Answer:
Option c. 17%
Step-by-step explanation:
Data provided in the question:
Amount paid on food last year = 12% of Annual after tax income
Amount paid on food for the current year = 10% of Annual before tax income
Now,
Let the after tax income be 'x'
and tax be 'y'
Therefore,
Income before tax = x + y
Amount paid on food = 12% of x
According to the question
12% of x = 10% of (x + y)
or
0.12x = 0.10 (x + y)
or
1.2x - x = y
0.2x = y
or
x = 5y
Thus,
percent of the family's annual before-tax income that was paid for tares last year
= [Tax ÷ Income before tax] × 100%
= [ y ÷ ( x + y )] × 100%
= [ y ÷ ( 5y + y )] × 100%
or
= 0.167 × 100% ≈ 17%
Hence,
Option c. 17%
Consider the number of loudspeaker announcements per day at school. Suppse thee snce of chance ofhaving 0 announcements, a 30% chance ofhaving i announcement, a 25% having 2 announcements, a 20% chance of having 3 announcements, and a \0 % chance announcements. Find the expected value of the number of announcements per day. of having A
Answer:
The expected value is 1.8
Step-by-step explanation:
Consider the provided information.
Suppose there’s a 15% chance of having 0 announcements, a 30% chance of having 1 announcement, a 25% chance of having 2 announcements, a 20% chance of having 3 announcements, and a 10% chance of having 4 announcements.
[tex]\text{Expected Value}=a \cdot P(a) + b \cdot P(b) + c \cdot P(c) + \cdot\cdot[/tex]
Where a is the announcements and P(a) is the probability.
[tex]\text{Expected Value}=0\cdot 15\% + 1 \cdot 30\% + 2 \cdot 25\% + 3\cdot20\%+4\cdot10[/tex]
[tex]\text{Expected Value}=1 \cdot 0.30+2 \cdot 0.25 +3 \cdot 0.2 + 4\cdot 0.10[/tex]
[tex]\text{Expected Value}=1.8[/tex]
Hence, the expected value is 1.8
A company wants to determine where they should locate a new warehouse. They have two existing production plants (i.e., Plant A and Plant B) that will ship units of a product to this warehouse. Plant A is located at the (X, Y) coordinates of (50, 100) and will have volume of shipping of 250 units a day. Plant B is located at the (X, Y) coordinates of (150, 200) and will have a volume of shipping of 150 units a day. Using the centroid method, which of the following are the X and Y coordinates for the new plant location?
Answer:
X = 87.5
Y = 137.5
Step-by-step explanation:
Let's X and Y be the xy-coordinates of the center warehouse.
We know that X is in between the x coordinates or the 2 plants:
50 < X < 150
Similarly Y is in between the y coordinates or the 2 plants:
100 < Y < 200
Using centroid method with the shipping units being weight we can have the following equations
250*50 + 150*150 = X*(250 + 150)
Hence X = (250*50 + 150*150)/(250+150) = 87.5
Similarly 250*100 + 150*200 = Y*(250 + 150)
Hence Y = (250*100 + 150*200)/(250+150) = 137.5
Over the past semester, you've collected the following data on the time it takes you to get to school by bus and by car:
• Bus:(15,10,7,13,14,9,8,12,15,10,13,13,8,10,12,11,14,11,9,12) • Car:(5,8,7,6,9,12,11,10,9,6,8,10,13,12,9,11,10,7)
You want to know if there's a difference in the time it takes you to get to school by bus and by car.
A. What test would you use to look for a difference in the two data sets, and what are the conditions for this test? Do the data meet these conditions? Use sketches of modified box-and-whisker plots to support your decision.
B. What are the degrees of freedom (k) for this test using the conservative method? (Hint: Don't pool and don't use your calculator.)
C. What are the sample statistics for this test? Consider the data you collected for bus times to be sample one and the data for car times to be sample two.D. Compute a 99% confidence interval for the difference between the time it takes you to get to school on the bus and the time it takes to go by car. Draw a conclusion about this difference based on this confidence interval using
E. Constructthesameconfidenceintervalyoudidinpartd,thistimeusingyour graphing calculator. Show what you do on your calculator, and what you put into your calculator, and give the confidence interval and degrees of freedom. (Hint: Go back to previous study materials for this unit if you need to review how to do this.)
F. How is the interval computed on a calculator different from the interval computed by hand? Why is it different? In this case, would you come to a different conclusion for the hypothesis confidence interval generated by the calculator?
Answer:
Step-by-step explanation:
Hello!
You have two study variables
X₁: Time it takes to get to school by bus.
X₂: Time it takes to get to school by car.
Data:
Sample 1
Bus:(15,10,7,13,14,9,8,12,15,10,13,13,8,10,12,11,14,11,9,12)
n₁= 20
Mean X[bar]₁= 11.30
S₁= 2.39
Sample 2
Car:(5,8,7,6,9,12,11,10,9,6,8,10,13,12,9,11,10,7)
n₂= 18
Mean X[bar]₂= 9.06
S₂= 2.29
A.
To test if there is any difference between the times it takes to get to school using the bus or a car you need to compare the means of each population.
The condition needed to make a test for the difference between means is that both the independent population should have a normal distribution.
The sample sizes are too small to use an approximation with the CLT. You can test if the study variables have a normal distribution using different methods, and hypothesis test, using a QQ-plot or using the Box and Whiskers plot. The graphics are attached.
As you can see both samples show symmetric distribution, the boxes are proportioned, the second quantile (median) and the mean (black square) are similar and in the center of the boxes. The whiskers have the same length and there are no outliers. Both plots show symmetry centered in the mean consistent with a normal distribution. According to the plots you can assume both variables have a normal distribution.
The next step to select the statistic to test the population means is to check whether there is other population information available.
If the population variances are known, you can use the standard normal distribution.
If the population variances are unknown, the distribution to use is a Student's test.
If the unknown population variances are equal, you can use a t-test with a pooled sample variance.
If the unknown population variances are not equal, the t-test to use is the Welch approximation.
Using an F-test for variance homogeneity the p-value is 0.43 so at a 0.01 level, you can conclude that the population variances are equal.
The statistic to use is a pooled t-test.
B.
Degrees of freedom.
For each study variable, you can use a t-test with n-1 degrees of freedom.
For X₁ ⇒ n₁-1 = 20 - 1 = 19
For X₂ ⇒ n₂-1 = 18 = 17
For X₁ + X₂ ⇒ (n₁-1) + (n₂-1)= n₁ + n₂ - 2= 20 + 18 - 2= 36
C.
See above.
D.
The formula for the 99% confidence interval is:
(X[bar]₁ - X[bar]₂) ± [tex]t_{n_1+n_2-2; 1- \alpha /2}[/tex] * [tex]Sa\sqrt{\frac{1}{n_1} + \frac{1}{n_2} }[/tex]
[tex]Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} }[/tex]
[tex]Sa= \sqrt{\frac{19*(2.39)^2+17*(2.29)^2}{36} }[/tex]
Sa= 2.34
[tex]t_{n_1+n_2-2; 1- \alpha /2}[/tex]
[tex]t_{36; 0.995}[/tex] = 2.72
(11.30 - 9.06) ± 2.72 * [tex]2.34\sqrt{\frac{1}{20} + \frac{1}{18} }[/tex]
[0.17;4.31]
With a 99% confidence level you'd expect that the difference between the population means of the time that takes to get to school by bus and car is contained in the interval [0.17;4.31].
E.
Couldn't find the original lesson to see what calculator is used.
F.
Same, no calculator available.
I hope it helps!
Answer:
this is nnot the answer i was looking for
Step-by-step explanation:
A bag contains 8 red marbles, 3 blue marbles and 6 green marbles. If three marbles are drawn out of the bag, what is the probability, to the nearest 1000th, that all three marbles drawn will be red?
Answer:
0.082
Step-by-step explanation:
There are a total of 17 marbles, 8 of which are red.
The probability that the first marble is red is 8/17.
The probability that the second marble is red is 7/16.
The probability that the third marble is red is 6/15.
Therefore, the probability that all three marbles are red is:
P = 8/17 × 7/16 × 6/15
P = 7/85
P = 0.082
The lifetime of a cheap light bulb is an exponential random variable with mean 36 hours. Suppose that 16 light bulbs are tested and their lifetimes measured. Use the central limit theorem to estimate the probability that the sum of the lifetimes is less than 600 hours.
Answer:
[tex] P(T<600)=P(Z< \frac{600-576}{144})=P(Z<0.167)=0.566[/tex]
Step-by-step explanation:
Previous concepts
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:
[tex]P(X=x)=\lambda e^{-\lambda x}, x>0[/tex]
And 0 for other case. Let X the random variable that represent "The number of years a radio functions" and we know that the distribution is given by:
[tex]X \sim Exp(\lambda=\frac{1}{16})[/tex]
Or equivalently:
[tex]X \sim Exp(\mu=16)[/tex]
Solution to the problem
For this case we are interested in the total T, and we can find the mean and deviation for this like this:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}=\frac{T}{n}[/tex]
If we solve for T we got:
[tex] T= n\bar X[/tex]
And the expected value is given by:
[tex] E(T) = n E(\bar X)= n \mu= 16*36=576[/tex]
And we can find the variance like this:
[tex] Var(T) = Var(n\bar X)=n^2 Var(\bar X)= n^2 *\frac{\sigma^2}{n}=n \sigma^2[/tex]
And then the deviation is given by:
[tex]Sd(T)= \sqrt{n} \sigma=\sqrt{16} *36=144[/tex]
And the distribution for the total is:
[tex] T\sim N(n\mu, \sqrt{n}\sigma)[/tex]
And we want to find this probability:
[tex] P(T< 600)[/tex]
And we can use the z score formula given by:
[tex]z=\frac{T- \mu_T}{\sigma_T}[/tex]
And replacing we got this:
[tex] P(T<600)=P(Z< \frac{600-576}{144})=P(Z<0.167)=0.566[/tex]
Using the central limit theorem, the probability that the sum of the lifetimes of 16 light bulbs is less than 600 hours is found to be approximately 0.2514 after calculating the mean, standard deviation, and z-score for the sum.
Explanation:To estimate the probability that the sum of the lifetimes of 16 light bulbs is less than 600 hours, we can use the central limit theorem. This theorem suggests that the sum (or average) of a large number of independent and identically distributed random variables will be approximately normally distributed, regardless of the original distribution of the variables. Here, each light bulb's lifetime is an exponential random variable with a mean of 36 hours.
First, we need to determine the mean (μ) and standard deviation (σ) of the sum of the lifetimes. For one light bulb, the mean is 36 hours, and since the standard deviation for an exponential distribution is equal to its mean, it is also 36 hours. For 16 light bulbs, the mean of the sum is 16 * 36 = 576 hours, and the standard deviation of the sum is √16 * 36 = 144 hours due to the square root rule for variances of independent sums.
To find the probability that the sum is less than 600 hours, we convert this to a standard normal distribution problem by calculating the z-score:
Z = (X - μ) / (σ/sqrt(n))
Z = (600 - 576) / (144/sqrt(16))
Z = 24 / 36
Z = 0.67
Now we look up the cumulative probability for a z-score of 0.67 using a standard normal distribution table or a calculator with normal distribution functions. The probability associated with a z-score of 0.67 is approximately 0.7486. Therefore, the probability that the sum of the lifetimes is less than 600 hours is 1 - 0.7486 = 0.2514.
Before lending someone money, banks must decide whether they believe the applicant will repay the loan. One strategy used is a point system. Loan officers assess information about the applicant, totalling points they award for the persons income level, credit history, current debt burden, and so on. The higher the point total, the more convinced the bank is that it’s safe to make the loan. Any applicant with a lower point total than a certain cut-off score is denied a loan. We can think of this decision as a hypothesis test. Since the bank makes its profit from the interest collected on repaid loans, their null hypothesis is that the applicant will repay the loan and therefore should get the money. Only if the persons score falls below the minimum cut-off will the bank reject the null and deny the loan. This system is reasonably reliable, but, of course, sometimes there are mistakes.a) When a person defaults on a loan, which type of error did the bank make?b) Which kind of error is it when the bank misses an opportunity to make a loan to someone who would have repaid it?c) Suppose the bank decides to lower the cut-off score from 250 points to 200. Is that analogous to choosing a higher or lower value of for a hypothesis test? Explain.d) What impact does this change in the cut-off value have on the chance of each type of error?
Answer:
(a) Type II error
(b) Type I error
(c) It is analogous to choosing a lower value for a hypothesis test
(d) There will be more tendency of making type II error and less tendency of making type I error
Step-by-step explanation:
(a) The bank made a type II error because they accepted the null hypothesis when it is false
(b) The bank made a type I error because they rejected the null hypothesis when it is true
(c) By lowering the value for the hypothesis test, they give applicants who do not meet the initial cut-off point the benefit of doubt of repaying the loan thus increasing their chances of making more profit
(d) There will be more tendency of making type II error because the bank accepts the null hypothesis though they are not fully convinced the applicants will repay the loan and less tendency of making type I error because the bank rejects the null hypothesis knowing the applicants might not be able to repay the loan
In hypothesis testing, a person defaulting on a loan represents a Type I error, while missing an opportunity to make a loan to someone who would have repaid it represents a Type II error. Lowering the cut-off score is analogous to increasing the value in a hypothesis test, accepting more risk. This increases the likelihood of Type I errors but decreases the likelihood of Type II errors.
Explanation:In the context of hypothesis testing in banking and the financial capital market, (a) when a person defaults on a loan, the bank made a Type I error: they lent money to an individual who failed to repay it. (b) If the bank does not lend money to someone who would have repaid it, it's a Type II error: they missed an opportunity to profit from interest because they incorrectly predicted the person would not pay back the loan. (c) Lowering the cut-off score from 250 points to 200 is analogous to choosing a higher value for a hypothesis test, which means the bank is willing to accept more risk. (d) Changing the cut-off value impacts the chance of each kind of error. By lowering the score, the bank is more likely to make Type I errors (lending to individuals who won't repay), but less likely to make Type II errors (not lending to individuals who would repay).
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An article reported that for a sample of 58 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 165.4.
(a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.) , ppm Interpret the resulting interval. We are 95% confident that the true population mean lies below this interval. We are 95% confident that this interval does not contain the true population mean. We are 95% confident that this interval contains the true population mean. We are 95% confident that the true population mean lies above this interval.
(b) Suppose the investigators had made a rough guess of 184 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 47 ppm for a confidence level of 95%?
Answer:
Step-by-step explanation:
Identify the type of observational study (cross-sectional, retrospective, or prospective) described below. A research company uses a device to record the viewing habits of about 2500 households, and the data collected over the past 2 years will be used to determine whether the proportion of households tuned to a particular children's program increased. Which type of observational study is described in the problem statement?
A. A prospective study
B. A retrospective study
C. A cross-sectional study
Answer:
B
Step-by-step explanation:
The retrospective or historic cohort story, is a longitudinal cohort story that considers a particular set of individuals that share the same exposure factor to ascertain its influence in the developments of an occurrence, which are compared with the other set or cohort which were not exposed to the same factors.
Retrospective studies have existed about the same time as prospective studies, hence their names.
Tyler has a baseball bat that weighs 28 ounces. Find this weight in kilograms and in grams. (Note 1 kilogram=35 ounces)
Answer:0.8 kilograms
800 grams
Step-by-step explanation:
The weight of Tyler's baseball bat is 28 ounces. We would convert the weight in ounces to kilogram and grams.
Let x represent the number of kilograms that is equal to 28 ounces. Therefore
1 kilogram = 35 ounces
x kilogram = 28 ounces
Cross multiplying, it becomes
35 × x = 28 × 1
35x = 28
x = 28/35 = 0.8 kilograms
We would convert 0.8 kilograms to grams
Let y represent the number of grams that is equal to 0.8 kilograms. Therefore,
1000 grams = 1 kilogram
y grams = 0.8 kilograms
Cross multiplying,
y × 1 = 0.8 × 1000
y = 800 grams
Answer:
0.2
Step-by-step explanation:
Identify the sampling technique used. In a recent television survey, participants were asked to answer "yes" or "no" to the question "Are you in favor of the death penalty?" Six thousand five hundred responded "yes" while 51 00 responded "no". There was a fifty- cent charge for the call.
Answer:
Convenience sampling. See explanation below.
Step-by-step explanation:
For this case they not use random sampling since all the individuals for the population are not included on the sampling frame, some individuals have probability of inclusion 0, because they are just using a charge for the call and some people would not answer the call.
Is not stratified sampling since we don't have strata clearly defined on this case, and other important thing is that in order to apply this method we need homogeneous strata groups and that's not satisfied on this case.
Is not systematic sampling since they not use a random number or a random starting point, and is not mentioned, they just use a call that is charge with 50 cents.
Is not cluster sampling since we don't have clusters clearly defined, and again in order to apply this method we need homogeneous characteristics on the clusters and an equal chance of being a part of the sample, and that's not satisfid again with the call charge used.
So then the only method that can be possible for this case is convenience sampling because they use a non probability sampling with some members of the potential population with probability of inclusion 0.
The sampling technique used in the given scenario is voluntary response sampling, where participants decide whether to take part in the survey. In this technique, participants chose to respond to the television survey by making a call. This method can be biased as the responses could lean towards those who hold strong views on the topic.
Explanation:The sampling technique used in this scenario is referred as voluntary response sampling or self-selection sampling. In this method, participants themselves decide to participate or not, usually by responding to a call for participants. This often happens when surveys are disseminated widely such as through television or online. Since there was a call to answer "yes" or "no" for the question with a charge, individuals chose to participate by making a call. It is important to note that the main drawback of this technique is that it tends to be biased, as the sample could be skewed in favor of those who felt strongly about the topic.
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You perform a X2 goodness-of-fit test to see if the number of birthdays occurring each month matches the expected number (assuming each month is equally likely to be the birth month for any given individual). You get 20.5 as your X2 value. What is the P-value for this test?
Answer:
[tex]p_v = P(\chi^2_{11} >20.5)=0.0389[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(20.5,11,TRUE)"
Step-by-step explanation:
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
We need to conduct a chi square test in order to check the following hypothesis:
H0: Each month is equally likely to be the birth month for any given individual
H1: Each month is NOT equally likely to be the birth month for any given individual
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
After calculate the statistic we got [tex]\chi^2 = 20.5[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=categories-1=12-1=11[/tex]
And we have categories =12 since we have 12 months in a year
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{11} >20.5)=0.0389[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(20.5,11,TRUE)"
A researcher developing scanners to search for hidden weapons at airports has concluded that a new scanner is significantly better than the current scanner. He made his decision based on a test using alpha equals 0.025 . Would he have made the same decision at alpha equals 0.10 question mark How about alpha equals 0.005 question mark Explain.
Step-by-step explanation:
Since the decision is made on the test based on the use of alpha equals 0.025, the p-value of the test would have been higher than the level of significance provided that is 0.025 since the test is not important.
p > 0.025
Now if we know that p > 0.025, this would not necessarily mean that p > 0.1 also, therefore we do not know with the given information that he would have made the same decision for 0.1 level of significance, ( we are not sure about his decision in that case ).
Now for the level of significance of 0.005, we would be sure that p > 0.005 as it is greater than 0.025, therefore the test is not significant at this level of significance as well. Therefore he would have made the same decision for 0.005 level of significance.
A box contains the following numbered tickets: 1,1,5,9,9
a) If I draw two tickets with replacement, what is the chance that the sum of the two tickets is greater than or equal to 10?
b) Drawing three tickets without replacement, what is the chance the first two tickets are not 5's, and the last ticket is a 5?
c) Calculate b) if the draws are made with replacement.
d) If I repeat the procedure in a) 8 times (ie draw 2 tickets and find their sum, and do this 8 times), what is the chance that I get a sum greater than or equal to 10 exactly 6 of the 8 times?
Answer:
Step-by-step explanation:
Feel free to ask if anything is unclear
If SSXY = −16.32 and SSX = 40.00 for a set of data points, then what is the value of the slope for the best-fitting linear equation? a. −0.41 b. −2.45 c. positive d. There is not enough information; you would also need to know the value of SSY.
Answer: a. −0.41
Step-by-step explanation:
The slope for the best-fitting linear equation is given by :-
[tex]b=\dfrac{SS_{xy}}{SS_x}[/tex]
where , [tex]SS_x[/tex] =sum of squared deviations from the mean of X.
[tex]SS_{xy}[/tex] = correlation between y and x in terms of the corrected sum of products.
As per given , we have
[tex]SS_x=10.00[/tex]
[tex]SS_{xy}=-16.32[/tex]
Then, the value of the slope for the best-fitting linear equation will be
[tex]b=\dfrac{-16.32}{40.00}=-0.408\approx -0.41[/tex]
Hence, the value of the slope for the best-fitting linear equation= -0.41
So the correct answer is a. −0.41 .
The value of the slope for the best-fitting linear equation is -0.41
The given parameters are:
[tex]SS_{xy} = -16.32[/tex] --- the correlation between y and x
[tex]SS_{x} = 40.00[/tex] --- the sum of squared deviations from the mean of X.
The slope (b) is calculated using the following formula
[tex]b = \frac{SS_{xy}}{SS_x}[/tex]
Substitute values for SSxy and SSx
[tex]b = \frac{-16.32}{40.00}[/tex]
Divide -16.32 by 40.00
[tex]b = -0.408[/tex]
Approximate
[tex]b = -0.41[/tex]
Hence, the value of the slope for the best-fitting linear equation is -0.41
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If n is a positive integer, which of following statement is individually sufficient to prove whether 289 is a factor of n?a. The greatest common divisor of n and 344 is 86. b. Least common multiple of n and 272 is 4624. c. The least common multiple of n and 289 is 289n.
Answer:
The statement b) is individually sufficient to prove than 289 is a factor of n
Step-by-step explanation:
The least common multiple of n and 272 is the smallest number that is a multiple of n and a multiple of 272. Therefore:
272 x X = 4624 ⇒ X = 17 but 272 = 17 · 16 and 289 = 17 · 17
Therefore 17·17 must be a factor of n. That means 289 is a factor of n
For the Data Set below, calculate the Variance to the nearest hundredth decimal place. (Do not use a coma in your answer) 175 349 234 512 638 549 500 611
Answer:
The variance of the data is 29966.3.
Step-by-step explanation:
The given data set is
175, 349, 234, 512, 638, 549, 500, 611
We need to find the variance to the nearest hundredth decimal place.
Mean of the data
[tex]Mean=\dfrac{\sum x}{n}[/tex]
where, n is number of observation.
[tex]Mean=\dfrac{3568}{8}=446[/tex]
The mean of the data is 446.
[tex]Variance=\dfrac{\sum (x-mean)^2}{n-1}[/tex]
[tex]Variance=\dfrac{(175-446)^2+(349-446)^2+(234-446)^2+(512-446)^2+(638-446)^2+(549-446)^2+(500-446)^2+(611-446)^2}{8-1}[/tex]
[tex]Variance=\dfrac{209764}{7}[/tex]
[tex]Variance=29966.2857[/tex]
[tex]Variance\approx 29966.3[/tex]
Therefore, the variance of the data is 29966.3.
Final answer:
The variance of the given data set is calculated by finding the mean, squaring the differences from the mean, summing these squares, and dividing by the count minus one. It results in a variance of 12790.54 when rounded to the nearest hundredth decimal place.
Explanation:
To calculate the variance of the data set, follow these steps:
First, find the mean (average) of the data set by adding all the numbers together and dividing by the total count.
Next, subtract the mean from each data point and square the result to get the squared differences.
Then, add up all of the squared differences.
Finally, divide the sum of the squared differences by the total number of data points minus one to get the variance (since this is a sample variance).
Data Set: 175, 349, 234, 512, 638, 549, 500, 611
Mean = (175 + 349 + 234 + 512 + 638 + 549 + 500 + 611) / 8 = 3793 / 8 = 474.125
Squared differences = (175 - 474.125)^2 + (349 - 474.125)^2 + (234 - 474.125)^2 + (512 - 474.125)^2 + (638 - 474.125)^2 + (549 - 474.125)^2 + (500 - 474.125)^2 + (611 - 474.125)^2
Sum of squared differences = 89533.78125
Variance = 89533.78125 / (8 - 1) = 12790.54
Therefore, the variance of the data set, to the nearest hundredth decimal place, is 12790.54.
Which of the following is used to determine the significance of predictions made by a best fitting linear equation?A. correlational analysisB. analysis of varianceC. analysis of regressionD. method of least squares
Answer:
D. method of least squares
Step-by-step explanation:
The Least Squares Method (LSM) is a mathematical method used to solve various problems, based on minimizing the sum of the squared deviations of some functions from the desired variables. It can be used to “solve” over-determined systems of equations (when the number of equations exceeds the number of unknowns), to find a solution in the case of ordinary (not redefined) linear or nonlinear systems of equations, to approximate the point values of a function. OLS is one of the basic regression analysis methods for estimating the unknown parameters of regression models from sample data.
Correlation analysis is a statistical method used to assess the strength of the relationship between two quantitative variables. A high correlation means that two or more variables have a strong relationship with each other, while a weak correlation means that the variables are hardly related. In other words, it is a process of studying the strength of this relationship with available statistics.
Analysis of Variance (or ANOVA) is a collection of statistical models used to analyze group averages and related processes (such as intra- and inter-group variation) in statistical science. When using Variance Analysis, the observed variance of a specified variable is divided into the variance component that can be based on different sources of change. In its simplest form, "Analysis of Variance" is a inferential statistical test to test whether the averages of several groups are equal or not, and this test generalizes the t-test test for two-groups to multiple-groups. If multiple two-sample-t-tests are desired for multivariate analysis, it is clear that this results in increased probability of type I error. Therefore, the variance analysis would be more useful to compare the statistical significance of three or more means (for groups or for variables) with the test.
Regression analysis is an analysis method used to measure the relationship between two or more variables. If analysis is performed using a single variable, it is called univariate regression, and if more than one variable is used, it is called multivariate regression analysis. With the regression analysis, the existence of the relationship between the variables, if there is a relationship between the strength of the information can be obtained. The logic here is that the variable to the left of the equation is affected by the variables to the right. The variables on the right are not affected by other variables. Not being influenced here means that when we put these variables into a linear equation in mathematical sense, it has an effect. Multiple linearity, sequential dependency problems are not meant.
A particle moves according to the law of motion s(t) = t^{3}-8t^{2}+2t, where t is measured in seconds and s in feet.
(a) Find the velocity at time t.
(b) What is the velocity after 3 seconds?
(c) When is the particle at rest?
Answer:
a) [tex]v(t) = 3t^{2} - 16t + 2[/tex]
b) The velocity after 3 seconds is -3m/s.
c) [tex]t = 0.13s[/tex] and [tex]t = 5.2s[/tex].
Step-by-step explanation:
The position is given by the following equation.
[tex]s(t) = t^{3} - 8t^{2} + 2t[/tex]
(a) Find the velocity at time t.
The velocity is the derivative of position. So:
[tex]v(t) = s^{\prime}(t) = 3t^{2} - 16t + 2[/tex].
(b) What is the velocity after 3 seconds?
This is v(3).
[tex]v(t) = 3t^{2} - 16t + 2[/tex]
[tex]v(3) = 3*(3)^{2} - 16*(3) + 2 = -19[/tex]
The velocity after 3 seconds is -3m/s.
(c) When is the particle at rest?
This is when [tex]v(t) = 0[/tex].
So:
[tex]v(t) = 3t^{2} - 16t + 2[/tex]
[tex]3t^{2} - 16t + 2 = 0[/tex]
This is when [tex]t = 0.13s[/tex] and [tex]t = 5.2s[/tex].
A linear enzyme is formed by four alpha and two beta protein subunits. How manydifferent arrangements are there?
Answer:
15
Step-by-step explanation:
We are given that
Number of alpha protein subunits=4
Number of beta protein subunits=2
Total number of protein sub-units=2+4=6
We have to find the number of different arrangements are there.
When r identical letters and y identical letters and total object are n then arrangements are
[tex]\frac{n!}{r!x!}[/tex]
n=6,r=2,x=4
By using the formula
Then, we get
Number of different arrangements =[tex]\frac{6!}{2!4!}[/tex]
Number of different arrangements=[tex]\frac{6\times 5\times 4!}{2\times 1\times 4!}[/tex]
Number of different arrangements=15
Hence, different arrangements are there= 15
Evaluate the integral Integral from nothing to nothing ∫ StartFraction 3 Over t Superscript 4 EndFraction 3 t4 sine left parenthesis StartFraction 1 Over t cubed EndFraction minus 6 right parenthesis sin 1 t3 −6dt
Answer:
[tex]\cos (\frac{1}{t^3}-6)} + c[/tex]
Step-by-step explanation:
Given function:
[tex]\int {\frac{3}{t^4}\sin (\frac{1}{t^3}-6)} \, dt[/tex]
Now,
let [tex]\frac{1}{t^3}-6[/tex] be 'x'
Therefore,
[tex]d(\frac{1}{t^3}-6)[/tex] = dx
or
[tex]\frac{-3}{t^4}dt[/tex] = dx
on substituting the above values in the equation, we get
⇒ ∫ - sin (x) . dx
or
⇒ cos (x) + c [ ∵ ∫sin (x) . dx = - cos (x)]
Here,
c is the integral constant
on substituting the value of 'x' in the equation, we get
[tex]\cos (\frac{1}{t^3}-6)} + c[/tex]
The price of a new computer is p dollars. The computer is on sale for 30% off. Which expression shows the savings that are being offered on the computer?
A. p - 0.3p B. 0.7 × p C. 0.3 × p D. p ÷ 30
Option C
Expression that shows the savings that are being offered on the computer is 0.3p
Solution:
Given that price of a new computer is p dollars
The computer is on sale for 30% offer
To find: Expression that shows the savings that are being offered on the computer
Computer is on sale for 30% offer which means 30 % offer on original price "p"
Original price = "p" dollars
offer price / saved price = 30 % of "p"
[tex]\text{ saved price } = 30 \% \times p\\\\\text{ saved price } = \frac{30}{100} \times p\\\\\text{ saved price } = 0.3p[/tex]
Thus the required expression is 0.3p
Thus option C is correct.
A marketing company is interested in the proportion of people that will buy a particular product. Match the vocabulary word with its corresponding example. The 380 randomly selected people who are observed to see if they will buy the product The proportion of the 380 observed people who buy the product fAll people in the marketing company's region The list of the 380 Yes or No answers to whether the person bought the product The proportion of all people in the company's region who buy the product Purchase: Yes or No whether a person bought the product a. Statistic b. Data Sample d. Variable e. Parameter f. Population Points possible: 6 License
The matching is as follow:
a -> Statistic
b -> Data Sample
d -> Variable
e -> Parameter
f -> Population
a. Statistic: The proportion of the 380 observed people who buy the product
b. Data Sample: The 380 randomly selected people who are observed to see if they will buy the product
d. Variable: Purchase - Yes or No whether a person bought the product
e. Parameter: The proportion of all people in the company's region who buy the product
f. Population: All people in the marketing company's region
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The 380 randomly selected people are the 'Data Sample', the proportion of these who buy is a 'Statistic', all people in the region are the 'Population', the list of 380 Yes/No answers is the 'Variable', proportion of all people in the region who buy the product is 'Parameter', and yes/no answer for each person's purchase is also deemed a 'Variable'.
Explanation:In this question, we are dealing with terms related to statistic studies. The 380 randomly selected people who are observed to see if they will buy the product represent the Data Sample. The proportion of the 380 observed people who buy the product is considered a Statistic. All people in the marketing company's region is the Population. The list of the 380 Yes or No answers to whether the person bought the product constitutes the Variable. The proportion of all people in the company's region who buy the product is an example of a Parameter. Lastly, the Purchase: Yes or No whether a person bought the product is the Variable.
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Suppose that ten bats were used in the experiment. For each trail, the zoo keeper pointed to one of two "feeders". Suppose that the bats went to the correct feeder (the one that the zoo keeper pointed at) 8 times. Find the 95% confidence interval for the population proportion of times that the bats would follow the point. (0.62, 1.0) (0.477, 0.951) (0.321, 0.831)
Answer: (0.477, 0.951)
Step-by-step explanation:
Given : Number of observations : n = 10
Number of successes : x = 8
Let p be the population proportion of times that the bats would follow the point.
Because the number of observation is not enough large , so we use plus four confidence interval for p.
Plus four estimate of p=[tex]\hat{p}=\dfrac{\text{No. of successes}+2}{\text{No. of observations}+4}[/tex]
[tex]\hat{p}=\dfrac{8+2}{10+4}=\dfrac{10}{14}\approx0.714[/tex]
We know that , the critical value for 95% confidence level : z* = 1.96 [By using z-table]
Now, the required confidence interval will be :
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}}[/tex] , where N= 14
[tex]0.714\pm (1.96)\sqrt{\dfrac{0.714(1-0.714)}{14}}[/tex]
[tex]0.714\pm (1.96)\sqrt{0.014586}[/tex]
[tex]0.714\pm (1.96)(0.120772513429)[/tex]
[tex]\approx0.714\pm0.237=(0.714-0.237,\ 0.714+0.237)[/tex]
[tex](0.477,\ 0.951)[/tex]
Hence, the 95% confidence interval for the population proportion of times that the bats would follow the point = (0.477, 0.951)
The 95% confidence interval for the proportion of the times that bats would follow the point is (0.552, 1.0). The result was adjusted because proportions cannot exceed 1.
Explanation:To calculate the 95% confidence interval for the population proportion, we follow these steps:
First, we calculate the sample proportion (p-hat) as the number of successes (bats going to the correct feeder) divided by the total number of observations. In this case, p-hat = 8 ÷ 10 = 0.8.Next, our goal is to construct the confidence interval using the formula p-hat ± Z * sqrt[p-hat(1 - p-hat) / n], where Z is the Z-value in the standard normal distribution corresponding to the desired confidence level (1.96 for 95% confidence level), n is the number of observations, and p-hat is the calculated sample proportion.Substituting all values into the formula, we get 0.8 ± 1.96 * sqrt[0.8(0.2) / 10] = 0.8 ± 1.96 * 0.126 = (0.552, 1.048).However, this interval contains value bigger than 1, which is not possible because proportion cannot exceed 1.Hence, we adjust our interval to (0.552, 1.0).Learn more about Confidence Interval here:https://brainly.com/question/34700241
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The weight of people on a college campus are normally distributed with mean 185 pounds and standard deviation 20 pounds. What's the probability that a person weighs more than 200 pounds? (round your answer to the nearest hundredth)
Answer:
0.23.
Step-by-step explanation:
We have been given that the weight of people on a college campus are normally distributed with mean 185 pounds and standard deviation 20 pounds.
First of all, we will find the z-score corresponding to sample score 200 using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=[/tex] Z-score,
[tex]x=[/tex] Sample score,
[tex]\mu=[/tex] Mean,
[tex]\sigma=[/tex] Standard deviation.
[tex]z=\frac{200-185}{20}[/tex]
[tex]z=\frac{15}{20}[/tex]
[tex]z=0.75[/tex]
Now, we need to find [tex]P(z>0.75)[/tex]. Using formula [tex]P(z>a)=1-P(z<a)[/tex], we will get:
[tex]P(z>0.75)=1-P(z<0.75)[/tex]
Using normal distribution table, we will get:
[tex]P(z>0.75)=1-0.77337 [/tex]
[tex]P(z>0.75)=0.22663 [/tex]
Round to nearest hundredth:
[tex]P(z>0.75)\approx 0.23[/tex]
Therefore, the probability that a person weighs more than 200 pounds is approximately 0.23.
Answer:the probability that a person weighs more than 200 pounds is 0.23
Step-by-step explanation:
Since the weight of people on a college campus are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = weight of people on a college campus
u = mean weight
s = standard deviation
From the information given,
u = 185
s = 20
We want to find the probability that a person weighs more than 200 pounds. It is expressed as
P(x greater than 200) = P(x greater than 200) = 1 - P(x lesser than lesser than or equal to 200).
For x = 200,
z = (200 - 185)/20 = 0.75
Looking at the normal distribution table, the probability corresponding to the z score is 0.7735
P(x greater than 200) = 1 - 0.7735 = 0.23
A random sample of 100 high school students was surveyed regarding their favorite subject. The following counts were obtained: Favorite Subject Number of Students English Math Science 30 Art/Music The researcher conducted a test to determine whether the proportion of students was equal for all four subjects. What is the value of the test statistic? O b. 25 OOOO d. -4 How many degrees of freedom does the chi-square test statistic for a goodness of fit have when there are 10 categories? a. 74 OOOO d. 62
Answer:
a) [tex]\chi^2 = \frac{(25-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(15-25)^2}{25}=6[/tex]
b) [tex]df=Categories-1=10-1=9[/tex]
Step-by-step explanation:
We assume the following info:
Favorite Subject Number of students
English 25
Math 30
Science 30
Art/Music 15
Total 100
Previous concepts
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Part a
The system of hypothesis on this case are:
H0: There is no difference with the distribution proposed
H1: There is a difference with the distribution proposed
The level os significance assumed for this case is [tex]\alpha=0.05[/tex]
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The table given represent the observed values, we just need to calculate the expected values are 25 for each category.
And the calculations are given by:
[tex]E_{English} =25[/tex]
[tex]E_{Math} =25[/tex]
[tex]E_{Science} =25[/tex]
[tex]E_{Music} =25[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(25-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(15-25)^2}{25}=6[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=Categories-1=4-1=3[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{3} >6)=0.112[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(6,3,TRUE)"
Part b
For this case we have this formula:
[tex]df=Categories-1=10-1=9[/tex]
A supervisor records the repair cost for 11 randomly selected refrigerators. A sample mean of $82.43 and standard deviation of $13.96 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
Final answer:
The critical value for constructing a 99% confidence interval is 2.576.
Explanation:
To determine the critical value for constructing the 99% confidence interval, we need to find the Z-value that represents the level of confidence. For a 99% confidence interval, the alpha level (1 - confidence level) is 0.01. Since the data is approximately normally distributed and the sample size is greater than 30, we can use the Z-distribution. Using a Z-table or calculator, we find that the Z-value for a 0.01 alpha level is approximately 2.576.
A psychologist wants to see if a certain company has fair hiring practices in an industry where 60% of the workers are men and 40% are women. She finds that the company has 55 women and 52 men. Test to see if these numbers are different from the industry numbers, and if so, how are they different? Use alpha -.05 and four steps. A) what is the null hypothesis? B) what is the alternative hypothesis? C) what is the critical value of the test statistic? D) what is the value of the test statistic? E) Reject or accept the null? And why?
The hypothesis test examines if the company's hiring distribution differs from industry standards. The null hypothesis represents no difference, while the alternative suggests a discrepancy.
The critical value for the test statistic at a 0.05 significance level is ±1.96 for a two-tailed test, and we either reject or fail to reject the null based on the comparison of the calculated Chi-square statistic to the critical value.
To determine if there is a significant difference between the hiring practices of a certain company and the industry standard, we use a hypothesis test for proportions.
A. Null Hypothesis (H₀)
The null hypothesis H0: P_(men) = 0.60 and P_(women) = 0.40, where P represents the proportion of men and women in the company, respectively.
B. Alternative Hypothesis (Ha)
The alternative hypothesis Ha: P_(men) ≠ 0.60 and P_(women) ≠ 0.40.
C. Critical Value of Test Statistic
The critical value for a two-tailed test at alpha = 0.05 is z = ±1.96.
D. Value of the Test Statistic
To calculate the test statistic, we use the formula for a test of proportions:
Calculate the expected counts based on industry proportions: expected men = 107 * 0.60 = 64.2, expected women = 107 * 0.40 = 42.8.
Compute the Chi-square test statistic: Χ2 = ((52-64.2)2/64.2) + ((55-42.8)2/42.8).
The resulting Χ₂ statistic can then be compared against the critical Χ₂ value with 1 degree of freedom at alpha = 0.05, which is 3.841.
E. Reject or Accept the Null Hypothesis
If the calculated Χ₂ is greater than 3.841, we reject the null hypothesis; if not, we fail to reject the null hypothesis. Without the actual calculation of the Χ₂, we cannot definitively conclude the action on the null hypothesis in this context.
The random variable X = the number of vehicles owned. Find the expected number of vehicles owned. Round answer to two decimal places.
Answer:
The expected number of vehicles owned to two decimal places is: 1.85.
Step-by-step explanation:
The table to the question is attached.
[tex]E(X) =[/tex]∑[tex]xp(x)[/tex]
Where:
E(X) = expected number of vehicles owned
∑ = Summation
x = number of vehicle owned
p(x) = probability of the vehicle owned
[tex]E(X) = (0 * 0.1) + (1 * 0.35) + (2 * 0.25) + (3 * 0.2) + (4 * 0.1)\\E(X) = 0 + 0.35 + 0.50 + 0.60 + 0.4\\E(X) = 1.85[/tex]
The expected number of vehicles owned is 1.85.
The expected number of vehicles owned, based on probability of ownership of 0 to 3 vehicles, is calculated by multiplying each possible number of vehicles by their corresponding probabilities and then summing up all the products. The calculated expected number is approximately 1.7 vehicles.
Explanation:To find the expected number of vehicles owned, we first need to multiply each possible number of vehicles someone could own by the probability of them owning that many vehicles. Then, sum up all of these products.
For instance, if they could own up to 3 cars and the probability for owning 0, 1, 2, or 3 cars is 0.1, 0.3, 0.4, and 0.2 respectively:
For 0 cars: 0 * 0.1 = 0
For 1 car: 1 * 0.3 = 0.3
For 2 cars: 2 * 0.4 = 0.8
For 3 cars: 3 * 0.2 = 0.6
Adding these together gives the expected number of cars:
0 + 0.3 + 0.8 + 0.6 = 1.7 (rounded to two decimal places).
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Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type.
11.8 7.7 6.5 6.8 9.7 6.8 7.3
7.9 9.7 8.7 8.1 8.5 6.3 7.0
7.3 7.4 5.3 9.0 8.1 11.3 6.3
7.2 7.7 7.8 11.6 10.7 7.0
a) Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion. [Hint: ?xi = 219.5.] (Round your answer to three decimal places.)
MPa
State which estimator you used.
x
p?
s / x
s
x tilde
Answer:
The point estimate for population mean is 8.129 Mpa.
Step-by-step explanation:
We are given the following in the question:
Data on flexural strength(MPa) for concrete beams of a certain type:
11.8, 7.7, 6.5, 6.8, 9.7, 6.8, 7.3, 7.9, 9.7, 8.7, 8.1, 8.5, 6.3, 7.0, 7.3, 7.4, 5.3, 9.0, 8.1, 11.3, 6.3, 7.2, 7.7, 7.8, 11.6, 10.7, 7.0
a) Point estimate of the mean value of strength for the conceptual population of all beams manufactured
We use the sample mean, [tex]\bar{x}[/tex] as the point estimate for population mean.
Formula:
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]\bar{x} = \dfrac{\sum x_i}{n} = \dfrac{219.5}{27} = 8.129[/tex]
Thus, the point estimate for population mean is 8.129 Mpa.
To estimate the mean flexural strength, the sum of strengths (219.5 MPa) is divided by the total number of beams measured (26), which yields a mean value of 8.442 MPa when rounded to three decimal places. The estimator used is the sample mean.
Explanation:To calculate a point estimate of the mean value for flexural strength (MPa) for a conceptual population of concrete beams, we use the sum of all measured strengths and divide by the number of measurements. The sum of the flexural strengths is provided as Σxi = 219.5 MPa.
Given the dataset:
11.87.76.56.89.76.87.37.99.78.78.18.56.37.07.37.45.39.08.111.36.37.27.77.811.610.77.0The number of measurements is the number of data points, which is 26. To find the mean:
mean = Sum of strengths / Number of measurements
mean = 219.5 MPa / 26
mean = 8.442 MPa (rounded to three decimal places)
The estimator used here is the sample mean (×).
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2. I Using the example { 2/3+4/3 X, explain why we add fractions the way we do. What is the logic behind the procedure? Make math drawings to support your explanation
Answer:
The procedure emphasizes the idea of the summation of one physical quantity. In this case, X.
Step-by-step explanation:
1. When we add fractions like these we do it simply by rewriting a new one, the summation of the numerators over the same denominator:
[tex]\frac{2}{3}X+\frac{4}{3})X=\frac{6}{3}X= 2X[/tex]
The procedure emphasizes the idea of the summation of one physical quantity, in this case, X.
2) This physical quantity x could be miles, oranges, gallons, etc.