An international polling agency estimates that 36 percent of adults from Country X were first married between the ages of 18 and 32, and 26 percent of adults from Country Y were first married between the ages of 18 and 32. Based on the estimates, which of the following is closest to the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y (Country X minus Country Y) who were first married between the ages of 18 and 32 is greater than 0.15?

(A) 0.1398
(B) 0.2843
(C) 0.4315
(D) 0.5685
(E) 0.7157

Answer: [tex]\$3108[/tex]

Step-by-step explanation:

Mr. Kim, his wife and two children are 4 persons. Now, he needs to buy 3 plane tickets for each one, which means he has to buy 12 plane tickets:

[tex](3 tickets)(4)=12 tickets[/tex]

On the other hand, we know each ticket costs [tex]\$259[/tex] per person. So, if we have 12 tickets to buy, we will have to multiply [tex]\$259[/tex] by 12:

[tex](12)(\$259)=\$3108[/tex] This is the amount Mr. Kim will spend in the plane tickets

Answer:

a) [tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2[/tex]

[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311[/tex]

b) [tex]63.331 < \mu_{left arm}-\mu_{right arm} <81.069[/tex]  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution

Let put some notation  

x=value for right arm , y = value for left arm

x: 102, 101,94,79,79

y: 175,169,182,146,144

The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:

d: 73, 68, 88, 67, 65

Part a

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2[/tex]

The third step would be calculate the standard deviation for the differences, and we got:

[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311[/tex]

Part b

The next step is calculate the degrees of freedom given by:

[tex]df=n-1=5-1=4[/tex]

Now we need to calculate the critical value on the t distribution with 4 degrees of freedom. The value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], so we need a quantile that accumulates on each tail of the t distribution 0.05 of the area.

We can use the following excel code to find it:"=T.INV(0.05;4)" or "=T.INV(1-0.05;4)". And we got [tex]t_{\alpha/2}=\pm 2.13[/tex]

The confidence interval for the mean is given by the following formula:  

[tex]\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)  

Now we have everything in order to replace into formula (1):  

[tex]72.2-2.13\frac{9.311}{\sqrt{5}}=63.331[/tex]  

[tex]72.2+2.13\frac{9.311}{\sqrt{5}}=81.069[/tex]  

So on this case the 90% confidence interval would be given by (63.331;81.069).

[tex]63.331 < \mu_{left arm}-\mu_{right arm} <81.069[/tex]  

Answer:

B. II only

Step-by-step explanation:

For this case the correct options would be:

B. II only

We analyze one by one the statements:

I.This study provides good evidence that walking is effective in controlling cholesterol

That's FALSE because there are many other factors that can influence the cholesterol and we just have two possible conditions (women who walk at least 10 miles and women who do not exercise at all), and we have just two averages obtained from two samples that we don't know if is a paired values or independent values. And is not appropiate to conclude this with the lack of info on this case.

II. This is an observational study, not an experiment

Correct we don't have a design for this experiment or factors selected in order to test the hypothesis of interest. So for this case we can conclude that we have an observational study for this case.

III. Although the study was conducted only on women, we can confidently generalize the results to men in the same age group.

False, we can't generalize results from women to men since they are different groups of people and with different characteristics.

Answer:

(b) a one-way between-subjects ANOVA

That's the correct option since we have one factor (class grade) and we have more than two groups.

Step-by-step explanation:

(a) a two-independent sample t test

We can't apply a two independnet t test since we are comparing more than two groups (Freshman, sophomore, Junior and senior). And for this case when we have more than two groups, the most powerful method is the one way ANOVA between subjects.

(b) a one-way between-subjects ANOVA

That's the correct option since we have one factor (class grade) and we have more than two groups.

One way Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:  

[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]

[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]

[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]

And we have this property

[tex]SST=SS_{between}+SS_{within}[/tex]

(c) a two-way between-subjects ANOVA

We can't apply a two way ANOVA since we have just one factor (or variable of interest) the class grades measured with a score. So then is not appropiate use this method for this case.

(d) both a two-independent sample t test and a one-way between-subjects ANOVA

False since we can't apply the two way ANOVA, that's not correct.

Answer:

[tex]P=\frac{40}{560}=0.0714[/tex]

Step-by-step explanation:

Notation

[tex]n_{sales}=4, n_{manufacturing}=7, n_{accounting}=5 [/tex]

Total = n= 4+7+5=16 people

We are going to select 3 people and will be given gift certificates to a local restaurant so then r =3.

Determine the probability that two of those selected will be from the accounting department and one will be from the sales department.

For this case we can use combinatory nCx, since the selection is without replacment.  

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

So then the definition of probability is given by :

[tex]P=\frac{Possible outcomes}{Total outcomes}[/tex]

Let's begin with the total outcomes, we have a total of n=16 people and we wan't to select 3 of them, so the possible outcomes are:

[tex]16C3= \frac{16!}{(16-3)! 3!}=560[/tex]

And now let's analyze the possible outcomes, we need that the group of 3 would be conformed by two people from the accounting department and one from the sales deparment. So then the possible outcomes are:

[tex](5C2)*(4C1)= \frac{5!}{(5-2)! 2!} \frac{4!}{(4-1)! 1!}=10*4=40[/tex]

And the reason is because we have a total of 5 people at the accounting and we want to select 2. And we have a total of 4 people at the sales department and we want to select just 1. And the multiplication it's because the order on the selection no matter (we assume this).

So then replacing into our formula of probability we got:

[tex]P=\frac{40}{560}=0.0714[/tex]

To calculate the probability, we need to determine the total number of ways to select 3 people out of all the employees and the total number of ways to select 2 people from the accounting department and 1 person from the sales department. The probability is then calculated by dividing the second calculation by the first.

To determine the probability that two of the selected people will be from the accounting department and one will be from the sales department, we need to first calculate the total number of ways to select 3 people out of the total number of employees. Then, we need to calculate the total number of ways to select 2 people from the accounting department and 1 person from the sales department. Finally, we divide the second calculation by the first calculation to get the probability.

Number of ways to select 3 people out of the total number of employees = 16C3 = 560

Number of ways to select 2 people from the accounting department and 1 person from the sales department = 5C2 * 4C1 = 10 * 4 = 40

Probability = Number of ways to select 2 people from the accounting department and 1 person from the sales department / Number of ways to select 3 people out of the total number of employees = 40 / 560 = <<40/560=0.0714>>0.0714

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Final answer:

The null and alternative hypotheses for testing the mean travel time on scenic and nonscenic routes.

Explanation:

a. In this case, the hypotheses to be tested are:

Null hypothesis (H0): The mean travel time for the nonscenic route is not shorter than 10 minutes compared to the scenic route, or in other words, u2 - u1 ≤ 10.

Alternative hypothesis (Ha): The mean travel time for the nonscenic route is shorter than 10 minutes compared to the scenic route, or in other words, u2 - u1 > 10.

b. If u1 is the mean for the nonscenic route and u2 for the scenic route, the hypotheses to be tested would be:

Null hypothesis (H0): The mean travel time for the scenic route is not shorter than 10 minutes compared to the nonscenic route, or in other words, u2 - u1 ≤ 10.

Alternative hypothesis (Ha): The mean travel time for the scenic route is shorter than 10 minutes compared to the non scenic route, or in other words, u2 - u1 > 10.

The correct options are : - (a) The correct hypothesis is 2. [tex]\(\mu_1 - \mu_2 < -10\)[/tex] and (b) The correct hypothesis is 5. [tex]\(\mu_1 - \mu_2 < 10\)[/tex]

Part (a): [tex]\(\mu_1\)[/tex] is the mean for the scenic route and [tex]\(\mu_2\)[/tex] is the mean for the non-scenic route

The individual decides that the non-scenic route is justified only if it reduces the mean travel time by more than [tex]10[/tex] minutes. In this context, we need to test whether the mean travel time for the scenic route [tex]\(\mu_1\)[/tex] minus the mean travel time for the non-scenic route [tex](\(\mu_2\))[/tex] is less than [tex]\(-10\)[/tex]

Null Hypothesis [tex](\(H_0\)) : \(\mu_1 - \mu_2 = -10\)[/tex]

Alternative Hypothesis[tex](\(H_a\)) : \(\mu_1 - \mu_2 < -10\)[/tex]

So, the correct option for (a) is: 2.[tex]\(\mu_1 - \mu_2 < -10\)[/tex]

Part (b): [tex]\(\mu_1\)[/tex] is the mean for the non-scenic route and [tex]\(\mu_2\)[/tex] is the mean for the scenic route

In this case, we need to test whether the mean travel time for the non-scenic route [tex](\(\mu_1\))[/tex] minus the mean travel time for the scenic route [tex](\mu_2\))[/tex] is less than [tex]\(-10\)[/tex]

Null Hypothesis [tex](\(H_0\)) : \(\mu_1 - \mu_2 = 10\)[/tex]

Alternative Hypothesis [tex](\(H_a\)) : \(\mu_1 - \mu_2 < 10\)[/tex]

So, the correct option for (b) is: 5. [tex]\(\mu_1 - \mu_2 < 10\)[/tex]

The complete Question is

An individual can take either a scenic route to work or a non-scenic route. She decides that use of the non-scenic route can be justified only if it reduces the mean travel time by more than 10 minutes.

(a) If μ1 is the mean for the scenic route and μ2 for the non-scenic route, what hypotheses should be tested?

1. μ1 − μ2 = −10

2. μ1 − μ2 < −10μ1 − μ2 = −10

3. μ1 − μ2 > −10 μ1 − μ2 = −10

4. μ1 − μ2 ≠ −10μ1 − μ2 = 10

5. μ1 − μ2 < 10μ1 − μ2 = 10

6. μ1 − μ2 > 10

(b) If μ1 is the mean for the non-scenic route and μ2 for the scenic route, what hypotheses should be tested?

1. μ1 − μ2 = −10

2. μ1 − μ2 < −10μ1 − μ2 = −10

3. μ1 − μ2 > −10 μ1 − μ2 = −10

4. μ1 − μ2 ≠ −10μ1 − μ2 = 10

5. μ1 − μ2 < 10μ1 − μ2 = 10

6. μ1 − μ2 > 10

If a box with a square base and open top must have a volume of 32,000 cm3. The dimensions of the box that minimize the amount of material used are:

(a) Sides of the base ≈ 40 cm

(b) Height ≈ 20 cm

Let's  the sides of the square base be x cm

Height of the box =h cm.

Volume of the box is given by:

Volume (V) = x² * h

We are given that the volume of the box is 32,000 cm³:

x²* h = 32,000

Surface area (A) of the box is the sum of the area of the base and the four sides:

Surface Area (A) = x² + 4xh

Express hin terms of x using the volume equation:

h = 32,000 / x²

Substitute

A = x² + 4x * (32,000 / x²)

A = x² + 128,000 / x

Let find the critical points of the surface area function

dA/dx = 2x - 128,000 / x^2 = 0

Solve for x

2x = 128,000 / x²

x³ = 64,000

x ≈ 40 cm

No, we can find the corresponding height

h = 32,000 / x²

h = 32,000 / (40²)

h ≈ 20 cm

Therefore the dimensions of the box that minimize the amount of material used are: Sides of the base ≈ 40 cm,  Height ≈ 20 cm

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To minimize the amount of material used, we need to minimize the surface area of the box while keeping the volume constant. The dimensions of the box that minimize the amount of material used are approximately x ≈ 252.98 cm and h ≈ 32000 / (x²) cm.

To find the dimensions of the box that minimize the amount of material used, we need to minimize the surface area of the box while keeping the volume constant. Let's assume the side length of the square base is x and the height of the box is h.

The volume of the box is given by [tex]V = x^2 * h = 32000 cm^3.[/tex]

The surface area of the box is given by [tex]A = x^2 + 4xh.[/tex]

To minimize A, we can use the volume equation to solve for h in terms of x: h = 32000 / (x^2). Substituting this into the surface area equation, we get:

[tex]A(x) = x^2 + 4x(32000 / x^2)[/tex]

To find the critical points of A(x), we differentiate A(x) with respect to x and set the result to 0: [tex]dA(x)/dx = 2x - (128000 / x^2) = 0.[/tex]

Solving this equation gives us x = sqrt(64000) or x ≈ 252.98 cm.

Since we are dealing with real-world dimensions, we can't have a negative value for x. Therefore, the dimensions of the box that minimize the amount of material used are approximately x ≈ 252.98 cm and h ≈ 32000 / (x²) cm.

Answer:

b. observing different participants one time in each group

[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]

If we analyze the formula for the sum of squares between we see that we are subtracting the mean for each group minus the grand mean. But in order to find the mean of each group we just need to observe just one time the dependent variable of interest for each group.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Solution to the problem

If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:  

[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]

[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]

If we analyze the formula for the sum of squares between we see that we are subtracting the mean for each group minus the grand mean. But in order to find the mean of each group we just need to observe just one time the dependent variable of interest for each group.

For this reason the best option on this case is:

b. observing different participants one time in each group

[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]

And we have this property :

[tex]SST=SS_{between}+SS_{within}[/tex]

Answer:

b. observing different participants one time in each group

Step-by-step explanation:

Between-subjects or groups design is a common experimental design used in psychology and other social science fields. Between-subjects is a type of experimental design in which the subjects of an experiment are assigned to different groups or conditions or participants in which each subject is being tested base on only one of the experimental conditions at a time. In Between-subjects experimental study, participants can be part of the treatment group or the control group, but cannot be part of both. A complete new group is required for each group, If more than one treatment is tested. The other way of assigning test to participants is Within-subjects design.

The major difference between Between-subjects design and Within-subjects design is that in Within-subjects design, the same participants test all the conditions of the experiment while in Between-subjects design different participants test each condition of the experiment.

Answer:

[tex]x_{n+1} = x_{n} - \frac{f(x_{n} )}{f^{'}(x_{n})}[/tex]

[tex]x_{1} = -10[/tex]

[tex]x_{2} = -3.95[/tex]

Step-by-step explanation:

Generally, the Newton-Raphson method can be used to find the solutions to polynomial equations of different orders. The formula for the solution is:

[tex]x_{n+1} = x_{n} - \frac{f(x_{n} )}{f^{'}(x_{n})}[/tex]

We are given that:

f(x) = [tex]x^{2} + 21[/tex]; [tex]x_{0} = -21[/tex]

[tex]f^{'} (x)[/tex] = df(x)/dx = 2x

Therefore, using the formula for Newton-Raphson method to determine [tex]x_{1}[/tex] and [tex]x_{2}[/tex]

[tex]x_{1} = x_{0} - \frac{f(x_{0} )}{f^{'}(x_{0})}[/tex]

[tex]f(x_{0}) = x_{0} ^{2} + 21 = (-21)^{2} + 21 = 462[/tex]

[tex]f^{'}(x_{0}) = 2*(-21) = -42[/tex]

Therefore:

[tex]x_{1} = -21 - \frac{462}{-42} = -21 + 11 = -10[/tex]

Similarly,

[tex]x_{2} = x_{1} - \frac{f(x_{1} )}{f^{'}(x_{1})}[/tex]

[tex]f(x_{1}) = (-10)^{2} + 21 = 100+21 = 121[/tex]

[tex]f^{'}(x_{1}) = 2*(-10) = -20[/tex]

Therefore:

[tex]x_{2} = -10 - \frac{121}{20} = -10+6.05 = -3.95[/tex]

The statements that are true are;

Step-by-step explanation:

Given that function 1 has the table;

x  y

0 -2

2  0

3   1

5   3

Finding the slope of the linear function gives the rate of change of the function. In this case,

m=Δy/Δx

m=3-1/5-3 = 2/2 =1

The equation of the linear function is given as;

m=Δy/Δx

y-3/x-5= 1

y-3=x-5

y=x-5+3

y=x-2

y-intercept is -2

In function 2, the line passes through points (-2,0) and (0,4)

Finding the slope of the line,

m₁=Δy/Δx

m₁=4-0/0--2

m₁= 4/2 =2

Rate of change is 2

Finding the equation of the line

y-4/x-0 = 2

y-4=2(x-0)

y-4 =2x

y=2x+4

The y-intercept is 4

The statements that are true are;

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Function 2 has a greater rate of change with a slope of 2 compared to Function 1's slope of 1. Additionally, Function 2 has a greater y-intercept at y = 4, while Function 1's y-intercept is at y = -2.

When comparing the rates of change for Function 1 and Function 2, we look at the slope of the lines representing these functions. The slope is the ratio of the rise to the run (change in y over change in x). For Function 1, considering the points (0, −2) and (2, 0), the slope is (0 - (-2)) / (2 - 0) = 2 / 2 = 1. Looking at Function 2, which passes through (-2, 0) and (0, 4), the slope is (4 - 0) / (0 - (-2)) = 4 / 2 = 2. Therefore, Function 2 has a greater rate of change than Function 1. Regarding y-intercepts, Function 1 starts at y = -2 (since the point (0, -2) is included), while Function 2 passes through y = 4 when x = 0, indicating the y-intercept is 4. This means that Function 2 has a greater y-intercept than Function 1.

A. 15917 ft   B.∠D=88.35°  C.0.0532 rad

Step-by-step explanation:

A. Given  that angle ∠ACB = 88.60° and the distance from C to B is 389 ft then triangle ABC is right ,90° at B. Applying the formula for tangent of an angle which is;

Tan of an angle = opposite side length/adjacent side length

Tan Ф = O/A =AB/389 ft

Tan 88.60°= AB/389

AB=389*tan 88.60° = 389×40.92 =15916.87 ft ⇒15917 ft

B.

The distance from B to D is given as 459 ft and the distance between the towers , AB, is 15917 ft. To get angle ∠D apply the formula for tangent of an angle where ;

Tan ∠D=O/A =15917/459 =34.6775599129

∠D =tan⁻(34.6775599129)

∠D=88.35°

C. To get angle ∠A subtract the sum of angle ∠C and ∠D from 180°. Apply the sum of angles in a triangle theorem

 ∠A =180° - (88.60°+88.35°)

   ∠A = 180°-(176.95°)=3.05°

    Changing degrees to radians you multiply the degree value with 0.0174533

3.05°=3.05*0.0174533=0.05323254 rad

To 4 decimal places

=0.0532 rad

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Answer:

Step-by-step explanation:

Answer

The answer and procedures of the exercise are attached in the following archives.

Explanation  

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

The pairs (C,r) where the expected profit from a warranty is zero are calculated by balancing the warranty cost to the store for stoves failing within the warranty period with the revenue from selling the warranties. Reasonable values should consider customer appeal, market competition, and business risk.

The question involves calculating the expected profit from a warranty which depends on an exponential random variable representing the lifetime of a stove. The lifetime of the stove, modeled as an exponential random variable, is typically used for modeling the lifespan of objects like mechanical or electronic devices whose failure rate is constant over time. This property is also known as the memoryless property. The distribution is defined by the parameter lambda (λ). In this case, λ equals 1/10, which suggests the average lifespan of the stove is 10 years.

The cost to replace the stove is $800, and the price charged for the warranty is denoted as C. The question asks for pairs of (C,r) where the expected profit is zero. This occurs when the cost of the warranty C is equal to the cost of replacing the stove, $800, over the time r in years for which the warranty is valid. It is also important to remember that not every stove will require a replacement, only those that fail within the warranty period r. The probability that a stove fails within the r years is computed using the exponential distribution's cumulative probability function as P(X

The expected profit is zero when the total warranty cost compensates for all the stoves replaced within their warranty periods i.e., C*P(X>r) = $800*P(X

Reasonable choices for C and r would depend on many factors such as the company's risk tolerance, competition in the warranty market, and customers' willingness to pay for extended warranties. A higher warranty price C increases profit but may discourage customers from buying the warranty. A longer warranty period r increases customer appeal but also the company's costs if more stoves fail and need to be replaced.

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Answer:

I get z = +1.8

x= 79

m= 73;s= 4;z= (x-m)/s= 1.5

hope it helps you

Trina's new z-score is calculated using the z-score formula z = (x-μ)/σ where the raw score, mean, and standard deviation are 79, 70, and 5, respectively. Therefore, Trina's z-score on the standardized distribution of the calculus exam is +1.80.

The subject you are asking about is related to calculating the z-score on a standardized distribution in a calculus exam. The z-score, in simple terms, tells you how many standard deviations a given value is from the mean. It's a way of standardizing scores on different distributions.

Now, we know Trina's original z-score was +1.50 when the mean was 73 with a standard deviation of 4. This score was moved to a new distribution that has a mean of 70 and standard deviation of 5. We can calculate the new z-score using the formula: z = (x-μ)/σ, where x is the raw score, μ is the mean, and σ is the standard deviation.

Plugging Trina's score into the formula we get z = (79-70)/5. Thus, Trina's new z-score on the standardized distribution of the calculus exam is +1.80.

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Answer:

a) Null hypothesis: [tex]\mu =0[/tex]

Alternative hypothesis: [tex]\mu \neq 0[/tex]

b) t =-7.44

c) [tex]p_v = 2*P(t_{98}<-7.44)<0.0001[/tex]

d) Reject Null hypothesis. The is enough evidence to conclude that the mean prediction error is not equal to 0.

We reject the null hypothesis because the [tex]p_v <\alpha[/tex]. So we can conclude that the difference is significantly different from 0 at 5% of significance.

Step-by-step explanation:

Assuming this output:

[tex]t_{difference}=-0.395[/tex]

t(observed value) =-7.44

t(Critical value )= 1.984

DF = 98

p value (two tailed) < 0.0001

[tex]\alpha =0.05[/tex]

a) What are the null and alternative hypothesis

Null hypothesis: [tex]\mu =0[/tex]

Alternative hypothesis: [tex]\mu \neq 0[/tex]

The reason is because the output says a bilateral test so for this case w eselect this option.

b) Identify the statistic

The correct formula for the statistic is given by:

[tex]t=\frac{\bar X_1 -\bar X_2 -0}{\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}}[/tex]

Based on the output the calculated value its t =-7.44

c) Identify the p value

We have the degrees of freedom given 98

Based on the alternative hypothesis the p value is given by:

[tex]p_v = 2*P(t_{98}<-7.44)<0.0001[/tex]

d) State the final conclusion that addresses the original claim

Reject Null hypothesis. The is enough evidence to conclude that the mean prediction error is not equal to 0.

We reject the null hypothesis because the [tex]p_v <\alpha[/tex]. So we can conclude that the difference is significantly different from 0 at 5% of significance.

Answer:

Upper Riemann Sum is 9/16

Step-by-step explanation:

Final answer:

To compute the upper Riemann sum for the function f(x) = x^2 over the interval x ∈ [-1, 1] with respect to the partition P = [-1, -1/4, 1/4, 3/4, 1], we need to find the value of the function at each partition point and multiply it by the width of the corresponding subinterval. The upper Riemann sum is obtained by summing up all these values.

Explanation:

To compute the upper Riemann sum, we need to find the value of the function at each partition point and multiply it by the width of the corresponding subinterval.

Given the function f(x) = x^2 and the partition P = [-1, -1/4, 1/4, 3/4, 1], we can calculate the upper Riemann sum as follows:

Calculate the width of each subinterval: Δx = (1 - (-1)) / 4 = 1/2.

Find the value of the function at each partition point:

Multiply the value of the function at each partition point by the width of the corresponding subinterval:

Sum up all the values obtained: 1/2 + 1/32 + 1/32 + 9/32 + 1/2 = 17/16

Therefore, the upper Riemann sum for the given function over the interval x ∈ [-1, 1] with respect to the partition P = [-1, -1/4, 1/4, 3/4, 1] is 17/16.

Answer:

B. 1635

Step-by-step explanation:

We have been given that the seating for an outdoor stage is arranged such that there are 11 seats in the first row. For each additional row after the first row, there are 3 more seats than there are in the previous row.

We can see that the seating order is in form of arithmetic sequence, whose first term is 11 and common difference is 3.

Since there are 30 rows altogether, so we need to find the sum of 30 first terms of the sequence using sum formula.

[tex]S_n=\frac{n}{2}[2a+(n-1)d][/tex], where,

[tex]S_n=[/tex] Sum of n terms,

n = Number of terms,

a = First term,

d = Common difference.

Upon substituting our given value is above formula, we will get:

[tex]S_n=\frac{30}{2}[2(11)+(30-1)3][/tex]

[tex]S_n=15[22+(29)3][/tex]

[tex]S_n=15[22+87][/tex]

[tex]S_n=15[109][/tex]

[tex]S_n=1635[/tex]

Therefore, there are 1635 seats in all and option B is the correct choice.

Answer: the total number of seats is 1635

Step-by-step explanation:

For each additional row after the first row, there are 3 more seats than there are in the previous row. This means that the seats in each row is increasing in arithmetic progression with a common difference of 3. The formula for determining the sum of n terms of an arithmetic sequence is expressed as

Sn = n/2[2a + (n - 1)d]

Where

n represents the number of terms in the sequence.

a represents the first term,

d represents the common difference.

From the information given,

a = 11

n = 30

d = 3

Therefore,

S30 = 30/2[2×11 + (30 - 1)3]

S30 = 15[22 + 29×3]

S30 = 15 × 109= 1635

Answer:

Step-by-step explanation:

Given than n = 25 , mean = 52800 sd = 4600

1) P(X<50000) , so please keep the z tables ready

we must first convert this into a z score so that we can look for probability values in the z table

using the formula

Z = (X-Mean)/SD

(50000- 52800)/4600 = -0.608

checkng the value in the z table

P ( Z>−0.608 )=P ( Z<0.608 )=0.7291

b)

again using the same formula and converting to z score

we need to calculate P(X<50000)

Z = (X-Mean)/SD

(50000- 52800)/4600 = -0.608

P ( Z<−0.608 )=1−P ( Z<0.608 )=1−0.7291=0.2709

proportion is 27% approx

c)

When your data points are measurements on a numerical scale you have variables data , here yield stress is numeric in nature , hence its a sampling by variable plan

Answer:

B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]

The correct answer is:

B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.

As regards using the normal model, the correct answer is D. No, there are not at least 10 people with weak bones and 10 people with strong bones in each group.

In sampling distributions, the normal model can be used if np ≥ 10 and n (1 - p) ≥ 10.

In this case, those with weak bones are:

= 8.5% x 82

= 6.97 people which is less than 10

= 1% x 593

= 5.93 people

We do not have 10 or more people for the sample sizes so the normal model will not be a good fit.

Find out more on the normal model at https://brainly.com/question/15399601.

Answer:0.05

Step-by-step explanation:

There are 5 different bank accounts

Probability that A will give maximum benefit i.e. Jake save most in account A

[tex]P_1=\frac{1}{5}[/tex]

so picking 1 bank we are left with 4 banks . So we Probability that he will save least into account B is [tex]P_2=\frac{1}{4}[/tex]

Probability that Jake save most into account A and save least into account B is

[tex]=P_1\times P_2[/tex]

[tex]=\frac{1}{5}\times \frac{1}{4}=\frac{1}{20}[/tex]

Answer:

Average value of temperature will be [tex]8.335^{\circ}C[/tex]

Step-by-step explanation:

We have given the temperature in degree Celsius [tex]T(x,y)=19-4x^2-y^2[/tex]

It is given that x varies between 0 to 2

So [tex]0\leq x\leq 2[/tex]

And y varies between 0 and 4

So [tex]0\leq y\leq 4[/tex]

So area between the region A = 4×2 = 8[tex]cm^2[/tex]

Now average temperature is given by

[tex]Average\ value=\frac{1}{A}\int \int T(x,y)dA[/tex]

[tex]Average\ value=\frac{1}{8}\int \int (19-4x^2-y^2)dxdy[/tex]

[tex]=\frac{1}{8}\int \int (19x-4\frac{x^3}{3}-y^2x)dy[/tex]

As limit of x is 0 to 2

[tex]=\frac{1}{8}\int  (19\times 2-4\times \frac{2^3}{3}-y^2\times 2)-(19\times 0-4\times \frac{0^3}{3}-y^2\times 0))dy=\frac{1}{8}\int(38-\frac{32}{3}-2y^2)dy[/tex]

=[tex]=\frac{1}{8}(38y-\frac{32}{3}y-\frac{2y^3}{3})[/tex]

As limit of y is 0 to 4

So [tex]Average\ value=\frac{1}{8}(38\times 4-\frac{16}{3}\times 4-\frac{2\times 4^3}{3})-0=8.335^{\circ}C[/tex]

Answer:

Answer explained below

Step-by-step explanation:

Let a be the working hours of plant A.

Let b be the working hours of plant H.

Let c be the working hours of plant M.

We have to minimize , Z = 60a + 92b + 140c

subject to constraint , 40a + 20b + 40c >=350

                20a + 40b + 40c >=240

where a>=0 , b>=0 ,c>=0

So ,by solving this , a = 7.67 , b= 2.17 , c = 0

So ,working hours of plant A = 8 hrs

working hours of plant H = 3 hrs

working hours of plant M = 0 hrs

Answer:

A = 8hrs  

H = 3hrs  

M = 0hrs  

Step-by-step explanation:

Let a be the working hours of plant A.  

Let b be the working hours of plant H.  

Let c be the working hours of plant M.  

∴ Hours of plants A,H,M are a,b,c respectively.  

We have to minimize the cost of production, Z = 60a + 92b + 140c  

Regular ice cream: 40a + 20b + 40c = 350  

Deluxe ice cream: 20a + 40b +40c =240  

Where a>=0, b>=0, c>=0  

So, by solving this, a= 7.67, b= 2.77,c= 0  

So working hours of plant A = 8 hrs  

Working hours of plant H = 3 hrs  

Working hours of plant M = 0 hrs  

Answer: A) .1587

Step-by-step explanation:

Given : The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.

i.e. [tex]\mu=12.30[/tex] and [tex]\sigma=0.20[/tex]

Let x denotes the amount of soda in any can.

Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.

Then, the probability that a randomly selected can will need to go through the mentioned process =  probability that a randomly selected can has more than 12.50 ounces of soda poured into it =

[tex]P(x>12.50)=1-P(x\leq12.50)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{12.50-12.30}{0.20})\\\\=1-P(z\leq1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.8413\ \ \ [\text{By z-table}]\\\\=0.1587[/tex]

Hence, the required probability= A) 0.1587

Final answer:

The probability that a randomly selected can will need a special cleaning process because it holds more than 12.50 ounces of soda is calculated using the Z-score. The correct answer is A) 0.1587.

Explanation:

The question asks us to calculate the probability that a randomly selected can will require a special cleaning process if it holds more than 12.50 ounces of soda, given a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce. To find this probability, we can use the standard normal distribution, also known as the Z-score.

First, we calculate the Z-score for 12.50 ounces using the formula: Z = (X - \\mu\) / \\sigma\, where X is the value we are evaluating (12.50 ounces), \mu is the mean (12.30 ounces), and \sigma is the standard deviation (0.20 ounce). Plugging in the values gives us Z = (12.50 - 12.30) / 0.20 = 1. The Z-score represents the number of standard deviations the value X is from the mean.

Next, to find the probability that a can will overflow (have more than 12.50 ounces), we look up the corresponding probability in the standard normal distribution table or use a calculator suitable for such statistical computations. The probability corresponding to a Z-score of 1 is approximately 0.8413. However, since we are interested in the probability of a can having more than 12.50 ounces, we need the area to the right of the Z-score, which is 1 - 0.8413 = 0.1587. Therefore, the probability that a randomly selected can will require a special cleaning process is 0.1587.

The correct answer to the question is A) 0.1587.

Answer: It's neither arithmetic or a geometric sequence

Explanation: For an arithmetic sequence, to get the common difference, we subtract the first term from the second, third term from second which should give the same value. While for a geometric sequence, to get the common ratio, divide the second term by the first term, third term by second term and so on, all of which should give the same answer. But in the above sequence, it doesn't follow this pattern

The sequence 45, 59, 65, 70, 85 is neither arithmetic nor geometric.

To determine if a sequence is arithmetic, one must check if the difference between consecutive terms is constant. For this sequence:

- The difference between the second and first terms is 59 - 45 = 14.

- The difference between the third and second terms is 65 - 59 = 6.

- The difference between the fourth and third terms is 70 - 65 = 5.

- The difference between the fifth and fourth terms is 85 - 70 = 15.

Since these differences are not the same, the sequence is not arithmetic.

To determine if a sequence is geometric, one must check if the ratio between consecutive terms is constant. For this sequence:

- The ratio between the second and first terms is 59/45.

- The ratio between the third and second terms is 65/59.

- The ratio between the fourth and third terms is 70/65.

- The ratio between the fifth and fourth terms is 85/70.

Simplifying these ratios:

- 59/45 = 1.3111

- 65/59 = 1.1017

- 70/65 = 1.0769

- 85/70 = 1.2143

Since these ratios are not the same, the sequence is not geometric.

The upper 90% confidence limit for the true proportion of firefighter masks of this type whose lenses would pop out at 250° is approximately 0.2949.

To construct a 90% upper confidence limit for the true proportion of firefighter's masks whose lenses would pop out at 250°, firstly, we need to calculate the sample proportion (p) which is the ratio of the number of masks that had lenses popping out (12) to the total number of masks tested (60). Thus, the sample proportion is 12/60 = 0.2.

Next, we use the formula for an upper confidence limit for proportions: p + z*sqrt((p*(1-p))/n), where z is the z-score associated with the desired level of confidence (for 90% confidence, z is 1.645), p is the sample proportion, and n is the sample size.

So the upper 90% confidence limit would be calculated as: 0.2 + 1.645*sqrt((0.2*0.8)/60) = 0.2 + 1.645*0.05774 = 0.2949322. Rounding to four decimal places, we get an upper confidence limit of 0.2949.

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The carrying capacity K for the logistic model is 45. In the context of the problem, the constant b, which refers to the growth rate divided by the carrying capacity, would be approximately 0.00444. The logistic model representing P vs. m under these conditions would be P(m) = 45/ (1+ (45/15-1) * e^(-0.00444m)).

In the field of mathematics, specifically in growth modeling, a logistic model incorporates a carrying capacity. The carrying capacity, denoted as K, is the maximum stable value of the population, in this case, the number of parents attending the PTA meeting. The carrying capacity is expected to be 45 in this instance.

The constant b in the logistic model can be found using the initial value (15 parents) and the growth rate (20%). However, the question does not specify whether this rate is relative or absolute. Assuming relative growth, the initial growth rate r is 0.2 and the constant b = r/K would be 0.00444. This is not the traditional definition of b in a logistic equation; typically, b would denote the initial population size, so the question seems to have a specific, non-standard usage in mind that we need to respect.

A good starting value of ther, in order to ensure convergence and stability of the numerical method, can be 15, the initial population size.

The logistic model would thus be represented as P(m) = K/ (1+ (K/P_0-1) * e^(-bm)), where P_0 is the initial number of parents, K is the carrying capacity, b is the constant/ growth rate, m is the number of months, and e is the mathematical constant approximated as 2.718.

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Answer:

Step-by-step explanation:

1) The diagram is a polygon with unequal sides. The number of sides and angles is 5. This means that it is an irregular Pentagon. The formula for the sum of interior angles of a polygon is expressed as

(n - 2)180

Where n is the number if sides of the polygon. Since the number of sides of the given polygon is 5, the sum if the interior angles would be

(5-2)×180 = 540 degrees. Therefore,

10x - 3 + 5x + 2 + 7x - 11 + 13x - 31 + 8x - 19 = 540

43x - 62 = 540

43x = 540 + 62 = 602

x = 602/43 = 14

Angle S = 13 - 31 = 13×14 - 31 = 182 - 31

Angle S = 151 degrees.

4) The diagram is a rectangle. Opposite sides are equal. Triangle JNM is an isosceles triangle. It means that its base angles, Angle NMJ and angle NJM are equal. Therefore

3x + 38 = 7x - 2

7x - 3x = 38 + 2

4x = 40

x = 40/4

x = 10

Angle NMJ = 7x - 2 = 7×10 - 2 = 68 degrees.

Angle JML = 90 degrees ( the four angles in a rectangle are right angles). Therefore,

Angle NML = 90 - 68 = 22 degrees

Answer:

[tex]t=\frac{20.49-20}{\frac{1.42}{\sqrt{63}}}=2.738[/tex]    

[tex]p_v =P(t_{(62)}>2.738)=0.0040[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can say that the true mean is significantly higher than 20 min .  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=20.49[/tex] represent the mean time for the sample  

[tex]s=1.42[/tex] represent the sample standard deviation for the sample  

[tex]n=63[/tex] sample size  

[tex]\mu_o =20[/tex] represent the value that we want to test

[tex]\alpha=0.01[/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 time is actually higher than 20 min, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 20[/tex]  

Alternative hypothesis:[tex]\mu > 20[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

[tex]t=\frac{20.49-20}{\frac{1.42}{\sqrt{63}}}=2.738[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=63-1=62[/tex]  

Since is a one side right tailed test the p value would be:  

[tex]p_v =P(t_{(62)}>2.738)=0.0040[/tex]  

And we can use the following excel code to find it:

"=1-T.DIST(2.738,62,TRUE)"

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can say that the true mean is significantly higher than 20 min .  

To perform the hypothesis test, set up null and alternative hypotheses, calculate the test statistic, and compare it to the critical value from the t-distribution table.

To perform a hypothesis test, we need to set up the null and alternative hypotheses. The null hypothesis, H0, states that the mean delivery time is 20 minutes or less. The alternative hypothesis, Ha, states that the mean delivery time is more than 20 minutes. We can perform a one-sample t-test using the sample mean, standard deviation, sample size, and the desired level of significance. Calculate the test statistic and compare it to the critical value from the t-distribution table. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is sufficient evidence to support the alternative hypothesis.

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Answer:

D. All of the above

Step-by-step explanation:

Linear regression models relate to some assumptions about the distribution of error terms. If they are violated violently, the model is not suitable for drawing conclusions. Therefore, it is important to consider the suitability of the model for information before further analysis can be performed based on this model.

The fit of the model is related to the remaining behavior complying with the basic assumptions for error values in the model. When a regression model is constructed from a series of data, it should be shown that the model responds to the standard statistical assumptions of the linear model because of conducting inference. Residual analysis is an effective tool to investigate hypotheses. This method is used to test the following statistical assumptions for a simple linear regression model:

-Regression function is linear in parameters,

-Error conditions have constant variance,

-Error conditions are normally distributed,

-Error conditions are independent.

If no statistical hypothesis of the model is fulfilled, the model is not suitable for data. The fourth hypothesis (independence of error conditions) relates to the regulation of time series data. It is now used some simple graphical methods to analyze analysis, the feasibility of a model, and some formal statistical tests. In addition, when a model fails to meet these assumptions, some data changes may be made to ensure that the assumptions are reasonable for the modified model.