Answer:
B. 0.3 and 0.022
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a sample proportion p in a sample of size n, the standard error is [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
135 of the 450 residents sampled live within one mile of a hazardous waste site.
So the sample proportion is [tex]p = \frac{135}{450} = 0.3[/tex]
Standard error
[tex]s = \sqrt{\frac{0.3*0.7}{450}} = 0.022[/tex]
So the correct answer is:
B. 0.3 and 0.022
Final answer:
The sample proportion of people who live within one mile of a hazardous waste site is 0.3, and the standard error for this sample proportion is 0.022. The correct answer is B. 0.3 and 0.022.
Explanation:
To estimate the proportion of people who live within one mile of a hazardous waste site using a sample, we divide the number of residents living within one mile by the total number of residents sampled. In this case, 135 residents live within one mile out of a sample of 450 residents, which makes the sample proportion 135/450 = 0.3.
To calculate the standard error (SE) for the sample proportion, we use the formula SE = √(p(1-p)/n), where p is the sample proportion and n is the sample size. Plugging in the values, we get SE = √(0.3(1-0.3)/450), which approximates to 0.022.
The correct answer is B. 0.3 and 0.022 for the sample proportion and its standard error, respectively.
implify the product. 2p(–3p2 + 4p – 5)
Answer:
=2p(-2p-5)
Step-by-step explanation:
2p(–3p2 + 4p – 5)
=2p(4p-6p-5)
=2p(-2p-5)
Select all the expression that
Answer:
2nd, 4th and last one are correct answers
Step-by-step explanation:
[tex] {16}^{ \frac{5}{2} } \\ = ( {4}^{2} )^{ \frac{5}{2} } \\ = {4}^{{2} \times \frac{5}{2} } \\ = {4}^{5} \\ [/tex]
Options: 2nd, 4th and last one are correct answers
A chord is 16 units from the center of a circle. The radius of the circle is 20 units. What is the length of the chord?
How is the measure of a central angle and the corresponding chord related to the measure of the arc intercepted by the chord?
Answer:
Length of the chord: 24 units
Angles are equal
Step-by-step explanation:
Drop a Perpendicular from the centre onto the chord. It will divide the chord into two equal parts, d units each
d² + 16² = r²
d² = 20² - 16²
d² = 144
d = 12
Chord = 2d = 24 units
Measure of central angle and the corresponding chord related to the measure of the arc intercepted by the chord are the same, they're equal
Answer:
The entire chord length is 12*2 = 24
The degree measure of a minor arc is equal to the measure of the central angle that intercepts it.
Step-by-step explanation:
We can make a right triangle to solve for 1/2 of the chord length. The hypotenuse is 20 and one of the legs is 16
a^2+b^2 = c^2
16^2 + b^2 = 20^2
256 +b^2 = 400
Subtract 256 from each side
b^2 = 400-256
b^2 =144
Take the square root of each side
b = 12
That means 1/2 of the chord length is 12
The entire chord length is 12*2 = 24
"A movie data base claims that the average length of movies is 117 minutes. A researcher collected a random sample of 160 movies released during 2010–2015. The mean length of those movies is 118.44 minutes and the standard deviation is 8.82. The researcher wonders if the actual mean length of movies released during 2010-2015 is more than the data base value and wants to carry out a hypothesis test. What are the null and alternative hypothesis?"
Answer:
We need to conduct a hypothesis in order to check if the mean the actual mean length of movies released during 2010-2015 is more than the data base value (a;ternative hypothesis) ,so then the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 117[/tex]
Alternative hypothesis:[tex]\mu > 117[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X=118.44[/tex] represent the sample mean
[tex]s=8.82[/tex] represent the sample standard deviation
[tex]n=160[/tex] sample size
[tex]\mu_o =117[/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)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean the actual mean length of movies released during 2010-2015 is more than the data base value (a;ternative hypothesis) ,so then the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 117[/tex]
Alternative hypothesis:[tex]\mu > 117[/tex]
Evaluate the expression. 23 [ 14 + 4(36 ÷ 12)]
Answer:
598
Step-by-step explanation:
Answer:
598
Step-by-step explanation:
[tex]23[14 + 4(36 \div 12)] \\ = 23[14 + 4(3)] \\ = 23[14 + 12] \\ = 23 \times 26 \\ = 598 \\ [/tex]
Suppose you are going to conduct a two tail test concerning the population mean. Suppose that you do not know what the population standard deviation is and that you have a sample of 55 observations. If you are going to conduct this test at the .01 level of significance what is the critical value?
Answer:
The critical value of t at 0.01 level of significance is 2.66.
Step-by-step explanation:
The hypothesis for the two-tailed population mean can be defined as:
H₀: μ = μ₀ vs. H₀: μ ≠ μ₀
It is provided that the population standard deviation is not known.
Since there is no information about the population standard deviation, we will use a t-test for single mean.
The test statistic is defined as follows:
[tex]t_{cal.}=\frac{\bar x-\mu}{s/\sqrt{n}}\sim t_{\alpha/2, (n-1)}[/tex]
The information given is:
n = 55
α = 0.01
Compute the critical value of t as follows:
[tex]t_{\alpha/2, (n-1)}= t_{0.01/2, (55-1)}=t_{0.005, 54}=2.66[/tex]
*Use a t-table for the value.
If the desired degrees of freedom are not provided consider he next highest degree of freedom.
Thus, the critical value of t at 0.01 level of significance is 2.66.
Ximena spent 72 of the day
shopping at the mall. She spent
74 of this time trying on jeans.
What fraction of the day did
Charlotte spend trying on jeans?
Answer:
Step-by-step explanation:
1/8
1/2 multiply 1/4
A political candidate has asked his/her assistant to conduct a poll to determine the percentage of people in the community that supports him/her. If the candidate wants a 6% margin of error at a 90% confidence level, what size of sample is needed? Be sure to round accordingly.
A sample size of 188 would be needed to achieve a 6% margin of error at a 90% confidence level.
Used the formula for sample size calculation:
[tex]n = \dfrac{z^2 \times p (1 - p)}{E^2}[/tex]
Where: n = sample size.
Z = Z-score corresponding to the desired confidence level.
p = estimated proportion of the population that supports the candidate.
E = desired margin of error.
Here we have to give that,
90% confidence level corresponds to a Z-score of approximately 1.645.
That is, z = 1.645
p = 0.5
E = 0.06
Substituting the given values into the formula, we get:
[tex]n = \dfrac{z^2 \times p (1 - p)}{E^2}[/tex]
[tex]n = \dfrac{(1.645)^2 \times 0.5(1 - 0.5)}{0.06^2}[/tex]
[tex]n = \dfrac{0.6765}{0.0036}[/tex]
[tex]n =187.9[/tex]
Rounded to whole numbers,
[tex]n = 188[/tex]
Therefore, the size of the sample would be 188.
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In order to achieve a 6% margin of error and a 90% confidence level in a poll, the correct sample size must be determined using the error bound formula. A larger and representative sample will minimize the margin of error and improve the poll's precision and accuracy. A number of key factors, like identification of the target population and select a random sample, should be considered while conducting a poll.
Explanation:The subject matter dealt with here involves conducting a poll to gauge public opinion, with a desired accuracy defined by a 6% margin of error and a 90% confidence level. This is primarily related to Statistics, a branch of Mathematics.
To determine an appropriate sample size for a given margin of error and confidence level, the error bound formula is utilized. Although the actual formula could appear complex for some, many online calculators can complete the task. The 90% confidence level means that if we were to take multiple sets of samples, roughly 90% of the calculated confidence intervals would contain the true population mean or proportion.
Furthermore, sample size influences the margin of error; a larger sample reduces the margin of error and makes the poll result more precise and accurate. As a rule of thumb, the sample should be random and representative of the population in question to ensure the accuracy and reliability of poll results.
Take note that many criteria must be met to conduct a valid and scientifically accurate poll, some of which include: identifying the desired population of respondents, ensuring the sample is both random and representative of the wider population, and interviewing a sufficient number of citizens to achieve a reasonable sample size.
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The "cold start ignition time" of an automobile engine is being investigated by a gasoline manufacturer. The following times, in seconds, were obtained for a test vehicle. 0.17, 1.94, 2.62, 2.35, 3.05, 3.15, 2.53, 4.81 and 1.92 Write answers as decimal values (i.e. 0.12) and round to two decimal places (if necessary)
Answer:
1. Sample Mean = 2.50
2. Sample Variance = 1.35
3. Standard Deviation = 1.16
Step-by-step explanation:
Note: As this question is not complete, similar can be found on internet where following are asked to calculate and here I will calculate these as well.
1. Sample Mean.
2. Sample Variance.
3. Sample Standard Deviation.
So, for sample mean to calculate from the given data points. We need to apply to following formula to calculate sample mean.
Sample Mean = (X1 + X2 + .... + XN) divided by Total number of data points.
Sample Mean = (0.17 + 1.94 + 2.62 + 2.35 + 3.05 + 3.15 + 2.53 + 4.81 + 1.92) divided by (9).
Sample Mean = 2.50
Likewise, from the answer of sample mean we can calculate sample variance by following steps.
Solution:
1. Subtract the obtained mean from each of the data point given.
(0.17 - 2.50) = -2.33
(1.94 - 2.50) = -0.56
(2.62 - 2.50) = 0.12
(2.35 - 2.50) = -0.15
(3.05 - 2.50) = 0.55
(3.15 - 2.50) = 0.65
(2.53 - 2.50) = 0.03
(4.81 - 2.50) = 2.31
(1.92 - 2.50) = -0.58
2. Square each of the differences obtained in step 1.
[tex]-2.33^{2}[/tex] = 5.43
[tex]-0.56^{2}[/tex] = 0.31
[tex]0.12^{2}[/tex] = 0.01
[tex]-0.15^{2}[/tex] = 0.02
[tex]0.55^{2}[/tex] = 0.30
[tex]0.65^{2}[/tex] = 0.42
[tex]0.03^{2}[/tex] = 0.01
[tex]2.31^{2}[/tex] = 5.34
[tex]-0.58^{2}[/tex] = 0.34
3. Sum all of these squares.
(5.43 + 0.31 + 0.01 + 0.02 + 0.30 + 0.42 + 0.01 + 5.34 + 0.34) = 12.18
4. Divide 12.18 by (n-1), where n = 9.
Sample Variance = 12.18/9 = 1.35
Now, by using sample variance, we can calculate standard deviation with easy simple steps.
So, in order to find standard deviation, we just need to take the square root of sample variance that we have already calculated above.
Standard Deviation = [tex]\sqrt{1.35}[/tex] = 1.16
Good Luck!
Confidence interval precision: We know that narrower confidence intervals give us a more precise estimate of the true population proportion. Which of the following could we do to produce higher precision in our estimates of the population proportion? Group of answer choices We can select a higher confidence level and increase the sample size. We can select a higher confidence level and decrease the sample size. We can select a lower confidence level and increase the sample size. We can select a lower confidence level and decrease the sample size.
Answer:
We can select a lower confidence level and increase the sample size.
Step-by-step explanation:
The precision of the confidence interval depends on the margin of error ME = Zcritical * Sqrt[(p(1-p)/n]
In this Zcritical value is in the numerator. Z critical decreases as Confidence level decreases. (Zc for 99% = 2.576, Zc for 95% is 1.96, Zc for 90% = 1.645). Therefore decreasing the Confidence level decreases ME.
Also we see that sample size n is in the denominator. So the ME decreases as sample size increases.
Therefore, We can select a lower confidence level and increase the sample size.
The best option that could be used to produce higher precision in our estimates of the population proportion is;
Option C; We can select a lower confidence level and increase the sample size.
Formula for confidence interval is given as;
CI = p^ ± z√(p^(1 - p^)/n)
Where;
p^ is the sample proportion
z is the critical value at given confidence level
n is the sample size
Now, the margin of error from the CI formula is:
MOE = z√(p^(1 - p^)/n)
Now, the lesser the margin of error, the narrower the confidence interval and thus the more precise is the estimate of the population proportion.
Now, looking at the formula for MOE, two things that could change aside the proportion is;
z and n.
Now, the possible values of z are;
At CL of 99%; z = 2.576
At CL of 95%; z = 1.96
At CL of 90%; z = 1.645
We can see that the higher the confidence level, the higher the critical value and Invariably the higher the MOE.
Thus, to have a narrow CI, we need to use a lower value of CL and increase the sample size.
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Determine the value of so that the area under the standard normal curve a. in the right tail is Round your answer to two decimal places. b. in the left tail is Round your answer to two decimal places. c. in the left tail is Round your answer to two decimal places. d. in the right tail is Round your answer to two decimal places. Click if you would like to Show Work for this ques
Answer:
a) 2.81 b)-2.33 c) -2.88 d)3.09
Step-by-step explanation:
The complete question is:
Determine the value of z so that the area under the standard normal curve
a. in the right tail is 0.025 Round your answer to two decimal places.
b. in the left tail is 0.01 Round your answer to two decimal places.
c. in the left tail is 0.002 Round your answer to two decimal places.
d. in the right tail is 0.01 Round your answer to two decimal places.
a)P( Z> ???)= 0.0025
P(Z> ???)= 1-P(Z<???)
P(Z<???)= 1-0.0025
P(Z<??)= 0.9975
From Z distribution table,
Z = 2.81
b) P(Z<???)= 0.01
From Z distribution table
Z= -2.33
c) P(Z< ??? ) = 0.002
From Z distribution table
Z= -2.88
d) P( Z> ???)= 0.001
P(Z> ???)= 1-P(Z<???)
P(Z<???)= 1-0.001
P(Z<??)= 0.999
From Z distribution table,
Z=3.09
The moods of U.S. Marines following a month-long training exercise conducted at cold temperature and at high altitudes were assessed. Negative moods, including fatigue and anger, increased substantially during the training and lasted up to three month after the training ended. The scores for 5 of the Marines were 14, 10, 13, 10, 11. The mean mood score was compared to population norms for college men; the population mean anger score for college men is 8.90. a) Test the null hypothesis that the population mean is 8.90 against the alternative that the population mean is greater than 8.90 at α=.05. Show all 6 steps. b) Interpret the results. What did we learn about the Marines’ negative moods?
Answer:
Step-by-step explanation:
Sample mean = (14 + 10 + 13 + 10 + 11)/5 = 11.6
Sample standard deviation,s = √(summation(x - mean)/n
Summation(x - mean) = (14 - 11.6)^2 + (10 - 11.6)^2 + (13 - 11.6)^2 + (10 - 11.6)^2 + (11 - 11.6)^2 = 13.2
s = √(13.2/5) = 1.62
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 8.9
For the alternative hypothesis,
µ > 8.9
it is a right tailed test because of >.
Since the number of samples is 5 and no population standard deviation is given, the distribution is a student's t.
Since n = 5,
Degrees of freedom, df = n - 1 = 5 - 1 = 4
t = (x - µ)/(s/√n)
Where
x = sample mean = 11.6
µ = population mean = 8.9
s = samples standard deviation = 1.62
t = (11.6 - 8.9)/(1.62/√5) = 3.73
We would determine the p value using the t test calculator. It becomes
p = 0.01
Since alpha, 0.05 > than the p value, 0.01, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that mean anger score of the marines is greater than that of college men.
Therefore, the marines's negative moods increased and it is higher than that of college men.
Which of the following is the equation of a line that passes through (-2, -1) and (-4, -3)?
Answer:
y = x+1
Step-by-step explanation:
You should use the point-slope form of y-y1 = m(x-x1) to solve this.
First you use the slope formula to find your slope. (y2-y1)/(x2-x1)
(-1 - -3)/(-2 - -4) = 2/2 = 1
With the slope of 1 found, just plug a point into the point-slope form.
y-(-1) = (1)(x-(-2)) ->
y+1 = x+2 ->
y = x+1
y=x+1 is the equation of a line that passes through (-2, -1) and (-4, -3)
What is Slope of Line?The slope of the line is the ratio of the rise to the run, or rise divided by the run. It describes the steepness of line in the coordinate plane.
The slope intercept form of a line is y=mx+b, where m is slope and b is the y intercept.
The slope of line passing through two points (x₁, y₁) and (x₂, y₂) is
m=y₂-y₁/x₂-x₁
The slope of line passing through (-2, -1) and (-4, -3)
m = -3+1/-4+2
=-2/-2= 1
Now let us find y intercept
-1=1(-2)+b
-1=-2+b
b=1
Hence, y=x+1 is the equation of a line that passes through (-2, -1) and (-4, -3)
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The diagram shows the number of dollars each child has in a family. How can you redistribute the money so that each child has the same amount? Check all that apply.
A balance diagram going from 1 to 9. 2 circles are above 4 and 2 circles are above 8.
Each child who has $8 must give away $4.
Each child who has $8 must give away $2.
Each child who has $4 must be given $4.
Each child who has $4 must be given $2.
When fairly balanced, each child has $6.
When fairly balanced, each child has $7.
Answer:
2,4,5
Step-by-step explanation:
Answer:2,4,and5
Step-by-step explanation:Common sense
What is the value for x 2x+3=x-4
Answer:
x = -7
Step-by-step explanation:
2x + 3 = x - 4
→ Minus x from both sides to isolate -4
x + 3 = -4
→ Minus 3 from both sides to isolate x and henceforth find the value of x
x = -7
Write an equation in slope-intercept form of the line with the given parametric equations.
x = 8t+3
y = 9t+ 3
The equation in slope-intercept from is [tex]y = \frac{9}{8}x - \frac{3}{8}[/tex].
Step-by-step explanation:
Given that,
x = 8t+3
y = 9t+ 3
Now, getting the value of t on both side to compare the equations.
[tex]x = 8t+3[/tex]
[tex]x -3 = 8t[/tex]
[tex]t = \frac{x-3}{8}[/tex]
also
[tex]y = 9t+ 3[/tex]
[tex]t= \frac{y-3}{9}[/tex]
Now, by comparing the values of t, we get
[tex]\frac{x-3}{8} = \frac{y-3}{9}[/tex]
[tex]9x - 27 = 8y -24[/tex]
[tex]9x - 3 = 8y[/tex]
[tex]y = \frac{9}{8}x - \frac{3}{8}[/tex]
Hence, the equation in slope-intercept from is [tex]y = \frac{9}{8}x - \frac{3}{8}[/tex].
Consider the following sets of matrices: M2(R) is the set of all 2 x 2 real matrices; GL2(R) is the subset of M2(R) with non-zero determinant: SL2(R) is the subset of GL2(R) with determinant 1. We know that multiplication is a binary operation on M2(R); show that it is an induced operation on the other 2 sets (You may freely use known facts from Math 3A for this).
Answer:
Step-by-step explanation:
REcall the following definition of induced operation.
Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.
So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.
For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).
Case SL2(R):
Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So
[tex]\text{det(AB)} = \text{det}(A)\text{det}(B)\neq 0[/tex].
So AB is also in SL2(R).
Case GL2(R):
Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So
[tex]\text{det(AB)} = \text{det}(A)\text{det}(B)=1\cdot 1 = 1[/tex].
So AB is also in GL2(R).
With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).
Please help me! I need this to complete my math hw
Answer:
1
Step-by-step explanation:
You "complete the square" by adding the square of half the x-term coefficient. Here, that is ...
((-2)/2)² = 1 . . . . value added to complete the square
If you want to keep 0 on the right, you must also subtract this value:
x² -2x -36 = 0
x² -2x +1 -36 -1 = 0 . . . . . . add and subtract 1 on the left
(x -1)² -37 = 0 . . . . . . . . . . . written as a square
How many 5 digit codes are possible if the first digit cannot be 0 and digits can be repeated?
Answer:
Step-by-step explanation:
Zip codes with 5 digits (all five digits can be repeated): 10^5=100,000 if first digit is 0, its no longer 5-digit right? Zip codes with no digit being repeated: 10*9*8*7*6=30,240 if first digit is 0, its no longer 5-digit right? Zip codes with at least one digit being repeated: 90000-27216 = 62784 ?N
3 to the power of 5 and (3) to the power of 5
Answer:
243
Step-by-step explanation:
3x3=9
9x3=27
27x3=81
81x3=243
help need full solution i &ii
Step-by-step explanation:
[tex]1)\:v = 2 {e}^{3t} + 5 {e}^{ - 3t} \\ differentiating \: w.r.t.t \: on \: both \: sides \\ acceleration =\\ \frac{dv}{dt} = \frac{1}{dt} (2 {e}^{3t} + 5 {e}^{ - 3t} ) \\ = 2 \times \frac{1}{dt} {e}^{3t} + 5 \times \frac{1}{dt} {e}^{ - 3t} \\ \\ = 2 \times {e}^{3t} \times 3 + 5 \times {e}^{ - 3t} \times ( - 3) \\ = 6{e}^{3t} - 15{e}^{ - 3t} \\ \therefore \frac{dv}{dt} = 6{e}^{3t} - 15{e}^{ - 3t} \\ \therefore \bigg(\frac{dv}{dt} \bigg) _{t=1} = 6 {e}^{3 \times 1} - 15 {e}^ { - 3 \times 1} \\ \bigg(\frac{dv}{dt} \bigg) _{t=1} = 6 {e}^{3} - 15 {e}^ { - 3 } \\ acceleration = \\ \purple{ \boxed{ \bold{\bigg(\frac{dv}{dt} \bigg) _{t=1} = \bigg(\frac{6 {e}^{6} - 15}{ {e}^{3} } \bigg) \: m {s}^{ - 2} }}} \\ \\ 2) \: let \:s \: be \: the \: total \: distance \: travelled \\ \therefore \: s = v \times t \\ \therefore \: s= (2 {e}^{3t} + 5 {e}^{ - 3t}) \times t \\ \therefore \: (s)_{t=2} = (2 {e}^{3 \times 2} + 5 {e}^{ - 3 \times 2}) \times 2 \\ \therefore \: (s)_{t=2} = (2 {e}^{6} + 5 {e}^{ - 6}) \times 2 \\ \therefore \: (s)_{t=2} = 4 {e}^{6} + 10{e}^{ - 6} \\ \red{ \boxed{ \bold{\therefore \: (s)_{t=2} = \bigg(\frac{4 {e}^{12} + 10}{{e}^{ 6}} \: \bigg)m}}}\\ [/tex]
The number of cars running a red light in a day, at a given intersection, possesses a distribution with a mean of 2.4 cars and a standard deviation of 4. The number of cars running the red light was observed on 100 randomly chosen days and the mean number of cars calculated. Describe the sampling distribution of the sample mean.
Answer:
The sampling distribution of the sample mean is:
[tex]\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)[/tex]
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
Then, the mean of the distribution of sample means is given by,
[tex]\mu_{\bar x}=\mu[/tex]
And the standard deviation of the distribution of sample means is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]
Let X = number of cars running a red light in a day, at a given intersection.
The information provided is:
[tex]E(X)=\mu=2.4\\SD(X)=\sigma=4\\n=100[/tex]
The sample selected is quite large, i.e. n = 100 > 30.
The Central limit theorem can be used to approximate the sampling distribution of the sample mean number of cars running a red light in a day, by the Normal distribution.
The mean of the sampling distribution of the sample mean is:
[tex]\mu_{\bar x}=\mu=2.4[/tex]
The standard deviation of the sampling distribution of the sample mean is:
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{4}{\sqrt{100}}=0.40[/tex]
The sampling distribution of the sample mean is:
[tex]\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)[/tex]
Final answer:
The sampling distribution of the sample mean is approximately normally distributed, with a mean of 2.4 cars and a standard deviation of 0.4 cars.
Explanation:
The sampling distribution of the sample mean can be described as follows:
The sampling distribution of the sample mean is approximately normally distributed.The mean of the sampling distribution of the sample mean is equal to the mean of the population, which is 2.4 cars in this case.The standard deviation of the sampling distribution of the sample mean, also known as the standard error, is equal to the standard deviation of the population divided by the square root of the sample size. In this case, the standard deviation of the sampling distribution of the sample mean is 4 / sqrt(100) = 0.4 cars.A sample of 900900 computer chips revealed that 49%49% of the chips do not fail in the first 10001000 hours of their use. The company's promotional literature states that 52%52% of the chips do not fail in the first 10001000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0H0, at the 0.020.02 level.
Answer:
The proportion of chips that do not fail in the first 1000 hours of their use is 52%.
Step-by-step explanation:
The claim made by the company is that 52% of the chips do not fail in the first 1000 hours of their use.
A quality control manager wants to test the claim.
A one-proportion z-test can be used to determine whether the proportion of chips do not fail in the first 1000 hours of their use is 52% or not.
The hypothesis can be defined as:
H₀: The proportion of chips do not fail in the first 1000 hours of their use is 52%, i.e. p = 0.52.
Hₐ: The proportion of chips do not fail in the first 1000 hours of their use is different from 52%, i.e. p ≠ 0.52.
The information provided is:
[tex]\hat p=0.49\\n=900\\\alpha =0.02[/tex]
The test statistic value is:
[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.49-0.52}{\sqrt{\frac{0.52(1-0.52)}{900}}}=-1.80[/tex]
The test statistic value is -1.80.
Decision rule:
If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.
Compute the p-value as follows:
[tex]p-value=2\times P(Z<-1.80)\\=2\times 0.03593\\=0.07186[/tex]
*Use a z-table.
The p-value = 0.07186 > α = 0.02.
The null hypothesis was failed to be rejected at 2% level of significance.
Conclusion:
The proportion of chips that do not fail in the first 1000 hours of their use is 52%.
3) A state policeman has a theory that people who drive red cars are more likely to drive too fast. On his day off, he borrows one of the department’s radar guns, parks his car in a rest area, and measures the proportion of red cars and non-red cars that are driving too fast (he decided ahead of time to define "driving too fast" as exceeding the speed limit by more than 5 miles per hour). To produce a random sample, he rolls a die and only includes a car in his sample if he rolls a 5 or a 6. He finds that 18 out of 28 red cars are driving too fast and 75 of 205 other cars are driving too fast. Is this convincing evidence that people who drive red cars are more likely to drive too fast, as the policeman has defined it?
Answer:
Yes, it is convincing evidence to conclude that the proportion of red cars that drive too fast on this
highway is greater than the proportion of non-red cars that drive too fast.
Step-by-step explanation:
From the, we wish to first test;
H0; P_r - P_o = 0
And; H0; P_r - P_o > 0
Where; P_r and P_o are the proportion of red cars and other cars, respectively, who are driving too fast.
We will use a significance level of a = 0.05.
Thus;
The procedure is a two-sample z-test for the difference of proportions.
For, Random Conditions: The policemen chose cars randomly by rolling a die.
10%: We can safely assume that the number of cars driving past the rest area is essentially infinite, so the 10% restriction does not apply.
Large counts: The number of successes and failures in the two groups are 18, 10, 75, and 130—all of which are at least 10.
So, P_r = 18/28 = 0.64
P_o = 75/205 = 0.37
P_c = (18 + 75)/(28 + 205) = 0.4
Thus:
z = [(0.64 - 0.37) - 0]/√[[(0.4 x 0.6)/28] + [(0.4 x 0.6)/205]]
z = 2.73
From the one tailed z-score calculator online, I got P value = 0.003167
Thus, the P-value of 0.0032 is less than a = 0.05, so we reject H0. We
have sufficient evidence to conclude that the proportion of red cars that drive too fast on this
highway is greater than the proportion of non-red cars that drive too fast.
Find the surface area of a triangular prism with measurements 8 cm 6 cm 5 cm and 3 cm
Answer:
86.96
Step-by-step explanation:
not sure what is height or base.
Answer:
86.96
Step-by-step explanation:
The plot shows the temperatures (in ºF) for a group of children who visited a doctor’s office.
A plot shows the temperature of children at a doctor's office. 1 child had a temperature of 96 degrees; 2, 97 degrees; 5, 98 degrees; 2, 99 degrees; 1, 100 degrees.
What conclusions can be drawn from the data set? Check all that apply.
Answer:
the answer is 1,3,4,5
Step-by-step explanation:
Does the sum of 1/5 + 1/5 equals 2/5 make sense in situations
Answer:
It equals 2/5.
Step-by-step explanation:
Basically you ignore the 5 because it stays the same. All you have to add are the numerators. 1+1=2. So it would be 2/5.
A scenario:
2 pizzas cut into 5th's. Everyone ate 4 in each and kept one slice. When you add those 2 together you would be 2 slices.
6. A survey for brand recognition is done and it is determined that 68% of consumers have heard of Dull Computer Company. A survey of 800 randomly selected consumers is to be conducted. For such groups of 800, would it be unusual to get 568 consumers who recognize the Dull Computer Company name? Find the mean and standard deviation to answer this question. Must show work and explain why or why not.
Given Information:
Population = n = 800
Probability = p = 68% = 0.68
Answer:
We can say with 68% confidence that 568 lies in the range of (518, 570) therefore, it would not be unusual to get 568 consumers who recognize the Dull Computer Company name.
Step-by-step explanation:
Let us first find out the mean and standard deviation.
mean = μ = np
μ = 800*0.68
μ = 544
standard deviation = σ = √np(1-p)
σ = √800*0.68(1-0.68)
σ = 13.2
we know that 68% of data fall within 2 standard deviations from the mean
μ ± 2σ = 544-2*13.2, 544+2*13.2
μ ± 2σ = 544-26.4 , 544+26.4
μ ± 2σ = 517.6, 570.4
μ ± 2σ = 518, 570
We can say with 68% confidence that 568 lies in the range of (518, 570) therefore, it would not be unusual to get 568 consumers who recognize the Dull Computer Company name.
Please help
Prove that the median to the hypotenuse of a right triangle is half the hypotenuse.
To compare lengths, use the _____ Formula.
Answer:
it is the midpoint formula because that is the only equation of that is being divided by 2. The formula is X1 +X2 divied by 2 then it is Y1 + Y2 dived by two which gives you half of something like a line
Step-by-step explanation:
Answer:
mid point
Step-by-step explanation:
A school bought pens, each costing $1, and pencils, each costing $0.5. The cost of the whole purchase was $220. How many pens and pencils were purchased if there were 80 more pencils than pens?
Answer:
200 pencils and 120 pens
Step-by-step explanation:
Let the number of pencils be x while pens be y.
Considering the cost of whole purchase then, we get the equation
0.5x+y=220
Considering the number of items bought, then the equation is
x-y=80
Adding the two equations then we have
1.5x=300
x=300/1.5=200
Consideeing that x-y=80 and x is 200 then
200-y=80
y=200-80=120
Therefore, the pencils were 200 and pens 120 pieces