compute P(5 ? X). (d) What is the probability that the number of anomalies does not exceed the mean value by more than one standard deviation?
Answer:
0.6284,0.4335,0.1953.0.9786
Step-by-step explanation:
Given that X the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk is following a poisson distribution with parameter 4
a) [tex]P(X\leq 4)=0.6284\\P(X<4)=0.4335[/tex]
b) [tex]P(4\leq x<5)\\=P(4)=0.1953\\[/tex]
c) P([tex]5\leq x)[/tex]=0.8046
d) the probability that the number of anomalies does not exceed the mean value by more than one standard deviation
=[tex]P(0\leq X\leq 8)\\=F(8)-F(0)\\=0.9786[/tex]
Salaries of 49 college graduates who took a statistics course in college have a mean, x overbar, of $ 65 comma 300. Assuming a standard deviation, sigma, of $17 comma 805, construct a 95% confidence interval for estimating the population mean mu.
Answer: [tex]60,540< \mu<70,060[/tex]
Step-by-step explanation:
The confidence interval for population mean is given by :-
[tex]\overline{x}-z^*\dfrac{\sigma}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\sigma[/tex] = Population standard deviation.
n= sample size
[tex]\overline{x}[/tex] = Sample mean
z* = Critical z-value .
Given : [tex]\sigma=\$17,000[/tex]
n= 49
[tex]\overline{x}= \$65,300[/tex]
Two-tailed critical value for 95% confidence interval = [tex]z^*=1.960[/tex]
Then, the 95% confidence interval would be :-
[tex]65,300-(1.96)\dfrac{17000}{\sqrt{49}}< \mu<65,300+(1.96)\dfrac{17000}{\sqrt{49}}[/tex]
[tex]=65,300-(1.96)\dfrac{17000}{7}< \mu<65,300+(1.96)\dfrac{17000}{7}[/tex]
[tex]=65,300-4760< \mu<65,300+4760[/tex]
[tex]=60,540< \mu<70,060[/tex]
Hence, the 95% confidence interval for estimating the population mean [tex](\mu)[/tex] :
[tex]60,540< \mu<70,060[/tex]
Compute the work done by the force F = sin(x + y), xy, (x^2)z> in moving an object along the trajectory that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (3, 6, 8) when force is measured in Newtons and distance in meters.
Parameterize the line segments by
[tex]\vec r(t)=(1-t)\langle1,1,1\rangle+t\langle2,2,2\rangle=\langle1+t,1+t,1+t\rangle[/tex]
and
[tex]\vec s(t)=(1-t)\langle2,2,2\rangle+t\langle3,6,8\rangle=\langle2+t,2+4t,2+6t\rangle[/tex]
both with [tex]0\le t\le1[/tex]. Then
[tex]\vec r'(t)=\langle1,1,1\rangle[/tex]
[tex]\vec s'(t)=\langle1,4,6\rangle[/tex]
so that the work done by [tex]\vec F(x,y,z)=\langle\sin(x+y),xy,x^2z\rangle[/tex] over the respective line segments is given by
[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r=\int_0^1\langle\sin(2+2t),(1+t)^2,(1+t)^3\rangle\cdot\langle1,1,1\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1\sin(2+2t)+(1+t)^2+(1+t)^3\,\mathrm dt=\boxed{\frac{73+6\cos2-6\cos4}{12}}[/tex]
and
[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec s=\int_0^1\langle\sin(4+5t),(2+t)(2+4t),(2+t)^2(2+6t)\rangle\cdot\langle1,4,6\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1\sin(4+5t)+4(2+t)(2+4t)+6(2+t)^2(2+6t)\,\mathrm dt=\boxed{\frac{3695+3\cos4-3\cos9}{15}}[/tex]
(both measured in Newton-meters)
To compute the work done by the force in this scenario, we need to apply the principle of line integrals of force with respect to displacement, calculating work for each segment of the trajectory separately and then adding these results.
Explanation:Computing work done by the given force while moving an object generally involves utilizing the line integral of the force with respect to displacement. The principle is simplistically shown in the equation W = F⋅d = Fd cos θ, where W represents work, F denotes force, and d symbolizes displacement, with '⋅' denoting the dot product, and cos θ representing the cosine of the angle between the force and displacement vectors. F can break down into its components, such as Fx, Fy, Fz, and similarly, d can be broken down to dx, dy, dz. Using these, we can express work for three dimensions as dW = Fxdx + Fydy + Fzdz which extends the notion of work done to three-dimensional space. This concept is categorically illustrated while calculating the infinitesimal work done by a variable force.
The trajectory (path) for this problem can be divided into two line segments, one from (1, 1, 1) to (2, 2, 2) and the other from (2, 2, 2) to (3, 6, 8). The work for each segment can be calculated independently, based on the variable force function and the displacement during each segment, after which the two results can be added up to determine the total work done.
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Suppose that in a bowling league, the scores among all bowlers are normally distributed with mean µ = 182 points and standard deviation σ = 14 points. A trophy is given to each player whose score is at or above the 97th percentile. What is the minimum score needed for a bowler to receive a trophy?
Answer:
209 points
Step-by-step explanation:
Mean points scored (μ) = 182 points
Standard deviation (σ) = 14 points
The z-score for any given game score 'X' is defined as:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
At, the 97th percentile of a normal distribution, the z-score, according to a z-score table, is 1.881.
Therefore, the minimum score, X, needed for a bowler to receive a trophy is:
[tex]1.881=\frac{X-182}{14}\\X=208.334[/tex]
Since only whole point scores are possible, X=209 points.
To find the minimum score at the 97th percentile for bowlers in a league, one must calculate the z-score for the 97th percentile and then apply the formula Score = μ + (z * σ) using the league's mean and standard deviation.
Explanation:To find the minimum score needed for a bowler to receive a trophy (which is at or above the 97th percentile), we need to use the normal distribution properties. With a mean (μ) of 182 points and a standard deviation (σ) of 14 points, we can find the z-score corresponding to the 97th percentile using a z-table or a calculator with normal distribution functions. Once we have the z-score, we can use the formula:
Score = μ + (z * σ)
to calculate the score that corresponds to the 97th percentile.
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Suppose a television news broadcast reports that the proportion of people in the United States who are living with a particular disease is 0.09. A team of biomedical students examined a random sample of 527 medical records and found that 34 of them had this disease. They constructed the following 95% z z‑confidence interval for the proportion, p p, of people in the United States who have this disease. 0.0435 < p < 0.0855 0.0435
Answer:
On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real proportion at least at 95% of confidence.
Step-by-step explanation:
1) Notation and definitions
[tex]X=34[/tex] number of people living at USA with a particular disease.
[tex]n=527[/tex] random sample taken
[tex]\hat p=\frac{34}{527}=0.0645[/tex] estimated proportion of people living at USA with a particular disease
[tex]p[/tex] true population proportion of people living at USA with a particular disease.
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 means 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.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.0645 - 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0435[/tex]
[tex]0.0645 + 1.96\sqrt{\frac{0.0645(1-0.0645)}{527}}=0.0855[/tex]
The 95% confidence interval would be given by (0.0435;0.0855)
On this case the 0.09 value is not included on the interval so we can say that the statement on the television news broadcast reports, is a value away from the real mean at least at 95% of confidence.
The area of a rectangular plot 24 feet long and 16 feet wide will be doubled by adding an equal width to each side of the plot. Which equation can be used to find this added width?
(2x + 24)(2x + 16) = 768
(x + 24)(x + 16) = 384
(x + 24)(x + 16) = 768
(2x + 24)(2x + 16) = 384
Answer:
B
Step-by-step explanation:
Stainless steels are frequently used in chemical plants to handle corrosive fluids, however, these steels are especially susceptible to stress corrosion cracking in certain environments. In a sample of 295 steel alloy failures that occurred in oil refineries and petrochemical plants in Japan over the last 10 tears, 118 were caused by stress corrosion cracking and corrosion fatigue (Materials Performance, 1981). Construct a 95% confidence interval for the true proportion of alloy failures caused by stress corrosion cracking.
Answer: 95% confidence interval would be (0.344,0.456).
Step-by-step explanation:
Since we have given that
n = 295
x = 118
so, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{118}{295}=0.4[/tex]
At 95% confidence, z = 1.96
So, margin of error would be
[tex]z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.4\times 0.6}{295}}\\\\=0.056[/tex]
so, 95% confidence interval would be
[tex]\hat{p}\pm \text{margin of error}\\\\=0.4\pm 0.056\\\\=(0.4-0.056,0.4+0.056)\\\\ =(0.344,0.456)[/tex]
Hence, 95% confidence interval would be (0.344,0.456).
Answer:
95% confidence interval would be (0.344,0.456).Step-by-step explanation:
a 14 ft long ladder is placed against a house with an angle of elevation of 72 degrees. How high above the ground is the top of the ladder?
Answer:
Answer is 13.3 ft
Step-by-step explanation:
i explained in the image below
The Martian Colonies elect their government through a lottery. There are100,000 people living on Mars, and every year, a council of 99 co-equalleaders is randomly selected from the population. In how many ways canthe leadership be elected? Give your answer in terms of permutations orcombinations and explain your choice. You do not have to evaluate.
Answer:
The answer is a 100,000-choose-99 (a combination.)
[tex]P(100000, 99) = \displaystyle \left( \begin{array}{c}100000\cr 99\end{array}\right)[/tex].
Step-by-step explanation:
A combination [tex]C(n, r)[/tex] or equivalently [tex]\displaystyle \left(\begin{array}{c}n \cr r\end{array}\right)[/tex] gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] elements.
A permutation [tex]P(n, r)[/tex] also gives the number of ways to choose [tex]r[/tex] out of [tex]n[/tex] elements. On top of that, it accounts for the order of the elements. Two elements in different order counts twice in a permutation, but only once in a combination.
The question emphasize that the council members are "co-equal." That implies that the order of the members don't really matter. Hence a combination with [tex]n = 100000[/tex] and [tex]r = 99[/tex] would be a more suitable choice.
A deck of cards contains 52 cards. They are divided into four suits: spades, diamonds, clubs and hearts. Each suit has 13 cards: ace through 10, and three picture cards: Jack, Queen, and King. Two suits are red in color: hearts and diamonds. Two suits are black in color: clubs and spades.
Use this information to compute the probabilities asked for below and leave them in fraction form. All events are in the context that three cards are dealt from a well-shuffled deck without replacement.
a. The first and second cards are both hearts.
b. The third card is an eight.
c. None of the three cards is an ace.
Answer:
a) 13/52 and the second 12/51
b) Two solutions:
b.1 if we did not picked up an eight in the first two cards 4/50
b.2 there is an eight i the two previous 3/59
c ) (48/52)*(47/51)*(46/50)
Step-by-step explanation:
Condition: Cards are taken out without replacement
a) Probability of first card is heart
There are 52 cards and 4 suits with the same probability , so you can compute this probability in two ways
we have 13 heart cards and 52 cards then probability of one heart card is 13/52 = 0.25
or you have 4 suits, to pick up one specific suit the probability is 1/4 = 0,25
Now we have a deck of 51 card with 12 hearts, the probability of take one heart is : 12/51
b) There are 4 eight (one for each suit ) P = 4/50 if neither the first nor the second card was an eight of heart, if in a) previous we had an eight, then this probability change to 3/50
c) The probability of the first card different from an ace is 48/52 , the probability of the second one different of an ace is 47/51 and for the thirsd card is 46/50. The probability of none of the three cards is an ace is
(48/52)*(47/51)*(46/50)
Nestor Milk Powder is sold in packets with an advertised mean weight of 1.5kgs. The standard deviation is known to be 184 grams. A consumer group wishes to check the accuracy of the advertised mean and takes a sample of 52 packets finding an average weight of 1.49kgs. What is the set of hypotheses that should be used to test the accuracy of advertised weight?
(a) oH:μ= 1.49; 1H: μ≠1.49
(b) oH:μ= 1.5; 1H: μ< 1.5
(c) oH: μ= 1.5; 1H: μ≠1.5
(d) oH: x = 1.5; 1H: x < 1.5
Answer: c) [tex]H_0:\mu= 1.5\\\\H_a:\mu \neq1.5[/tex]
Step-by-step explanation:
Given: Nestor Milk Powder is sold in packets with an advertised mean weight of 1.5 kgs.
i.e. [tex]\mu= 1.5[/tex]
A consumer group wishes to check the accuracy of the advertised mean and takes a sample of 52 packets finding an average weight of 1.49 kgs.
So 1.49 is sample mean but hypothesis is the statement about the parameter which is [tex]\mu[/tex].
i.e. he wanted to check whether [tex]\mu= 1.5[/tex] or [tex]\mu \neq1.5[/tex]
Since null hypothesis[tex](H_0)[/tex] contains equality and alternative hypothesis[tex](H_a)[/tex] is against it.
Therefore, the set of hypotheses that should be used to test the accuracy of advertised weight would be :
[tex]H_0:\mu= 1.5\\\\H_a:\mu \neq1.5[/tex]
Hence, the correct answer is option c) oH: μ= 1.5; 1H: μ≠1.5
A news vendor sells newspapers and tries to maximize profits. The number of papers sold each day is a random variable. However, analysis of the past month's data shows the distribution of daily demand in Table 16. A paper costs the vendor 20C. The vendor sells the paper for 30C. Any unsold papers are returned to the publisher for a credit of IOC. Any unsatisfied demand is estimated to cost IOC in goodwill and lost profit. If the policy is to order a quantity equal to the preceding day's demand, determine the average daily profit of the news vendor by simulating this system. Assume that the demand for day 0 is equal to 32. Demand per day Probability30 .0531 .1532 .2233 .3834 .1435 .06
Find the P-value for the indicated hypothesis test. An airline claims that the no-show rate for passengers booked on its flights is less than 6%. Of 380 randomly selected reservations, 19 were no-shows. Find the P-value for a test of the airline's claim.
A. 0.3508
B. 0.2061
C. 0.0746
D. 0.1230
P-value for the airline's claim that the no-show rate for passengers booked on its flights is less than 6% is 0.2061. Hence, option (B) is correct.
Given that, an airline claims that the no-show rate for passengers booked on its flights is less than 6%.
From this claim, it is clear that:
Null hypothesis : Proportion of no-show rate for passengers booked on its flights, P = 6%. i.e. H₀: P = 0.06
Alternative hypothesis: Proportion of no-show rate for passengers booked on its flights P < 6%, i.e. P < 0.06.
Out of 380 randomly selected reservations, 19 were no-shows
Sample proportion, [tex]\hat{p} = \frac{19}{380}[/tex] = 0.05
Then, the standard error for the sample size of 380 is:
[tex]\text{S.E.} = \sqrt{{\frac{p(1-p)}{n}[/tex]
[tex]\text{S.E.} = \sqrt{{\frac{0.06(1-0.06)}{380}[/tex]
[tex]\text{S.E.} = 0.012[/tex]
Now calculating the test statistic
[tex]z = \frac{( \hat{p} - P)}{S.E}[/tex]
[tex]z = \frac{(0.05 - 0.06)}{0.012 }[/tex]
z = -0.833
p value for z = -0.083 is 0.2061 (From the normal table).
Hence, the p-value for a test of the airline's claim is 0.2061. Option (B) is correct.
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Final answer:
To find the P-value for the hypothesis test, calculate the sample proportion and standard error, then use the standard normal distribution to calculate the z-score and find the corresponding P-value.
Explanation:
To find the P-value for the hypothesis test, we need to use the given data and perform calculations.
First, we need to calculate the sample proportion, which is the number of no-shows divided by the total number of reservations: 19/380 = 0.05.
Next, we calculate the standard error of the sample proportion using the formula: √((p' * (1 - p')) / n), where p' is the sample proportion and n is the sample size. In this case, the standard error is √((0.05 * (1 - 0.05)) / 380) = 0.014.
Finally, we use the standard normal distribution to calculate the z-score and find the corresponding P-value. In this case, the observed proportion is less than the claimed proportion, so we use a one-tailed test and calculate the z-score as (observed proportion - claimed proportion) / standard error = (0.05 - 0.06) / 0.014 = -0.714. Looking up the P-value for a z-score of -0.714, we find that it is approximately 0.4714.
Therefore, the P-value for the test of the airline's claim is approximately 0.4714.
Which of the following may be used to check the conditions needed to perform a two sample test for mean (independent samples)?
I. Both populations are approximately normally distributed
II. Both sample sizes greater than 30
III. Population of differences is approximately normally distributed
A) I or III
B) II or III
C) I or II
D) I, II, or III
Answer:
Option D is right
Step-by-step explanation:
given that a two sample test for mean of independent samples to be done.
We create hypotheses as:
[tex]H_0: \ bar x = \bar y[/tex] vs alternate suitably right or left or two tailed according to the needs.
The conditions needed for conducting this test would be
I. Both populations are approximately normally distributed
II. Both sample sizes greater than 30
III. Population of differences is approximately normally distributed
i.e. either i, ii or III
Option D is right.
A poll showed that 48 out of 120 randomly chosen graduates of California medical schools last year intended to specialize in family practice. What is the width of a 90 percent confidence interval for the proportion that plan to specialize in family practice? Select one:a. ± .0736b. ± .0447c. ± .0876d. ± .0894
Answer:
± 0.0736
Step-by-step explanation:
Data provided in the question:
randomly chosen graduates of California medical schools last year intended to specialize in family practice, p = [tex]\frac{48}{120}[/tex] = 0.4
Confidence level = 90%
sample size, n = 120
Now,
For 90% confidence level , z-value = 1.645
Width of the confidence interval = ± Margin of error
= ± [tex]z\times\sqrt\frac{p\times(1-p)}{n}[/tex]
= ± [tex]1.645\times\sqrt\frac{0.4\times(1-0.4)}{120}[/tex]
= ± 0.07356 ≈ ± 0.0736
Hence,
The correct answer is option ± 0.0736
The correct option is a. [tex]\±0.0736.[/tex] The width of a [tex]90\%[/tex] confidence interval for the proportion that plan to specialize in family practice
To find the width of a confidence interval for a proportion, we can use the formula:
[tex]\[ \text{Width} = Z \times \sqrt{\frac{p(1-p)}{n}} \][/tex]
where:
[tex]\( Z \)[/tex] is the Z-score corresponding to the desired confidence level,
[tex]\( p \)[/tex] is the sample proportion,
[tex]\( n \)[/tex] is the sample size.
Given:
[tex]Sample\ proportion \( p = \frac{48}{120} = \frac{2}{5} = 0.4 \)[/tex],
[tex]Sample\ size \( n = 120 \)[/tex]
[tex]Confidence \ level = 90\%[/tex], which corresponds to a Z-score of approximately [tex]1.645.[/tex]
Substitute the values into the formula:
[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.4 \times 0.6}{120}} \][/tex]
[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.24}{120}} \][/tex]
[tex]\[ \text{Width} = 1.645 \times \sqrt{0.002} \][/tex]
[tex]\[ \text{Width} = 1.645 \times 0.0447 \][/tex]
[tex]\[ \text{Width} = 0.0736 \][/tex]
Which function has (2,8) on its graph
Answer:
y = 2x^2
Step-by-step explanation:
The the coordinates (2,8) are expressed in terms of (x,y). Substitute the x with 2 and y with 8 and plug them into the formulas. The one that has the two sides equal to each other is the function that has (2,8) on its line.
A manager is interested in determining if the population standard deviation has dropped below 134. Based on a sample of n=27 items selected randomly from the population, conduct the appropriate hypothesis test at a 0.05 significance level. The sample standard deviation is 126. Determine the Null and alternative hypothese
Answer:
H₀: σ² ≥ 17956
H₁: σ² < 17956
Step-by-step explanation:
Hello!
You are asked to test if the population standard deviation of a certain population. Now keep in mind that wherever you want to make a hypothesis test for a population parameter, you have to have a known distribution that includes this parameter. Since the population standard deviation is no parameter of any distribution, what you have to do is test the population variance. Any decision you make about the population variance can be extrapolated to the population standard deviation.
To make a hypothesis test for the population variance, you need a variable with normal distribution and the statistic to use is a Chi-square statistic.
The hypothesis is that the population standard deviation is less than 134, symbolically: σ < 134
Translated in terms of the population variance: σ² < 17956
H₀: σ² ≥ 17956
H₁: σ² < 17956
α: 0.05
χ²= (n-1)S² ~χ²[tex]_{n-1}[/tex]
σ²
χ²= (27-1)(126)² = 22.988
17956
The test is one tailed (left)
χ²[tex]_{n-1;α}[/tex] =χ²[tex]_{26;0.05}[/tex] = 15.379
If the calculated Chi-square value is ≤ than the critical value, you reject the null hypothesis.
If the calculated Chi-square value is > than the critical value, you don't reject the null hypothesis.
Since the value is greater than the critical value, you do not reject the null hypothesis. So at a 5% level, there is not enough evidence to reject the null hypothesis, this means the population variance is at least 17956. On the same level, you can conclude that the population standard deviation is at least 134.
I've made the test so that you have an example of how to do it.
I hope it helps!
Complete parts (a) through (c) below.
(a) Determine the critical value(s) for a right-tailed test of a population mean at the alphaequals0.01 level of significance with 10 degrees of freedom.
(b) Determine the critical value(s) for a left-tailed test of a population mean at the alphaequals0.10 level of significance based on a sample size of nequals15.
(c) Determine the critical value(s) for a two-tailed test of a population mean at the alphaequals0.01 level of significance based on a sample size of nequals12.
Answer:
a) t = 2.7638
b) t = - 2.6245
c) t = 3.1058 on the right side and
t = -3.1058 on the left
Step-by-step explanation:
a)Determine critical value for a right-tail test for α = 0.01 level of significance and 10 degrees of fredom
From t-student table we find:
gl = 10 and α = 0.01 ⇒ t = 2.7638
b)Determine critical value for a left-tail test for α = 0.01 level of significance and sample size n = 15
From t-student table we find:
gl = 14 and α = 0.01 gl = n - 1 gl = 15 - 1 gl = 14
t = - 2.6245
c) Determine critical value for a two tails-test for α = 0.01 level of significance the α/2 = 0.005 and sample size n = 12
Then
gl = 11 and α = 0.005
t = 3.1058 on the right side of the curve and by symmetry
t = - 3.1058
From t-student table we find:
One method for measuring air pollution is to measure the concentration of carbon monoxide, or CO, in the air. Suppose Nina, an environmental scientist, wishes to estimate the CO concentration in Budapest, Hungary. She randomly selects 48 locations throughout the city measures the CO concentration at each location. Based on her 48 samples, she computes the margin of error for a 95% t-confidence interval for the mean concentration of CO in Budapest, in g/m3, to be 4.28 What would happen to the margin of error if Nina decreases the confidence level to 90%? Nina increases the confidence level to 99%? Nina decreases the sample size to 34 locations? Nina increases the sample size to 70 locations? Answer Bank Decrease Stay the sameIncrease
Answer:
a) Nina decreases the confidence level to 90%? (Decrease)
b) Nina decreases the sample size to 34 locations? (Increase)
c) Nina increases the sample size to 70 locations? (Decrease)
Step-by-step explanation:
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 means 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.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n=48 represent the original sample size
Confidence =95% or 0.95
ME=4.28 represent the margin of error.
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
And the margin of error is given by the following expression:
[tex]ME= t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (4)
Based on the formula (4) we can answer all the questions involved:
a) Nina decreases the confidence level to 90%?
On this case the value for [tex]t_{\alpha/2}[/tex] will also decrease so the margin of error would decrease.
b) Nina decreases the sample size to 34 locations?
If we analyze the original sample size of 48 we see that if we reduce the value of n to 34, the margin of error would increase, because n is on the denominator of the margin of error.
c) Nina increases the sample size to 70 locations?
If we analyze the original sample size of 48 we see that if we increase the value of n to 70, the margin of error would decrease, because n is on the denominator of the margin of error.
Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integral ∫C7y2xdx+8x2ydy.
Answer:
1/2
Step-by-step explanation:
The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that
[tex]\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy[/tex]
Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.
Mx(x,y) = d/dx 8x²y = 16xyLy(x,y) = d/dy 7y²x = 14xyThus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral
[tex] \int\limits^1_0\int\limits^1_0 {2xy} \, dxdy[/tex]
We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows
[tex]\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2[/tex]
We conclude that the line integral is 1/2
To evaluate the line integral using Green's Theorem, we need to find the curl of the vector field and the area enclosed by the square. The line integral of the vector field along the square is equal to the double integral of the curl over the region enclosed by the square. Using this method, we can find the value of the line integral to be 1/3.
Explanation:To evaluate the line integral using Green's Theorem, we first need to find the curl of the vector field. In this case, the vector field is F(x, y) = 7y^2x i + 8x^2y j. Taking the partial derivatives of its components with respect to x and y, we get curl(F) = (8x^2 - 14xy^2) k.
Next, we need to find the area enclosed by the square C, which is 1 unit^2. Using Green's Theorem, the line integral of F along C is equal to the double integral of curl(F) over the region D enclosed by C. Integrating curl(F) with respect to y, we get -7xy^2 + 6x^2y. Integrating this with respect to x over the given limits, we find the value of the line integral to be 1/3.
Learn more about Green's Theorem here:https://brainly.com/question/30763441
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An electrical engineer wishes to compare the mean lifetimes of two types of transistors in an application involving high-temperature performance. A sample of 60 transistors of type A were tested and were found to have a mean lifetime of 1827 hours and a standard deviation of 176 hours. A sample of 180 transistors of type B were tested and were found to have a mean lifetime of 1658 hours and a standard deviation of 246 hours. Let μX represent the population mean for transistors of type A and μY represent the population mean for transistors of type B. Find a 95% confidence interval for the difference μX−μY . Round the answers to three decimal places.
Answer:
add the decimals thats all
Step-by-step explanation:
In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In the Cozumel region about 48% of strikes (while trolling) resulted in a catch. Suppose that on a given day a fleet of fishing boats got a total of 20 strikes. Find the following probabilities. (Round your answers to four decimal places.) (a)-12 or fewer fish were caught (b)-5 or more fish were caught (c)-between 5 and 12 fish were caught
Answer:
a) [tex]P(X\leq 12)=0.8586[/tex]
b) [tex]P(X\geq 5)=0.9802[/tex]
c) [tex]P(5\leq X\leq 12)=0.8389[/tex]
Step-by-step explanation:
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=20, p=0.48)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
We need to check the conditions in order to use the normal approximation.
[tex]np=20*0.48=9.6 \approx 10 \geq 10[/tex]
[tex]n(1-p)=20*(1-0.48)=10.4 \geq 10[/tex]
So we see that we satisfy the conditions and then we can apply the approximation.
If we appply the approximation the new mean and standard deviation are:
[tex]E(X)=np=20*0.48=9.6[/tex]
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{20*0.48(1-0.48)}=2.234[/tex]
Part a
We want this probability:
[tex]P(X\leq 12)[/tex]
We can use the z score given by this formula [tex]Z=\frac{x-\mu}{\sigma}[/tex].
[tex]P(X\leq 12)=P(\frac{X-\mu}{\sigma}\leq \frac{12-9.6}{2.234})=P(Z\leq 1.074)=0.8586[/tex]
Part b
We want this probability:
[tex]P(X\geq 5)[/tex]
We can use again the z score formula and we have:
[tex]P(X\geq 5)=1-P(X<5)=1-P(\frac{X-\mu}{\sigma}< \frac{5-9.6}{2.234})=1-P(Z<- 2.059)=0.9802[/tex]
Part c
We want this probability:
[tex]P(5\leq X\leq 12)=P(\frac{5-9.6}{2.234}\leq \frac{X-\mu}{\sigma}\leq \frac{12-9.6}{2.234})=P(-2.059\leq Z \leq 1.074)[/tex]
[tex]=P(Z<1.074)-P(Z<-2.059)=0.8586-0.0197=0.8389[/tex]
A researcher selects a sample from a population with μ = 30 and uses the sample to evaluate the effect of a treatment. After treatment, the sample has a mean of M = 32 and a variance of s2 = 6. Which of the following would definitely increase the likelihood of rejecting the null hypothesis?
Question options:
a.
Decrease the sample variance
b.
Increase the sample mean
c.
Increase the sample size
d.
All of the other options will increase the likelihood of rejecting the null hypothesis
Answer:
Option b) Increase the sample mean
Step-by-step explanation:
Given that a researcher selects a sample from a population with μ = 30 and uses the sample to evaluate the effect of a treatment. After treatment, the sample has a mean of M = 32 and a variance of s2 = 6.
This is a paired test with test statistic
=mean diff/std error
Mean difference would increase if sample mean increases.
This would increase the test statistic
Or otherwise decrease in variance will increase the test statistic
Or Increase in sample size would also increase test statistic
Of all these the II option is definite in increasing the likelihood of rejecting the null hypothesis because this would definitely increase the chances of rejecting H0.
Others may also have effect but not as much direct as sample mean difference.
Because variance and sample size have influence only upto square root of the difference.
Compute the lower Riemann sum for the given function f(x)=x2 over the interval x∈[−1,1] with respect to the partition P=[−1,− 1 2 , 1 2 , 3 4 ,1].
Answer:
21/64
Step-by-step explanation:
First, we need to note that the function f(x) = x² is increasing on (0, +∞), and it is decreasing on (-∞,0)
The first interval generated by the partition is [-1, -1/2], since f is decreasing for negative values, we have that f takes its minimum values at the right extreme of the interval, hence -1/2.
The second interval is [-1/2, 1/2]. Here f takes its minimum value at 0, because f(0) = 0, and f is positive otherwise.
Since f is increasing for positive values of x, then, on the remaining 2 intervals, f takes its minimum value at their respective left extremes, in other words, 1/2 and 3/4 respectively.
We obtain the lower Riemman sum by multiplying this values evaluated in f by the lenght of their respective intervals and summing the results, thus
LP(f) = f(-1/2) * ((-1/2) - (-1)) + f(0) * (1/2 - (-1/2)) + f(1/2)* (3/4 - 1/2) + f(3/4) * (1- 3/4)
= 1/4 * 1/2 + 0 * 1 + 1/4 * 1/4 + 9/16 * 1/4 = 1/8 + 0 + 1/16 + 9/64 = 21/64
As a result, the lower Riemann sum on the partition P is 21/64
Final answer:
To compute the lower Riemann sum for the given function f(x) = x² over the interval x ∈ [-1,1] with respect to the partition P = [-1, -1/2, 1/2, 3/4 ,1], we need to find the height of each subinterval and multiply it by the width of the subinterval. The lower Riemann sum is 5/8.
Explanation:
To compute the lower Riemann sum for the given function f(x) = x² over the interval x ∈ [-1,1] with respect to the partition P = [-1, -1/2, 1/2, 3/4 ,1], we need to find the height of each subinterval and multiply it by the width of the subinterval.
First, let's calculate the width of each subinterval:
width of subinterval 1: (-1) - (-1) = 0
width of subinterval 2: (-1/2) - (-1) = 1/2
width of subinterval 3: (1/2) - (-1/2) = 1
width of subinterval 4: (3/4) - (1/2) = 1/4
width of subinterval 5: (1) - (3/4) = 1/4
Next, let's calculate the height of each subinterval by substituting the left endpoint of each subinterval into the function:f((-1/2))^2 = 1/4, f(1/2)² = 1/4, f(3/4)² = 9/16, f(1)² = 1
Finally, we compute the lower Riemann sum by multiplying the height by the width for each subinterval and summing them up:
(0 * 0) + (1/2 * 1/4) + (1 * 9/16) + (1/4 * 1) = 5/8.
help please
1 though 5
Answer:
5/8,-5√2,{(2x+3)(x-5)}
Step-by-step explanation:
1) 5x-10/8x-16
=5(x-2)/8(x-2)
=5/8
2) √32-3√18
=4√2-3√18
=4√2-9√2
=(4-9)√2
= -5√2
4) 2x^2-7x-15
=2x^2+3x-10x-15
=(2x+3),(x-5)
A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal. (a) Determine the sample mean in cents (Round to 3 decimal places)
Answer:
53¢
Step-by-step explanation:
First, I'll put these in order.
20¢;30¢; 30¢;75¢;40¢;40¢;40¢;40¢;50¢;55¢55¢65¢;65¢; $1.50;
Then, I'll combine like terms.
30+30=60
40+40+40+40(or 40 x 4)=160
55+55=110
65+65=130
60+160+110+130+20+75+50+$1.50=$7.55/14=53¢
PLZ correct me if i'm wrong :-D
A queue with a single server receives 50 requests per second on average. The average time for the server to address a request is 10 milliseconds. What is the probability that there are exactly k requests in this system?
Answer:
[tex]P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}[/tex]
Step-by-step explanation:
This is a typical example where the Poisson distribution is a good choice to model the situation.
In this case we have an interval of time of 50 milliseconds as average time for the server to address one request and 50 requests per second.
By cross-multiplying we determine the expected value of requests every 50 milliseconds.
We know 1 second = 1,000 milliseconds
50 requests __________ 1000 milliseconds
x requests __________ 50 milliseconds
50/x = 1000/50 ===> x = 2.5
and the expected value is 2.5 requests per interval of 50 milliseconds.
According to the Poisson distribution, the probability of k events in 50 milliseconds equals
[tex]\bf P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}[/tex]
A farmer wants to fence a rectangular garden next to his house which forms the northern boundary. The fencingfor the southern boundary costs $6 per foot, and the fencing for the east and west sides costs $3 per foot. If hehas a budget of $120 for the project, what are the dimensions of the largest area the fence can enclose?
Answer:
10 ft x 10 ft
Area = 100 ft^2
Step-by-step explanation:
Let 'S' be the length of the southern boundary fence and 'W' the length of the eastern and western sides of the fence.
The total area is given by:
[tex]A=S*W[/tex]
The cost function is given by:
[tex]\$ 120 = \$3*2W+\$6*S\\20 = W+S\\W = 20-S[/tex]
Replacing that relationship into the Area function and finding its derivate, we can find the value of 'S' for which the area is maximized when the derivate equals zero:
[tex]A=S*(20-S)\\A=20S-S^2\\\frac{dA}{dS} = \frac{d(20S-S^2)}{dS}\\0= 20-2S\\S=10[/tex]
If S=10 then W =20 -10 = 10
Therefore, the largest area enclosed by the fence is:
[tex]A=S*W\\A=10*10 = 100\ ft^2[/tex]