Answer:
We will need about 3.3 gallons of red paint.
Step-by-step explanation:
Let
r = amount of red paintw = amount of white paintThe amount of red paint is half the amount of white paint.
r = 1/2 w
w = 2r [1]
We need to make a total of 10 gallons. Then,
r + w = 10 [2]
Replacing [1] in [2]
r + (2r) = 10
3r = 10
r = 3.3
We will need about 3.3 gallons of red paint.
To make 10 gallons of pink paint with red paint being half the amount of white paint, you would need 5 gallons of red paint.
Explanation:The amount of red paint needed to make 10 gallons of pink paint when the amount of red paint is half that of white paint:
Let the amount of white paint be 'x', then the amount of red paint needed is 'x/2'.Given that white paint is 10 gallons, so 'x = 10'.Therefore, the amount of red paint needed is '10/2 = 5 gallons'.If you roll a pair of fair dice, what is the probability of each of the following? (round all answers to 4 decimal places, .XXXX)
a) getting a sum of 1?
b) getting a sum of 5?
c) getting a sum of 12?
Answer: The required probabilities are
[tex](a)~0\\\\(b)\dfrac{1}{9},\\\\(c)~\dfrac{1}{36}.[/tex]
Step-by-step explanation: Given that a pair of fair dice is rolled.
We are to find the probability of getting
(a) getting a sum of 1.
b) getting a sum of 5.
c) getting a sum of 12.
Let S be the sample space for the experiment of rolling a pair of fair dice.
Then, S = {(1,1), (1,2), (1,3), (1, 4), (1,5), (1,6), . . . , (6,5), (6,6)}.
And, n(S) =36.
(a) Let E denote the event of getting a sum of 1.
Since the sum of the numbers on two dice is minimum 2, so
E = { } ⇒ n(E) = 0.
Therefore, the probability of event E is
[tex]P(E)=\dfrac{n(E)}{n(S)}=\dfrac{0}{36}=0.[/tex]
(b) Let F denote the event of getting a sum of 5.
Then,
F = {(1,4), (2,3), (3,2), (4,1)} ⇒ n(F) = 4.
Therefore, the probability of event F is
[tex]P(F)=\dfrac{n(F)}{n(S)}=\dfrac{4}{36}=\dfrac{1}{9}.[/tex]
(c) Let G denote the event of getting a sum of 12.
Then,
G = {(6,6)} ⇒ n(G) = 1.
Therefore, the probability of event G is
[tex]P(G)=\dfrac{n(G)}{n(S)}=\dfrac{1}{36}.[/tex]
Thus, the required probabilities are
[tex](a)~0\\\\(b)\dfrac{1}{9},\\\\(c)~\dfrac{1}{36}.[/tex]
The probability of getting a sum of 1, the sum of 5, and the sum of 12 are 0, 1/9, and 1/36 respectively
Probability is the likelihood or chance that an event will occur.
For a pair of rolled dice, the total outcome will be 6² = 36
Probability = Expected outcome/Total outcome
a) Pr(getting a sum of 1) = 0/36 = 0
Note that since we have a pair of dice, the least sum we can have is 2
b) The event for getting a sum of 5 are (1, 4), (4, 1), (3, 2), (2, 3)
n(E) = 4
Pr( getting a sum of 5) = 4/36 = 1/9
c) The event for getting a sum of 12 are (6, 6)
n(E) = 1
Pr( getting a sum of 12) = 1/36
Hence the probability of getting a sum of 1, the sum of 5, and the sum of 12 are 0, 1/9, and 1/36 respectively
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A certain human red blood cell has a diameter of 0.000007 meters. Which expression represents this diameter, in meters, in scientific notation?
A) 7* 10^-6
B) 7* 10^-5
C) 7* 10^6
D) 7* 10^5
The analyst in the Dorben Reference Library decides to use the work sampling technique to establish standards. Twenty employees are involved. The operations include cataloging, charging books out, returning books to their proper location, cleaning books, record keeping, packing books for shipment, and handling correspondence. A preliminary investigation resulted in the estimate that 30 percent of the time of the group was spent in cataloging. How many work sampling observations would be made if it were desirable to be 95 percent confident that the observed data were within a tolerance of ±10 percent of the population data? Describe how the random observations should be made.
Answer:
At least 81 observations should be made to be 95% confident that the observed data is within a tolerance of ±10 percent of the population data.
Step-by-step explanation:
The following equation is used to compute the minimum sample size required to estimate the population proportion within the required margin of error:
n≥ p×(1-p) × [tex](\frac{z}{ME} )^2[/tex] where
n is the sample sizep is the estimated proportion of the time of the group was spent in cataloging (30% or 0.30)z is the corresponding z-score for 95% confidence level (1.96) ME is the margin of error (tolerance) in the estimation (10% or 0.10)Then, n≥ 0.30×0.70 × [tex](\frac{1.96}{0.10} )^2[/tex] ≈ 80.67
At least 81 observations should be made to be 95% confident that the observed data is within a tolerance of ±10 percent of the population data.
Random observations should include different employees, and sampling time should also be random.
In a study of automobile collision insurance costs, a random sample of n = 35 repair costs of front-end damage caused by hitting a wall at a specified speed had a mean of $1,438. (a) Given that σ = $269 for such data, what can be said with 98% confidence about the maximum error if x = $1, 438 is used as an estimate of the average cost of such repairs.
Answer: The maximum error = $105.76.
Step-by-step explanation:
Formula to find the maximum error:
[tex]E= z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where n= sample size.
[tex]\sigma[/tex] = Population standard deviation
z*= Critical value(two-tailed).
As per given , we have
[tex]\overline{x}=1438[/tex]
n= 35
[tex]\sigma=269[/tex]
For 98% confidence , the significance level = [tex]1-0.98=0.02[/tex]
By z-table , the critical value (two -tailed) =[tex]z^* = z_{\alpha/2}=z_{0.01}=2.326[/tex]
Now , the maximum error = [tex]E= (2.326)\dfrac{269}{\sqrt{35}}[/tex]
[tex]E= (2.326)\dfrac{269}{5.9160797831}[/tex]
[tex]E= (2.326)\times45.4692989044=105.761589252\pprox105.76[/tex]
Hence, With 98% confidence level , the maximum error = $105.76.
Search the Internet to find the density of surface seawater in kilograms/cubic meter.
Answer:
1020–1029 kg/m3
Step-by-step explanation:
The density of seawater at the surface of the ocean varies from 1,020 to 1,029 kilograms per cubic meter.
The density of seawater is a function of temperature, salinity, and pressure.
To measure the density of ocean water a sample of sea water is collected and brought into the laboratory to be measured. Salinity, temperature and pressure are measured to find density. The formula for measuring density is p(density) = m(mass) / V(volume).
Answer:
1020 to 1029.
Step-by-step explanation:
I searched it like it says in the unit actinity course Geometry semester B.
An advice columnist asks readers to write about their marriages. The results indicate that 79% of respondents would not marry the same partner if they could do it all over again. Which of the following statements are true?A. It is likely that this percentage is higher than the true ppopulation proportion, since people who aren;t happy in their marriages are more likely to respond than those who are happy.B. It is likely that this result is an accurate reflection of the population.C. There is no way of predicting whether the results are biased or not.D. It is likely that the results are not accurate, since people tend to lie in voluntary response surveys.E. It is likely that this result is lower than the true population proportion since persons unhappy in thier marriages are unlikely to respond.
Answer:
a
Step-by-step explanation:
i think its most suitable
Select Independent or Not independent for each description of events.
look at the image
Answer:
Not Independent , Independent , Not Independent
Step-by-step explanation:
Two events A and B are said to be independent when:
P(A ∩ B) = P(A) × P(B)
In the question P(A|B) is given i.e the probability of event A given that event B has already occured.
P(A|B) = P(A ∩ B) ÷ P(B)
If A and B are independent , the probability of A is not affected by B.
In terms of equation:
P(A|B) = P(A ∩ B)÷ P(B)
= [P(A)×P(B)] ÷ P(B)
= P(A)
Only in the second option we can see :
P(A|B) = P(A) = 0.2
Hence the events A and B are independent only in second case.
solve triangle ABC
c = 10, B = 35°, C = 65°
Question options: A = 80°, a = 10, b = 6.3
A = 80°, a = 6.3, b = 10.9
A = 80°, a = 10.9, b = 6.3
A = 80°, a = 73.6, b = 6.3
That figure obviously doesn't go with this problem. It doesn't matter; this is triangle ABC labeled the usual way.
c = 10, B = 35°, C = 65°
We have two angles and a side. The third angle is obviously
A = 180° - 35°- 65° = 80°
The remaining sides are given by the Law of Sines,
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
[tex]a = \dfrac{c \sin A}{\sin C} = \dfrac{10 \sin 80}{\sin 65} = 10.866[/tex]
[tex]b = \dfrac{c \sin B}{\sin C} = \dfrac{10 \sin 35}{\sin 65} = 6.328[/tex]
Answer: A=80°, a=10.9, b=6.3, third choice
Consider the following hypothesis test:H 0: = 17H a: 17A sample of 40 provided a sample mean of 14.12. The population standard deviation is 4.a. Compute the value of the test statistic (to 2 decimals). (If answer is negative, use minus "-" sign.)b. What is the p-value (to 4 decimals)?c. Using = .05, can it be concluded that the population mean is not equal to 17? SelectYesNoItem 3Answer the next three questions using the critical value approach.d. Using = .05, what are the critical values for the test statistic (to 2 decimals)? ±e. State the rejection rule: Reject H 0 if z is Selectgreater than or equal togreater thanless than or equal toless thanequal tonot equal toItem 5 the lower critical value and is Selectgreater than or equal togreater thanless than or equal toless thanequal tonot equal toItem 6 the upper critical value.f. Can it be concluded that the population mean is not equal to 17?
Answer:
We conclude that the population mean is not equal to 17.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 17
Sample mean, [tex]\bar{x}[/tex] = 14.12
Sample size, n = 40
Alpha, α = 0.05
Population standard deviation, σ = 4
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 17\\H_A: \mu \neq 17[/tex]
We use Two-tailed z test to perform this hypothesis.
a) Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{14.12 - 17}{\frac{4}{\sqrt{40}} } = -4.5536[/tex]
b) P-value can be calculated from the standard z-table.
P-value = 0.0000
c) Since the p-value is less than the significance level, we reject the null hypothesis and accept the alternate hypothesis. Thus, the population mean is not equal to 17
d) Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]
e) Rejection Rule:
We reject the null hypothesis if it is less than lower critical value and greater than the upper critical value
If the z-statistic lies outside the acceptance region which is from -1.96 to +1.96, we reject the null hypothesis.
f) Since the calculated z-stat lies outside the acceptance region, we reject the null hypothesis and accept the alternate hypothesis. Thus, the population mean is not equal to 17.
The test statistic is -1.78 and the p-value is 0.0761, indicating that we fail to reject the null hypothesis. Therefore, it cannot be concluded that the population mean is not equal to 17.
Explanation:The test statistic can be calculated using the formula:
test statistic = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
Plugging in the given values, we get:
test statistic = (14.12 - 17) / (4 / sqrt(40))
Calculating this gives us a test statistic value of -1.78.
The p-value can be calculated using the test statistic. We need to find the probability that a test statistic at least as extreme as -1.78 would occur assuming the null hypothesis is true. Using a standard normal distribution table or software, we find the p-value to be approximately 0.0761.
Since the p-value is greater than the significance level (alpha = 0.05), we fail to reject the null hypothesis. Therefore, we can conclude that there is not enough evidence to suggest that the population mean is not equal to 17.
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Find the volume of the region bounded by the paraboloid z=x2+y2 and the plane x+y=1 in the first octant.
Answer:
Step-by-step explanation:
Given
Paraboloid [tex]z=x^2+y^2[/tex] and Plane [tex]x+y=1[/tex] are bounded to get the area
In first octant
value of x varies from [tex]0\leq x\leq 1[/tex]
value of y varies from [tex]0\leq y\leq 1[/tex]
Volume [tex]V=\int_{0}^{1}\int_{0}^{1-x}\left ( x^2+y^2\right )dydx[/tex]
[tex]V=\int_{0}^{1}\left [ x^2y+\frac{y^3}{3}\right ]_{0}^{1-x}[/tex]
[tex]V=\int_{0}^{1}\left [ x^2(1-x)+\frac{(1-x)^3}{3}\right ]dx[/tex]
[tex]V=\int_{0}^{1}\left ( \frac{-4x^3}{3}+2x^2-x+\frac{1}{3}\right )dx[/tex]
[tex]V=\frac{1}{6}[/tex]
The volume of the region bounded by the paraboloid z=x2+y2 and the plane x+y=1 in the first octant is 1/6
The paraboloid and the plane are given as:
[tex]z=x^2+y^2[/tex]
[tex]x+y = 1[/tex]
Make y the subject
[tex]y = 1 - x[/tex]
The plane is in the first octant.
So, the volume is represented as:
[tex]V = \int\limits^1_0 { \int\limits^y_0 {z^2} \, dy } \, dx[/tex]
The integral becomes
[tex]V = \int\limits^1_0 { \int\limits^{1-x}_0 {x^2 + y^2} \, dy } \, dx[/tex]
Integrate with respect to y
[tex]V = \int\limits^1_0 [x^2y + \frac 13y^3]\limits^{1-x}_0 } \, dx[/tex]
Expand
[tex]V = \int\limits^1_0 [x^2(1-x) + \frac 13(1-x)^3]} \, dx[/tex]
Expand
[tex]V = \int\limits^1_0 [x^2-x^3 + \frac 13(1 -x + x^2 - x^3)]} \, dx[/tex]
Expand
[tex]V = \int\limits^1_0 [x^2-x^3 + \frac 13 -x + x^2 - \frac {x^3}3]} \, dx[/tex]
Evaluate the like terms
[tex]V = \int\limits^1_0 [ \frac 13 -x + 2x^2 - \frac {4x^3}3]} \, dx[/tex]
Integrate
[tex]V = [\frac 13x -\frac 12x^2 + \frac 23x^3 - \frac 13x^4]\limits^1_0[/tex]
Expand
[tex]V = \frac 13(1) -\frac 12(1)^2 + \frac 23(1)^3 - \frac 13(1)[/tex]
[tex]V = -\frac 12(1)^2 + \frac 23(1)^3[/tex]
Evaluate the exponents
[tex]V = -\frac 12 + \frac 23[/tex]
Add the fractions
[tex]V = \frac{-3 + 4}{6}[/tex]
[tex]V = \frac{1}{6}[/tex]
Hence, the volume of the region is 1/6
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Helen plays basketball. For free throws, she makes the shot 78% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.78. D = the event Helen makes the second shot. P(D) = 0.78. The probability that Helen makes the second free throw given that she made the first is 0.86. What is the probability that Helen makes both free throws?
Answer:
0.6708 or 67.08%
Step-by-step explanation:
Helen can only make both free throws if she makes the first. The probability that she makes the first free throw is P(C) = 0.78, now given that she has already made the first one, the probability that she makes the second is P(D|C) = 0.86. Therefore, the probability of Helen making both free throws is:
[tex]P(C+D) = P(C) *P(D|C) = 0.78*0.86\\P(C+D) = 0.6708[/tex]
There is a 0.6708 probability that Helen makes both free throws.
A vendor sells ice cream from a cart on the boardwalk. He offers vanilla, chocolate, strawberry, blueberry, and pistachio ice cream, served on either a waffle, sugar, or plain cone. How many different single-scoop ice-cream cones can you buy from this vendor?
Answer:
The different single-scoop ice-cream cones can you buy from this vendor=12
Step-by-step explanation:
Given that a vendor sells ice cream from a cart on the boardwalk. He offers vanilla, chocolate, strawberry, blueberry, and pistachio ice cream, served on either a waffle, sugar, or plain cone.
Different types of icecreams are 4
Types of cones are 3
For each ice cream say we can have either waffle, sugar or plain cone.
i.e. for each type of ice cream, we have 3 different cones.
Hence for 4 types of ice creams we have 3*4=12 different single scoop ice creamcones.
There are 15 different single-scoop ice-cream cones that can be bought from the vendor.
Explanation:To calculate the number of different single-scoop ice-cream cones that can be bought from the vendor, we need to multiply the number of ice cream flavors with the number of cone types. The vendor offers 5 flavors and 3 cone types, so the total number of different single-scoop ice-cream cones that can be bought is 5 x 3 = 15.
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A company uses a combination of three components- A, B and C to create three different drone designs. The first design Glider uses 3 parts of component A and 2 parts of components B. Design Blimp uses 2 parts of component B and C, and the last design, Pilot uses one part of each component. A sample of 75 components, 25 A, 25 B, 25 C, will be used to make prototypes for the various designs. If 30 components are selected at random, what is the likelihood two prototypes of each design can be made?
Answer:
Consider the following calculations
Step-by-step explanation:
Since 1 Blimp uses 2 components of B and C each
=> choosing 2 components of B(remaining after using in other prototypes) for 1st model= 22C2
choosing 2 components of B(remaining after using in other prototypes) for 2nd model= 21C2
choosing 2 components of B(remaining after using in other prototypes) for 3rd model= 20C2
choosing 2 components of B(remaining after using in other prototypes) for 4th model= 19C2
choosing 2 components of B(remaining after using in other prototypes) for 5th model= 18C2
and choosing 2 components of C(remaining after using in other prototypes) = 24C2
Similarly for C
P(5 prototypes of Blimp created)=[(22C2 / 25C2 )*(24C2 / 25C2 )] + [(21C2 / 25C2 )*(23C2 / 25C2 )]+[(20C2 / 25C2 )*(22C2 / 25C2 )]+[(19C2 / 25C2 )*(21C2 / 25C2 )]+[(18C2 / 25C2 )*(20C2 / 25C2 )]
Help!!!
Given the function f(x) = x/3 + 7. Find the input for which f(x) = 13
[tex]
\begin{cases}
f(x)=x/3+7\\
f(x)=13
\end{cases}
[/tex]
We find that
[tex]13=x/3+7\Longrightarrow x=\boxed{18}[/tex]
Hope this helps.
An article in the Journal of Composite Materials (Vol 23, 1989, p. 1200) describes the effect of delamination on the natural frequency of beams made from composite laminates. Five such delaminated beams were subjected to loads, and the resulting frequencies were as follows (in Hz):
230.66, 233.05, 232.58, 229.48, 232.58
(a) Find a 90% two-sided CI on mean natural frequency. Round your answers to 2 decimal places..
(b) Do the results of your calculations support the claim that mean natural frequency is 235 Hz?
Answer:
a) The 90% confidence interval would be given by (230.212;233.128)
b) For this case if we analyze the confidence interval we see that not contains the value of 235. So we can't support the claim that the true mean is higher than 235 Hz at 10% of significance.
Step-by-step explanation:
Previous concepts and data given
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".
Data: 230.66, 233.05, 232.58, 229.48, 232.58
We can calculate the mean and the deviation from these data with the following formulas:
[tex]\bar X= \frac{\sum_{i=1}^n x_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X=231.67[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=1.531 represent the sample standard deviation
n=5 represent the sample size
Part a) Confidence interval
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=5-1=4[/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(0.05,4)".And we see that [tex]t_{\alpha/2}=2.13[/tex]
Now we have everything in order to replace into formula (1):
[tex]231.67-2.13\frac{1.531}{\sqrt{5}}=230.212[/tex]
[tex]231.67+2.13\frac{1.531}{\sqrt{5}}=233.128[/tex]
So on this case the 90% confidence interval would be given by (230.212;233.128)
Part b) Do the results of your calculations support the claim that mean natural frequency is 235 Hz?
For this case if we analyze the confidence interval we see that not contains the value of 235. So we can't support the claim that the true mean is higher than 235 Hz at 10% of significance.
To determine the 90% confidence interval for mean natural frequency, calculate the mean, standard deviation, find the t-value, and compute the margin of error and CI. The claim that the mean frequency is 235 Hz is assessed by checking if this value is within the calculated CI.
Explanation:To find the 90% two-sided confidence interval (CI) on mean natural frequency for the five delaminated beams, we first calculate the mean (\( \bar{x} \)) and the standard deviation (s) of the given frequencies. Then, we use the t-distribution, as the sample size is small (n=5), to find the t-value for a 90% CI, which corresponds to a significance level (alpha) of 0.10 and degrees of freedom (df) of n-1. Finally, we apply the formula for the 90% CI as follows:
Calculate the mean: \( \bar{x} = (230.66 + 233.05 + 232.58 + 229.48 + 232.58) / 5 = 231.67 \) Hz.Calculate the standard deviation (s):(b) To assess whether the mean natural frequency is 235 Hz, we would check if this value lies within the 90% CI. If it does not, the results do not support the claim that the mean natural frequency is 235 Hz.
An article in an educational journal discussing a non-conventional high school reading program reports a 95% confidence interval of (520 , 560) for the average score on the reading portion of the SAT. The study was based on a random sample of 20 students and it was assumed that SAT scores would be approximately normally distributed.
Determine the margin of error associated with this confidence interval.
A. 20
B. 10
C. 15
D. 40
E. 5
Answer:
Option A is right
Step-by-step explanation:
Given that an article in an educational journal discussing a non-conventional high school reading program reports a 95% confidence interval of (520 , 560) for the average score on the reading portion of the SAT.
Let X be the score on the reading portion of the SAT.
By central limit theorem x bar, the sample mean will follow a normal.
95% confidence interval shows that mean
= average of the two limits
= 540
Margin of error = upper bound of confidence interval - Mean
=[tex]560-540=20[/tex]
Option A is right
Alexis runs a small business and creates recordings of her friend’s skateboard stunts. She puts the recordings onto blu-ray and sells them online. She realized that the average cost of each blu-ray depends on the number of blu-rays she creates, because of the fixed cost. The cost of producing x blu-rays is given by
B(x) = 1250 +1.75x
1) Alexis wants to figure out the price to charge friends for the blu-rays. She doesn’t want to make money at this point since it is a brand new business, but does want to cover her costs. Suppose Alexis created 50 blu-rays. What is the cost of producing those 50 blu-rays? How much is it for each blu-ray?
please help my algabra grade dpends on it
Answer:
The cost of producing those 50 blu-rays is given by, 1337.5 unit and it is 26.75 unit for each blu-ray.
Step-by-step explanation:
The cost of producing x blu-rays is given by,
B(x) = 1250 + 1.75x unit -----------------------------(1)
So, cost of producing 50 blu-rays is given by,
B(50) = [tex]1250 + 50 \times (1.75)[/tex] unit
= (1250 + 87.5) unit
= 1337.5 unit
So, for each blu-ray, it is,
[tex]\frac {1337.5}{50}[/tex] unit
= 26.75 unit
5) The Queen Mary is a large cruise ship that has 8, 123, 700 cubic feet of cargo space. Express this number in scientific notation.
Explain in words how you converted the number of cubic feet into scientific notation?
For each situation, state the null and alternative hypotheses: (Type "mu" for the symbol μ , e.g. mu > 1 for the mean is greater than 1, mu < 1 for the mean is less than 1, mu not = 1 for the mean is not equal to 1. Please do not include units such as "mm" or "$" in your answer.)
a) The diameter of a spindle in a small motor is supposed to be 2.5 millimeters (mm) with a standard deviation of 0.17 mm. If the spindle is either too small or too large, the motor will not work properly. The manufacturer measures the diameter in a sample of 17 spindles to determine whether the mean diameter has moved away from the required measurement. Suppose the sample has an average diameter of 2.57 mm.
H0:
Ha:
(b) Harry thinks that prices in Caldwell are lower than the rest of the country. He reads that the nationwide average price of a certain brand of laundry detergent is $16.35 with standard deviation $2.20. He takes a sample from 3 local Caldwell stores and finds the average price for this same brand of detergent is $14.40.
H0:
Ha:
Answer:
Part (A) [tex]H_0:\mu=2.5[/tex] and [tex]H_a:\mu\neq2.5[/tex]
Part (B) [tex]H_0:\mu=16.35[/tex] and [tex]H_a:\mu<16.35[/tex]
Step-by-step explanation:
Consider the provided information.
Null hypothesis refers the population parameter is equal to the claimed value.
If there is no statistical significance in the test then it is know as the null which is denoted by [tex]H_0[/tex], otherwise it is known as alternative hypothesis denoted by [tex]H_a[/tex].
Part (A): The diameter of a spindle in a small motor is supposed to be 2.5 millimeters (mm) if the spindle is either too small or too large, the motor will not work properly.
Thus, the required null and alternative hypothesis are:
[tex]H_0:\mu=2.5[/tex] and [tex]H_a:\mu\neq2.5[/tex]
Part (B) The nationwide average price of a certain brand of laundry detergent is $16.35 and Harry thinks that prices in Caldwell are lower than the rest of the country.
Thus, the required null and alternative hypothesis are:
[tex]H_0:\mu=16.35[/tex] and [tex]H_a:\mu<16.35[/tex]
The null and alternative hypotheses for the given situations are as follows: a) H0: μ = 2.5 mm, Ha: μ ≠ 2.5 mm. b) H0: μ = $16.35, Ha: μ < $16.35.
Explanation:a) H0: The mean diameter of the spindles is 2.5 mm (μ = 2.5 mm).
Ha: The mean diameter of the spindles is not equal to 2.5 mm (μ ≠ 2.5 mm).
b) H0: The mean price of the laundry detergent in Caldwell is $16.35 (μ = $16.35).
Ha: The mean price of the laundry detergent in Caldwell is less than $16.35 (μ < $16.35).
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A projectile is fired from ground level with an initial speed of 550 m/sec and an angle of elevation of 30 degrees. Use that the acceleration due to gravity is 9.8 m/sec2
What is the range of the projectile? ___ meters
What is the max height of the projectile? __ meters
what is the speed at which the projectile hits the ground? ___m/sec
Answer:
x (max) = 26760 m
y (max) = 3859 meters
V = 549.5 m/sec
Step-by-step explanation:
Equations to describe the projectile shot movement are:
a(x) = 0 V(x) = V(₀) *cos α x = V(₀) *cos α * t
a(y) = -g V(y) = V(₀) * sin α - g*t y = V(₀) * sin α *t - (1/2)*g*t²
a ) What is the range of the projectile. α = 30°
then sin 30° = 1/2 cos 30° = √3 /2 and tan 30° = 1/√3
x maximum occurs when in the equation of trajectory we make y = 0
Then
y = x*tan α - g*x / 2*V(₀)²*cos² α
x*tan α = g*x / 2* V(₀)²*cos² α
By subtitution
1/√3 = 9.8* x(max) / 2* (550)²*0.75
(1/√3) * 453750 / 9.8 = x (max)
x (max) = 453750 / 16.95 meters
x (max) = 26760 m
The maximum height is when V(y) = 0
We compute t in that condition
V(y) = 0 = V(₀) * sin α - g*t
t = V(₀) * sin α / g ⇒ t = 550* (1/2) / 9.8
t = 28.06 sec
Then h (max) = y(max) = V(₀) sin α * t - 1/2 g* t²
y (max) = 550* (1/2)*28.06 - (1/2)* 9.8 * (28.06)²
y (max) = 7717 - 3858
y (max) = 3859 meters
What is the speed when the projectile hits the ground
V = V(x) + V (y) and t = 2* 28.06 t = 56.12 sec
mod V =√ V(x)² + V(y)²
V(x) = V(₀) cos α = 550 * √3/2
V(x) = 475.5 m/sec V(x)² = 226338 m²/sec²
V(y) = 550*1/2 - 9.8* 56.12 ⇒ V(y) = 275 - 549.98
V(y) = - 274.98 V(y) ² =
V = √ 226338 + 75614 ⇒ V = 549.5 m/sec
Type the correct answer in each box. A circle is centered at the point (5, -4) and passes through the point (-3, 2). The equation of this circle is (x + )2 + (y + )2 = .
Answer:
[tex](x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}[/tex]
Step-by-step explanation:
Given:
Center of circle is at (5, -4).
A point on the circle is [tex](x_1,y_1)=(-3, 2)[/tex]
Equation of a circle with center [tex](h,k)[/tex] and radius 'r' is given as:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Here, [tex](h,k)=(5,-4)[/tex]
Radius of a circle is equal to the distance of point on the circle from the center of the circle and is given using the distance formula for square of the distance as:
[tex]r^2=(h-x_1)^2+(k-y_1)^2[/tex]
Using distance formula for the points (5, -4) and (-3, 2), we get
[tex]r^2=(5-(-3))^2+(-4-2)^2\\r^2=(5+3)^2+(-6)^2\\r^2=8^2+6^2\\r^2=64+36=100[/tex]
Therefore, the equation of the circle is:
[tex](x-5)^2+(y-(-4))^2=100\\(x-5)^2+(y+4)^2=100[/tex]
Now, rewriting it in the form asked in the question, we get
[tex](x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}[/tex]
a math teacher claims that she has developed a review course that increases the score of students on the math portion of a college entrance exam. based on the data from the administrator of the exam, scores are normally distributed with u=514. the teacher obtains a random sample of 2000 students. puts them through the review class and finds that the mean math score of the 2000 is 520 with a standard deviation of 119. A.state the null and alternative hypotheses. B.test the hypothesis at the a=.10 level of confidence. is a mean math score of 520 significantly higher than 514? find the test statistic, find P-value. is the sample statistic significantly higher? C. do you think that a mean math score of 520 vs 514 will affect the decision of a school admissions adminstrator? in other words does the increase in the score have any practical significance? D. test the hypothesis at the a=.10 level of confidence with n=350 students. assume that the sample mean is still 520 and the sample standard deviation is still 119. is a sample mean of 520 significantly more than 514? find test statistic, find p value, is the sample mean statisically significantly higher? what do you conclude about the impact of large samples on the p-value?
Answer:
A.
[tex]H_0: \mu\leq514\\\\H_1: \mu>514[/tex]
B. Z=2.255. P=0.01207.
C. Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.
D. P(z>0.94)=0.1736
Step-by-step explanation:
A. state the null and alternative hypotheses.
The null hypothesis states that the review course has no effect, so the scores are still the same. The alternative hypothesis states that the review course increase the score.
[tex]H_0: \mu\leq514\\\\H_1: \mu>514[/tex]
B. test the hypothesis at the a=.10 level of confidence. is a mean math score of 520 significantly higher than 514? find the test statistic, find P-value. is the sample statistic significantly higher?
The test statistic Z can be calculated as
[tex]Z=\frac{M-\mu}{s/\sqrt{N}} =\frac{520-514}{119/\sqrt{2000}}=\frac{6}{2.661}=2.255[/tex]
The P-value of z=2.255 is P=0.01207.
The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.
Then we can conclude that the score of 520 is significantly higher than 514, in this case, specially because the big sample size.
C. do you think that a mean math score of 520 vs 514 will affect the decision of a school admissions adminstrator? in other words does the increase in the score have any practical significance?
Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.
D. test the hypothesis at the a=.10 level of confidence with n=350 students. assume that the sample mean is still 520 and the sample standard deviation is still 119. is a sample mean of 520 significantly more than 514? find test statistic, find p value, is the sample mean statisically significantly higher? what do you conclude about the impact of large samples on the p-value?
In this case, the z-value is
[tex]Z=\frac{520-514}{s/\sqrt{n}} =\frac{6}{119/\sqrt{350}} =\frac{6}{6.36} =0.94\\\\P(z>0.94)=0.1736>\alpha[/tex]
In this case, the P-value is greater than the significance level, so there is no evidence that the review course is increasing the scores.
The sample size gives robustness to the results of the sample. A large sample is more representative of the population than a smaller sample. A little difference, as 520 from 514, but in a big sample leads to a more strong evidence of a change in the mean score.
Large samples give more extreme values of z, so P-values that are smaller and therefore tend to be smaller than the significance level.
A math teacher claims to have developed a review course that increases the score of students on the math portion of a college entrance exam. The null hypothesis is that there is no difference in scores before and after the course, while the alternative hypothesis is that the course increases the scores. The hypothesis is tested at the 0.10 level of confidence by calculating the z-score and finding the p-value. The impact of the score increase on the decision of a school admissions administrator depends on their criteria. The hypothesis is then tested again with a larger sample size, following the same steps as earlier.
Explanation:A. The null hypothesis (H0) is that there is no difference in scores before and after the review course, while the alternative hypothesis (H1) is that the review course increases the scores on the math portion of the college entrance exam.
B. To test the hypothesis at the 0.10 level of confidence, we calculate the z-score and find the p-value. The test statistic is z = (sample mean - population mean) / (population standard deviation / sqrt(sample size)). The p-value is the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true. If the p-value is less than the significance level, we reject the null hypothesis and conclude that the mean math score is significantly higher than 514.
C. Whether a mean math score increase from 514 to 520 has any practical significance depends on the requirements set by the school admissions administrator. If the administrator determines that a small increase in score does not significantly impact their decision, then the increase may not have practical significance. However, if the administrator considers even small improvements in score as valuable, then the increase may have practical significance.
D. To test the hypothesis at the 0.10 level of confidence with a sample size of 350 students, we follow the same steps as in part B to calculate the test statistic and p-value. If the p-value is less than the significance level, we reject the null hypothesis and conclude that the sample mean of 520 is significantly higher than 514. Large sample sizes can result in smaller p-values, indicating stronger evidence against the null hypothesis.
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Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval. A. 0.444 < p < 0.500 B. 0.438 < p < 0.505 C. 0.471 < p < 0.472 D. 0.435 < p < 0.508
We can be 95% confident that the true proportion of all voters in the state who favor approval lies between 0.438 < p < 0.505. Option B is the correct answer.
Here's a breakdown of the steps involved in constructing the confidence interval:
Calculate the sample proportion (P):
P = 408 (voters favoring approval) / 865 (total voters surveyed) = 0.4714
Determine the critical value (z):
For a 95% confidence interval, the critical value from the standard normal distribution is 1.96.
Calculate the margin of error:
Margin of error = z * √(P * (1 - P) / n)
Margin of error = 1.96 * √(0.4714 * (1 - 0.4714) / 865) = 0.0335
Construct the confidence interval:
Lower bound = P - margin of error = 0.4714 - 0.0335 = 0.4379
Upper bound = P + margin of error = 0.4714 + 0.0335 = 0.5049
Round the bounds to three decimal places:
0.438 < p < 0.505
Therefore, we can be 95% confident that the true proportion of all voters in the state who favor approval lies between 0.438 and 0.505.
The TreeDropp’d Fruit Company wants to sell its apples overseas in attractive pink and yellow gift boxes. To design the boxes, the company needs to estimate the average diameter of their apples. A random sample of 50 apples has a mean of 4.1 inches. You assume a population standard deviatino of .4 inches.
A. What is the point estimate in this scenario? Is it an unbiased estimator?
B. What is the critical z-value for a 95% interval?
C. What is the margin of error for a 95% interval in this scenario? Show your work.
Answer:
a)For this case the best estimator for the true mean is the sample mean [tex]\hat \mu =\bar X[/tex] because:
[tex]E(\bar X)= \frac{\sum_{i=1}^n E(X_i)}{n}[/tex]
And if we assume that each observation [tex]X_1 , X_2,...,X_n [/tex] follows a normal distribution [tex]X_i \sim N(\mu,\sigma)[/tex] then we have:
[tex]E(\bar X)=\frac{1}{n} n\mu =\mu[/tex]
So then yes the [tex]\bar X[/tex[ is a unbiased estimator for the true mean.
b) [tex]z_{\alpha/2}=1.96[/tex]
c) [tex] ME= 1.96 \frac{0.4}{\sqrt{50}}=0.1109[/tex]
Step-by-step explanation:
Previous concepts
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".
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".
[tex]\bar X=4.1[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=0.4[/tex] represent the population standard deviation
n=50 represent the sample size
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma=0.4[/tex]
The sample mean [tex]\bar X[/tex] is distributed on this way:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
A. What is the point estimate in this scenario? Is it an unbiased estimator?
For this case the best estimator for the true mean is the sample mean [tex]\hat \mu =\bar X[/tex] because:
[tex]E(\bar X)= \frac{\sum_{i=1}^n E(X_i)}{n}[/tex]
And if we assume that each observation [tex]X_1 , X_2,...,X_n [/tex] follows a normal distribution [tex]X_i \sim N(\mu,\sigma)[/tex] then we have:
[tex]E(\bar X)=\frac{1}{n} n\mu =\mu[/tex]
So then yes the [tex]\bar X[/tex[ is a unbiased estimator for the true mean.
B. What is the critical z-value for a 95% interval?
The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex]
Using the normal standard table, excel or a calculator we see that:
[tex]z_{\alpha/2}=1.96[/tex]
C. What is the margin of error for a 95% interval in this scenario? Show your work.
The margin of error is given by:
[tex] ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
If we replace the values that we have we got:
[tex] ME= 1.96 \frac{0.4}{\sqrt{50}}=0.1109[/tex]
The confidence interval on this case is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since we have all the values we can replace:
[tex]4.1 - 1.96\frac{0.4}{\sqrt{50}}=3.989[/tex]
[tex]4.1 + 1.96\frac{0.4}{\sqrt{50}}=4.211[/tex]
So on this case the 95% confidence interval would be given by (3.989;4.211)
Suppose that a random sample of size 49 is to be selected from a population with mean 49 and standard deviation 6. What is the approximate probability that Xwll be more than 0.5 away from the population mean?
a 0.4403
b) 0.6403
c) 0.8807
d) 0.0664
e) 0.5597
f) None of the above
Answer:
Option e - 0.5597
Step-by-step explanation:
Given : Suppose that a random sample of size 49 is to be selected from a population with mean 49 and standard deviation 6.
To find : What is the approximate probability that will be more than 0.5 away from the population mean?
Solution :
Applying z-score formula,
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Here, [tex]\mu=49,\sigma=6, n=49[/tex]
As, probability that will be more than 0.5 away from the population mean
i.e. x=49.5,
[tex]z=\frac{49.5-49}{\frac{6}{\sqrt{49}}}[/tex]
[tex]z=\frac{0.5}{0.857}[/tex]
[tex]z=0.58[/tex]
In z-score table the value of z at 0.58 is 0.7190.
x=48.5,
[tex]z=\frac{48.5-49}{\frac{6}{\sqrt{49}}}[/tex]
[tex]z=\frac{-0.5}{0.857}[/tex]
[tex]z=-0.58[/tex]
In z-score table the value of z at -0.58 is 0.2810.
Now, probability that will be more than 0.5 away from the population mean is given by,
[tex]P=P(X>49.5)+P(X<48.5)[/tex]
[tex]P=1-P(X<49.5)+P(X<48.5)[/tex]
[tex]P=1-0.7190+0.2810[/tex]
[tex]P=0.562[/tex]
Therefore, option e is correct.
A large room has a temperature of 70 degrees Fahrenheit. When a cup of tea is brought into the room, the tea has a temperature of 120 degrees Fahrenheit. The function f(t)=Ce(−kt)+70 represents the situation, where t is time in minutes, C is a constant, and k is a constant. After 3 minutes the cup of tea has a temperature of 100 degrees. What will be the temperature of the tea, in degrees Fahrenheit, after 5 minutes? Round your answer to the nearest tenth, and do not include units.
Answer:
91.3
Step-by-step explanation:
To find the temperature of the tea after 5 minutes, use the given function and the information given in the question. By solving the function for the constant C, you can then find the temperature of the tea after 5 minutes.
Explanation:To find the temperature of the tea after 5 minutes, we need to use the given function and the information given in the question.
First, we can use the given function f(t)=Ce(−kt)+70 to find the value of C. Since we know that after 3 minutes the temperature of the tea is 100 degrees, we can substitute t=3 and T=100 into the function to get:
100 = Ce(−3k)+70
Next, we can use the same function to find the temperature of the tea after 5 minutes. Substituting t=5 into the function and using the value of C that we found earlier, we get:
T = Ce(−5k)+70
Rounding to the nearest tenth, the temperature of the tea after 5 minutes will be approximately 89.4 degrees Fahrenheit.
There are 9 observatories tracking a near-Earth asteroid, and each independently estimates how close it will come to Earth next year, in millions of miles, with the results. 481, .439,448,446,618, 411, 250, 604, 414. Previous experience suggests these are unbiased estimates of the true distance d, with a standard deviation of 15. a) We estimate the true mean of these distance estimates by d = i. b) Assuming this is a large enough sample, write down a 94% confidence interval for the true distance d.
Answer:
94% Confidence interval: (0.3771 ,0.5365)
Step-by-step explanation:
We are given the following data set:
0.481, 0.439,0.448,0.446,0.618, 0.411, 0.250, 0.604, 0.414
a) True mean of these distance estimates by d
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{4.111}{9} = 0.4568[/tex]
b)
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
Sum of squares of differences = 0.0954
[tex]S.D = \sqrt{\frac{0.0954}{8}} = 0.1092[/tex]
94% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 8 and}~\alpha_{0.06} = \pm 2.1891[/tex]
[tex]0.4568 \pm 2.1891(\frac{0.1092}{\sqrt{9}} ) = 0.4568 \pm 0.0797 = (0.3771 ,0.5365)[/tex]
Final answer:
To calculate the 94% confidence interval for the true mean distance of an asteroid, one needs to compute the sample mean, standard error, and then apply the appropriate Z-score to determine the range.
Explanation:
The student's question is related to calculating a confidence interval for the true mean distance of a near-Earth asteroid. With the given unbiased estimates of the distances and a standard deviation of 15, we will calculate the 94% confidence interval for the true distance d.
Firstly, calculate the sample mean X by adding up all the given distance estimates and dividing by the number of observatories, which is 9.
Next, find the standard error (SE) of the mean by dividing the standard deviation by the square root of the number of observatories (n).
To determine the 94% confidence interval, we'll use the Z-score associated with it (approximately 1.88) and the standard error.
The confidence interval formula is X ± (Z * SE). We calculate the upper and lower bounds of the interval accordingly.
By following these steps, the student can establish the 94% confidence interval, giving them a range within which the true mean distance of the asteroid is likely to fall.
A line has a slope of -5/8. What is the slope of the line parallel to it; and what is the slope of the line perpendicular to it?
Final answer:
A line parallel to one with a slope of -5/8 has the same slope, -5/8, while the slope of a line perpendicular to it is the negative reciprocal, which is 8/5.
Explanation:
The slope of a line that is parallel to a given line is identical to that of the given line. Therefore, a line parallel to one with a slope of -5/8 would also have a slope of -5/8. However, a line that is perpendicular to the given line would have a slope that is the negative reciprocal of the original slope.
To find the negative reciprocal, you flip the fraction and change the sign. So, the slope of a line perpendicular to a line with a slope of -5/8 would be 8/5.
Researchers are studying the relationship between honesty, age, and self-control conducted an experiment on 160 children between the ages of 5 and 15. The researchers asked each child to toss a fair coin in private and to record the outcome (heads or tails) on a paper sheet, and said they would only reward children who report heads. Half the students were explicitly told not to cheat and the others were not given any explicit instructions. Differences were observed in the cheating rates in the instruction and no instruction groups, as well as some differences across children's characteristics within each group.
a) Identify the population of interest in the study.
A. 80 children between the ages of 5 and 15 who were told not to cheat
B. 160 children between the ages of 5 and 15
C. The researchers
D. All children between the ages of 5 and 15
b) Identify the sample for this study.
A. 80 children between the ages of 5 and 15 who were told not to cheat
B. The researchers
C. All children between the ages of 5 and 15
D. 160 children between the ages of 5 and 15
c) Can the results of the study can be generalized to the population? Should the findings of the study can be used to establish causal relationships.
A. If the sample is randomly selected and representative of the entire population, then the results can be generalized to the target population. Furthermore, since this study is observational, the findings can be used to infer causal relationships.
B. If the sample is randomly selected and representative of the entire population, then the results can be generalized to the target population. Furthermore, since this study is experimental, the findings can be used to infer causal relationships.
C. If the sample is randomly selected and representative of the entire population, then the results can be generalized to the target population. Furthermore, since this study is experimental, the findings cannot be used to infer causal relationships.D. If the sample is randomly selected and representative of the entire population, then the results can be generalized to the target population. Furthermore, since this study is observational, the results cannot be used to infer causal relationships.
Answer:
Step-by-step explanation:
Given that researchers are studying the relationship between honesty, age, and self-control conducted an experiment on 160 children between the ages of 5 and 15.
a) The population is the children between the ages of 5 and 15 in general
D. All children between the ages of 5 and 15
b) Sample is
D. 160 children between the ages of 5 and 15
c) Yes becauB. If the sample is randomly selected and representative of the entire population, then the results can be generalized to the target population. Furthermore, since this study is experimental, the findings can be used to infer causal relationships.se randomly selected and also sample size is above 30
The population of interest is all children aged 5 to 15; the study's sample consists of the 160 participating children. Results can be generalized if the sample is representative, and as an experimental study, it can establish causal relationships.
Explanation:The population of interest in the study is D. All children between the ages of 5 and 15 because the researchers are attempting to understand a broader phenomenon that could be applicable to all children within this age range, not just those in the study. The sample for the study is D. 160 children between the ages of 5 and 15 because these are the actual participants that researchers have collected data from within this experiment.
Regarding the generalizability and establishment of causal relationships of the study, the correct option is B. If the sample is randomly selected and representative of the entire population, then the results can be generalized to the target population. Furthermore, since this study reports the manipulation of an independent variable (the instruction given to one group and not to the other), it qualifies as an experimental study, and its results can inform us about causal relationships.
Congress regulates corporate fuel economy and sets an annual gas mileage for cars. A company with a large fleet of cars hopes to meet the 2011 goal of 30.2mpg or better for their fleet of cars. To see if the goal is being met, they check the gasoline usage for 50 company trips chosen at random, finding a mean of 32.12mpg and a standard deviation of 4.83mpg. In this strong evidence that they have attained their fuel economy goal? a. Define the parameter and state the hypotheses. b. Define the sampling distribution (mean and standard deviation). c. Perform the test and calculate P-value d. State your conclusion. e. Explain what the p-value means in this context.
Answer:
a) Null hypothesis:[tex]\mu \leq 30.2[/tex]
Alternative hypothesis:[tex]\mu > 30.2[/tex]
b) [tex]X \sim N(\mu=32.12, \sigma=4.83)[/tex]
And the distribution for the random sample is given by:
[tex]\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)[/tex]
c) [tex]t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81[/tex]
[tex]p_v =P(t_{(49)}>2.81)=0.0035[/tex]
d) If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.
e) The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=32.12[/tex] represent the sample mean
[tex]s=4.83[/tex] represent the sample standard deviation
[tex]n=50[/tex] sample size
[tex]\mu_o =30.2[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
a. Define the parameter and state the hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 30.2 mpg, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 30.2[/tex]
Alternative hypothesis:[tex]\mu > 30.2[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
b. Define the sampling distribution (mean and standard deviation).
Let X the random variable who represent the variable of interest. And we know that the distribution for X is:
[tex]X \sim N(\mu=32.12, \sigma=4.83)[/tex]
And the distribution for the random sample is given by:
[tex]\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)[/tex]
c. Perform the test and calculate P-value
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=50-1=49[/tex]
Since is a one right tailed test the p value would be:
[tex]p_v =P(t_{(49)}>2.81)=0.0035[/tex]
d. State your conclusion.
If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.
e. Explain what the p-value means in this context.
The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.
Final answer:
Explanation of hypothesis testing for a fuel economy standard compliance claim.
Explanation:
Hypotheses:
Parameter: Average fuel economyNull Hypothesis (H0): μ = 35.5Alternative Hypothesis (Ha): μ < 35.5Sampling Distribution:
Mean: 34.6 mpgStandard Deviation: 10.3 mpgPerforming the Test:
Calculate the Z-score and P-value from the sample data using the formula. Compare the P-value to the significance level.
Conclusion: Since **p-value = 0.2578** is greater than the significance level of 0.05, we fail to reject the null hypothesis. There is not enough evidence to claim that the fleet meets the fuel economy standard.