Answer:
[tex] \epsilon = Y -X\beta[/tex]
And the expected value for [tex] E(\epsilon) = 0[/tex] a vector of zeros and the covariance matrix is given by:
[tex] Cov (\epsilon) = \sigma^2 I[/tex]
So we can see that the error terms not have a variance of 0. We can't assume that the errors are assumed to have an increasing mean, and we other property is that the errors are assumed independent and following a normal distribution so then the best option for this case would be:
The regression model assumes the errors are normally distributed.
Step-by-step explanation:
Assuming that we have n observations from a dependent variable Y , given by [tex] Y_1, Y_2,....,Y_n[/tex]
And for each observation of Y we have an independent variable X, given by [tex] X_1, X_2,...,X_n[/tex]
We can write a linear model on this way:
[tex] Y = X \beta +\epsilon [/tex]
Where [tex]\epsilon_{nx1}[/tex] i a matrix for the error random variables, and for this case we can find the error ter like this:
[tex] \epsilon = Y -X\beta[/tex]
And the expected value for [tex] E(\epsilon) = 0[/tex] a vector of zeros and the covariance matrix is given by:
[tex] Cov (\epsilon) = \sigma^2 I[/tex]
So we can see that the error terms not have a variance of 0. We can't assume that the errors are assumed to have an increasing mean, and we other property is that the errors are assumed independent and following a normal distribution so then the best option for this case would be:
The regression model assumes the errors are normally distributed.
Final answer:
The best statement is that the regression model assumes the errors are normally distributed. In regression analysis, it is essential that the errors are independent, normally distributed, and have constant variance, which supports the validity of the model's predictions.
Explanation:
The correct statement among the provided options is that the regression model assumes the errors are normally distributed. This is a fundamental assumption of linear regression analysis, where it's assumed that the residuals or errors of the regression model are randomly distributed about an average of zero. These error terms must be independent, normal, and have constant variance (homoscedasticity) across all levels of the independent variables.
According to the theoretical foundation of regression, it is not assumed that errors have an increasing mean, nor that they have zero variance, as some diversity in errors is expected. Additionally, the assumption that errors are indeed dependent would violate the principles of ordinary least squares (OLS) regression, making the model invalid.
Normality, independence, and equal variance are key premises in regression analysis to ensure the validity of the model's inferences. Indeterminate errors that affect the dependent variable 'y' are assumed to be normally distributed and independent of the independent variable 'x'. This maintains the integrity of the regression model.
Using the common denominator what is the equivalent fraction of 1/8
Answer:
2/16
Step-by-step explanation:
1/8 x 2 = 2/16
Both 2 and 16 divide by 8, so one can know this is correct.
Also, both 8 and 16 are divisible by 8, so one can see this is the common denominator.
A bag of marbles contains 3 yellow, 6 blue, 1 green, 8 red, 2 orange. Find the probability: P(yellow, then orange WITHOUT replacement)
Answer:
P(yellow, then orange WITHOUT replacement) = 0.0158
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes
This problem asks:
The probability of first getting a yellow marble, then an orange marble, without replacemtn.
Yellow marble first:
3 yellow marbles out of 3+6+1+8+2 = 20 marbles.
So P(a) = 3/20.
Then the orange marble:
Now the yellow marble is retired.
2 orange marbles out of 19.
So P(b) = 2/19
Probability:
P(yellow, then orange WITHOUT replacement) = P(a)P(b) = (3/20)*(2/19) = 0.0158
A sample of 16 elements is selected from a population with a reasonably symmetrical distribution. The sample mean is 100 and the sample standard deviation is 40. We can say that we are 90% confident that is between what two numbers? 12. What is the upper number? Maintain three decimal points in your calculations and give at least two decimal points in your answer.
Answer:
We can say that we are 90% confident that the population mean is between 29.876 and 170.124.
The upper number is 170.124.
Step-by-step explanation:
We have the sample's standard deviation, so we use the students' t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 16 - 1 = 15
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95([tex]t_{95}[/tex]). So we have T = 1.7531
The margin of error is:
M = T*s = 40*1.7531 =70.124
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 100 - 70.124 = 29.876
The upper end of the interval is the sample mean added to M. So it is 100 + 70.124 = 170.124
We can say that we are 90% confident that the population mean is between 29.876 and 170.124.
The upper number is 170.124.
Answer:
[tex]100-1.753\frac{40}{\sqrt{16}}=82.470[/tex]
[tex]100+ 1.753\frac{40}{\sqrt{16}}=117.530[/tex]
So on this case the 90% confidence interval would be given by (82.470;117.530)
And the upper value would be 117.530 for this case
Step-by-step explanation:
Previous concepts
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=100[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=40 represent the sample standard deviation
n=16 represent the sample size
Solution to the problem
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 critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=16-1=15[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,15)".And we see that [tex]t_{\alpha/2}=1.753[/tex]
Now we have everything in order to replace into formula (1):
[tex]100-1.753\frac{40}{\sqrt{16}}=82.470[/tex]
[tex]100+ 1.753\frac{40}{\sqrt{16}}=117.530[/tex]
So on this case the 90% confidence interval would be given by (82.470;117.530)
You spin the spinner shown below once. Each sector shown has an equal area.
What is p (not squirrel)
Answer: 0.8
Step-by-step explanation:
There are 5 possible equally sized sections meaning there is a 0.2 or 20% of landing on each section. We can subtract the probability of getting a squirrel from the whole thing (1 as probability always equals 1) to determine the probability of not squirrel.
P(squirrel) = 0.2
1 - P(squirrel) = P(not a squirrel)
1 - 0.2 = 0.8
Answer: the anserw is 0.8
Step-by-step explanation:
The trucking company has a policy that any truck with a mileage more than 2.5 standard deviations above the mean should be removed from the road and inspected. What is the probability that a randomly selected truck from the fleet will have to be inspected? ROUND YOUR ANSWER TO 4 DECIMAL PLACES
Answer:
0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected
Step-by-step explanation:
Z-score
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
What is the probability that a randomly selected truck from the fleet will have to be inspected?
Values of Z above 2.5. So this probability is 1 subtracted by the pvalue of Z = 2.5.
Z = 2.5 has a pvalue of 0.9938
1 - 0.9938 = 0.0062
0.0062 = 0.62% probability that a randomly selected truck from the fleet will have to be inspected
Answer:
For this case we want this probability:
[tex] P(X > \mu +2.5 \sigma) [/tex]
And we can use the z score formula given by:
[tex] z = \frac{ X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]
We want this probability:
[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]
And using the normal standard table or excel we got:
[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/tex]
Step-by-step explanation:
Previous concepts
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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the mileage of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
For this case we want this probability:
[tex] P(X > \mu +2.5 \sigma) [/tex]
And we can use the z score formula given by:
[tex] z = \frac{ X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z = \frac{\mu +2.5 \sigma -\mu}{\sigma}= 2.5[/tex]
We want this probability:
[tex] P(Z>2.5) = 1-P(Z<2.5)[/tex]
And using the normal standard table or excel we got:
[tex] P(Z>2.5) = 1-P(Z<2.5)= 1-0.9938= 0.0062[/tex]
We survey a random sample of American River College students and ask if they drink coffee on a regular basis. The 90% confidence interval for the proportion of all American River College students who drink coffee on a regular basis is (0.262, 0.438). What will be true about the 95% confidence interval for these data? Group of answer choices The 95% confidence interval is narrower than the 90% confidence interval. The 95% confidence interval is wider than the 90% confidence interval. The two intervals will have the same width. It is impossible to say which interval will be wider.
Answer:
Correct option:
The 95% confidence interval is wider than the 90% confidence interval.
Step-by-step explanation:
The (1 - α)% confidence interval for the population proportion is:
[tex]CI=\hat p\pm z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The width of this interval is:
[tex]W=UL-LL\\=2\times z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The width of the interval is directly proportional to the critical value.
The critical value of a distribution is based on the confidence level.
Higher the confidence level, higher will be critical value.
The z-critical value for 95% and 90% confidence levels are:
[tex]90\%: z_{\alpha /2}=z_{0.05}=1.645\\95\%: z_{\alpha /2}=z_{0.025}=1.96[/tex]
*Use a z-table.
The critical value of z for 95% confidence level is higher than that of 90% confidence level.
So the width of the 95% confidence interval will be more than the 90% confidence interval.
Thus, the correct option is:
"The 95% confidence interval is wider than the 90% confidence interval."
The diameters of a batch of ball bearings are known to follow a normal distribution with a mean 4.0 in and a standard deviation of 0.15 in. If a ball bearing is chosen randomly, find the probability of realizing the following event: (a) a diameter between 3.8 in and 4.3 in, (b) a diameter smaller than 3 9 in, (c) a diameter larger than 4.2 in
Answer:
a) 88.54% probability of a diameter between 3.8 in and 4.3 in
b) 25.14% probability of a diameter smaller than 3.9in
c) 90.82% probability of a diameter larger than 4.2 in
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 4, \sigma = 0.15[/tex]
(a) a diameter between 3.8 in and 4.3 in,
This is the pvalue of Z when X = 4.3 subtracted by the pvalue of Z when X = 3.8.
X = 4.3
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4.3 - 4}{0.15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
X = 3.8
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3.8 - 4}{0.15}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a pvalue of 0.0918
0.9772 - 0.0918 = 0.8854
88.54% probability of a diameter between 3.8 in and 4.3 in
(b) a diameter smaller than 3 9 in,
This is the pvalue of Z when X = 3.9. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3.9 - 4}{0.15}[/tex]
[tex]Z = -0.67[/tex]
[tex]Z = -0.67[/tex] has a pvalue of 0.2514
25.14% probability of a diameter smaller than 3.9in
(c) a diameter larger than 4.2 in
This is 1 subtracted by the pvalue of Z when X = 4.2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4.2 - 4}{0.15}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082
90.82% probability of a diameter larger than 4.2 in
Imagine you conducted a study to look at the association between whether an expectant mother eats breakfast (or not) and the gender of her baby. Cramér's V = .22. How would you interpret this value? (Hint: Cramér's V can range between 0 and 1.) There was a small to medium association between baby gender and whether the mother ate breakfast every day. 22% of the variation in frequency counts of baby gender (boy or girl) can be explained by whether or not the mother ate breakfast every day. 2.2% of the variation in frequency counts of baby gender (boy or girl) can be explained by whether or not the mother ate breakfast every day. There is a medium to large association between the gender of the baby and whether or not the mother ate breakfast every day.
Answer:
: The percentage of newborn infants who are exclusively breastfed at the time of hospital discharge.
Data Source: Rhode Island Department of Health, Center for Health and Data Analysis, Newborn Developmental Risk Screening Program Database and Maternal and Child Health Database.
Footnotes: The year indicated is the mid-year of a five-year period of data.
Step-by-step explanation:
Cramér's V value of .22 in the study indicates a small to medium association between whether the mother ate breakfast and the gender of her baby.
Explanation:In the study, Cramér's V measuress the strength of association between two nominal variables, in this case, the mother's breakfast-eating habits and the gender of her baby. The value of Cramer’s V ranges from 0 (indicating no association) to 1 (indicating a perfect association)). Given the value of Cramér's V at .22 for this study, there is a small to medium association between whether a mother ate breakfast and the gender of her baby. It does not mean that 22% of the variation in baby gender can be explained by the mother's breakfast-eating habits. Rather, it quantifies the extent of the association between these two factors.
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A recipe requires 3/4 cups of sugar for every 14 cookies how many cups of sugar will be needed to bake 168 cookies
Answer:
9 cups
Step-by-step explanation:
3/4 cups of sugar for 14 cookies
There are 168÷14 = 12 set of 14 cookies in 168
So we need 12 times 3/4 sugar
12 × 3/4 = 9 cups of sugar needed
A random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph. From past information, it is known that the standard deviation of the population is 22 mph. The 95 percent confidence interval for μ is determined as (61.08, 68.92). If we are to reduce the sample size to 100 (other factors remain unchanged), the 95 percent confidence interval for μ would: Question 6 options: be the same. become wider. become narrower. be impossible to determine.
Answer:
95 percent confidence interval for μ is determined by
(60.688 , 69.312)
a) be the same
Step-by-step explanation:
Step (i):-
Given data a random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph.
The sample size is n= 121
The mean of the sample x⁻ = 65mph.
The standard deviation of the population is 22 mph
σ = 22mph
Step (ii) :-
Given The 95 percent confidence interval for μ is determined as
(61.08, 68.92)
we are to reduce the sample size to 100 (other factors remain unchanged) so
given The sample size is n= 100
The mean of the sample x⁻ = 65mph.
The standard deviation of the population is 22 mph
σ = 22mph
Step (iii):-
95 percent confidence interval for μ is determined by
[tex](x^{-} - 1.96\frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} } )[/tex]
[tex](65 - 1.96\frac{22}{\sqrt{100} } , 65 + 1.96 \frac{22}{\sqrt{100} } )[/tex]
(65 - 4.312 ,65 +4.312)
(60.688 , 69.312) ≅(61 ,69)
This is similar to (61.08, 68.92) ≅(61 ,69)
Conclusion:-
it become same as (61.08, 68.92)
In the given scenario, the 95 percent confidence interval for μ would become wider if the sample size is reduced from 121 to 100. This is due to the increased uncertainty in estimation as a result of a smaller sample size.
Explanation:In statistics, a confidence interval is a range in which a population parameter is estimated to lie. The calculation of a confidence interval involves several factors, one of which is the sample size. The width of the confidence interval decreases as the sample size increases, assuming that all other factors remain the same.
In this case, the sample size is proposed to decrease from 121 to 100. As a result, the confidence interval for μ would become wider. This is because as the sample size decreases, the standard error increases, which in turn widens the confidence interval, allowing for more uncertainty in estimations.
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The graph of a line passes through the points (0,5) and (-10,0). What is the equation
of the line?
CLEAR
CHECK
y = 2x - 10
y = 5x + 5
y= 1 / 8x 10
y = -2x + 5
Answer:
Y=1/2x +5
Step-by-step explanation:
(0,5) (-10,0)
0-5/-10-0=1/2
(0,5) (x, y)
Y-5/x-10=1/2(crossmultiply)
2y-10=x
2y= x +10(divide all sides by 2)
Y=1/2x+5
Cuál es el valor de la expresión -2+3(8+-13)?
Answer:
-17
Step-by-step explanation:
-2+3(8+-13)
-2+3(8-13)
-2+3(-5)
-2-15
-17
[tex]\text{Solve:}\\\\-2+3(8+-13)\\\\\text{Solve the variables inside of the parenthesis}\\\\-2+3(-5)\\\\-2-15\\\\\boxed{-17}[/tex]
How do the areas of triangle ABC and DEF compare?
O The area of ABC is 1 square unit less than the area of
ADEF
O The area of AABC is equal to the area of ADEF.
O The area of ABC is 1 square unit greater than the
area of ADEF.
O The area of ABC is 2 square units greater than the
area of ADEF.
Answer:
C: The area of ABC is 1 square unit greater than the
area of ADEF
Step-by-step explanation:
Due to the area of triangle ABC and triangle DEF, the ABC triangle has dimensions that are 1 square unit greater that the area of ADEF. This is because the lines have different lengths. Good Luck on the Rest of Your Questions!!
Answer: C
Step-by-step explanation:
A deck of cards contains 52 cards.two decks are put together and divided among eight players.how many cards does each person get
Answer:
13 cards
Step-by-step explanation:
One deck has 52 cards.
2 decks have 2 * 52 cards, or 104 cards.
104 cards are divided among 8 players.
104/8 = 13
Answer: 13 cards
sara pays $4100 tuition and fees for each of the two semesters, plus an additional $41" for textbooks each semester. what is the monthly cost?
Answer:
The monthly cost is $690.17.
Step-by-step explanation:
Each semester costs $4100.
In each semester, there are $41 spent on textbooks.
In a year:
2*4100 + 2*41 = $8282
Monthly cost
A year has 12 months. So
8282/12 = 690.17.
The monthly cost is $690.17.
49w = −8
what is the value of w
Answer:
w= -0.163265306122449
Step-by-step explanation:
49w = -8
divide both sides by 49
w= -8/49 = -0.163265306122449
check answer by multiplying -0.163265306122449 by 49
49 * -0.163265306122449 = -8
Answer:
w = -8/49
Step-by-step explanation:
49w = −8
Divide each side by 49
49w/49 = -8/49
w = -8/49
Suppose the selling price of homes is skewed right with a mean of 350,000 and a standard deviation of 160000 If we record the selling price of 40 randomly selected US homes what will be the shape of the distribution of sample means what will be the mean of this distribution what will be the standard deviation of this distribution
Answer:
The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Population:
Suppose the selling price of homes is skewed right with a mean of 350,000 and a standard deviation of 160000
Sample of 40
Shape approximately normal
Mean 350000
Standard deviation [tex]s = \frac{160000}{\sqrt{40}} = 25298[/tex]
The distribution will be approximately normal, with mean 350,000 and standard deviation 25,298.
Answer:
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The mean would be:
[tex] \mu_{\bar X} =350000[/tex]
And the standard deviation would be:
[tex]\sigma_{\bar X} =\frac{160000}{\sqrt{40}}= 25298.221[/tex]
Step-by-step explanation:
Previous concepts
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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the selling price of a population, and for this case we know the following info
Where [tex]\mu=350000[/tex] and [tex]\sigma=160000[/tex]
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".
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The mean would be:
[tex] \mu_{\bar X} =350000[/tex]
And the standard deviation would be:
[tex]\sigma_{\bar X} =\frac{160000}{\sqrt{40}}= 25298.221[/tex]
1. Walking to improve health. In a study investigating a link between walking and improved health (Social Science & Medicine, Apr. 2014), researchers reported that adults walked an average of 5.5 days in the past month for the purpose of health or recreation. a) Specify the null and alternative hypotheses for testing whether the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days. b) Is the test one-tailed or two-tailed? c) Using α = .01, specify the critical value (CV), the rejection region, and draw the distribution.
Answer:
a) We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 5.5[/tex]
Alternative hypothesis:[tex]\mu < 5.5[/tex]
b) For this case since we want to determine if the true mean is lower than an specified value is a one tailed test
c) For this case the significance level is given [tex]\alpha=0.01[/tex] and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:
[tex]z_{cric}= -2.326[/tex]
And the rejection zone would be: [tex] (\infty ,-2.326)[/tex]
And the region is on the figure attached
Step-by-step explanation:
Part a: State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 5.5[/tex]
Alternative hypothesis:[tex]\mu < 5.5[/tex]
Part b
For this case since we want to determine if the true mean is lower than an specified value is a one tailed test
Part c
For this case the significance level is given [tex]\alpha=0.01[/tex]and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:
[tex]z_{cric}= -2.326[/tex]
And the rejection zone would be: [tex] (\infty ,-2.326)[/tex]
And the region is on the figure attached
An archer is able to hit the bull's-eye 53% of the time. If she shoots 10 arrows, what is the probability that she gets exactly 4 bull's-eyes?
To find the probability that the archer gets exactly 4 bull's-eyes out of 10 shots, use the binomial probability formula. The probability is approximately 0.221.
Explanation:To find the probability that the archer gets exactly 4 bull's-eyes out of 10 shots, we can use the binomial probability formula. The formula is:
P(x) = nCx * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of getting exactly x successesn is the number of trialsx is the number of successes we want to find the probability forp is the probability of a single successIn this case, n = 10, x = 4, and p = 0.53 (since the archer hits the bull's-eye 53% of the time).
Plugging these values into the formula:
P(4) = 10C4 * (0.53)^4 * (1-0.53)^(10-4)
Calculating this gives us P(4) ≈ 0.221
Final answer:
To find the probability of exactly 4 bull's-eyes out of 10 shots for an archer with a 53% success rate, use the binomial probability formula, resulting in approximately a 20.6% chance.
Explanation:
The probability that an archer who hits the bull's-eye 53% of the time getting exactly 4 bull's-eyes out of 10 shots can be calculated using the binomial probability formula: P(X = k) = nCk × p^k × (1 - p)^(n - k), where:
n is the number of trials (10 arrows)
k is the number of successes (4 bull's-eyes)
p is the probability of success on a single trial (0.53)
nCk is the number of combinations of n things taken k at a time
Applying these values:
Calculate the number of combinations for getting 4 successes out of 10 trials:
10C4 = 210.
Calculate the probability of getting exactly 4 bull's-eyes:
P(X = 4) = 210 × (0.53)^4 × (0.47)^6.
Compute the result:
P(X = 4) ≈ 0.206 (rounded to three decimal places).
The archer has approximately a 20.6% chance of hitting exactly 4 bull's-eyes out of 10 shots.
The same researcher wants to see if recently widowed spouses leave the house more after they perform some activity (e.g., Bingo, Support Group, etc.). She wants to compare participants before and after a group activity. Which type of t-test should she use?
Answer:
paired t-test
Step-by-step explanation:
Paired t-test is used to measure before and after effect. For example, a researcher wants to check the effect of new teaching method and he wants to compares marks of students before new teaching method and marks of students before new teaching method.
In the given example there are two groups participants before group activity and participants after group activity. A researcher wants to investigate that whether recently widowed spouses leave house after performing group activity or not. The observations in these two groups are paired naturally and we take the differences of these two groups are considered as random sample. This type of t-test is known as paired t-test.
Find equations of the tangent plane and the normal line to the given surface at the specified point. x + y + z = 8exyz, (0, 0, 8) (a) the tangent plane (b) the normal line (x(t), y(t), z(t)) =
Let [tex]f(x,y,z)=x+y+z-8e^{xyz}[/tex]. The tangent plane to the surface at (0, 0, 8) is
[tex]\nabla f(0,0,8)\cdot(x,y,z-8)=0[/tex]
The gradient is
[tex]\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)[/tex]
so the tangent plane's equation is
[tex](1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8[/tex]
The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by [tex]t[/tex], then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation
[tex](1,1,1)t+(0,0,8)=(t,t,t+8)[/tex]
or [tex]x(t)=t[/tex], [tex]y(t)=t[/tex], and [tex]z(t)=t+8[/tex].
(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)
Final answer:
This detailed answer explains how to find the equation of the tangent plane and the normal line to a given surface at a specified point. The parametric equations of the normal line are [tex]\( (x(t), y(t), z(t)) = (t, t, 8 + t) \).[/tex]
Explanation:
To find the equations of the tangent plane and the normal line to the given surface x + y + z = 8exyz at the specified point 0, 0, 8) we first need to find the partial derivatives of x + y + z - 8exyz with respect to x, y, and z.
Given the surface equation:
[tex]\[ x + y + z = 8exyz \][/tex]
Let's denote the function as [tex]\(F(x, y, z) = x + y + z - 8exyz\).[/tex]
(a) To find the equation of the tangent plane, we use the gradient vector of F at the point 0, 0, 8 as the normal vector to the tangent plane.
The gradient vector of F is given by:
[tex]\[ \nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \][/tex]
Let's calculate the partial derivatives:
[tex]\[ \frac{\partial F}{\partial x} = 1 - 8eyz \][/tex]
[tex]\[ \frac{\partial F}{\partial y} = 1 - 8exz \][/tex]
[tex]\[ \frac{\partial F}{\partial z} = 1 - 8exy \][/tex]
Now, evaluate these partial derivatives at the point (0, 0, 8):
[tex]\[ \frac{\partial F}{\partial x} = 1 - 8e(0)(0) = 1 \][/tex]
[tex]\[ \frac{\partial F}{\partial y} = 1 - 8e(0)(0) = 1 \][/tex]
[tex]\[ \frac{\partial F}{\partial z} = 1 - 8e(0)(0) = 1 \][/tex]
So, the gradient vector of [tex]\(F\) at \((0, 0, 8)\) is \( \nabla F = (1, 1, 1) \).[/tex]
Therefore, the equation of the tangent plane is:
[tex]\[ x + y + z = 8 \][/tex]
(b) To find the equation of the normal line, we use the same gradient vector [tex]\( \nabla F = (1, 1, 1) \)[/tex] as the direction vector for the line.
Given the point \((0, 0, 8)\) on the surface, the parametric equations of the normal line are:
[tex]\[ x(t) = 0 + t(1) = t \][/tex]
[tex]\[ y(t) = 0 + t(1) = t \][/tex]
[tex]\[ z(t) = 8 + t(1) = 8 + t \][/tex]
So, the parametric equations of the normal line are [tex]\( (x(t), y(t), z(t)) = (t, t, 8 + t) \).[/tex]
please help me ASAP!!
Answer:
x = 16
Step-by-step explanation:
The sum of measures of these three angles of any triangle is equal to 180 °
54 + 90 + (2x + 4) = 180
144 + 2x + 4 = 180
148 + 2x = 180
2x = 180 - 148
2x = 32
[tex]x= \frac{32}{2} \\\\x=16[/tex]
Focus on triangle KLM. Since JKLM is a rectangle, angle KLM is right. Since the sum of the interior angles of every triangle is 180, we deduce that
[tex]\hat{K}+\hat{L}+\hat{M}=180[/tex]
Now plug the values for each angle and you have
[tex]54+90+(2x+4)=180[/tex]
Move 54 and 90 to the right side to get
[tex]2x+4=36 \iff 2x=32 \iff x=16[/tex]
The two dot plots represent a sample of the number of
people in households in two towns. Which statements
are true about the data sets?
Check all that apply.
Both have the same number of data points.
Both means are between 3 and 4.
O Both have the same median.
Both have the same range.
Westwood has less variability than Middleton.
The answer is:
Both have the same number of data points
Both means are between 3 and 4
Westwood has less variability than Middleton
Answer:
The right anwers are : A B and E
Step-by-step explanation:
right on edg!
The average THC content of marijuana sold on the street is 10%. Suppose the THC content is normally distributed with standard deviation of 2%. Let X be the THC content for a randomly selected bag of marijuana that is sold on the street. Round all answers to two decimal places and give THC content in units of percent. For example, for a THC content of 11%, write "11" not 0.11.
Answer:
(a) [tex]X\sim N(0.10,\ 0.02^{2})[/tex].
(b) The probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.
(c) The 60th percentile is 11%.
Step-by-step explanation:
The complete question is:
The average THC content of marijuana sold on the street is 10%. Suppose the THC content is normally distributed with standard deviation of 2%. Let X be the THC content for a randomly selected bag of marijuana that is sold on the street. Round all answers to two decimal places and give THC content in units of percent. For example, for a THC content of 11%, write "11" not 0.11.
A. X ~ N( , )
B. Find the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11%.
C.Find the 60th percentile for this distribution.
Solution:
The random variable X is defined as the THC content for a bag of marijuana that is sold on the street.
(a)
The random variable X is Normally distributed with mean, μ = 0.10 and standard deviation, σ = 0.02.
Thus, [tex]X\sim N(0.10,\ 0.02^{2})[/tex].
(b)
Compute the value of P (X > 0.11) as follows:
[tex]P(X>0.11)=P(\frac{X-\mu}{\sigma}>\frac{0.11-0.10}{0.02})\\=P(Z>0.50)\\=1-P(Z<0.50)\\=1-0.69146\\=0.30854\\\approx 0.31[/tex]
*Use a z-table.
Thus, the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 11% is 31%.
(c)
The pth percentile is a data value such that at least p% of the data-set is less than or equal to this data value and at least (100 - p)% of the data-set are more than or equal to this data value.
Let x represent the 60th percentile value.
The, P (X < x) = 0.60.
⇒ P (Z < z) = 0.60
The value of z for this probability is,
z = 0.26.
Compute the value of x as follows:
[tex]z=\farc{x-\mu}{\sigma}\\0.26=\farc{x-0.10}{0.02}\\x=0.10+(0.26\times 0.02)\\x=0.1052\\x\approx 0.11[/tex]
Thus, the 60th percentile is 11%.
A worker has asked her supervisor for a letter of recommendation for a new job. She estimates that there is an 80 percent chance that she will get the job if she receives a strong recommendation, a 40 percent chance if she receives a moderately good recommendation, and a 10 percent chance if she receives a weak recommendation. She further estimates that the probabilities that the recommendation will be strong, moderate, and weak are 0.7, 0.2, and 0.1, respectively.
a. How certain is she that she will receive the new job offer?
b. Given that she does receive the offer, how likely should she feel that she received a strong recommendation?
i. a moderate recommendation?
ii. a weak recommendation?
c. Given that she does not receive the job offer, how likely should she feel that she received a strong recommendation?
i. a moderate recommendation?
ii. a weak recommendation?
Answer:
a) P(Job offer) = 0.65
b) P(S|J) = 0.862
P(M|J) = 0.143
P(W|J) = 0.0179
c) P(S|J') = 0.400
P(M|J') = 0.343
P(W|J') = 0.257
Step-by-step explanation:
- Let the event that she gets a job be J
- Let the event that she does not get the job be J'
- Let the event that she receives a strong recommendation be S
- Let the event that she receives a moderate recommendation be M
- Let the event that she receives a weak recommendation be W.
Given in the question,
P(J|S) = 80% = 0.8
P(J|M) = 40% = 0.4
P(J|W) = 10% = 0.1
P(S) = 0.70
P(M) = 0.20
P(W) = 0.10
a) How certain is she that she will receive the new job offer?
P(J) = P(J n S) + P(J n M) + P(J n W) (since S, M and W are all of the possible outcomes that lead to a job)
But note that the conditional probability, P(A|B) is given mathematically as,
P(A|B) = P(A n B) ÷ P(B)
P(A n B) is then given as
P(A n B) = P(A|B) × P(B)
So,
P(J n S) = P(J|S) × P(S) = 0.80 × 0.70 = 0.56
P(J n M) = P(J|M) × P(M) = 0.40 × 0.20 = 0.08
P(J n W) = P(J|W) × P(W) = 0.10 × 0.10 = 0.01
P(J) = P(J n S) + P(J n M) + P(J n W)
P(J) = 0.56 + 0.08 + 0.01 = 0.65
b) Given that she does receive the offer, how likely should she feel that she received a strong recommendation?
This probability = P(S|J)
P(S|J) = P(J n S) ÷ P(J) = 0.56 ÷ 0.65 = 0.862
i. a moderate recommendation?
P(M|J) = P(J n M) ÷ P(J) = 0.08 ÷ 0.65 = 0.143
ii. a weak recommendation?
P(W|J) = P(J n W) ÷ P(J) = 0.01 ÷ 0.65 = 0.0179
c) Probability that she doesn't get job offer, given she got a strong recommendation = P(J'|S)
P(J'|S) = 1 - P(J|S) = 1 - 0.80 = 0.20
Probability that she doesn't get job offer, given she got a moderate recommendation = P(J'|M)
P(J'|M) = 1 - P(J|M) = 1 - 0.40 = 0.60
Probability that she doesn't get job offer, given she got a weak recommendation = P(J'|S)
P(J'|W) = 1 - P(J|W) = 1 - 0.10 = 0.90
Total probability that she doesn't get job offer
P(J') = P(J' n S) + P(J' n M) + P(J' n W)
P(J' n S) = P(J'|S) × P(S) = 0.20 × 0.70 = 0.14
P(J' n M) = P(J'|M) × P(M) = 0.60 × 0.20 = 0.12
P(J' n W) = P(J'|W) × P(W) = 0.90 × 0.10 = 0.09
Total probability that she doesn't get job offer
P(J') = P(J' n S) + P(J' n M) + P(J' n W)
= 0.14 + 0.12 + 0.09 = 0.35
Given that she does not receive the job offer, how likely should she feel that she received a strong recommendation?
This probability = P(S|J')
P(S|J') = P(J' n S) ÷ P(J') = 0.14 ÷ 0.35 = 0.400
i. a moderate recommendation?
P(M|J') = P(J' n M) ÷ P(J') = 0.12 ÷ 0.35 = 0.343
ii. a weak recommendation?
P(W|J') = P(J' n W) ÷ P(J') = 0.09 ÷ 0.35 = 0.257
Hope this Helps!!!
She has 62% certainty of getting the job. If she gets it, there are high chances (~90%) that the recommendation was strong. In case of rejection, she's slightly more likely to have received a strong recommendation than a moderate or weak one.
Explanation:The subject of this problem involves the understanding of conditional probabilities and the principle of total probability. Let's denote the events as follows:
A1: the recommendation is strong A2: the recommendation is moderate A3: the recommendation is weak B: she gets the new job offer
a. The total probability that she will receive the job offer is calculated as follows:
P(B)=P(A1)P(B|A1)+P(A2)P(B|A2)+P(A3)P(B|A3) = 0.7*0.8+0.2*0.4+0.1*0.1=0.62, so she is 62% certain that she will receive the job offer.
b. If she does receive the offer, the probabilities that she got a strong, moderate and weak recommendation are calculated using Bayes' Theorem:
P(A1|B)=P(B|A1)P(A1)/P(B)=0.8*0.7/0.62 =~ 0.9 P(A2|B)=P(B|A2)P(A2)/P(B)=0.4*0.2/0.62 =~ 0.13 P(A3|B)=P(B|A3)P(A3)/P(B)=0.1*0.1/0.62 =~ 0.016
c. If she doesn't get the job, the probabilities are calculated similarly:
P(A1|Not B)=P(Not B|A1)P(A1)/P(Not B)=0.2*0.7/0.38 =~ 0.37 P(A2|Not B)=P(Not B|A2)P(A2)/P(Not B)=0.6*0.2/0.38 =~ 0.32 P(A3|Not B)=P(Not B|A3)P(A3)/P(Not B)=0.9*0.1/0.38 =~ 0.24Learn more about Conditional Probability here:
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If Sergio has 16 quarters and dimes in his pocket, and they have a combined value of 265 cents, how many of each coin does he have?
The price-earnings (PE) ratios of a sample of stocks have a mean value of 10.5 and a standard deviation of 3. If the PE ratios have a normal distribution, use the Empirical Rule (also called the 68-95-99.7 Rule) to estimate the percentage of PE ratios that fall between:
Answer:
a) 68% falls within 7.5 and 13.5
b) 95% falls within 4.5 and 16.5
c) 99.7% falls within 1.5 and 19.5
Step-by-step explanation:
Given that:
mean (μ) = 10.5, Standard deviation (σ) = 3.
The Empirical Rule (also called the 68-95-99.7 Rule) for a normal distribution states that all data falls within three standard deviations. That is 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
Therefore using the empirical formula:
68% falls within the first standard deviation (µ ± σ) = (10.5 ± 3) = (7.5, 13.5)
95% within the first two standard deviations (µ ± 2σ) = (10.5 ± 2(3)) = (10.5 ± 6) = (4.5, 16.5)
99.7% within the first three standard deviations (µ ± 3σ) = (10.5 ± 3(3)) = (10.5 ± 9) = (1.5, 19.5)
Solve the following:
1)If 32x=2, find x
2) simplify 9-1\2
Answer:
the answers are
1) 1/16
2) -18
Answer:
[tex]1)32x = 2 \\ \frac{32x}{32} = \frac{2}{32} \\ x = \frac{1}{16} [/tex]
[tex]2)9 - \frac{1}{2} \\ \frac{9}{1} \frac{ \times }{ \times } \frac{2}{2} - \frac{1}{2} \\ \frac{18}{2} - \frac{1}{2} \\ = \frac{17}{2} \\ = 8 \frac{1}{2} [/tex]
Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book. From a random sample of 50 students, she found that 32 students read at least 1 book last month. Assuming all conditions for Inference are met, which of the following defines a 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month? 32 +1.645,8908 32 + 1.9620 0.64 + 1.282,16.620.) © 264 265 CHAT 0.04 + 1.00, layanan
Answer:
0.64 ± 0.1117 or
[tex]0.64\pm 1.645*\sqrt{\frac{0.64*(0.36)}{50}}[/tex]
Step-by-step explanation:
Sample size (n) = 50
Z-score for a 90% confidence interval (z) = 1.645
Proportion of students that read at least one book (p):
[tex]p=\frac{32}{50}=0.64[/tex]
The confidence interval is given by:
[tex]p\pm z*\sqrt{\frac{p*(1-p)}{n} }[/tex]
Applying the given data:
[tex]0.64\pm 1.645*\sqrt{\frac{0.64*(1-0.64)}{50}}\\ 0.64\pm 0.1117[/tex]
The confidence interval is 0.64 ± 0.1117
The 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month is [tex]0.64\pm 0.11178[/tex].
Given information:
From a random sample of 50 (n) students, Alma found that 32 students read at least 1 book last month.
Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book.
It is required to find the 90 percentage confidence interval for the proportion of all students in her district who read at least 1 book last month.
Now, from tha table, the z-score of 90 percent confidence level is,
[tex]z=1.645[/tex]
The probability p for the proportion of students who read atleast one book is,
[tex]p=\dfrac{32}{50}\\p=0.64[/tex]
So, the 90 percent confidence interval will be calculated as,
[tex]p\pm z\sqrt{\dfrac{p(1-p)}{n}}=0.64\pm 1.645\times \sqrt{\dfrac{0.64(1-0.64)}{50}}\\=0.64\pm 0.11178[/tex]
Therefore, the 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month is [tex]0.64\pm 0.11178[/tex].
For more details, refer to the link:
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Find the derivative.
StartFraction d Over dt EndFraction Integral from 0 to t Superscript 8 StartRoot u cubed EndRoot du
a. by evaluating the integral and differentiating the result.
b. by differentiating the integral directly.
Answer:
Using either method, we obtain: [tex]t^\frac{3}{8}[/tex]
Step-by-step explanation:
a) By evaluating the integral:
[tex]\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du[/tex]
The integral itself can be evaluated by writing the root and exponent of the variable u as: [tex]\sqrt[8]{u^3} =u^{\frac{3}{8}[/tex]
Then, an antiderivative of this is: [tex]\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}[/tex]
which evaluated between the limits of integration gives:
[tex]\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}[/tex]
and now the derivative of this expression with respect to "t" is:
[tex]\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}[/tex]
b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then
[tex]g(x)=\int\limits^x_a {f(t)} \, dt[/tex]
is continuous on [a,b], differentiable on (a,b) and [tex]g'(x)=f(x)[/tex]
Since this this function [tex]u^{\frac{3}{8}[/tex] is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:
[tex]\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}[/tex]