Answer:
(1) The probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.
(2) The two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.
(3) The two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.
Step-by-step explanation:
Let X = number of senior professionals who thought that global warming is having a significant impact on the environment.
The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.65.
But the sample selected is too large and the probability of success is close to 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of p if the following conditions are satisfied:
np ≥ 10 n(1 - p) ≥ 10Check the conditions as follows:
[tex]np= 100\times 0.65=65>10\\n(1-p)=100\times (1-0.65)=35>10[/tex]
Thus, a Normal approximation to binomial can be applied.
So, [tex]\hat p\sim N(p, \frac{p(1-p)}{n})=N(0.65, 0.002275)[/tex].
(1)
Compute the value of [tex]P(0.64<\hat p<0.69)[/tex] as follows:
[tex]P(0.64<\hat p<0.69)=P(\frac{0.64-0.65}{\sqrt{0.002275}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{0.69-0.65}{\sqrt{0.002275}})[/tex]
[tex]=P(-0.20<Z<0.80)\\=P(Z<0.80)-P(Z<-0.20)\\=0.78814-0.42074\\=0.3674[/tex]
Thus, the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.
(2)
Let [tex]p_{1}[/tex] and [tex]p_{2}[/tex] be the two population percentages that will contain the sample percentage with probability 90%.
That is,
[tex]P(p_{1}<\hat p<p_{2})=0.90[/tex]
Then,
[tex]P(p_{1}<\hat p<p_{2})=0.90[/tex]
[tex]P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}})=0.90[/tex]
[tex]P(-z<Z<z)=0.90\\P(Z<z)-[1-P(Z<z)]=0.90\\2P(Z<z)-1=0.90\\2P(Z<z)=1.90\\P(Z<z)=0.95[/tex]
The value of z for P (Z < z) = 0.95 is
z = 1.65.
Compute the value of [tex]p_{1}[/tex] and [tex]p_{2}[/tex] as follows:
[tex]-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.65=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.65\times 0.05)\\p_{1}=0.5675\\p_{1}\approx0.57[/tex] [tex]z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.65=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.65\times 0.05)\\p_{1}=0.7325\\p_{1}\approx0.73[/tex]
Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.
(3)
Let [tex]p_{1}[/tex] and [tex]p_{2}[/tex] be the two population percentages that will contain the sample percentage with probability 95%.
That is,
[tex]P(p_{1}<\hat p<p_{2})=0.95[/tex]
Then,
[tex]P(p_{1}<\hat p<p_{2})=0.95[/tex]
[tex]P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}})=0.95[/tex]
[tex]P(-z<Z<z)=0.95\\P(Z<z)-[1-P(Z<z)]=0.95\\2P(Z<z)-1=0.95\\2P(Z<z)=1.95\\P(Z<z)=0.975[/tex]
The value of z for P (Z < z) = 0.975 is
z = 1.96.
Compute the value of [tex]p_{1}[/tex] and [tex]p_{2}[/tex] as follows:
[tex]-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.96=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.96\times 0.05)\\p_{1}=0.552\\p_{1}\approx0.55[/tex] [tex]z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.96=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.96\times 0.05)\\p_{1}=0.748\\p_{1}\approx0.75[/tex]
Thus, the two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.
The questions concern calculating the probability of a sample proportion falling within a given range and constructing confidence intervals around the population proportion. These questions are related to statistical concepts such as the normal approximation to the binomial distribution and the use of z-scores for interval estimation, which are typically covered in college-level statistics courses.
Explanation:The student has asked about determining the probability that a sample percentage will fall within certain ranges, given a known proportion from a survey conducted by the Institute of Management Accountants (IMA) regarding the impact of global warming on their companies. Specifically, the student is looking to estimate probabilities related to the sample proportion and construct confidence intervals around a population percentage. These are statistical concepts typically covered in college-level courses in probability and statistics, specifically in chapters related to sampling distributions and confidence interval estimation.
To answer the first question, we would need to use the normal approximation to the binomial distribution, since the sample size is large (n=100). The sample proportion p = 0.65, and we can calculate the standard error for the sampling distribution of the sample proportion. However, since the full calculations are not provided here, a specific numerical answer cannot be given.
For the second and third questions, constructing confidence intervals at 90% and 95% requires using the standard error and the appropriate z-scores that correspond to these confidence levels. Again, the specific limits are not calculated here, but the process involves multiplying the standard error by the z-score and adding and subtracting this product from the sample percentage.
A 2010 estimate of Australia's population is 21,515,754.
Which is the best estimate of the number of Australians with type A blood in 2010?
Blood Types in Australia
Answer:
[tex]\mathrm{Number\:of\:Australians\:with\:type\:A\:Blood\:Group\:in\:2010}\:\approx\:8175987[/tex]
Step-by-step explanation:
[tex]\mathrm{Percent}:\\\\\mathrm{A\:ratio\:expressed\:as\:a\:fraction\:out\:of\:a\:hundred.}\\\\\mathrm{In\:order\:to\:convert\:a\:percent\:to\:a\:ratio,\:divide\:it\:by\:100}:\\\\38\%\:=\frac{38}{100}\\\\\mathrm{Percentage\:of\:population\:with\:A\:Blood\:Type}\:=\frac{38}{100}\times 21515754\\\\\approx8175987[/tex]
An estimated 8,175,987 Australians have Type A blood.
There were 21,515,754 in Australia in 2010. In the same year, it was estimated that 38% of people had Type A blood.
This means that the best estimate of people in Australia with Type A blood is:
= Percentage of people with type A blood x Number of people in Australia
= 38% x 21,515,754
= 8,175,986.52
= 8,175,987 people
In conclusion, approximately 8,175,987 Australians have Type A blood.
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The demand function for a certain make of replacement cartridges for a water purifier is given by the following equation where p is the unit price in dollars and x is the quantity demanded each week, measured in units of a thousand.
p = −0.01x2 − 0.1x + 32
Determine the consumers' surplus if the market price is set at $2/cartridge. (Round your answer to two decimal places.)
Demand function: p = -0.01x^2-0.2x+13 (1).
The equation (2) is: CS = ∫ D(p) dp, where D is the demand curve expressed in terms of the price. The bottom limit of the integrand is the market price given as $5 and the top limit is $13 found by setting x = 0 in (1) above. Equation (2) above requires we solve the original equation, (1) above, for x in terms of p. The first step is (a) to multiply both sides of equation by -100 to clear decimals, then (b) place equation in standard quadratic form, namely, x^2 + 20x + 100p – 1300 = 0. Step (c): Solve for x by applying quadratic formula, namely, x = (-b +- √(b^2 – 4ac)) / 2a to solve for x. Use a = 1, b = 20, c = 100p – 1300. The new demand equation is: x = -10 + 10√(14 – p). Now calculate ∫ (-10 + 10(14 – p)^(1/2)) dp and evaluate at 5 and 13. This gives: -10(13 – 5) – (2/3) (10) ((14 - 13)^(3/2) – (14 – 5)^(3/2)) . This evaluates to: -80 – (20/3) + 180. So, CS = $93.33.
The consumer surplus demanded will be 50 per week if the market price is set at $2/cartridge.
What do you mean by domain and range of a function?For any function y = f(x), Domain is the set of all possible values of [x] for which [y] exists. Range is the set of all values of [y] that exists for the given domain.
Given is the demand function for a certain make of replacement cartridges for a water purifier is given by the following equation where [p] is the unit price in dollars and [x] is the quantity demanded each week, measured in units of a thousand.
The demand function is -
p = − 0.01x² − 0.1x + 32
For p = 2, we can write -
− 0.01x² − 0.1x + 32 = 2
− 0.01x² − 0.1x + 30 = 0
Graph the equation and note the [x] intercepts. Taking the positive value of [x] intercept, we get 50.
Hence, the consumer surplus demanded will be 50 per week if the market price is set at $2/cartridge.
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Let X and Y denote the values of two stocks at the end of a five-year period. X is uniformly distributed on the interval (0, 12). Given X = x, Y is uniformly distributed on the interval (0, x). Determine Cov(X, Y) according to this model.
Answer:
Cov (X,Y) = 6
Step-by-step explanation:
hello,
Cov(X,Y) = E(XY) - E(X)E(Y)
we must first find E(XY), E(X), and E(Y).
since X is uniformly distributed on the interval (0,12), then E(X) = 6.
next we find the joint density f(x,y) using the formula
[tex]f(x,y) = g(y|x)f_{X}(x)[/tex]
[tex]f_{X} (x) = \frac{1}{12} \ $for$\ 0<x<12[/tex] this is because f is uniformly distributed on the the interval (0,12)
also since the conditional probability density of Y given X=x, is uniformly distributed on the interval [0,x], then
[tex]g(y|x)=\frac{1}{x}[/tex] for 0≤y≤x≤12
thus
[tex]f(x,y)=\frac{1}{12x}[/tex].
hence,
[tex]E(X,Y)= \int\limits^{12}_{x=0} \int\limits^x_{y=o} xy\frac{1}{12x} \,dy dx[/tex]
[tex]E(X,Y)=\frac{!}{24} \int\limits^{12}_{x=0} x^2 \, dx = 24[/tex]
also,
[tex]E(Y) = \int\limits^{12}_{x=0} \int\limits^x_{y=0} y\frac{1}{12x} \, dydx[/tex]
[tex]E(Y)=\frac{1}{24}\int\limits^{12}_{x=0} {x} \, dx =3[/tex]
thus Cov(X,Y) = E(XY) - E(X)E(Y)
= 24 - (6)(3)
= 6
The Wall Street Journal recently published an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked of both men and women was: "Do you think sexual harassment is a major problem in the American workplace?" 24% of the men compared to 62% of the women responded "Yes". Suppose that 150 women and 200 men were interviewed.
a) What are the null and alternative hypotheses that The Wall Street Journal should test in order to show that its claim is true?
b) For a 0.01 level of significance, what is the critical value for the rejection region?
c) What is the value of the test statistic? (hint: you need to find pbar)
d) What is the p-value of the test?
e) What conclusion should be reached?
Answer:
(a) H₀: P₂ - P₁ = 0 vs. Hₐ: P₂ - P₁ > 0.
(b) The critical value for the rejection region is 2.33.
(c) The calculated z-statistic value is, z = 7.17.
(d) The p-value of the test is 0.
(e) The proportion of women who view sexual harassment on the job is more than that for men.
Step-by-step explanation:
Here we need to test whether the proportion of women who view sexual harassment on the job is more than that for men.
(a)
Our hypothesis will be:
H₀: The difference between the proportions of men and women who view sexual harassment on the job as a problem is same, i.e. P₂ - P₁ = 0
Hₐ: The difference between the proportions of men and women who view sexual harassment on the job as a problem is more than 0, i.e. P₂ - P₁ > 0.
(b)
The significance level of the test is:
α = 0.01
The rejection region is defined as:
If test statistic value, z[tex]_{t}[/tex] > z₀.₀₁ then then null hypothesis will be rejected.
Compute the critical value of the test as follows:
[tex]z_{\alpha}=z_{0.01}=2.33[/tex]
*Use z-table.
Thus, the critical value for the rejection region is 2.33.
(c)
The z-statistic for difference of proportions is,
[tex]z=\frac{\hat p_{2}-\hat p_{1}}{\sqrt{P(1-P)\times (\frac{1}{n_{2}}+\frac{1}{n_{1}})}}[/tex]
[tex]\hat p_{i}[/tex] = ith sample proportion,
P = population proportion
[tex]n_{i}[/tex] = ith sample size.
The given information is:
[tex]n_{1}=200\\n_{2}=150\\\hat p_{1}=0.24\\\hat p_{2}=0.62[/tex]
Since, there is no data about the population proportion the unbiased estimate of P is given by,
[tex]P=\frac{n_{1}\hat p_{1}+n_{2}\hat p_{2}}{n_{1}+n_{2}}=\frac{200\times 0.24+150\times 0.62}{200+150}=0.4029[/tex]
Using the given data we compute the z-statistic as:
[tex]z=\frac{\hat p_{2}-\hat p_{1}}{\sqrt{P(1-P)\times (\frac{1}{n_{2}}+\frac{1}{n_{1}})}}[/tex]
[tex]=\frac{0.62-0.24}{\sqrt{0.4029(1-0.4029)\times (\frac{1}{150}+\frac{1}{200})}}[/tex]
[tex]=7.17[/tex]
Thus, the calculated z-statistic value is, z = 7.17.
(d)
Compute the p-value of the test as follows:
[tex]p-value=P(Z>z_{t})[/tex]
[tex]=P(Z>7.17)\\=1-P(Z<7.17)\\=1 -(\approx1)\\=0[/tex]
Thus, the p-value of the test is 0.
(e)
As stated in part (b), if z₀.₀₁ > z[tex]_{t}[/tex] then then null hypothesis will be rejected.
z[tex]_{t}[/tex] = 7.17 > z₀.₀₁ = 2.33
Thus, the null hypothesis will be rejected at 1% level of significance.
Conclusion:
The proportion of women who view sexual harassment on the job is more than that for men.
A researcher wants to study exercise as a way of reducing anger expression. He plans to throw water balloons at participants, measure their anger using behavioral cues (e.g., number of swear words, redness of face, etc.), make them do 20 pushups, then measure their anger again. He wants to see if there are differences before and after 20 pushups. Which t-test should he use? a. Single sample b. Paired samples c. Independent samples
Answer:
Option B: Paired sample t-test, since there is a "before 20 push ups" and "after 20 push up " score for each participant.
Step-by-step explanation:
In this question, we have a case where each participant undergoes different exercises to see differences before and after 20 push ups.
Now, it's obvious that the same condition applies to all participants i.e. differences before and after 20 push ups.
Thus, this will be a case of paired sample t-test or dependent t test because conditions for all participants in both tests are dependent or same.
So, the only option that corresponds with my explanation to use paired sample is option B
Is 26, 31, 36, 41, 46 an arithmetic sequence?
Answer:
Yes, it is arithmetic sequence.
Step-by-step explanation:
31-26=5
36-31=5
41-36=5
46-41=5
Difference between consecutive numbers is the same, so we have arithmetic sequence.
Answer:
Yes it is an arithmetic sequence
26+5=31
31+5=36
36+5=41
41+5=46
Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim.The owner of a football team claims that the average attendance at games is over 694, and he is therefore justified in moving the team to a city with a larger stadium. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is failure to reject the null hypothesis, state the conclusion in nontechnical terms.There is sufficient evidence to support the claim that the mean attendance is greater than than 694.There is sufficient evidence to support the claim that the mean attendance is less than 694.There is not sufficient evidence to support the claim that the mean attendance is less than 694.There is not sufficient evidence to support the claim that the mean attendance is greater than 694.
Answer:
Option D is correct.
Failure to reject the null hypothesis means that there is not sufficient evidence to support the claim that the mean attendance is greater than 694.
Step-by-step explanation:
For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
It usually maintains that random chance is responsible for the outcome or results of any experimental study/hypothesis testing.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
It usually maintains that other than random chance, there are significant factors affecting the outcome or results of the experimental study/hypothesis testing.
For this question, the owner of a football team claims that the average attendance at games is over 694, and he is therefore justified in moving the team to a city with a larger stadium.
So, the null hypothesis would be that there isn't enough evidence to support the claim that the mean attendance is greater than 694.
That is, the mean isn't greater than 694; the mean is less than or equal to 694.
While the alternative hypothesis would be that there is sufficient evidence to support the claim that the mean attendance is greater than 694.
Mathematically,
The null hypothesis is represented as
H₀: μ₀ ≤ 694
The alternative hypothesis is represented as
Hₐ: μ₀ > 694
Failure to reject the null hypothesis means that the null hypothesis is true. Hence, the answer choice picked is the obvious correct answer.
Hope this Helps!!!
How tall is a building that casts a 74-ft shadow when the angle of elevation of the sun is 36°?
Answer:
101.85ft
Step-by-step explanation:
We can use tan here
[tex]tan\theta=\frac{opp}{adj}\\adj = \frac{opp}{tan \theta} \\adj = \frac{74}{tan(36)} = 101.85[/tex]
The height should be 101.85 ft.
Calculation of the height:Since it casts a 74-ft shadow when the angle of elevation of the sun is 36°
So, the height should be
Here we can use tan
[tex]= 74\div tan\ 36[/tex]
= 101.85 ft
Therefore, we can conclude that The height should be 101.85 ft.
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You and a friend go to the movies. You each buy a ticket for x dollars and a popcorn-and-drink combo for $5.25. Use the distributive property to find the total amount of money you and your friend spent using the following expression 2(x + 5.25)
Answer:
2x + 10.5.
Step-by-step explanation:
Multiply 2 by x and 5.25.
Answer:
.
Step-by-step explanation:
Josh wants to download music online. He must buy a membership for $14.
Then he can download songs for 99 cents each. Josh has $20. How many
songs can Josh buy without spending more money than he has?
Answer:
6 songs
Step-by-step explanation:
It's best to start by setting up an equation equal to $20.
$14 + $.99s = $20
s being songs in this case.
First we add like terms by subtracting $14 from both sides.
$.99s = $6
Now we solve for s by dividing both sides by $.99
s = $6.06
Therefore, Josh can by 6 songs.
Final answer:
Josh can buy 6 songs with his remaining $6 after purchasing a $14 membership, since each song costs 99 cents.
Explanation:
Josh wants to download music online. With $20 to spend, he first needs to buy a membership for $14, leaving him with $6 to spend on songs. Each song costs 99 cents, so to determine how many songs Josh can purchase without exceeding his budget, we have to divide the remaining money by the cost per song.
Step 1: Calculate the remaining budget after membership: $20 - $14 = $6.
Step 2: Divide the remaining budget by the cost per song: $6 \(\div\) $0.99 ≈ 6 songs.
Thus, Josh can buy 6 songs without spending more than he has.
The point (-3,1) is on the terminal side of the angle in standard position. What is tanθ
Answer:
[tex] \tan \theta = - \frac{1}{3}[/tex]
Step-by-step explanation:
Since, point (-3, 1) is on the terminal side of the angle in standard position.
[tex] \therefore \: ( - 3, \: 1) = (x, \: y) \\ \therefore \:x = - 3 \: \: and \: \: y = 1 \\ \because \tan \theta = \frac{y}{x} \\ \therefore \:\tan \theta = \frac{1}{ - 3} \\ \\ \huge \red{ \boxed{\therefore \:\tan \theta = - \frac{1}{3} }}[/tex]
Write the word sentence as an inequality. A number h is at least −12.
Answer:
h ≥ -12
Step-by-step explanation:
at least means greater than or equal to
h ≥ -12
Answer:
h>-12
Step-by-step explanation:
Ali is riding his bicycle. He rides at a Speed of 12.8 kilometers per hour for 2 hours how many kilometers did he ride
Answer:
25.6 kilometers
Step-by-step explanation:
12.8 km in an hour
he is riding for 2 hours
12.8 x 2 = 25.6
Using the formula 'Distance equals Speed times Time', we find that Ali rode a distance of 25.6 kilometers given that he rode his bicycle at a speed of 12.8 kilometers per hour for 2 hours.
Explanation:To answer this question, you can use the formula to calculate distance, which is Speed multiplied by Time. In Ali's case, he's riding his bicycle at a speed of 12.8 kilometers per hour and for 2 hours. Using the formula, you'll multiply 12.8(km/h) by 2(h). Thus, Ali's distance covered would be 25.6 kilometers.
Here's how it works in a step-by step format:
Use the formula: Distance = Speed x TimePlug in the given values: Distance = 12.8 km/h x 2 hCompute: Distance = 25.6 kmTherefore, Ali rode his bicycle for a distance of 25.6 kilometers.
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A survey was conducted two years ago asking college students their top motivations for using a credit card. To determine whether this distribution has changed, you randomly select 425 college students and ask each one what the top motivation is for using a credit card. Can you conclude that there has been a change in the claimed or expected distribution? Use alpha = 0.10.
Response Old Survey % New Survey Frequency
Reward 27 112
Low rate 23 96
Cash back 21 109
Discount 9 48
Others 20 60
Answer:
See explanation
Step-by-step explanation:
Solution:-
- A survey was conducted among the College students for their motivations of using credit cards two years ago. A randomly selected group of sample size n = 425 college students were selected.
- The results of the survey test taken 2 years ago and recent study are as follows:
Old Survey ( % ) New survey ( Frequency )
Reward 27 112
Low rate 23 96
Cash back 21 109
Discount 9 48
Others 20 60
- We are to test the claim for any changes in the expected distribution.
We will state the hypothesis accordingly:
Null hypothesis: The expected distribution obtained 2 years ago for the motivation behind the use of credit cards are as follows: Rewards = 27% , Low rate = 23%, Cash back = 21%, Discount = 9%, Others = 20%
Alternate Hypothesis: Any changes observed in the expected distribution of proportion of reasons for the use of credit cards by college students.
( We are to test this claim - Ha )
We apply the chi-square test for independence.
- 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 other.
- We will compute the chi-square test statistics ( X^2 ) according to the following formula:
[tex]X^2 = Sum [ \frac{(O_i - E_i)^2}{Ei} ][/tex]
Where,
O_i : The observed value for ith data point
E_i : The expected value for ith data point.
- We have 5 data points.
So, Oi :Rewards = 27% , Low rate = 23%, Cash back = 21%, Discount = 9%, Others = 20% from a group of n = 425.
Ei : Rewards = 112 , Low rate = 96, Cash back = 109, Discount = 48, Others = 60.
Therefore,
[tex]X^2 = [ \frac{(112 - 425*0.27)^2}{425*0.27} + \frac{(96 - 425*0.23)^2}{425*0.23} + \frac{(109 - 425*0.21)^2}{425*0.21} + \frac{(48 - 425*0.09)^2}{425*0.09} + \frac{(60 - 425*0.20)^2}{425*0.20}]\\\\X^2 = [ 0.06590 + 0.03132 + 4.37044 + 2.48529 + 7.35294]\\\\X^2 = 14.30589[/tex]
- Then we determine the chi-square critical value ( X^2- critical ). The two parameters for evaluating the X^2- critical are:
Significance Level ( α ) = 0.10
Degree of freedom ( v ) = Data points - 1 = 5 - 1 = 4
Therefore,
X^2-critical = X^2_α,v = X^2_0.1,4
X^2-critical = 7.779
- We see that X^2 test value = 14.30589 is greater than the X^2-critical value = 7.779. The test statistics value lies in the rejection region. Hence, the Null hypothesis is rejected.
Conclusion:-
This provides us enough evidence to conclude that there as been a change in the claimed/expected distribution of the motivations of college students to use credit cards.
Final answer:
To test for changes in credit card usage motivations among college students, calculate the expected frequencies from the old survey percentages, apply the chi-square goodness-of-fit test, and compare the statistic to the chi-square critical value at alpha = 0.10.
Explanation:
Understanding the Hypothesis Test for Changed Distribution
To determine whether there has been a shift in the distribution of college students' motivations for using credit cards since the previous survey, we conduct a hypothesis test for goodness of fit. The test will compare the observed frequencies from the new survey with the expected frequencies based on the old survey percentages.
First, calculate the expected frequency for each category using the formula: Expected Frequency = (Old Survey Percentage / 100) × Total Number of Students. Then, employ the chi-square goodness-of-fit test to analyze the data:
Reward: Expected = 0.27 × 425 = 114.75
Low rate: Expected = 0.23 × 425 = 97.75
Cash back: Expected = 0.21 × 425 = 89.25
Discount: Expected = 0.09 × 425 = 38.25
Others: Expected = 0.20 × 425 = 85.00
With the expected frequencies calculated, the chi-square statistic will be computed:
χ² = Σ((Observed - Expected) ² / Expected)
This statistic will be compared to a critical value from the chi-square distribution table with (n - 1) degrees of freedom, where n is the number of categories, at alpha = 0.10 to determine whether to reject the null hypothesis (no change in distribution).
A study was done on the timeliness of flights (on-time vs. delayed) of two major airlines: StatsAir and AirMedian. Data were collected over a period of time from five major cities and it was found that StatsAir does better overall (i.e., has a smaller percentage of delayed flights). However, in each of the five cities separately, AirMedian does better.
Which of the following is correct?
(a) This situation is mathematically impossible.
(b) This is an example of Simpson's Paradox.
(c) "City" is a lurking variable in this example.
(d) This is an example of a negative association between variables.
(e) Both (b) and (c) are correct.
Answer:
The answer is Option E; both B and C are correct
Step-by-step explanation:
Both (b) and (c) are correct. Simpson's paradox is a paradox in which a trend that appears in different groups of data disappears when these groups are combined, and the reverse trend appears for the aggregate data. This result is often encountered in social-science and medical-science statistics, and is particularly confounding when frequency data are unduly given causal interpretations. Simpson's Paradox disappears when causal relations are brought into consideration.
Now, this question is an example of Simpsons paradox because the groups of collected data over a period of time from five major cities showed a trend that StatsAir does better overall, but this trend is reversed when the groups are studied separately to show that air median does better.
So, option B is correct.
Also, City is a variable that influences both the dependent variable and independent variable, causing a spurious association. That is it is the cause of why the 2 results are biased. Thus, city is a lurking variable.
So, option C is also correct
A pair of snow boots at an equipment store in Big Bear that originally cost $60 is on sale for 40% off. Then, you have a coupon for 10% off. What is the final cost?
Answer:
$30 i think
Step-by-step explanation:
because %40+%10=%50
and %50 is half of %100 so 60-(60×(50%)=30
sorry is its wrong
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:
See below.
Step-by-step explanation:
1) the measure of center are the same
3) there is little variability in the data
4) the average temperature is 98
5) the data is clustered around the mean
Write the equation for g(x)
Answer:
g-x=3
Step-by-step explanation:
The number of undergraduates at Johns Hopkins University is approximately 2000, while the number at Ohio State University is approximately 40,000. A simple random sample of 50 undergraduates at Johns Hopkins University will be obtained to estimate the proportion of all Johns Hopkins students who feel that drinking is a problem among college students. A simple random sample of 50 undergraduates at Ohio State University will be obtained to estimate the proportion of all Ohio State students who feel that drinking is a problem among college students. Answer Questions 4-5 below.
The remaining part of Question:
4) what can we conclude about sampling variability in the sample proportion calculated in the sample at John Hopkins as compared to that calculated in the sample at Ohio State.
5) The number of undergraduates at Johns Hopkins is approximately 2000 while the number at Ohio State is approximately 40000, suppose instead that at both schools, a simple random sample of about 3% of the undergraduates Will be taken.
Answer:
4) The sample proportion from Johns Hopkins will have about the same sampling variability as that from Ohio State
5) The sample proportion from John Hopkins will have more sampling variability than that from Ohio State
Step-by-step explanation:
Note: The sampling variability in the sample proportion decreases with increase in the sample size.
4) since the sample size at both Johns Hopkins and Ohio State is the same (i.e. n = 50), the sample variability of the sample proportion will be the same for both cases.
5) 3% of the population are selected for the observation in both cases.
At Johns Hopkins, sample size, n = 3% * 2000
n = 60
At Ohio State, sample size, n = 3% * 40000
n = 1200
Since sampling variability in the sample proportion decreases with increase in the sample size, the sampling variability in sample proportion will be higher at Johns Hopkins than at Ohio State.
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. Of the 29 tires surveyed, the mean lifespan was 45,800 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly consistent with the claim?
Answer:
Step-by-step explanation:
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 ≥ 50000
For the alternative hypothesis,
µ < 50000
Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is
z = (x - µ)/(σ/√n)
Where
x = lifetime of the tyres
µ = mean lifetime
σ = standard deviation
n = number of samples
From the information given,
µ = 50000 miles
x = 45800 miles
σ = 8000
n = 29
z = (50000 - 45800)/(8000/√29) = - 2.83
Looking at the normal distribution table, the probability corresponding to the z score is 0.9977
Since alpha, 0.05 < than the p value, 0.9977, then we would accept the null hypothesis. Therefore, At a 5% level of significance, the data is not highly consistent with the claim.
What is 3 to the 3rd power
Answer:
27
Step-by-step explanation:
3 x 3 x 3=27
Answer:
27
Step-by-step explanation:
you can just do 3x3x3 or 3^3 to figure it out.. 3x3=9 and 9x3= 27.. hope this helped..
What percentage of the data values falls between the values of 27 and 45 in the data set shown?
A box-and-whisker plot. The number line goes from 25 to 50. The whiskers range from 27 to 48, and the box ranges from 32 to 45. A line divides the box at 36.
25%
50%
75%
100%
Answer:
the answer is 75% hope that helps
Step-by-step explanation:
i did the test and i got it right
The required percentage of the data values falls between the values of 27 and 45 in the data set shown is 75%.
What is a box plot?A straightforward method of expressing statistical data on a plot in which a rectangle is drawn to represent the second and third quartiles, with a vertical line inside to indicate the median value. Horizontal lines on both sides of the rectangle show the lower and upper quartiles.
Based on the box-and-whisker plot, we can see that the majority of the data values fall between the values of 32 and 45, and the median value is 36.
To find the percentage of data values that fall between 27 and 45, we can estimate by visually analyzing the plot. Since the whiskers go from 27 to 48, we can assume that all data values between 27 and 48 are included. Then, we need to estimate how much of the data falls between 32 and 45, which is the range of the box.
Since the box takes up most of the range of values between 32 and 45, we can estimate that around 75% of the data values fall within this range. Therefore, the percentage of data values that fall between 27 and 45 can be estimated as around 75%.
Learn more about box plots here:
brainly.com/question/1523909
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A university planner is interested in determining the percentage of spring semester students who will attend summer school. She takes a pilot sample of 160 spring semester students discovering that 56 will return to summer school. (7 points) What is the proportion of semester students returning to summer school
Answer:
The proportion of semester students returning to summer school
p = 0.35
Step-by-step explanation:
Explanation:-
Given data A university planner is interested in determining the percentage of spring semester students who will attend summer school.
She takes a pilot sample of 160 spring semester students discovering that 56 will return to summer school
Given the sample size 'n' = 160
let 'x' = 56
Sample proportion of semester students returning to summer school
[tex]p = \frac{x}{n} = \frac{56}{160}[/tex]
p = 0.35
Carmen needs $3560 for future project. She can invest $2000 now at annual rate of 7.8%, compound monthly. Assuming that no withdrawals are made, how long will it take for her to have enough money for her project?
Answer:
7.42 years (7 years 5 months)
Step-by-step explanation:
The future value of Carmen's account can be modeled by
FV = P(1 +r/12)^(12t)
where P is the principal invested, r is the annual rate, and t is the number of years.
Solving for t, we have ...
FV/P = (1 +r/12)^(12t)
log(FV/P) = 12t·log(1 +r/12)
t = log(FV/P)/(12·log(1 +r/12))
For FV = 3560, P=2000, r = 0.078, the time required is ...
t = log(3560/2000)/(12·log(1 +.078/12))
t ≈ 7.42
It will take Carmen about 7 years 5 months to reach her savings goal.
multiply 3/4 x 16/19
To multiply fractions, multiply the numerators and denominators separately. For [tex]\frac{3}{4} \times \frac{16}{19}[/tex], the result is [tex]\frac{12 }{ 19}[/tex] after simplifying.
To multiply fractions, you simply multiply the numerators together and the denominators together. In this case, [tex]\frac{3}{4} \times \frac{16}{19}[/tex] would be calculated as: [tex]\frac{3 \times 16}{4 \times 19} = \frac{48 }{ 76}[/tex] This fraction can be simplified by dividing both the numerator and denominator by their greatest common factor, which in this case is 4: [tex]\frac{48 \div 4 }{ 76 \div 4 }= \frac{12 }{ 19}[/tex].
8. A well-balanced stock market portfolio will often experience an exponential growth. A
particular investor with a well-balanced stock market portfolio records the portfolio balance
every month, in thousands of dollars, from the date of investment. The roughly exponential
growth can be transformed to a linear model by plotting the natural log of the balances
versus time, in months, where t - 0 represents the date the money was invested. The linear
regression equation for the transformed data is
In (balance)-5.550 + 0.052t.
Using this equation, what is the predicted balance of the portfolio after 2 years (24 months)?
(A) $5,654
(B) $6,798
(C) $285,431
(D) $896,053
(E) $948,464
Answer:
D) $896,053
Step-by-step explanation:
Suppose samples of size 100 are drawn randomly from a population that has a mean of 20 and a standard deviation of 5. What are the values of the mean and the standard deviation of the distribution of the sample means
Answer:
a) The mean of the sampling distribution of means μₓ = μ = 20
b) The standard deviation of the sample σₓ⁻ = 0.5
Step-by-step explanation:
Explanation:-
Given sample size 'n' =100
Given mean of the Population 'μ' = 20
Given standard deviation of population 'σ' = 5
a) The mean of the sampling distribution of means μₓ = μ
μₓ = 20
b) The standard deviation of the sample σₓ⁻ = [tex]\frac{S.D}{\sqrt{n} }[/tex]
Given standard deviation of population 'σ' = 5
= [tex]\frac{5}{\sqrt{100} } = 0.5[/tex]
Final answer:
The mean of the distribution of the sample means drawn from a population with a mean of 20 and a standard deviation of 5, using samples of size 100, is 20. The standard deviation of this distribution, also known as the standard error, is 0.5.
Explanation:
The question involves understanding the concept of the distribution of sample means, also known as the sampling distribution. When samples of size 100 are drawn from a population with a mean (μ) of 20 and a standard deviation (σ) of 5, the mean of the distribution of the sample means will be the same as the population mean, which is 20. However, the standard deviation of the distribution of sample means, known as the standard error (SE), will be the population standard deviation divided by the square root of the sample size (√n), which in this case is 5/√100 = 0.5.
Therefore, the mean of the distribution of the sample means is 20 and the standard deviation (standard error) of this distribution is 0.5. This is based on the central limit theorem which states that, for a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normally distributed regardless of the population’s distribution, with these exact parameters of mean and standard deviation.
g Exercise 6. Let X be a Gaussian random variable with X ∼ N (0, σ2 ) and let U be a Bernoulli random variable with U ∼ Bern(?) independent of X. Define V as V = XU. (a) Find the characteristic function of V , ϕV = E(e jsV ) = RfV (v)e jsv. Hint: use iterated expectation. (b) Find the mean and variance of V .
Answer:
Step-by-step explanation
question solved below
How many centimeter cubes will fill a box 20cm by 6cm by 3cm
Answer:
360
Step-by-step explanation:
formula is L x B x H so first layer would be 6 rows of 20 cubes which is 120cubes
H is 3cm so 120x3 = 360
or find out the volume of the box, 20x6x3 = 360
The U.S. Department of Agriculture claims that the mean cost of raising a child from birth to age 2 by husband-wife families in the United States is $13,120. A random sample of 500 children (age 2) has a mean cost of $12,925 with a standard deviation of $1745. At α=.10, is there enough evidence to reject the claim? (Adapted from U.S. Department of Agriculture Center for Nutrition Policy and Promotion).
Answer:
Step-by-step explanation:
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 = 13120
For the alternative hypothesis,
µ ≠ 13120
This is a 2 tailed test
Since the population standard deviation is not given, the t test would be used to determine the test statistic. The formula is
t = (x - µ)/(s/√n)
Where
s = sample standard deviation
x = sample mean
µ = population mean
n = number of samples
From the information given,
µ = $13120
x = $12925
n = 500
s = $1745
t = (12925 - 13120)/(1745/√500) = - 2.5
Degree of freedom = n - 1 = 500 - 1 = 499
Using the t score calculator to find the probability value,
p = 0.012
Since α = 0.10 > p = 0.012, it means that there is enough evidence to reject the claim