Answer:
c. y? =+4.198 x
Step-by-step explanation:
Hello!
Using the given data you need to estimate the equation of linear regression.
Dependent variable:
Y: Annual percentage change in stock price for a company.
Independent variable:
X: Annual percentage change in profits for the company.
The population regression line equation is:
[tex]Y_i= \alpha + \beta X_i + E_i[/tex]
To estimate the equation you need to find the point estimator for α and β.
The following formulas are the ones to use:
α ⇒ a= y[bar] - bX[bar]
β ⇒ b= (∑xy - [(∑x)(∑y)]/n)/(∑X²-(∑x)²/n)
As you can see you need to make several summatories before calculating the values of a and b:
n= 7
∑x= 36.30
∑x²= 667.09
∑y= 51.60
∑y²= 747.98
∑xy= 560.74
Sample mean of de dependent variable Y[bar]= ∑y/n= (51.60/7)= 7.37
Sample mean of the independent variable X[bar]= ∑x/n= (36.30/7)= 5.19
b= [tex]\frac{560.74-\frac{(36.3*51.6)}{7} }{667.09-\frac{(51.60)^2}{7} }[/tex]
b= 0.6122
a= [tex]7.37-(0.61*5.19)[/tex]
a= 4.196
The estimated regression equation is:
Y= 4.196 + 0.6122x
I hope it helps!
An advertisement for a word-processing class claims that students who complete the class will, on average, be able to type 60 words per minute (wpm) with a standard deviation of 6 wpm. At the end of the class, 49 students are tested and their mean score is 58.5 wpm.
A. Is this evidence that the true mean is different from 60 wpm? Give a complete answer, using a significance level of .05. (12 points)
B. Suppose the person conducting the research had believed, before collecting data, that the graduates weren't as good as claimed. Would this belief have changed the analysis you performed in part (A)? (12 points)
C. Construct a 99% confidence interval for the true mean of the population. What conclusion can you draw from this interval, and do you have evidence to reject the claim that the average graduate can type 60 wpm? (6 points)
Answer:
Reject at 5%, accept at 1% the null hypothesis
Step-by-step explanation:
Set up hypotheses as
[tex]H_0: \bar x = 60\\H_a: \bar x < 60[/tex]
(Left tailed test)
Population std dev = 6
Sample std error = [tex]\frac{6}{\sqrt{49} } \\=0.8555[/tex]
Mean difference = -1.5
Since sigma is known we can use Z test
Z = mean diff/std error = -1.7533
p value = 0.039
a) Since p value <0.05 we reject H0. There is evidence that the true mean is different from 60 wpm
b) Yes, because his sample would have been biased since he may want to prove his belief so slow or inefficient persons he would have selected in the sample.
c) For 99% confidence interval critical value = 2.58
Confidence interval for population mean = 58.5±2.58*std error
=(56.2928, 60.7072)
Since this contains 60, the hypothesized mean, we accept null hypothesis.
we do not have evidence to reject the claim that the average graduate can type 60 wpm at 1% level of significance.
You are assigned to the jury of a paternity case; determining whether the the child’s guardian father is actually his biological father. After listening to all the witnesses regarding the child’s family, you are 75% convinced that the guardian father is the child’s biological father. Additionally, you have been presented with laboratory blood tests indicating that the child has blood type B. The laboratory further provided population statistics stating that(a) If the guardian father is assumed to be the biological father, the child has 50% chance of having blood type B.(b) If the guardian father is assumed to NOT be the biological father, the child has 0.91% chance to have a blood type OTHER than B.How confident are you(what is the probability) that the guarding father is the child’s biological father
Answer:
Answer: 0.6022
Consider the following calculation
Step-by-step explanation:
Let F shows the event that guardian father is biological father. So
P(F) = 0.75
By the complement rule,
P(F') = 1 - P(F) =1 - 0.75 = 0.25
Let B shows the event that child has blood type B. So we have
P(B|F) = 0.50, P(B' |F') = 0.0091
By the complement rule we have
P(B|F') = 1 - P(B' |F') = 0.9909
The probability that the guarding father is the child’s biological father given that child have blood type B is
P(BFPF) P(F|B) = PRI P(BF)P(F) + P(BF)P(F) 0.50 -0.75 0.50 -0.75 +0.9909 - 0.25
0.375 0.622725 = 0.6022
Answer: 0.6022
Ten experts rated a newly developed chocolate chip cookie on a scale of 1 to 50. Their ratings were:
34, 35, 41, 28, 26, 29, 32, 36, 38, and 40.
1. What is the mean deviation of the ratings?
Select one:
a. 8.00
b. 4.12
c. 12.67
d. 0.75
Answer:
Option B.
Step-by-step explanation:
The given data set is
34, 35, 41, 28, 26, 29, 32, 36, 38, 40
We need to find the mean deviation of the given data.
Number of observations, n = 10
Mean of the data is
[tex]Mean=\dfrac{\sum x}{n}[/tex]
[tex]Mean=\dfrac{34+35+41+28+26+29+32+36+38+40}{10}[/tex]
[tex]Mean=\dfrac{339}{10}[/tex]
[tex]Mean=33.9[/tex]
Formula for mean deviation is
[tex]\text{Mean deviation}=\dfrac{\sum |x-mean|}{n}[/tex]
[tex]\sum |x-mean|=|34-33.9|+|35-33.9|+|41-33.9|+|28-33.9|+|26-33.9|+|29-33.9|+ |32-33.9|+|36-33.9|+|38-33.9|+|40-33.9|=41.2[/tex]
[tex]\text{Mean deviation}=\dfrac{41.2}{10}[/tex]
[tex]\text{Mean deviation}=4.12[/tex]
The mean deviation of the ratings is 4.12.
Therefore, the correct option is B.
Answer:
b. 4.12
Step-by-step explanation:
We have been given that 10 experts rated a newly developed chocolate chip cookie on a scale of 1 to 50. Their ratings were:
34, 35, 41, 28, 26, 29, 32, 36, 38, and 40.
First of all, we will find the mean of the ratings.
[tex]\text{Mean of ratings}=\frac{34+35+41+28+26+29+32+36+38+40}{10}[/tex]
[tex]\text{Mean of ratings}=\frac{339}{10}[/tex]
[tex]\text{Mean of ratings}=33.9[/tex]
Let us find absolute deviation of each point from mean.
[tex]|34-33.9|=0.1[/tex]
[tex]|35-33.9|=1.1[/tex]
[tex]|41-33.9|=7.1[/tex]
[tex]|28-33.9|=5.9[/tex]
[tex]|26-33.9|=7.9[/tex]
[tex]|29-33.9|=4.9[/tex]
[tex]|32-33.9|=1.9[/tex]
[tex]|36-33.9|=2.1[/tex]
[tex]|38-33.9|=4.1[/tex]
[tex]|40-33.9|=6.1[/tex]
Now we will use mean deviation formula.
[tex]\text{Absolute mean deviation}=\frac{\Sigma |x-\mu|}{N}[/tex], where,
[tex]\mu=\text{Mean}[/tex] and N = Number of data points.
[tex]MD=\frac{0.1+1.1+7.1+5.9+7.9+4.9+1.9+2.1+4.1+6.1}{10}[/tex]
[tex]MD=\frac{41.2}{10}[/tex]
[tex]MD=4.12[/tex]
Therefore, the mean deviation of the ratings is 4.12 and option 'b' is the correct choice.
The mean time taken to design a house plan by 38 architects was found to be 22 hours with a standard deviation of 3.70 hours. a. Construct a 95% confidence interval for the population mean μ. Round your answers to two decimal places.
To construct a 95% confidence interval for the population mean of the time taken to design a house plan, use the formula which states 21.08 to 22.92 hours.
Explanation:To construct a 95% confidence interval for the population mean, we can use the formula:
Confidence Interval = mean ± (critical value) * (standard deviation/sqrt(sample size))
Given that the mean time taken to design a house plan is 22 hours, the standard deviation is 3.70 hours, and the sample size is 38, we can calculate the confidence interval:
Confidence Interval = 22 ± (1.96) * (3.70/sqrt(38))
Calculating this gives us a confidence interval of approximately 21.08 to 22.92 hours.
Learn more about Confidence Intervals here:https://brainly.com/question/34700241
#SPJ11
Find the area of the following regions, expressing your results in terms of the positive integer n ≥ 2. The region bounded by f(x)=x and g(x)=x^1/n, for x≥0
Answer:
The area of the searched region is [tex]A= a+b+ \frac{2n}{n+1}- \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1}-2[/tex]
Step-by-step explanation:
If you want to find the area of a region bounded by functions f(x) and G(x) between two limits (a,b), you have to do a double integral. you must first know which of the functions is greater than the other for the entire domain.
In this case, for 0<x<1, f(x)<g(x)
while for 1<x, g(x)<f(x).
Therefore if our domain is all real numbers superior to 0 (where the limit 0<a<1 and 1<b), we have to do 2 integrals:
A=A(a<x<1)+A(1<x<b)
[tex]A(a<x<1)=\int\limits^1_a {\int\limits^{x^{\frac{1}{n}} }_{x} } {} \, dy } \, dx = \int\limits^1_a {x^{\frac{1}{n} } -x \, dx = a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1}[/tex]
[tex]A(1<x<b)=\int\limits^b_1 {\int\limits^{x}_{x^{\frac{1}{n} } } {} \, dy } \, dx = \int\limits^b_1 {x-x^{\frac{1}{n} } \, dx =b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1}[/tex]
[tex]A=a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1} + b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1} = a+b+ \frac{2n}{n+1} - \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1} -2[/tex]
When individuals in a sample of 150 were asked whether or notthey supported capital punishment, the following information wasobtained.
Doyousupport Numberof
capitalpunishment? individuals
Yes 40
No 60
No Opinion 50
We are interested in determining whether or not the opinionsof the individuals (as to Yes, No and No Opinion) are uniformlydistributed.
The expected frequency for each group is?
a. 0.333
b. 0.50
c. 1/3
d. 50
Answer:
They are not uniformly distributed.
The expected frequency of each group is 50
Step-by-step explanation:
In probability distributions, uniform distribution refers to a probability distribution for which all of the values that a random variable can take on occur with equal probability.
In other words, for n number of events, the probability of occurrence 1,2,3,4......n is 1/n
There are 3 possible occurrence in the question above
1. Yes
2. No
3. No Opinion.
For the above events to have a uniform distribution, then they must have a probability of ⅓ each.
The expected frequency of each would then be ⅓ of n where n = 150
⅓ of 150 = 50
Sophia buys a certain brand of cereal that costs $5 per box. Yani changes to a super-saving brand of the same size. The equation shows the price, y, as a function of the number of boxes, x, for the new brand.
y = 4.35x
Part A: How many more dollars is the price of a box Sophia's original brand of cereal than the price of a box of the super-saving cereal? Show your work.
Part B: How much money does she save each month with the change in cereal brand if he buys 5 cereal boxes each month? Show your work.
$ 0.65 more dollars is the price of a box Sophia's original brand of cereal than the price of a box of the super-saving cereal
Amount saved each month with the change in cereal brand if he buys 5 cereal boxes each month is $ 3.25
Solution:
Given that Sophia buys a certain brand of cereal that costs $5 per box
The equation shows the price, y, as a function of the number of boxes, x, for the new brand:
y = 4.35x
Part A:
New brand, y = 4.35x where y is the price and x is the number of boxes
Original brand, y = 5x since given that cereal that costs $5 per box
If Sophia old cereal preference was $5, and the equation shows that the new cereal preference is $4.35, if I subtract the amount of the new one from the old,
we get , 5 - 4.35 = 0.65
Therefore, $ 0.65 more dollars is the price of a box Sophia's original brand of cereal than the price of a box of the super-saving cereal
Part B:
Given that if he buys 5 cereal boxes, let us calculate price for old and new brand
New brand, y = 4.35x
New brand, y = 4.35(5) = 21.75
Original brand, y = 5x = 5(5) = 25
Amount saved = $ 25 - $ 21.75 = $ 3.25
Thus amount saved each month with the change in cereal brand if he buys 5 cereal boxes each month is $ 3.25
Use the given data to find the best predicted value of the response variable. Ten pairs of data yield requals0.003 and the regression equation ModifyingAbove y with caret equals2plus3x. Also, y overbarequals5.0. What is the best predicted value of y for xequals2?
The best predicted value of 'y' when 'x' is 2, using the linear regression equation ŷ = 2 + 3x, is 8. However, the correlation coefficient of 0.003 indicates this prediction may not be accurate due to the weak linear relationship between the variables.
Explanation:The question is about predicting a value using a given linear regression equation. Given the regression equation ŷ = 2 + 3x, to predict 'y' when x = 2, we just replace 'x' with '2' in the regression equation. The equation becomes ŷ = 2 + 3*2 = 2 + 6 = 8. Therefore, the best predicted value of 'y' when 'x' is 2 is 8.
Note that the provided correlation coefficient (r) of 0.003 indicates a very weak linear relationship between the variables, hence this prediction might not be very reliable.
We use the regression line equation to make the prediction, this line of best fit has been calculated using the data provided. These predictions are most reliable when there is a strong correlation between the variables used.
Learn more about Linear Regression here:
https://brainly.com/question/36829492
#SPJ12
A restaurant owner is interested in the proportion of his customers who order dessert. He looks at 65 randomly selected receipts. Match the vocabulary word with its corresponding example.
The proportion of the 65 randomly selected customers who ordered dessert.
The list of the 65 Yes or No answers for whether each customer ordered dessert.
The 65 restaurant patrons whose receipts were observed by the owner.
The proportion of all customers who order dessert.
The answer: Yes or No to whether a customer ordered dessert.
All customers who come to the restaurant.
a. Data
b. Sample
c. Population
d. Statistic
e. Parameter
f. Variable
The context of a restaurant owner examining dessert orders involves several statistical concepts: Data is the list of Yes or No answers, the Sample is the 65 restaurant patrons, Population is all restaurant customers, the Statistic is the proportion of the sample who ordered dessert, the Parameter is the proportion of all customers who order dessert, and the Variable is whether an individual customer ordered dessert.
Explanation:The question involves the concepts of statistics in Mathematics, particularly focusing on how different terminologies are used. Here is the matching:
a. Data: This is the list of the 65 Yes or No answers for whether each customer ordered dessert.b. Sample: This applies to the 65 restaurant patrons whose receipts were observed by the owner.c. Population: 'All customers who come to the restaurant' is the complete set, or population in statistical terms.d. Statistic: The proportion of the 65 randomly selected customers who ordered dessert is a statistic because it provides an estimate of a particular characteristic, obtained from the sample.e. Parameter: The proportion of all customers who order dessert is the parameter. It's a numerical characteristic of the population.f. Variable: The answer: Yes or No to whether a customer ordered dessert is the variable in this context, because it can change for each customer or change over time.Learn more about Statistics Vocabulary here:https://brainly.com/question/33544588
#SPJ12
The test statistic of z equals=2.17 is obtained when testing the claim that pnot equals≠0.2170.217. a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed. b. Find the P-value. c. Using a significance level of alphaαequals=0.010.01, should we reject Upper H 0H0 or should we fail to reject Upper H 0H0?
Answer:
a) Two tailed test
Null hypothesis:[tex]p=0.217[/tex]
Alternative hypothesis:[tex]p \neq 0.217[/tex]
b) [tex]p_v =2*P(Z>2.17)=0.03[/tex]
c) If we compare the p value obtained and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
We Fail to reject the null hypothesis H0
Step-by-step explanation:
Data given and notation
n represent the random sample taken
X represent the outcomes desired in the sample
[tex]\hat p[/tex] estimated proportion of interest
[tex]p_o[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.99
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion is 0.217 or no:
a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.
Two tailed test
Null hypothesis:[tex]p=0.217[/tex]
Alternative hypothesis:[tex]p \neq 0.217[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
For this case the calculated value is given z =2.17
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
b. Find the P-value
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(Z>2.17)=0.03[/tex]
c. Using a significance level of alphaαequals=0.01, should we reject Upper H 0 or should we fail to reject Upper H 0?
If we compare the p value obtained and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
We Fail to reject the null hypothesis H0
Water is being pumped continuously from a pool at a rate proportional to the amount of water left in the pool. Initially there was 15,000 gallons of water in the pool; six minutes later there was 13,800 gallons.
At what rate was the amount of water in the pool decreasing when there were 14,000 gallons remaining and when will there be 5,000 gallons remaining?
Please show all steps.
Answer:
194.6 gpm at 14,000 gallons69.5 gpm at 5,000 gallons.Step-by-step explanation:
When a value is decreasing at a rate proportional to that value, it can be modeled by the formula
a = a0·e^(-kt)
where k is the constant of proportionality.
Alternatively, we can write the exponential function describing the pool volume* as ...
a = 15000·(138/150)^(t/6) = 15000·((138/150)^(1/6))^t
Comparing these, we see that ...
e^(-kt) = (138/150)^(t/6)
or ...
k = -ln(138/150)/6 ≈ 0.0138969
__
So, when 14000 gallons remain, the rate of decrease is ...
14000·0.0138969 ≈ 194.6 . . . gallons per minute
When 5000 gallons remain, the rate of decrease is ...
5000·0.0138969 ≈ 69.5 . . . gallons per minute
_____
* The generic form of this is ...
(initial value) · (multiplier per interval)^(number of intervals)
Here, the multiplier over a 6-minute period is 13800/15000 = 138/150, and the number of 6-minute intervals is t/6 when t is in minutes.
_____
Effectively, we make use of the fact that for ...
a = a0·e^(-kt)
the derivative is ...
da/dt = -k(a0·e^(-kt)) = -k·a
That is, k is the constant of proportionality mentioned in the first sentence of the problem statement.
Concerns about climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g., from renewable sources) and petrodiesel (from fossil fuel). Random blended fuel samples of size 35 are tested in a lab to ascertain the bio/total carbon ratio (X). If the true (i.e. population) mean is 0.948 and the true (i.e. population) standard deviation is 0.006, what is the distribution of Xbar?
Answer:
[tex]\bar X \sim N(\mu=0.948, \sigma_{\bar X}=\frac{0.006}{\sqrt{35}}=0.00101)[/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".
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".
Let X the random variable who represents the bio/total carbon ratio. We know from the problem that the distribution for the parameters for the random variable X are:
[tex]\mu = 0.948[/tex]
[tex]\sigma=0.006[/tex]
We select a sample of n=35 nails. That represent the sample size.
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]
So on this case :
[tex]\bar X \sim N(\mu=0.948, \sigma_{\bar X}=\frac{0.006}{\sqrt{35}}=0.00101)[/tex]
As reported on the package of seeds, the mean time until maturity of a certain variety of tomato plant is 70 days. It is also known that the maturity of this variety of tomato follows a normal distributed with standard deviation σ = 2.4. A researcher believes that it will indeed take more time in a given condition. To test his belief, he selects a sample of 36 plants of this variety under the given condition and measure the time until maturity. The sample mean is found to be 72 days. The value of the test statistic and p-value for testing H0: μ= 70, Ha: μ>70 are:
Group of answer choices
A. 1.00
B. 0.00
C. 0.786
A symbol used to name one or more parts of a whole or a set, or a location on a number line is a?
Answer:
Fraction
Step-by-step explanation:
Fraction is a symbol that represents a part of a whole. It consists of a numerator and a denominator. The numerator is the number above the fraction bar (also known as "Vinculum), while the denominator is the number below the fraction bar. The denominator is the total number of equal parts in a whole.
Examples of Fraction: [tex]\frac{1}{2}[/tex], [tex]\frac{2}{7}[/tex] and [tex]\frac{5}{8}[/tex].
In the first example, [tex]\frac{1}{2}[/tex], 1 is the numerator, while 2, is the denominator.
Additional Information
When the numerator is smaller than the denominator, the fraction is called "proper fraction." On the contrary, when the numerator is bigger than the denominator, the fraction is called "improper fraction."
Suppose that an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline that will have the best fuel economy. The four grades are: regular, below regular, premium, and super premium. The test car made three trial runs on the test track using each of the four grades and the miles per gallon recorded. At the 0.05 level, what is the critical value of F used to test the hypothesis that the miles per gallon for each fuel is the same. Show your work.
Miles Per Gallon
Regular: Below Regular Premium Super Premium
$39.31 36.69 38.99 40.04
39.87 40.00 40.02 39.89
39.87 41.01 39.99 39.93
A.) 3.49
B.) 4.07
C.) 2.33
D.) 3.86
E.) 3.26
Answer:
B) 4.07
Step-by-step explanation:
First we need to calculate the mean of all the data, which is the same as the mean of the means of each grade of gasoline:
Regular BelowRegular Premium SuperPremium
39.31 36.69 38.99 40.04
39.87 40.00 40.02 39.89
39.87 41.01 39.99 39.93
X1⁻=39.68 X2⁻= 39.23 X3⁻= 39.66 X4⁻= 39.95
Xgrand⁻ = (39.68+39.23+39.66+39.95)/4 = 39.63
Next we need to calculate the sum of squares within the group (SSW) and the sum of squares between the groups (SSB), and the respective degrees of freedom):
SSW = [ (39.31-39.68)² + (39.87-39.68)² + (39.87-39.68)² ] + [ (36.69-39.23)² + (40.00-39.23)² + (41.01-39.23)² ] + [ (38.99-39.66)² + (40.02-39.66)² + (39.99-39.66)² ] + [ (40.04-39.95)² + (39.89-39.95)² + (39.93-39.95)² ] = [0.2091] + [10.2129] + [0.6874] + [0.0121] = 11.12
SSW = 11.12
Degrees of freedom in this case is calculated by m(n-1), with m being the number of grades of gasoline (4) and n being the number of trial results for each one (3), so we would have 4(3-1) = 8 degrees of freedom
SSB = [ (39.68-39.63)² + (39.68-39.63)² + (39.68-39.63)²] + [ (39.23-39.63)² + (39.23-39.63)² + (39.23-39.63)² ] + [ (39.66-39.63)² + (39.66-39.63)² + (39.66-39.63)² ] + [ (39.95-39.63)² + (39.95-39.63)² +(39.95-39.63)² ] = [0.0075] + [0.48] + [0.0027] + [0.3072] = 0.7974
SSB = 0.80
For this case, the degrees of freedom are m-1, so we would have 4-1 = 3 degrees of freedom
Now we can establish the hypothesis for the test:
H0: μ1 = μ2 = μ3 = μ4
The null hypothesis states that the means of miles per gallon for each fuel are the same, indicating that the drade of gasoline does not make a difference, therefore our alternative hypothesis will be:
H1: the grade of gasoline does makes a difference
We will use the F statistic to test the hypothesis, which is calculated like follows:
F - statistic = (SSB/m-1) / (SSW/m(n-1)) = (0.80/3) / (11.12/8) = 0.19
We know that the level of significance we are using is α = 0.05, so to find the critical value F we need to look at some table of critical values for the F distribution for the 0.05 significance level (like the attached image). Then we just need to look fot the value that is located in the intersection between the degrees of freedom we have in the numerator (horizontal) and the denominator (vertical) of the statistic (3 and 8). That critical value is:
Fc = 4.07
Final answer:
The critical value of F used to test the hypothesis that the miles per gallon for each fuel is the same is 3.49.
Explanation:
To test the hypothesis that the miles per gallon for each fuel is the same, we can use an ANOVA test. The critical value of F at the 0.05 level can be found using the F-distribution table or by using statistical software. Since we have three trial runs for each grade of gasoline, we have a total of 12 observations. At a significance level of 0.05 and with 3 degrees of freedom for the numerator and 8 degrees of freedom for the denominator, the critical value of F is 3.49.
Design specifications for filling a bottled soda claim that bottles should contain 350-360 milliliters of liquid. Sample data indicate that the bottles contain an average of 355 milliliters of liquid, with a standard deviation of 2 milliliters. Is the filling operation capable of meeting the design specifications?
Answer:
It is high likely that the filling operation is capale of meeting design specifications.
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 = 355, \sigma = 2[/tex]
Is the filling operation capable of meeting the design specifications?
It will be capable if it is highly likely that the specifications will be met. A probability is said to be high likely when it is of at least 95%.
In this case, the probability of containing between 350 and 360 ml of liquid is the pvalue of Z when X = 360 subtracted by the pvalue of Z when X = 350.
X = 360
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{360 - 355}{2}[/tex]
[tex]Z = 2.5[/tex]
[tex]Z = 2.5[/tex] has a pvalue of 0.9938.
X = 350
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{350 - 355}{2}[/tex]
[tex]Z = -2.5[/tex]
[tex]Z = -2.5[/tex] has a pvalue of 0.0062.
This means that there is a 0.9938 - 0.0062 = 0.9876 = 98.76% probability that the filling operation is capable of meeting the design specifications. It is high likely that the filling operation is capale of meeting design specifications.
By determining f prime left parenthesis x right parenthesis equals ModifyingBelow lim With h right arrow 0 StartFraction f left parenthesis x plus h right parenthesis minus f left parenthesis x right parenthesis Over h EndFractionf′(x)=limh→0 f(x+h)−f(x) h, find f prime left parenthesis 7 right parenthesisf′(7) for the given function. f left parenthesis x right parenthesis equals 5 x squaredf(x)=5x2 f prime left parenthesis 7 right parenthesisf′(7)equals=nothing (Simplify your answer.)
Answer:
70 is answer
Step-by-step explanation:
Given that a function in x is
[tex]f(x) = 5x^2[/tex]
we have to find f'(7)
we know by derivative rule derivative of a function is
[tex]f'(x) = lim_({h-->0}) \frac{f(x+h)-f(x)}{h}[/tex]
For finding out at 7 we replace x by 7
[tex]f'(7) = lim_({h-->0}) \frac{f(7+h)-f(7)}{h}[/tex]
=[tex]lim\frac{5(7+h)^2-5*7^2}{h} \\= lim \frac{10h*7+h^2}{h} \\= 70+h = 70[/tex]
So f'(7) = 70
answer is 70
Answer:
f'(7)=70
Step-by-step explanation:
We have the definition of the derivative as:
[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]
Now we have a function [tex]f(x)=5x^2[/tex] and we want to approximate the first derivative around x=7, that is [tex]f'(7)[/tex].
We can replace this in the first formula as:
[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}= \lim_{h \to 0} \dfrac{5(x+h)^2-5x^2}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{5(x^2+2xh+h^2-x^2)}{h}\\\\f'(x)=\lim_{h \to 0}\dfrac{5(2xh+h^2)}{h}\\\\f'(x)=\lim_{h \to 0}5(2x+h)\\\\f'(x)=10x+lim_{h \to 0}h=10x+0=10x[/tex]
Then, the value for f'(7) is:
[tex]f'(7)=10\cdot 7=70[/tex]
Consider a manufacturing process with a quality inspection station. In the past, 15% of parts are defective. As soon as one defective part is found, the process is stopped. If 8 parts have been inspected without finding a defective part, what is the probability that at least 11 total parts will be inspected before the process is stopped?
Answer:
0.614125
Step-by-step explanation:
Given that a manufacturing process with a quality inspection station has on an average 15% of parts are defective.
As soon as one defective part is found, the process is stopped.
We find that number of defectives would be binomial because each part randomly selected has a constant probability of 0.15 being defective
Probability that at least 11 total parts will be inspected before the process is stopped/8 parts have been inspected without finding a defective part
=[tex]P(x\geq 11)/P(x=8)\\[/tex]
= Probability of 9th, 10th, 11th should not be defective
= [tex](1-0.15)^3\\= 0.614125[/tex]
A journalist reported that the average amount of time that a French person spends eating lunch at a restaurant is 22 minutes. Perform a hypothesis test to determine if a difference exists between the average time an American spends eating lunch when compared to a person from France. The following data represents the time, in minutes, that random French and American diners spent at lunch. Assume that the population variances are equal. Assume Population 1 is defined as French diners and Population 2 is defined as American diners. What is the test statistic for this hypothesis test?
American
21
17
17
20
25
16
20
16
French
24
18
20
28
18
29
17
Answer:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]
Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:
[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]
[tex]t=\frac{19 -22)-(0)}{4.095\sqrt{\frac{1}{8}+\frac{1}{7}}}=-1.416[/tex]
Step-by-step explanation:
Data given
American: 21,17,17,20,25,16,20,16 (Sample 1)
French: 24,18,20,28,18,29,17 (Sample 2)
When we have two independent samples from two normal distributions with equal variances we are assuming that
[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]
And the statistic is given by this formula:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]
Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:
[tex]S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]
This last one is an unbiased estimator of the common variance [tex]\sigma^2[/tex]
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 = \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2 \neq 0[/tex]
Our notation on this case :
[tex]n_1 =8[/tex] represent the sample size for group 1
[tex]n_2 =7[/tex] represent the sample size for group 2
[tex]\bar X_1 =19[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =22[/tex] represent the sample mean for the group 2
[tex]s_1=3.117[/tex] represent the sample standard deviation for group 1
[tex]s_2=5.0[/tex] represent the sample standard deviation for group 2
First we can begin finding the pooled variance:
[tex]S^2_p =\frac{(8-1)(3.117)^2 +(7 -1)(5.0)^2}{8 +7 -2}=16.770[/tex]
And the deviation would be just the square root of the variance:
[tex]S_p=4.095[/tex]
And now we can calculate the statistic:
[tex]t=\frac{19 -22)-(0)}{4.095\sqrt{\frac{1}{8}+\frac{1}{7}}}=-1.416[/tex]
Now we can calculate the degrees of freedom given by:
[tex]df=8+7-2=13[/tex]
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =2*P(t_{13}<-1.416) =0.1803[/tex]
So with the p value obtained and using the significance level assumed [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance we don't have significant differences between the two means.
The lumen output was determined for each of I = 3 different brands of lightbulbs having the same wattage, with J = 8 bulbs of each brand tested. The sums of squares were computed as SSE = 4776.3 and SSTr = 599.5. State the hypotheses of interest (including word definitions of parameters).
Answer:
Step-by-step explanation:
Hello!
The study variable is
X: Lumen of a bulb of the i brand. i=3
There are 3 populations of bulbs, Brand 1, Brand 2 and brand 3.
The objective is to test if the population means are equal.
The study parameters are:
μ₁: population mean lumen of the population of light bulbs of brand 1.
μ₂: population mean lumen of the population of light bulbs of brand 2.
μ₃: population mean lumen of the population of light bulbs of brand 3.
The hypothesis is:
H₀:μ₁= μ₂= μ₃= μ
H₁: At least one of the population means is different.
To test this hypothesis, considering the given information, I'll use an ANOVA test, then the statistic is defined as:
[tex]F= \frac{MSTr}{MSerror}[/tex]~[tex]F_{(I-1)(J-1)}[/tex]
Rejection region
This region is always one-tailed (right), the statistic is constructed as the mean square of the treatments divided by the mean square of the error, if the number of F is big, this means that the treatments have more effect over the populations. If the value of F is small, this means that there is no difference between the variability caused by the treatments and the one caused by the residues.
Since there is no significance level specified, I'll use α: 0.05
[tex]F_{(I-1);(J-1); 1 - \alpha } = F_{2; 7; 0.95} = 19.35[/tex]
You will reject the null hypothesis when F[tex]_{H_0}[/tex] ≥ 19.35
To calculate the statistic value you need to calculate the Mean Square of Treatments and the Mean Square of errors:
MSTr= SSTr/DfTr = 599.5/2= 299.75
MSerror= SSerror/Dferror= 4776.3/5= 955.26
F[tex]_{H_0}[/tex]= [tex]\frac{299.75}{955.26}[/tex]= 0.31
At this level the decision is to not reject the null hypothesis.
I hope it helps!
Use an Addition or Subtraction Formula to find the exact value of the expression, as demonstrated in Example 1. cos 41π 12.
Answer: [tex]\frac{\sqrt{6}-\sqrt{2}}{2} [/tex]
Step-by-step explanation:
We apply the formula [tex]\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y) [/tex].
Note that [tex]\cos(\frac{41}{12}\pi)=\cos((\frac{36}{12}+\frac{7}{12})\pi)=\cos(3\pi + \frac{7}{12})\pi) [/tex]. Take [tex]x=3\pi[/tex] and [tex]y=\frac{7}{12}\pi[/tex] in the formula above to get
[tex]\cos(\frac{41}{12}\pi)=\cos(3\pi)\cos(\frac{7}{12}\pi)-\sin(3\pi)\sin(\frac{7}{12}\pi)=(-1)\cdot \cos(\frac{7}{12}\pi)-0\cdot\sin(\frac{7}{12}\pi)=-\cos(\frac{7}{12}\pi)[/tex]
Then the value of this expression is [tex]-\cos(\frac{7}{12}\pi) [/tex]
We can use the cosine addition formula again to simplify further. Decompose the fraction in the argument as:
[tex]\cos(\frac{7}{12}\pi)=\cos((\frac{3}{12}+\frac{4}{12})\pi)=\cos((\frac{1}{4}\pi + \frac{1}{3})\pi) [/tex]
Applying the formula with [tex]x=\frac{1}{4}\pi[/tex] and [tex]y=\frac{1}{3}\pi[/tex] we obtain
[tex]\cos(\frac{7}{12}\pi)=\cos(\frac{1}{4}\pi)\cos(\frac{1}{3}\pi)-\sin(\frac{1}{4}\pi)\sin(\frac{1}{3}\pi)=\frac{\sqrt{2}}{2}\cdot\frac{1}{2} -\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{2}-\sqrt{6}}{2} [/tex]
We conclude that this expression has the value [tex]-\frac{\sqrt{2}-\sqrt{6}}{2}=\frac{\sqrt{6}-\sqrt{2}}{2} [/tex]
A professor is trying to determine if her students guessed on a certain multiple choice question. She expects that if the students guessed, the distribution of answers would be uniform for that question. She compares the observed distribution of answers with the uniform distribution. The professor conducts a chi-square Goodness-of-Fit hypothesis test at the 5% significance level.
a) The null and alternative hypotheses are: H0: The student answers have the uniform distribution. Ha: The student answers do not have the uniform distribution.
b) χ20=13.167.
c) χ20.05=7.815.
d) What conclusions can be made?
Select all that apply:
A. We should reject H0.
B. We should not reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
D. At the 5% significance level, there is not enough evidence to conclude that the students were not guessing.
Answer:
A. We should reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
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: The student answers have the uniform distribution.
H1: The student answers do not have the uniform distribution.
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 with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]
On this case we assume that the calculated statistic is given by:
Statistic calculated
[tex]\chi^2_{calc}=13.167[/tex]
P value
Assuming the we have 2 rows and 4 columns on the contingency table.
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(rows-1)(cols-1)=(2-1)(4-1)=3[/tex]
We can calculate the critical value with this formula in excel:" =CHISQ.INV(0.95,3)" On this case we got that the critical value is:
[tex]\chi^2_{crit}=7.815[/tex]
Since our calculated value is higher than the cirtical value we have enough evidence to reject the null hypothesis at the significance level of 5%.
And we can also calculate the p value given by:
[tex]p_v = P(\chi^2_{3} >13.167)=0.0043[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(13.167,3,TRUE)"
Since the p value is lower than the significance level we reject the null hypothesis at 5% of significance.
A. We should reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
The reason why we select option C is because if we reject the null hypothesis of uniform distribution then we are rejecting the claim that the students are guessing.
Juan roller skates at the constant speed of 8 miles per hour. How far can he travel in 2.4 hours.
Answer: Juan can travel 19 miles in 2.4 hours at a speed of 8 miles per hour
Step-by-step explanation:
Juan roller skates at the constant speed of 8 miles per hour. Distance travelled is expressed as
Distance = speed × time
Therefore, the distance that Juan can travel in 2.4 hours is
Distance = 2.4 × 8 = 19.2 miles
Approximating to the nearest whole number, it becomes 19 miles
Determine whether the random variable is discrete or continuous. In each case, state the possible values of the random variable.
(a)The number of points scored during a basketball game.
(b) The weight of a T-bone steak.
(A) Is the number of points scored during a basketball game discrete or continuous?
a.The random variable is continuous. The possible value are x=0,1,2,
b.The random variable is discrete. The possible values are x ? 0
Answer:
a) Discrete
b) continuous
Step-by-step explanation:
Given that there are two random variables
1) The number of points scored during a basketball game
2) The weight of a
T bone steak.
The number of points scored during a basketball game
-- Can take values as 0,1,2.....
This can be counted i.e. we can have a one to one correspondence with natural numbers.
So this is a discrete variable
The weight of a T-bone steak
-- This can take any value in decimal or fraction. This can be between an interval comprising all values over the interval. Hence we cannot set one to one correspondence with set of natural numbers.
So continuous variable.
The concept of aging as a result of cellular duplication errors is based on the fact that the body's ability to make new cells that are exact copies of the original ones: a. stops at about age 80. b. continues unchanged throughout life. c. becomes less able to repair duplication errors. d. stops at about age 18.
Answer:
Becomes less able to repair duplication errors
Step-by-step explanation:
This is premised on the fact that aging has been connected with the deterioration of DNA maintenance and repair machinery, which tends to lose its ability to replicate new cell as a person age with time.
Suppose a 95% confidence interval for the average amount of weight loss on a diet program for males is between 13.0 and 18.0 pounds. These results were based on a sample of 42 male participants who were deemed to be overweight at the start of the 4-month study. What is the margin of error for this study?
Answer:
The margin of error for this study is 2.5 pounds.
Step-by-step explanation:
The margin of error is the subtraction of the mean by the lower end of the confidence interval, and this must be equal to the subtraction of the upper end to the mean.
In this problem, we have that:
M - 13 = 18 - M
2M = 31
M = 15.5
The mean is 15.5 pounds.
So the margin of error for this study is 15.5 - 13 = 18 - 15.5 = 2.5 pounds.
Which of the following is the upper critical value of z (z*) for an 80% confidence interval?
a. 1.96
b. .84
c. 2.33
d. 1.45
e. 1.28
Answer:
e. 1.28
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.8}{2} = 0.1[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex]. This is our critical value.
So it is z with a pvalue of [tex]1-0.1 = 0.9[/tex], so [tex]z = 1.28[/tex]
The correct answer is:
e. 1.28
Great Lakes Health Care common stock offers an expected total return of 9.2 percent. The last annual dividend was $2.10 a share. Dividends increase at a constant 2.6 percent per year. What is the dividend yield?
A. 3.75 percent
B. 4.20 percent
C. 4.55 percent
D. 5.25 percent
E. 6.60 percent
Answer:
E. 6.60 percent
Step-by-step explanation:
We have been given that Great Lakes Health Care common stock offers an expected total return of 9.2 percent. The last annual dividend was $2.10 a share. Dividends increase at a constant 2.6 percent per year.
We will use total return formula to answer our given problem.
[tex]\text{Total return}=\text{Dividend yield}+\text{Growth rate}[/tex]
Upon substituting our given values in above formula, we will get:
[tex]9.2\%=\text{Dividend yield}+2.6\%[/tex]
[tex]\text{Dividend yield}=9.2\%-2.6\%[/tex]
[tex]\text{Dividend yield}=6.6\%[/tex]
Therefore, the dividend yield would be 6.60% and option E is the correct choice.
Assume that you have a sample of n 1 equals 8n1=8, with the sample mean Upper X overbar 1 equals 42X1=42, and a sample standard deviation of Upper S 1 equals 4S1=4, and you have an independent sample of n 2 equals 15n2=15 from another population with a sample mean of Upper X overbar 2 equals 34X2=34 and a sample standard deviation of Upper S 2 equals 5S2=5. What assumptions about the two populations are necessary in order to perform thepooled-variance t test for the hypothesis Upper H 0 : μ 1 equals μ 2H0: μ1=μ2 against the alternative Upper H 1 : μ 1 >μ 2H1: μ1>μ2 and make a statistical decision?
Answer:
Check the explanation below
Step-by-step explanation:
Hello!
To make a pooled variance t-test you have to make the following assumptions:
The study variables X₁ and X₂ must be independent.
Both variables should have a normal distribution, X₁~N(μ₁; σ₁²) and X₂~N(μ₂; σ₂²)
The population variances should be equal but unknown, σ₁² = σ₂² = ?.
You have the information of two samples:
Sample 1
n₁=8
sample mean X[bar]₁= 42
sample standard deviation S₁=4
Sample 2
n₂=15
sample mean X[bar]₂= 34
sample standard deviation S₂= 5
For the hypothesis:
H₀: μ₁ = μ₂
H₁: μ₁ > μ₂
The statistic is:
t= (X[bar]₁ - X[bar]₂) - (μ₁ - μ₂) ~[tex]t_{n_1 + n_2 - 2}[/tex]
Sa[tex]\sqrt{\frac{1}{n_1} + \frac{1}{n_2} }[/tex]
Sa²= [tex]\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}[/tex]
Sa²= 22
Sa= 4.69
[tex]t_{H0}[/tex]= 3.8962 ≅ 3.9
The critical region is one-tailed, for example for α: 0.05
[tex]t_{n_1 + n_2 - 2; 1 - \alpha } = t_{21; 0.95} = 1.721[/tex]
Since [tex]t_{H0}[/tex] > 1.721, then the decision is to reject the null hypothesis.
I hope it helps!
A leprechaun places a magic penny under a girl's pillow. The next night there are 2 magic pennies under her pillow. The following morning she finds four pennies. Apparently, while she sleeps each penny turns into two magic pennies. The total number of pennies seen under the pillow each day is the grand total; that is, the pennies from each of the previous days are not being stored away until more pennies magically appear. How many days would elapse before she has a total of more than $2 billion ? (Proceed by trial and error.)
nothing days
(Type a whole number.) PLEASE HURRY
Answer: 31
Step-by-step explanation:
2^x=2 000 000 000
log2^x=log2 000 000 000
xlog2 = log 2 000 000 000
x= log (2000 000 000)/log 2
x= 30.897352854
round to 31
gotchu bro