Answer:
There is 95% confidence that the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week is between 55.9% and 74.7%.
Step-by-step explanation:
We have to answer the population proportion for Vermont.
We can only do it by a confidence interval, as we only have information from a sample.
This sample, of size n=100, has a proportion p=0.653.
The degrees of freedom are:
[tex]df=n-1=100-1=99[/tex]
We will calculate a 95% confidence interval, which for df=99 has a critical value of t of t=1.984.
The margin of error can be calculated as:
[tex]E=t*\sigma_p=t\sqrt{\dfrac{p(1-p)}{n}}=1.984\sqrt{\dfrac{0.653*0.347}{100}}\\\\\\E=1.984*\sqrt{0.00226}=1.984*0.0476=0.094[/tex]
Then, the upper and lower bounds of the confidence interval are:
[tex]LL=p-E=0.653-0.094=0.559\\\\UL=p+E=0.653+0.094=0.747[/tex]
Then, we can say that there is 95% confidence that the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week is between 55.9% and 74.7%.
Final answer:
The estimated population proportion of Vermont residents who exercised for at least 30 minutes a day 3 days a week is 65.3%, which is based on the sample proportion from the Gallup survey.
Explanation:
The student is asking about the population proportion for people from Vermont who exercised for at least 30 minutes a day 3 days a week based on the Gallup survey results. The survey indicated that 65.3% of the Vermont respondents exercised at the mentioned rate.
To find the value of the population proportion (population proportion), we typically use the sample proportion as an estimate. From the survey, we have that the sample proportion (p-hat) for Vermont is 65.3%, which we express as a decimal, 0.653. Assuming the sample is representative, we would estimate the population proportion to also be 0.653 or 65.3%.
It's important to note that this is an estimate based on the sample and that to infer more confidently about the entire population of Vermont, a larger sample size or additional statistical methods such as confidence intervals or hypothesis testing may be applied. Nevertheless, with the information provided, the best estimate for the population proportion of Vermont residents who exercised according to the guidelines is the sample proportion of 65.3%.
Use the given information to bound the p-value of the F statistic for a one-tailed test with the indicated degrees of freedom. F = 4.23, df1 = 4, df2 = 5 p-value < 0.005 0.005 < p-value < 0.010 0.010 < p-value < 0.025 0.025 < p-value < 0.050 0.050 < p-value < 0.100 p-value > 0.100
Answer:
The range of the p-value is: 0.050 < p-value < 0.100.
Step-by-step explanation:
For checking the equivalence of two population variances of independent samples, we use the f-test.
The test statistic is given by:
[tex]F=\frac{S_{1}^{2}}{S_{2}^{2}}\sim F_{\alpha, (n_{1}-1)(n_{2}-1)}[/tex]
It is provided that the hypothesis test is one-tailed.
The computed value of the test statistic is:
F = 4.23.
The degrees of freedom of the numerator and denominator are:
[tex]df_{1}=4\\df_{2}=5[/tex]
Use MS-Excel to compute the p-value as follows:
Step 1: Select function fX → F.DIST.RT.
Step 2: A dialog box will open. Enter the values of f-statistic and the two degrees of freedom.
*See the attachment below.
Step 3: Press OK.
The p-value is, 0.0728.
The range of the p-value is:
0.050 < p-value < 0.100
A survey of top executives revealed that 35% of them regularly read Time magazine, and 40% read U.S. News & World Report. Ten percent read both Time and U.S. News & World Report. What is the probability that a particular top executive reads either Time or U.S. News & World Report regularly? Select one: a. 0.85 b. 0.75 c. 1 d. 0.65
Answer:
Option D) 0.65
Step-by-step explanation:
We are given the following in the question:
Percentage of executives who read Time magazine = 35%
[tex]P(M) = 0.35[/tex]
Percentage of executives who read U.S. News & World Report = 40%
[tex]P(N) = 0.4[/tex]
Percentage of executives who read both Time magazine and U.S. News & World Report = 10%
[tex]P(M\cap N) = 0.1[/tex]
We have to find the probability that a particular top executive reads either Time or U.S. News & World Report regularly.
Thus, we have to evaluate,
[tex]P(M\cup N) = P(M) + P(N) -P(M\cap N)[/tex]
Putting values, we get,
[tex]P(M\cup N) = 0.35 + 0.4 - 0.1=0.65[/tex]
0.65 is the probability that a particular top executive reads either Time or U.S. News & World Report regularly.
Thus, the correct answer is
Option D) 0.65
D. Now calculate resting Vm given the following relative permeability ratios. Assume that there is no meaningful permeability to Ca2+. (1pt) PK : PNa : PCl = 1.0 : 0.04 : 0.45 E. Let’s say that the K+ permeability increases, for example because of expression of more K+ channels. Calculate Vm with new the permeability values and discuss how the membrane potential has changed in relation to the potassium equilibrium potential EK. (roughly 1 sentence) (1pt)+
Answer:EX = RT ln [X]o
.........zF.....[X]i
EX = (1.987 cal/deg.mol)(293 deg) ln [X]o
.........z(23,062 cal/volt.mol)................[X]i
OR
EX = (8.315 joules/deg.mol)(293 deg) ln [X]o
.........z(96,485 joules/volt.mol)................[X]i
EK+ = 0.025 ln(12/400) = -0.088 V = -88 mV
ENa+ = 0.025 ln(450/55) = 0.053 V = 53 mV
ECa+2 = 0.0126 ln(10/0.0001) = 0.145 V = 145 mV
ECl- = -0.025 ln(550/56) = -0.058 V = -58 mV
Step-by-step explanation:
The Goldman equation is used to calculate resting membrane potential (Vm) considering the relative permeabilities of different ions. Given the permeabilities, the resting Vm can be estimated. If the K+ permeability increases, Vm will move closer to the Potassium equilibrium potential (EK).
Explanation:The resting membrane potential (Vm) can be calculated using the Goldman equation, which considers the relative permeabilities and concentrations of different ions. The equation is: Vm = 61.5 log ((PK[K+]out + PNa[Na+]out + PCl[Cl-]in) / (PK[K+]in + PNa[Na+]in + PCl[Cl-]out)), where Px indicates the relative permeability of each ion & the square brackets contain the ion concentrations inside (in) or outside (out) the cell.
Given the permeabilities PK : PNa : PCl = 1.0 : 0.04 : 0.45, assuming concentrations inside and outside the cell in a balanced condition with no net movement of any ion, you might estimate the resting Vm.
If the K+ permeability increases, Vm would reportedly move closer towards the Potassium equilibrium potential (EK). This is because the membrane is becoming more permeable to K+ and less responsive to the influences of other ions.
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The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $490 and standard deviation $10. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1?
Answer:
[tex]z=1.28<\frac{a-490}{10}[/tex]
And if we solve for a we got
[tex]a=490 +1.28*10=502.8[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 502.8.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the weeknly amount of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(490,10)[/tex]
Where [tex]\mu=490[/tex] and [tex]\sigma=10[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.1[/tex] (a)
[tex]P(X<a)=0.9[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.9[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.9[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28<\frac{a-490}{10}[/tex]
And if we solve for a we got
[tex]a=490 +1.28*10=502.8[/tex]
So the value of height that separates the bottom 90% of data from the top 10% is 502.8.
Final answer:
The company should budget approximately $502.80 for weekly repairs and maintenance to ensure that the probability of exceeding this amount is only 0.1.
Explanation:
We want to find how much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1.
Since the weekly amount of money spent is normally distributed with a mean of $490 and a standard deviation of $10, we can find the amount by looking up the z-score that corresponds to the 90th percentile
(since 100% - 10% = 90%) in a standard normal distribution table or using a calculator.
Let the z-score for the 90th percentile be denoted as z.
Looking up the standard normal distribution table or using a calculator, we find that z ≈ 1.28 for 0.9 cumulative probability.
We then use the z-score formula:
z = (X - mean) / standard deviation
Plugging in our z-score and the parameters, we can solve for X:
1.28 = (X - 490) / 10
X - 490 = 12.8
X = $502.80
Therefore, the company should budget approximately $502.80 for weekly repairs and maintenance to ensure that the probability of exceeding this amount is only 0.1.
Add together 8.03 m 1.26 m 0.5 m 4.09 m 3.5 m
Answer:
17.38m
Step-by-step explanation:
Answer:
17.38m
Step-by-step explanation:
Need help with this please thank you!
Answer:
4
Step-by-step explanation:
AC = AE = 4 cm
factor the expression completely 40-5x
Answer:
5(8-x)
Step-by-step explanation:
Fact using the GCF (Greatest Common Factor) of 40 and 5
Factors of 5:
1,5
Factors of 40:
1,2,4,5,8,10,20,40
The greatest factor they both have is 5
Factor by distributing out a 5
40-5x
5(8-x)
The GCF (Greatest Common Factor) of 40 and 5 is 5!
Divide both numbers by 5.
40 / 5 = 8
-5x / 5 = -x
Rewrite the expression using the distributive property.
5(8 - x)
Best of Luck!
When rolling two fair 6 sided dice, what is the probability that the total is at most 10?
Answer:
[tex]\frac{33}{36}[/tex]
Step-by-step explanation:
Combinations greater than a 10
5 - 5
5 - 6
6 - 5
There are total 36 combinations (6 * 6).
3 of these combinations are higher than 10.
So 36 - 3 combinations are less than 10.
Use this information to answers Questions 1 through 7. The U.S. government provides money to each state to maintain the interstate highway system in the state. The U.S. can revoke or reduce the money if the states do not safely maintain the highways. The U.S. government is particularly concerned about the speed of traffic on Kansas highways. If there is convincing evidence that the average speed of all interstate highway vehicles in Kansas exceeds the posted speed limit of 70 mph, the federal government will reduce the amount of funding it provides. If there is not convincing evidence, then the government will not reduce the funding.
Kansas Highway Patrol recorded the speed of 450 interstate vehicles and found the mean speed of 70.2 mph with standard deviation 1.6 mph. The U.S. government will use this information to conduct a hypothesis test at significance level 0.01 to decide whether or not to reduce to money sent to Kansas.
Suppose that mu is the true mean speed of all vehicles on the Kansas interstate highway system.
1. What are the null and alternative hypotheses that the U.S. government should test?
2. What is the value of the test statistic?
Answer:
1. Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 70 mph
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 70 mph
2. Value of test statistics is 2.652.
Step-by-step explanation:
We are given that Kansas Highway Patrol recorded the speed of 450 interstate vehicles and found the mean speed of 70.2 mph with standard deviation 1.6 mph.
We have to conduct a hypothesis test at significance level 0.01 to decide whether or not to reduce the money sent to Kansas.
Let [tex]\mu[/tex] = true mean speed of all vehicles on the Kansas interstate highway system.
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 70 mph {means that the federal government will not reduce the amount of funding it provides as the speed limit is less than or equal to 70 mph}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 70 mph {means that the federal government will reduce the amount of funding it provides as the speed limit exceed 70 mph}
The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean speed limit of 450 interstate vehicles = 70.2 mph
s = sample standard deviation = 1.6 mph
n = sample of vehicles = 450
So, test statistics = [tex]\frac{70.2-70}{\frac{1.6}{\sqrt{450} } }[/tex] ~ [tex]t_4_4_9[/tex]
= 2.652
Hence, the value of test statistics is 2.652.
What inequality is represented by this graph? A number line going from 1 to 9. An open circle is at 6. Everything to the left of the circle is shaded. x greater-than 6 x less-than-or-equal-to 6 x less-than 6 x greater-than-or-equal-to 6
x less-than-or-equal-to 6 is the inequality which is represented by the graph.
How to solve the problem?The problem can be solved by following steps
It is given that
A number line going from 1 to 9.An open circle is at 6.Everything to the left of the circle is shaded.According to above statement a graph is drawn below
The shaded part of the left side describes the value of x which are less than 6 or equal to 6
Hence , The inequality that is represented in the graph is x is less than or equal to 6
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Answer: B
Step-by-step explanatii got it right on edge
I don’t know how to do this. Care to explain?
Answer:
y=6x+7
Step-by-step explanation:
−6x+y=7
Step 1: Add 6x to both sides
y=6x+7
Answer:
y=6x+7
Step-by-step explanation:
whenever you solve for a letter you have to get it by itself. That means it has to be alone on the opposite side of the equation
move the -6x to the other side of the equation
that -6 changes to a positive 6 because, your moving it to the other side of the equation
your left with y=6x+7
pls mark me brainliest
A study of bulimia among college women studied the connection between childhood sexual abuse and a measure of family cohesion (the higher the score, the greater the cohesion). The sample mean on the family cohesion scale was 1.9 for 13 sexually abused students (sequals2.1) and 5.2 for 17 nonabused students (sequals3.5). a. Find the standard error for comparing the means. b. Construct a 95% confidence interval for the difference between the mean family cohesion for sexually abused students and non-abused students. Interpret.
Answer: a) 1.029, b) (-5.318, -1.282).
Step-by-step explanation:
Since we have given that
[tex]n_1=13\\\\n_2=17\\\\\bar{x_1}=1.9\\\\\bar{x_2}=5.2[/tex]
and
[tex]s_1=2.1\\\\s_2=3.5[/tex]
So, the standard error for comparing the means :
[tex]SE=\sqrt{\dfrac{s^2_1}{n_1}+\dfrac{s^2_2}{n_2}}\\\\SE=\sqrt{\dfrac{2.1^2}{13}+\dfrac{3.5^2}{17}}\\\\SE=\sqrt{1.0598}\\\\SE=1.029[/tex]
At 95% confidence interval, z = 1.96
So, Confidence interval would be
[tex]\bar{x_1}-\bar{x_2}\pm z\times SE\\\\=(1.9-5.2)\pm 1.96\times 1.0294\\\\=-3.3\pm 2.017624\\\\=(-3.3-2.018,-3.3+2.018)\\\\=(-5.318,-1.282)[/tex]
Hence, a) 1.029, b) (-5.318, -1.282).
Thirty-four college students were asked how much money they spent on textbooks for the current semester. Their responses are shown in the following stemplot.
1 2 3 3 4 5 5 6 7 8
2 1 2 3 4 5 6 8 8 9 9 9
3 1 2 2 7 8 9
4 1 4 5 7
5 1 3
6 2
7
8 1
Key: 1|2 = $120
a. Describe a procedure for identifying potential outliers, and use the procedure to decide whether there are outliers among the responses for the money spent on textbooks.
b. Based on the stemplot, write a few sentences describing the distribution of money spent on textbooks for the 34 students.
Answer:
(a) The outlier in the data is $810.
(b) The distribution of money spent on textbooks for the 34 students is right skewed.
Step-by-step explanation:
The data provided for the amount of money 34 college students spent on books is:
S = {120, 130, 130, 140, 150, 150, 160, 170, 180, 210, 220, 230, 240, 250, 260, 280, 280, 290, 290, 290, 310, 320, 320, 370, 380, 390, 410, 440, 450, 470, 510, 530, 620, 810}
(a)
An outlier of a data set is a value that is very different from the other values of a data set. It is either too large or too small.
The most common way to determine whether a data set consists of any outliers of not is,
Data value that less than Q₁ - 1.5 IQR are outliers.Data values that are more than Q₃ + 1.5 IQR are outliers.Here
Q₁ = first quartile
Q₃ = third quartile
IQR = Inter-quartile range = Q₃ - Q₁.
The first quartile is the value that is more than 25% of the data values. The first quartile is the median of the first half of the data.
Compute the value of first quartile as follows:
First half of data: {120, 130, 130, 140, 150, 150, 160, 170, 180, 210, 220, 230, 240, 250, 260, 280, 280}
There are 17 values.
The median of an odd data set is the middle value.
The middle value is: 180
The first quartile is Q₁ = 180.
The third quartile is the value that is more than 75% of the data values.
Compute the value of first quartile as follows:
Second half of data: {290, 290, 290, 310, 320, 320, 370, 380, 390, 410, 440, 450, 470, 510, 530, 620, 810}
There are 17 values.
The median of an odd data set is the middle value.
The middle value is: 390
The third quartile is Q₃ = 390.
Compute the inter-quartile range as follows:
IQR = Q₃ - Q₁
= 390 - 180
= 210
Compute the value of [Q₁ - 1.5 IQR] as follows:
[tex]Q_{1}-1.5\ IQR =180-(1.5\times 210)=-135[/tex]
Compute the value of [Q₃ + 1.5 IQR] as follows:
[tex]Q_{3}+1.5\ IQR =390+(1.5\times 210)=705[/tex]
There are no values that are less than [Q₁ - 1.5 IQR]. But there is one value that is more than [Q₃ + 1.5 IQR].
X = 810 > [Q₃ + 1.5 IQR] = 705
Thus, the outlier in the data is $810.
(b)
A distribution is known as to be skewed to the right, or positively skewed, when maximum of the data are collected on the left of the distribution.
In the stem plot above, it is shown that maximum of the data values are collected on the left of the chart. This implies that the distribution is positively skewed.
Thus, the distribution of money spent on textbooks for the 34 students is right skewed.
Outliers are data elements that are relatively far from other data elements
The outlier of the dataset is $810The distribution of the stem-plot is right skewed.(a) The outlier
From the plot, the stems are given as:
Stems: 1 2 3 4 5 6 7 8
However, stem 7 does not have any leaf
Using the above highlight, the next entry after 620 is 810
810 is relatively far from the other datasets.
Hence, 810 is an outlier
(b) The distribution
The data on the stem-plot is more concentrated at the top, and it reduces as the stem increases.
When there are more data elements at the top or left, then the distribution is right skewed.
Hence, the distribution is right skewed.
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Qual o valor que Z assume?
-Cz+ 6z = Tz+ 83
Answer:
[tex]\large \boxed{z = \dfrac{83}{6 - C - T}}[/tex]
Step-by-step explanation:
[tex]\begin{array}{rccl}-Cz + 6z & = & Tz + 83 & \\6z -Cz - Tz &= & 83 & \text{Subtracted Tz from each side}\\z(6 - C - T) & = & 83 & \text{Removed the common factor}\\z& = & \mathbf{\dfrac{83}{6 - C - T}} &\text{Divided each side by 6 - C - T}\\\end{array}\\\\\large \boxed{\mathbf{z = \dfrac{83}{6 - C - T}}}[/tex]
Jay was reaching into her purse and accidentally spilled her coin purse. 10 pennies fell on the floor. Jay noticed that only 2 of the pennies landed on heads. What is the theoretical probability of this happening?
Answer:
The theoretical probability of landing on 2 heads, when 10 coins are tossed is 0.0439 or 4.39%.
Step-by-step explanation:
Number of coins that fell on the floor = 10
Number of coins that landed on heads = 2
We have to find the theoretical probability of getting 2 coins landing of heads when 10 coins are tossed.
Notice that there are only 2 possible outcomes: Either that coin will land on head or it won't. Landing of each coin is independent of the others coins. Probability of each coin landing on head is constant i.e. 0.5 or 1/2. Number of trials, i.e. the number of times the experiment will be done is fixed, which is 10.
All the 4 conditions for an experiment to be considered a Binomial Experiment are satisfied. So we will use Binomial Probability to solve this problem.
Probability of success = Probability of coin landing on head = 0.5
Number of trials = n = 10
Number of success = r = 2
The formula for Binomial Probability is:
[tex]P(X = x) =^{n}C_{r}(p)^{r}(1-p)^{n-r}[/tex]
Using the values, we get:
[tex]P(X=2)=^{10}C_{2}(0.5)^2(0.5)^8=0.0439[/tex]
Thus, the theoretical probability of landing on 2 heads, when 10 coins are tossed is 0.0439 or 4.39%.
What do common multiples of 4 and 6 mean
Answer:
so the common multiple of 4 and 6 is 2 i hope this helps! :)
Step-by-step explanation:
well the common multiples of 4 and 6 is 2
2 * 2 = 4
2 * 3 = 6
so there has to be a certain number that goes into both of the numbers you are working with
Answer:
Common multiples of 4 and 6 are 12, 24, 36, 48, 60, 72, 84, 96...
x g(x)
−2 1/4
-1 1/2
0 1
1 2
3 8
Consider that f(x) = x + 2, while the table represents y = g(x). Which statement is true when comparing the rate of change for the functions?
A) The rate of increase for the functions is the same.
B) f(x) has a greater rate of increase than function g(x).
C) g(x) has a greater rate of increase than function f(x).
D) g(x) has a greater rate of decrease than function f(x).
Answer:
A
Step-by-step explanation:
Cause I know
Answer: C!!
Step-by-step explanation:
USA Test Prep told me! :)
You would like to determine if there is a higher incidence of smoking among women than among men in a neighborhood. Let men and women be represented by populations 1 and 2, respectively. The relevant hypotheses are constructed as....a) h0: p1-p2 ≥ 0 h1: p1-p2 < 0b) h0: p1-p2≤ 0 h1: p1-p2 > 0
Answer:
For this case we want to test if there is a higher incidence of smoking among women than among men in a neighborhood (alternative hypothesis). And we define p1 for men and p2 for women, so for this case the best system of hypothesis are:
Null hypothesis:[tex] p_1- p_2 \geq 0[/tex]
Alternative hypothesis: [tex]p_1 -p_2 <0 [/tex]
And the best option would be:
a) h0: p1-p2 ≥ 0 h1: p1-p2 < 0
Step-by-step explanation:
Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Solution to the problem
For this case we want to test if there is a higher incidence of smoking among women than among men in a neighborhood (alternative hypothesis). And we define p1 for men and p2 for women, so for this case the best system of hypothesis are:
Null hypothesis:[tex] p_1- p_2 \geq 0[/tex]
Alternative hypothesis: [tex]p_1 -p_2 <0 [/tex]
And the best option would be:
a) h0: p1-p2 ≥ 0 h1: p1-p2 < 0
A study is conducted to investigate whether customer satisfaction is greater among computer companies that offer tech support versus those that do not offer tech support. A random sample of 50 customers are selected from among those that purchased computers that offer tech support. A separate random sample of 40 customers are selected from among those that purchased computers that do not offer tech support.
The study found that the mean satisfaction rating was significantly greater among customers that purchased computers that offer tech support.
Which of the following is the best description of this study?
(A) An experiment using a completely randomized design.
(B) An experiment using a randomized block design.
(C) An experiment using a matched pairs design
(D) An observational study using a simple random sample.
Answer:
The correct answer is (E).
This is not an experiment because no treatment is being imposed upon the customers. Additionally, this study used a stratified sample because independent random samples were selected from two distinct populations of customers.
Step-by-step explanation:
The correct answer is (E) An observational study using a stratified sample.
What is stratified sample?Stratified sampling is also known as stratified random sampling. The stratified sampling process starts with researchers dividing a diverse population into relatively homogeneous groups called strata, the plural of stratum. Then, they draw a random sample from each group (stratum) and combine them to form their complete representative sample.
Given that a data of a survey, the study found that the mean satisfaction rating was significantly greater among customers that purchased computers that offer tech support.
We need to find which is the best description of this study,
This study used a stratified sample because independent random samples were selected from two distinct populations of customers.
This is not an experiment because no treatment is being imposed upon the customers.
Hence, the best description is an observational study using a stratified sample.
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The complete question is attached.
Two dice: Find the probability of rolling an odd number on the first die and an even number on the second die.
Answer:
1/4; 25%
Step-by-step explanation:
Both events happen with probability 1/2: there are 3 even numbers and 3 odd numbers in a die.
Since the two events are also independent (the outcome of the first die doesn't affect the outcome of the second), we have to multiply those probability.
So, you roll an odd number on the first die and an even number on the second die with probability
[tex]\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}[/tex]
Show that (2, 1) is a solution of the system of equations.
x + 3y = 5,
y = –x + 3
1. Substitute (2, 1) into x + 3y = 5 to get .____
Options: 1 + 32 = 5, 1 + 3(2) = 5, 2 + 31 = 5, 2 + 3(1) = 5
2. Simplify the equation to get ____
Options: 5 = 5 is true, 7 = 7 is true, 7 = 5 is false, 33 = 5 is false
3. Substitute (2, 1) into y = –x + 3 to get ____
Options: 1 = -(2) + 3, 1 = -(2 + 3), 2 = -(1) + 3, 2 = -(1 + 3)
4. Simplify the equation to get .____
Options: 1 = 1 is true, 1 = -1 is false, 2 = 2 is true, 2 = -2 is false
Answer: The answer is 1. Substitute (2,1) into x+3y=5
Step-by-step explanation: Remember, (2,1) 2 is substituted for x and 1 is used as the y substitute
Answer:
1. 2+3(1)=5
2. 5=5 is true
3. 1=-(2)+3
4. 1=1 is true
What is the slope of this line?
(1,4) (6,-1)
Answer:
The slope is -1
Step-by-step explanation:
Let's find the slope between your two points.
(1,4);(6,−1)
(x1,y1)=(1,4)
(x2,y2)=(6,−1)
Use the slope formula:
m= y2−y1/x2−x1 = −1−4/6−1
= −5/5
= −1
Hope this is a better explanation :)
Hope it helps!!!!!!!!!
According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones, etc.) in back-to-college spending per student. Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54. If a family of a returning college student is randomly selected, what is the probability that: (a) They spend less than $160 on back-to-college electronics
Answer:
7.64% probability that they spend less than $160 on back-to-college electronics
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 237, \sigma = 54[/tex]
Probability that they spend less than $160 on back-to-college electronics
This is the pvalue of Z when X = 160. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{160 - 237}{54}[/tex]
[tex]Z = -1.43[/tex]
[tex]Z = -1.43[/tex] has a pvalue of 0.0763
7.64% probability that they spend less than $160 on back-to-college electronics
HELP ME PLEASE HURRY: Wade has claims that quadrilateral ABCD is a square because he has found that all four sides are congruent as shown below:
AB = 8.3 units
BC = 8.3 units
CD= 8.3 units
AD = 8.3 units
Explain in at least two sentences why Wade is incorrect and what else he needs to show for ABCD to be a square. Be specific!
Answer:
At first we should know that:
The properties of the square are:
It has four equal sides.All angles are right angles or equal to 90º.The sum of its all angles is 360ºIt has two pairs of perpendicular lines.It has two pairs of parallel lines.The properties of Rhombus
It has equal four sides.The opposite sides are of the same length.It has two acute angles and two obtuse angles.The sum of its all angles is 360ºIt has zero pairs of perpendicular lines.It has two pairs of parallel lines.So, Wade is incorrect because the quadrilateral may be Rhombus
And the quadrilateral to be a square, she needs to show that It has two pairs of perpendicular lines.
If the drawing has dimensions 7.5 cm × 5 cm, what is the size of the soccer field?
Answer: 35.5cm^2
Step-by-step explanation:
I believe we're talking about a rectangular. So it would be length * width or 7.5cm * 5cm = 35.5cm^2
Look at the number line below. The letters f, g, h, and i all represent integers. Write two inequalities to compare f and g.
Without the provided number line, we cannot determine the exact relationship between f and g. Inequalities f < g or f > g represent f being less than or greater than g, respectively. To write the correct inequality, one must refer to the positions of f and g on the number line.
Explanation:Since the number line is not provided, we cannot see the exact positions of f and g. However, we can discuss how to write inequalities to compare two integers based on their positions on a number line. If f is located to the left of g on the number line, it means that f is less than g. The inequality for this scenario would be f < g. On the other hand, if f is located to the right of g, then f is greater than g, and the corresponding inequality would be f > g.
You can use an inequality symbol to show how two metric measurements are related. If two numbers are the same, the inequality symbol would be the equal sign, representing they are equivalent. However, without the number line, we cannot determine the exact relationship between f and g, so one must look at the number line to ascertain the correct inequality to use.
The following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5.
a. What is the point estimate of the population mean?10 b. What is the point estimate of the population standard deviation (to 2 decimals)?3.46 c. With confidence, what is the margin of error for the estimation of the population mean (to 1 decimal)?2.9 d. What is the confidence interval for the population mean (to 1 decimal)?
Answer:
10
3.4641
2.8965
12.8965
Step-by-step explanation:
Given: 10, 8, 12, 15, 13, 11, 6, 5
c = 95%
a. The point estimate of the population mean is the sample mean. The mean is the sum of all values divided by the number of values:
x = 10 + 8 + 12 + 15 + 13 + 11 + 6 + 5 /8
= 80/8
= 10
b. The point estimate of the population standard deviation is the sample standard deviation. The variance is the sum of squared deviations from the mean divided by n - 1. The standard deviation is the square root of the variance:
s = /(10 – 10)^2 +.... + (5– 10)^2/8 – 1
s = 3.4641
c. Determine the t-value by looking in the row starting with degrees of freedom df = n-1 = 8 –1 = 7 and in the column with [tex]\alpha[/tex] = (1 – c)/2 = 0.025 in table :
t_[tex]\alpha[/tex]/2 = 2.365
The margin of error is then:
E = t_[tex]\alpha[/tex]/2 * s/√n
= 2.365 x s 3.4641/ √8
= 2.8965
d. The confidence intent)] then becomes:
7.1035 = 10 – 2.8965 = x – E <u<x +E= 10 + 2.8965 = 12.8965
The point estimate of the population mean will be 10.
How to calculate the point estimate?The point estimate of the population mean will be calculated thus:
= (10 + 8 + 12 + 15 + 13 + 11 + 11 + 6 + 5) / 8
= 80/8
= 10
Also, the margin of error will be:
= 2.365 × 3.4641/✓8
= 2.8965
In conclusion, the margin of error is 2.8965.
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Penalty Shots in World Cup Soccer A study1 of 138 penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper correctly guessed the direction of the kick only 41% of the time. The article notes that this is ‘‘slightly worse than random chance." We use these data as a sample of all World Cup penalty shots ever. Test at a 5% significance level to see whether there is evidence that the percent guessed correctly is less than 50%. The sample size is large enough to use the normal distribution. The standard error from a randomization distribution under the null hypothesis is SE=0.043. 1St.John, A., ‘‘Physics of a World Cup Penalty-Kick Shootout - 2010 World Cup Penalty Kicks," Popular Mechanics, June 14, 2010.
Final answer:
To test whether the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer, we can use a one-sample proportion test. Using a significance level of 0.05, we find that the test statistic is -2.09. Comparing this to the critical value from the standard normal distribution (-1.645), we reject the null hypothesis and conclude that there is evidence to suggest that the percentage of correctly guessed penalty shots is less than 50%.
Explanation:
To test whether there is evidence that the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer, we can use a one-sample proportion test. We will assume that the null hypothesis is true and that the goalkeeper's correct guesses are no better than random chance. The alternative hypothesis would be that the goalkeeper's correct guesses are significantly less than 50%. Using a significance level of 0.05, we can calculate the test statistic and compare it to the critical value from the standard normal distribution.
Null hypothesis (H0): The percentage of correctly guessed penalty shots is 50%.
Alternative hypothesis (Ha): The percentage of correctly guessed penalty shots is less than 50%.
Test statistic: We can use the z-test statistic since the sample size is large enough. The formula for the z-test statistic is z = (p - P0) / SE, where p is the sample proportion, P0 is the hypothesized proportion, and SE is the standard error. In this case, since the standard error is given as 0.043, we can plug in the values to calculate the test statistic.
Calculate the z-test statistic: z = (0.41 - 0.5) / 0.043 = -2.09
Find the critical value: Since our alternative hypothesis is that the percentage is less than 50%, we will use a one-tailed test. With a significance level of 0.05, the critical value from the standard normal distribution is -1.645.
Compare the test statistic to the critical value: Since the test statistic (-2.09) is less than the critical value (-1.645), we can reject the null hypothesis. There is evidence to suggest that the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer.
An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm exam. After all the students had turned in their exam pa- pers, the instructor randomly ordered them before grading. Consider the first 15 graded exam papers. (a) Find the probability that exactly 10 of these are from the second section. Find the probability that at least 10 of these are from the second section. Find the probability that at least 10 of these are from the same section?
Answer:
The answers are for option (a) 0.2070 (b)0.3798 (c) 0.3938
Step-by-step explanation:
Given:
Here Section 1 students = 20
Section 2 students = 30
Here there are 15 graded exam papers.
(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070
(b) Here if x is the number of students copies of section 2 out of 15 exam papers.
here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15
Then,
Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798
(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)
so,
Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938
Note : Here the given distribution is Hyper-geometric distribution
where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.
Evaluate \dfrac p4 +pq
4
p
+pqstart fraction, p, divided by, 4, end fraction, plus, p, q when p=8p=8p, equals, 8 and q=6q=6q, equals, 6.
The value is 50
what is fraction?A fraction represents a part of a whole or, more generally, any number of equal parts.
Given:
p/4 + pq
As, p= 8, q= 6
We have,
8/4 + 8*6
= 2 + 48
= 50
Hence, p/4 + pq= 50
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The equation results to get 50.
To evaluate the expression p/4 + pq when p = 8 and q = 6, follow these steps:
First, substitute the given values of p and q into the expression.Calculate p/4: p/4 = 8/4 = 2.Next, calculate pq: pq = 8 × 6 = 48.Add the two results together: p/4 + pq = 2 + 48 = 50.Therefore, the value of the expression when p = 8 and q = 6 is 50.
The complete question is:
Evaluate p/4 +pq when p=8 and q=6