A company wants to determine where they should locate a new warehouse. They have two existing production plants (i.e., Plant A and Plant B) that will ship units of a product to this warehouse. Plant A is located at the (X, Y) coordinates of (50, 100) and will have volume of shipping of 250 units a day. Plant B is located at the (X, Y) coordinates of (150, 200) and will have a volume of shipping of 150 units a day. Using the centroid method, which of the following are the X and Y coordinates for the new plant location?

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is [tex]\bold{W=-3e^{7x}}[/tex]

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

[tex]W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|[/tex]

Thus replacing the functions of the exercise we get:

[tex]W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|[/tex]

Working with the determinant we get

[tex]W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}[/tex]

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

Final answer:

The Wronskian of [tex]y = e^{5x[/tex] and [tex]z = e^{2x[/tex] is calculated as [tex]-3e^{7x[/tex], which is nonzero for all x, thereby showing that these solutions are linearly independent. This property is vital in forming a solution space for differential equations.

Explanation:

Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions:  [tex]y = e^{5x[/tex] and  [tex]z = e^{2x[/tex]. To demonstrate that these solutions are linearly independent, we consider the corresponding vector solutions y1 and y2, and use the Wronskian for this purpose.

Calculating the Wronskian

The Wronskian of two functions f and g is defined as:

[tex]W(f,g) = det [[ f, g ],[ f', g' ]][/tex]

For our functions  [tex]y = e^{5x[/tex] and  [tex]z = e^{2x[/tex], the derivatives are [tex]5e^{5x} and 2e^{2x[/tex], respectively. Plugging these into the Wronskian formula, we get:

[tex]W(y,z) = det [[ e^{5x}, e^{2x} ],[ 5e^{5x}, 2e^{2x} ]][/tex]=[tex](e^{5x})(2e^{2x}) - (5e^{5x})(e^{2x}) = -3e^{7x[/tex]

This result is nonzero for all values of x, indicating that the solutions are linearly independent. The concept of linear independence is crucial in the study of differential equations, as it ensures that the solutions can form a basis for the solution space of the differential equation.

By deriving and equating to zero the revenue and profit functions, the maximum revenue is $900,000 achieved at 3000 units sold. The maximum profit is $384,000 when 2400 units are sold at $180 each. If taxed $55/unit, the maximum profit is $302,400 at 1840 units with a price of $208/unit.

To solve this, we'll first calculate the revenue function R(x) which is the product of the number of units sold and the price per unit, i.e., R(x) = x*p(x). Then, we'll find the profit function P(x), which is the difference between the revenue and the cost, i.e., P(x) = R(x) - C(x). Next, to maximize revenue and profit, we'll get the derivative of both R(x) and P(x), set them equal to zero, and solve.

For part (A), R(x) = x*(300 - x/20) = 300x - x^2/20. Its derivative is R'(x) = 300 - x/10. Setting R'(x) = 0, we get x = 3000 units, which leads to the maximum revenue of $900,000.

For part (B), P(x) = R(x) - C(x) = 300x - x^2/20 - (72000 + 60x). Its derivative, P'(x), when equal to zero gives us x = 2400 units. Thus, the maximum profit is $384,000 and the price per unit is $180.

For part (C), the new cost function very becomes C(x) = 72,000 + 115x. Setting the derivative of the new profit equation P'(x) = 0, we get x = 1840 units, which leads to the maximum profit of $302,400. The company should charge $208 per unit.

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(A) Maximum revenue: $450,000

(B) Maximum profit: $216,000 at production level of 2400 sets, price per set: $180

(C) With a $55 tax per set, maximum profit is $84,000 at production level of 2400 sets, price per set remains $180.

(A) To find the maximum revenue, we first need to maximize the revenue function:

[tex]\[ R(x) = (300 - \frac{x}{20}) \times x \]\[ R(x) = 300x - \frac{1}{20}x^2 \][/tex]

Now, let's find the critical points by taking the derivative of R(X) and setting it equal to zero:

[tex]\[ \frac{dR}{dx} = 300 - \frac{1}{10}x \]\[ \frac{1}{10}x = 300 \]\[ x = 3000 \][/tex]

So, the maximum revenue occurs at \( x = 3000 \).

Plugging \( x = 3000 \) back into the revenue function:

[tex]\[ R(3000) = (300 - \frac{3000}{20}) \times 3000 \]\[ R(3000) = (300 - 150) \times 3000 \]\[ R(3000) = 150 \times 3000 = 450000 \][/tex]

The maximum revenue is $450,000.

(B) To find the maximum profit, we need to maximize the profit function P(x) :

P(x) = R(x) - C(x)

Given C(x) = 72,000 + 60x, we have:

[tex]\[ P(x) = (300 - \frac{x}{20}) \times x - (72,000 + 60x) \]\[ P(x) = (300x - \frac{1}{20}x^2) - (72,000 + 60x) \]\[ P(x) = 300x - \frac{1}{20}x^2 - 72,000 - 60x \]\[ P(x) = -\frac{1}{20}x^2 + 240x - 72,000 \][/tex]

Now, let's find the critical points by taking the derivative of P(x)  and setting it equal to zero:

[tex]\[ \frac{dP}{dx} = -\frac{1}{10}x + 240 \][/tex]

Setting dP/dx = 0:

-1/10x + 240 = 0

1/10x = 240

x = 2400

Now, we need to check the endpoints of the interval [tex]\( 0 \leq x \leq 6000 \)[/tex] for potential maximum profit.

[tex]\[ P(0) = -\frac{1}{20}(0)^2 + 240(0) - 72,000 = -72,000 \]\[ P(6000) = -\frac{1}{20}(6000)^2 + 240(6000) - 72,000 = -132,000 \][/tex]

So, the maximum profit occurs at  x = 2400.

Plugging x = 2400  back into the profit function:

P(2400) = [tex]-\frac{1}{20}(2400)^2[/tex] + 240(2400) - 72,000  

P(2400) = -288,000 + 576,000 - 72,000

P(2400) = 216,000

The maximum profit is $216,000.

To find the price per set at the optimal production level, we use the price-demand equation:

p(2400) = 300 - 2400/20

p(2400) = 300 - 120

p(2400) = 180

So, the company should charge $180 per television set to realize the maximum profit.

(C) If the government decides to tax the company $55 for each set it produces, the cost function becomes:

C(x) = 72,000 + 60x + 55x

C(x) = 72,000 + 115x

To find the new maximum profit, we repeat the steps from part (B) with the updated cost function. We already know that the optimal production level is x = 2400, so we can directly plug this value into the new profit function.

P(x) = R(x) - C(x)

P(2400) = R(2400) - C(2400)

Plugging in x = 2400  into the revenue function:

R(2400) = (300 - 2400/20 * 2400

R(2400) = (300 - 120) * 2400

R(2400) = 180 * 2400

R(2400) = 432,000

Plugging in  x = 2400 into the updated cost function:

C(2400) = 72,000 + 115 * 2400

C(2400) = 72,000 + 276,000

C(2400) = 348,000

Now, calculate the new profit:

P(2400) = R(2400) - C(2400)

P(2400) = 432,000 - 348,000

P(2400) = 84,000

So, the company should manufacture 2400 sets each month to maximize its profit. The maximum profit is $84,000.

To find the price per set at the optimal production level with the tax:

p(2400) = 300 - 2400/20

p(2400) = 300 - 120

p(2400) = 180

The company should still charge $180 per television set to realize the maximum profit, even with the tax.

Therefore,

(A) Maximum revenue: $450,000

(B) Maximum profit: $216,000 at production level of 2400 sets, price per set: $180

(C) With a $55 tax per set, maximum profit is $84,000 at production level of 2400 sets, price per set remains $180.

Answer

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

Step-by-step 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.  

Answer:

[tex]SE=\sqrt{\frac{0.505 (1-0.505)}{832}+\frac{0.0311(1-0.0311)}{145}}=0.0226[/tex]

Step-by-step explanation:

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".  

1) Data given and notation  

[tex]X_{1}=420[/tex] represent the number of cats diagnosing malignant lymphoma from homes where herbicides were used​ regularly

[tex]X_{2}=17[/tex] represent the number of cats diagnosing malignant lymphoma from homes where NO herbicides were used​ regularly

[tex]n_{1}=832[/tex] sample 1 selected

[tex]n_{2}=145[/tex] sample 2  selected

[tex]\hat p_{1}=\frac{420}{832}=0.505[/tex] represent the proportion of of cats diagnosing malignant lymphoma from homes where herbicides were used​ regularly

[tex]\hat p_{2}=\frac{17}{145}=0.0311[/tex] represent the proportion of cats diagnosing malignant lymphoma from homes where NO herbicides were used​ regularly

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the value for the test (variable of interest)  

[tex]p_1 -p_2[/tex] parameter of interest

2) Solution to the problem

We are interested on the standard error for the difference of proportions and is given by this formula:

[tex]SE=\sqrt{\frac{\hat p_1 (1-\hat p_1)}{n_{1}}+\frac{\hat p_2 (1-\hat p_2)}{n_{2}}[/tex]

And if we replace the values given we got:

[tex]SE=\sqrt{\frac{0.505 (1-0.505)}{832}+\frac{0.0311(1-0.0311)}{145}}=0.0226[/tex]

The expected population of the Midwest city in 2036, based on the given parameters of an initial population (P₀) of 307,000 people in 2014, a yearly growth rate (r) of 0.673%, and a time period (t) of 22 years, is approximately 355,992.07.

Find the expected population of the Midwest city in 2036:

1. Define the variables:

P: Population in the year 2036 (unknown)

P₀: Initial population in 2014 (307,000 people)

t: Number of years since 2014 (2036 - 2014 = 22 years)

r: Yearly growth rate (0.673% = 0.00673 as a decimal)

2. Apply the formula:

The formula for exponential population growth is:

P = P₀ * e^(r * t)

where:

e is the base of the natural logarithm (approximately 2.71828)

3. Plug in the values:

P = 307,000 * e^(0.00673 * 22)

4. Calculate the result:

Using a calculator or spreadsheet, we get:

P ≈ 355,992.07

A circle centered at [tex]O(a,b)[/tex] with radius [tex]R[/tex] (the length of [tex]OP[/tex]) has equation

[tex](x-a)^2+(y-b)^2=R^2[/tex]

which can be parameterized by

[tex]\vec c(t)=\langle x(t),y(t)\rangle=\langle a+R\cos t,b+R\sin t\rangle[/tex]

with [tex]0\le t\le2\pi[/tex].

The tangent line to [tex]\vec c(t)[/tex] at a point [tex]P(x_0,y_0)[/tex] is [tex]\frac{\mathrm dy}{\mathrm dx}[/tex] with [tex]x=x_0[/tex] and [tex]y=y_0[/tex]. By the chain rule (and this is where we use implicit differentiation),

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\dfrac{R\cos t}{-R\sin t}=-\dfrac{\cos t}{\sin t}[/tex]

At the point [tex]P[/tex], we have

[tex]x_0=a+R\cos t\implies\cos t=\dfrac{x_0-a}R[/tex]

[tex]y_0=b+R\sin t\implies\sin t=\dfrac{y_0-b}R[/tex]

so that the slope of the line tangent to the circle at [tex]P[/tex] is

[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{\frac{x_0-a}R}{\frac{y_0-b}R}=-\dfrac{x_0-a}{y_0-b}[/tex]

Meanwhile, the slope of the line through the center [tex]O(a,b)[/tex] and the point [tex]P(x_0,y_0)[/tex] is

[tex]\dfrac{b-y_0}{a-x_0}[/tex]

Recall that perpendicular lines have slopes that are negative reciprocals of one another; taking the negative reciprocal of this slope gives

[tex]-\dfrac1{\frac{b-y_0}{a-x_0}}=-\dfrac{a-x_0}{b-y_0}=-\dfrac{x_0-a}{y_0-b}[/tex]

which is exactly the slope of the tangent line.

Answer:

The 80% confidence interval is given by (13.622;16.378)  

Step-by-step explanation:

1) 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".  

Assuming the X the random variable that represent the number of feet it takes the brakes to stop a car going 50 miles per hour follows a normal distribution  

[tex]X \sim N(\mu, \sigma)[/tex]  

[tex]\bar X =15[/tex] represent the sample mean

[tex]s=2.28[/tex] represent the sample deviation

n=6 sample size selected

Confidence =0.8 or 80%

2) Confidence interval

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 80% of confidence, our significance level would be given by [tex]\alpha=1-0.80=0.2[/tex] and [tex]\alpha/2 =0.1[/tex]. The degrees of freedom are given by:  

[tex]df=n-1=6-1=5[/tex]  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.1,5)" for [tex]t_{\alpha/2}=-1.48[/tex]  

"=T.INV(1-0.1,5)" for [tex]t_{1-\alpha/2}=1.48[/tex]  

The critical value [tex]tc=\pm 1.48[/tex]  

Now we can calculate the margin of error (m)  

The margin of error for the sample mean is given by this formula:  

[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]  

[tex]m=1.48 \frac{2.28}{\sqrt{6}}=1.378[/tex]  

The interval for the mean is given by this formula:  

[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]  

And calculating the limits we got:  

[tex]15 -1.48 \frac{2.28}{\sqrt{6}}=13.622[/tex]  

[tex]15 +1.48 \frac{2.28}{\sqrt{6}}=16.378[/tex]  

The 80% confidence interval is given by (13.622;16.378)  

Answer:

Option A) reject null hypothesis if z is greater than 1.645

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 250

p = 30% = 0.3

Alpha, α = 0.05

Number of women belonging to union , x = 75

First, we design the null and the alternate hypothesis  

[tex]H_{0}: p = 0.3\\H_A: p > 0.3[/tex]

This is a one-tailed(right) test.

Rejection Region:

[tex]z_{critical} \text{ at 0.05 level of significance } = 1.645[/tex]

So, the rejection region will be

[tex]z > 1.64[/tex]

That is we will reject the null hypothesis if the calculated z-statistic is greater than 1.645

Option A) reject null hypothesis if z is greater than 1.645

Final answer:

The rejection region for the given hypothesis test with a significance level of 0.05 is when the z-score is greater than 1.645.

Explanation:

To find the rejection region for this hypothesis test, we need to use the given significance level (alpha, a) of 0.05 to determine the critical z-value. In a one-tailed test, because we are looking for the proportion that exceeds 30%, we focus on the right tail of the normal distribution. Referencing the normal distribution table, a z-value with 0.05 to its right is approximately 1.645. Hence, we reject the null hypothesis if our test statistic z is greater than 1.645.

Utilizing the sample data where 75 out of 250 business students have PCs at home, we would calculate the test statistic and compare it to the critical value. If our calculated z-score exceeds 1.645, then we would reject the null hypothesis and conclude that more than 30% of business students have PCs at home, leading the lab to reconsider its proposed expansion.

Answer:

84.13%

Step-by-step explanation:

Population mean (μ) = 500 hours

Standard deviation (σ) = 50 hours

Assuming a normal distribution, for any given number of hours 'X', the z-score is determined by:

[tex]z=\frac{X-\mu}{\sigma}[/tex]

For X=550

[tex]z=\frac{550-500}{50}\\z=1[/tex]

For a z-score of 1, 'X' corresponds to the 84.13-th percentile of a normal distribution.

Therefore, the probability of that a randomly chosen light bulb lasts less than 550 hours, P(X<550), is 84.13%.

Answer:

a) 3⁵5³.

b) 1

c) 23³

d) 41·43·53

e) 1

f) 1111

Step-by-step explanation:

The greatest common divisor of two integers is the product of their common powers of primes with greatest exponent.

For example, to find gcd of 2⁵3⁴5⁸ and 3⁶5²7⁹ we first identify the common powers of primes, these are powers of 3 and powers of 5. The greatest power of 3 that divides both integers is 3⁴ and the greatest power if 5 that divides both integers is 5², then the gcd is 3⁴5².

a) The greatest common prime powers of 3⁷5³7³ and 2²3⁵5⁹ are 3⁵ and 5³ so their gcd is 3⁵5³.

b) 11·13·17 and 2⁹3⁷5⁵7³ have no common prime powers so their gcd is 1

c) The only greatest common power of 23³ and 23⁷ is 23³, so 23³ is the gcd.

d) The numbers 41·43·53 and 41·43·53 are equal. They both divide themselves (and the greatest divisor of a positive integer is itself) then the gcd is 41·43·53

e) 3³5⁷ and 2²7² have no common prime divisors, so their gcd is 1.

f) 0 is divisible by any integer, in particular, 1111 divides 0 (1111·0=0). Then 1111 is the gcd

Greatest common divisors were calculated for each of the given pairs. Many pairs had no common factors and their GCD was 1, while others had a GCD equal to a shared prime factor or to one of the pair elements.

In the subject of mathematics, specifically number theory, the greatest common divisor (GCD) is the largest number that divides two or more numbers without a remainder. Let's determine the greatest common divisors of the given pairs:

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Answer:

(A) It is less than the contribution predicted by the regression line.

Step-by-step explanation:

The residual is the subtraction of the observed value by the predicted value.

So, if James has a negative residual, it means that his contribution is less than what was expected by the regression line.

The correct answer is:

(A) It is less than the contribution predicted by the regression line.

James' actual contribution must be less than the contribution predicted by the regression line.

To determine which statement must be true about James' actual contribution, we need to understand the concept of residuals in regression analysis. A residual is the difference between the actual observed value and the predicted value. Since James' residual is negative, it means that his actual contribution is less than the contribution predicted by the regression line. (A) It is less than the contribution predicted by the regression line.

However, we cannot determine if James' actual contribution is less than the average contribution made by the 50 alumni solely based on the fact that his residual is negative. The regression line only predicts individual contributions based on the number of years since graduation, and the average contribution is not directly related to the regression line. Therefore, (B) It is less than the average contribution made by the 50 alumni. cannot be concluded.

Therefore, the correct answer is (A) It is less than the contribution predicted by the regression line.

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Answer:

1.8 cycles

Step-by-step explanation:

The average number of clock cycles per instruction is given by the sum of the product of each possible number of cycles by its likelihood.

1 cycle: 50%

2 cycles : 25%

3 cycles : 20%

4 cycles : 5%

[tex]Avg = 0.5*1+0.25*2+0.20*3+0.05*4\\Avg= 1.8\ cycles[/tex]

The average number of clock cycles per instruction is 1.8.

In a program with varying instruction types requiring different clock cycles to execute, the average number of clock cycles per instruction is calculated as 1.8. The calculation includes 25% of instructions requiring 2 cycles, 20% requiring 3 cycles, 5% requiring 4 cycles, with the rest needing only 1 cycle. For a Pentium chip which executes 100 million instructions per second with each instruction requiring one clock cycle, the actual execution rate for the average instruction mix would be over 55 million instructions per second.

To calculate the average number of clock cycles per instruction, we first multiply each type of instruction by the number of clock cycles they require, and then sum these up. The calculation would go as follows:

We then sum up these results: 0.5 + 0.6 + 0.2 + 0.5 = 1.8 clock cycles per instruction on average.

In relation to the Pentium chip information, this would mean that a single Pentium chip could execute an average instruction mix at a rate of over 55 million instructions per second, given that it can execute 100 million instructions per second if each instruction only required one clock cycle.

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Answer:

d. 99%

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]p = 0.30, n = 50[/tex]

Let's start from the higher confidence levels, since the higher the confidence level, the higher the width of the interval.

d. 99%

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.905[/tex], so [tex]Z = 2.575[/tex].

The lower limit is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.30 - 2.575\sqrt{\frac{0.3*0.7}{50}} = 0.133[/tex]

The upper limit is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.30 + 2.575\sqrt{\frac{0.3*0.7}{50}} = 0.467[/tex]

This is very close to the interval found, so d. is the correct answer

Answer:

Step-by-step explanation:

Given that 34​% of U.S. adults say they are more likely to make purchases during a sales tax holiday.

You randomly select 10 adults.

Let X be the no of adults in the selection of 10 who say they are more likely to make purchases during a sales tax holiday.

Each person is independent of the other and also there are two outcomes

Hence X is binomial with n =10 and p = 0.34

q = 0.66

P(X=r) [tex]=10Cr (0.34)^r(0.68)^{10-r}[/tex]

The probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is​

(a) exactly​ two,

=[tex]P(X=2)\\= 0.1873[/tex]

(b) more than​ two,

=[tex]P(X>2)\\= 1-F(2)\\=1-0.2838\\= 0.7162[/tex]

(c) between two and​ five, inclusive.

=[tex]P(2\leq x\leq 5)\\= F(5)-F(1)\\=0.9164-0.0965\\=0.8199[/tex]

The question is about binomial probability involving a scenario of U.S adults making a purchase during a tax sale. Depending on the number of adults to be considered (2, more than 2, between 2 & 5), the probabilities are calculated using the formula for binomial probability taking into account the probability of single trial success and the number of trials.

This question for the subject of Mathematics involves the concept of Binomial Probability

. To solve this question, we use the formula for binomial probability which is: P(k;n,p) = C(n, k) * (p^k) * ((1-p)^(n-k)) where C(n,k) refers to the combination of n things taken k at a time, n is the number of trials, k is the number of successes, and p is the probability of success on a single trial.

Given in the question, p = 0.34 (probability of an adult making a purchase during a tax sale), n = 10 (number of adults selected).

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Answer:

So on this case the 95% confidence interval would be given by [tex]-1.4485 \leq \mu_G -\mu_B \leq -0.9515[/tex]  

Since the confidence interval not contains the 0 we can say that we have significant differences between the mean of girls and boys.

1. What is the value of the lower end of the confidence interval?

b. – 1.4485

2. What is the value of the upper end of the confidence interval?

c. – 0.9515

What kind of distribution ( central limit theorem or T) and why?

We can use the t distribution but since the sample size is large enough we will have a distribution similar to the normal standard distribution. Because when the degrees of freedom  of the t distribution increases we have a normal distribution.

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".  

[tex]\bar X_1 =5.1[/tex] represent the sample mean 1 (girls)

[tex]\bar X_2 =6.3[/tex] represent the sample mean 2  (boys)

n1=250 represent the sample 1 size  

n2=280 represent the sample 2 size  

[tex]s_1 =1.2[/tex] sample standard deviation for sample 1

[tex]s_2 =1.7[/tex] sample standard deviation for sample 2

[tex]\mu_1 -\mu_2[/tex] parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:

[tex]\bar X_1 -\bar X_2 =5.1-6.3=-1.2[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:  

[tex]df=n_1 +n_2 -1=250+280-2=528[/tex]  

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,528)".And we see that [tex]t_{\alpha/2}=1.964[/tex]  

The standard error is given by the following formula:

[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex]

And replacing we have:

[tex]SE=\sqrt{\frac{1.2^2}{250}+\frac{1.7^2}{280}}=0.127[/tex]

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

[tex]-1.2-1.96\sqrt{\frac{1.2^2}{250}+\frac{1.7^2}{280}}=-1.4485[/tex]  

[tex]-1.2+1.96\sqrt{\frac{1.2^2}{250}+\frac{1.7^2}{280}}=-0.9515[/tex]  

So on this case the 95% confidence interval would be given by [tex]-1.4485 \leq \mu_G -\mu_B \leq -0.9515[/tex]  

Since the confidence interval not contains the 0 we can say that we have significant differences between the mean of girls and boys.

1. What is the value of the lower end of the confidence interval?

b. – 1.4485

2. What is the value of the upper end of the confidence interval?

c. – 0.9515

What kind of distribution ( central limit theorem or T) and why?

We can use the t distribution but since the sample size is large enough we will have a distribution similar to the normal standard distribution. Because when the degrees of freedom  of the t distribution increases we have a normal distribution.

So lets write down what we have a(t) = 66 and v(0) = 0 and s(0) = 0

a) Determine the position function.

To do this we have to integrate our acceleration function twice, or we have to integrate the acceleration function to get our velocity function and then integrate that to get our position function.

So:

[tex]\int\limits {a(t)} \, dt = \int\limits {66} \, dt[/tex]

= 66t + c

This means that v(t) = 66t + c

We know that v(0) = 0, so:

v(0) = 66(0) + c

c = 0

So v(t) = 66t (This will be helpful for us later)

Now we have to integrate again.

[tex]\int\limits {66t} \, dt[/tex]

[tex]= 33t^2 + c[/tex]

*note that both of these integrands are done with the reverse power rule for integration*

So we can say that [tex]s(t) = 33t^2 + c[/tex]

But we know that s(0) = 0

So

s(0) = 33(0)^2 + c

c = 0

So... s(t) = 33t^2

b) Now we can answer this question using our position function!

All we have to do is plug in t=5 for s(t)

So...

s(5) = 33(5)^2

s(5) = 825 ft

c) So this is essentially the same problem as b except now we are solving for the time instead of the distance.

We know that in 1 mile there are 5280 ft

So in 1/3 a mile there are 5280(1/3) ft or 1760 ft

Now we can set 1760 equal to s(t) and solve for t

1760 = 33t^2

t^2 = 53.33

[tex]t = \sqrt{53.33}[/tex]

t = 7.30s (we only consider positive answers for time)

d) This is the same question as c just a different distance so:

Setting 300 for s(t)

300 = 33t^2

t^2 = 9.091

[tex]t = \sqrt{9.091}[/tex]

t = 3.015s

e) So for this question we have to approach it in terms of the velocity function not the position function. Then we will solve for the time it took to travel with that velocity and then plug that time value into the position function.

So:

v(t) = 66t

We know that in this case v(t) = 178 ft/s

So: 178 = 66t

t = 2.6969 // t = 2.7s

Now we can use this time in our position function to solve for the distance traveled.

s(t) = 33t^2

s(2.7) = 33(2.7)^2

s(2.7) - 240.57 ft

Hope this helped!

To solve the problem we must know about the concept of Acceleration.

Acceleration is defined as the rate of change of velocity of an object with respect to time.

[tex]a = \dfrac{dv}{dt}[/tex]

Velocity is defined as the rate of change of position of an object with respect to time.

[tex]v=\dfrac{dx}{dt}[/tex]

As we know that acceleration is written as,

[tex]a = \dfrac{dv}{dt}\\\\\int dv =\int a\ dt[/tex]

substitute the value a(t) = 66,

[tex]\int dv =\int (66)\ dt[/tex]

[tex]v = 66t + c[/tex]

As the condition given v(0) = 0,

[tex]v = 66t + c\\\\0 = 66(0)+c\\\\c = 0[/tex]

Therefore, the velocity of the racer can be written as v=66t.

The velocity is written as,

[tex]v=\dfrac{dx}{dt}\\\\\int dx= \int v\ dt \\\\\int dx= \int (66t)\ dt\\\\x = 33t^2+c[/tex]

Substitute the given value x(0) = 0

[tex]0 = 33(0)^2+c\\\\c=0[/tex]

Thud, the position of the racer can be given by the function s = 33t².

A.) The position function t is greater than or equal to 0.

The position function t greater than or equal to 0 can be given by the function s = 33t².

B.) Distance traveled by the racer in 5 seconds.

To find the distance traveled by the racer in 5 seconds substitute the value of t in the position function,

s = 33(5)²

s = 825 m

C.) Time is taken by the racer to travel 1/3 mi.

We already have the position function,

s = 1/3 mi = 536.448 m

s = 33t²

536.448 = 33t²

t = 4.0319 second

D.) Time is taken by the racer to travel 300 ft

We already have the position function,

s = 300 ft = 91.44 m

s = 33t²

91.44 = 33t²

t = 1.6646 second

E.) Distance traveled by the race when it reaches a speed of 178 ft/s.

178 ft/s = 54.2544 m/s

v = 66t

54.2544 = 66t

t = 0.822 sec

Distance traveled in t = 0.8220 sec,

s = 33t²

s = 33(0.822)²

s = 22.2975 m

Learn more about Acceleration:

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Answer:

If (a and b )≤ 0 then [tex]log(ab)=log(a)+log(b)[/tex] is disproved

Step-by-step explanation:

If a and b are positive real numbers then:

[tex]log(ab)=log(a)+log(b)[/tex]

But if a and b are negative then this axiom is not true as log is not defined

[tex]log_{c}(x)[/tex]= undefined   [tex]for \quad x\leq 0[/tex]

So if (a and b )≤ 0 then [tex]log_{c}(a)[/tex] and  [tex]log_{c}(b)[/tex] are undefined but [tex]log_{c}(-a*-b)[/tex] is defined.

The statement 'If a and b are any two real numbers, then log(ab) = log(a) + log(b)' is a basic property of logarithms and it is true. It represents the concept of 'logarithms of products' and falls under the wider subject of exponentials and logarithms.

The provided statement 'If a and b are any two real numbers, then log(ab) = log(a) + log(b)' is actually a basic property of logarithms. If a and b are any two positive real numbers, then the logarithm of the product of these two numbers (ab) is indeed equal to the sum of the logarithm of the first number (a) and the logarithm of the second number (b). This is mathematically represented as log(ab) = log(a) + log(b).

This property of logarithms comes under the concept of 'logarithms of products', which is a part of the wider topic of exponentials and logarithms. Using similar properties, we can say that the logarithm of the number resulting from the division of two numbers is the difference between the logarithms of the two numbers. Also, the logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number.

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Answer: 12

Step-by-step explanation:

Given : Students at a liberal arts college study for an average of 10 hours per week with a standard deviation of 2 hours per week.

[tex]\mu=10\text{ hours}[/tex] and [tex]\sigma=2\text{ hours}[/tex]

The distribution of their study time happens to be uni-modal, symmetric and bell shaped i.e. Normally distributed.

According to the Empirical rule , about 68% of the population lies within one standard deviation from mean .

i.e. Approximately 68% of students study between [tex]\mu-\sigma[/tex] and  [tex]\mu+\sigma[/tex]  hours a week.

i.e. Approximately 68% of students study between [tex]10-2[/tex] and  [tex]10+2[/tex]  hours a week.

i.e. Approximately 68% of students study between 8 and 12 hours a week.

Hence, the value of B = 12.

The question is about a normal distribution in statistics where 68% of data falls within one standard deviation from the mean. Given the mean study, hours is 10 per week and the standard deviation is 2 hours, the value of B representing the upper limit of the 68% range is 12 hours per week.

The subject matter of this problem is based on the principles of statistics, particularly the concept of a normal distribution which is characterized by being uni-modal, symmetric, and bell-shaped. In a normal distribution, approximately 68% of data falls within one standard deviation from the mean.

In this question, we are given that the average study hours are 10 per week (the mean), and the standard deviation is 2 hours. A study time of 8 hours a week represents one standard deviation below the mean. Therefore, one standard deviation above the mean will represent 12 hours a week (mean + standard deviation: i.e. 10 + 2).

Thus, in terms of the normal distribution of study hours, the value of B is 12 hours. That means that approximately 68% of students at this particular liberal arts college study between 8 and 12 hours a week.

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The difference of proportions of high-school and college students reading newspapers regularly can be assessed using a 95% confidence interval, and the significance of this difference can be verified using a z-test considering appropriately formulated null and alternative hypotheses. The questions involve application of fundamental concepts of statistics and hypothesis testing.

Given the data, the proportions of high-school students (p₁) and college students (p₂) who read newspapers regularly are 287/500 (0.574) and 252/420 (0.6) respectively.

A: The 95% confidence interval for the difference between the proportions is calculated by [p₁-p₂ ± Z*√((p₁*(1-p₁)/n₁) + (p₂*(1-p₂)/n₂))], where Z is the z-value (1.96 for 95% confidence), n₁ and n₂ are the sample sizes. Plug in the given numbers to get the interval.

B: Based on the 95% confidence interval, we can judge whether 0 is in this interval. If so, we can't conclude that there's a significant difference between the two proportions. If not, there is a significant difference.

C: The null hypothesis (H₀) is p₁ - p₂ = 0, indicating no difference. And the alternative hypothesis (Hₐ) is p₁ - p₂ < 0, suggesting there's a significant difference and p₁ is less than p₂.

D: To find p ˆ, the pooled estimate is (x₁+x₂) / (n₁+n₂), where x₁ and x₂ are the counts of successes (those who read newspapers) in each group. Plug in the numbers to do the math.

E: Preconditions for a z-procedure: 1. The samples are random. 2. Both sample sizes are sufficiently large (n>30) to apply the Central Limit Theorem. 3. The events are independent.

F: SEp ˆ, the pooled estimate of the standard errors, is √(p ˆ*(1-p ˆ)*(1/n₁+1/n₂))

G: The test statistic is (p₁-p₂) / SEp ˆ, and this z-value can be used to find the P-value in a standard normal distribution table. If the P-value is small, we reject the null hypothesis in favor of the alternative.

H: If the P-value<0.05, then the conclusion would be that the proportion of high school students who read the newspaper regularly is significantly less than the proportion of college students.

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The confidence interval and hypothesis testing don't provide enough evidence to suggest a significant difference in the proportions of high-school and college students who read newspapers regularly. The requirements for using a z-interval and z-procedure were satisfied, suggesting that the statistical methods used were appropriate. The test statistic and P-value confirm that the null hypothesis cannot be rejected.

This question pertains to comparing proportions using hypothesis testing and confidence interval estimation. Before we perform these statistical analyses, we need to ensure that the conditions for each are satisfied.

A. To construct a 95% confidence interval, and to confirm that the conditions for a z-interval are satisfied, we consider the following:

Calculations work out to a confidence interval of ±3.83%. Therefore, according to our analysis, we are 95% confident that the true difference in proportions of high-school students and college students who read newspapers regularly lies in this range.

B. Given our confidence interval, we don't have enough evidence to suggest a significant difference between the two proportions of high-school and college students who read newspapers regularly.

C. The null hypothesis (H0) would be: p1=p2, indicating no difference between the proportions. The alternative hypothesis (Ha), would be: p1

D. To calculate pooled estimate of the population proportions (denoted as p ˆ), the formulas and calculations result in p ˆ = 0.574.

E. Sample size for both samples are larger than 30, implying that we can safely use a z-procedure for the hypothesis test. Also, the samples are independent and randomly selected, satisfying the necessary conditions.

F. The pooled estimate of the standard errors (SEp ˆ), works out to 0.028.

G. The calculated test statistic (z) is 0.83 and the P-value is 0.20.

H. Given a significance level, α = .05, the P-value > α. Therefore, we fail to reject the null hypothesis. This means we do not have enough evidence to suggest that the proportion of high-school students who read newspapers regularly is less than the proportion of college students who do.

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Answer:

Type I: 1.9%, Type II: 1.6%

Step-by-step explanation:

given null hypothesis

H0=the individual has not taken steroids.

type 1 error-falsely rejecting the null hypothesis

⇒   actually the null hypothesis is true⇒the individual has not taken steroids.

 but we rejected it ⇒our prediction is the individual has taken steroids.

typr II error- not rejecting null hypothesis when it has to be rejected

⇒actually null hypothesis is false ⇒the individual has taken steroids.

but we didnt reject⇒the individual has not taken steroids.

let us denote

the individual has taken steroids by 1

the individual has not  taken steroids.by 0

                            predicted

                              1       0

   actual          1    98.4%  1.6%

                        0   1.9%   98.1%

so for type 1 error  

 actual-0

predicted-1

therefore from above table we can see that probability of Type I error is 1.9%=0.019

so for type II error

   actual-1

predicted-0

therefore from above table we can see that probability of Type I error is 1.6%=0.016

Final answer:

The probability of a Type I error in the described scenario is 1.9%, and the probability of a Type II error is 1.6%.

Explanation:

The question asks about the probabilities of Type I and Type II errors in the context of a screening process for detecting a rare disease. A Type I error occurs when the test incorrectly indicates that the disease is present when it actually is not. The probability given for a Type I error is 1.9%, since this is the rate at which the test wrongly gives a positive reaction in individuals without the disease. A Type II error occurs when the test fails to indicate that the disease is present when it actually is. The probability given for a Type II error is the complement of the test's sensitivity, which means 100% - 98.4%, equaling 1.6%.

Therefore:

Probability of a Type I error: 0.019 (or 1.9%)

Probability of a Type II error: 0.016 (or 1.6%)

The estimated slope is approximately -4.028 and the estimated y-intercept is approximately 92.919.

The estimated value of y when x=5 is approximately 72.779, and the error prediction when x=2 is approximately 5.863.

The estimated slope (b1) and the estimated y-intercept (bo) in the regression equation y = bo + b1x, you can use the following formulas:

[tex]\[b1 = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}\][/tex]

[tex]\[bo = \frac{\sum y - b1(\sum x)}{n}\][/tex]

Where:

n is the number of data points,

[tex]\(\sum xy\)[/tex] is the sum of the product of x and y,

[tex]\(\sum x\)[/tex] is the sum of x,

[tex]\(\sum y\)[/tex] is the sum of y,

[tex]\(\sum x^2\)[/tex] is the sum of the squared x values.

Now let's calculate these values step by step:

a. Find the estimated slope (b1):

[tex]\[n = 10\][/tex]

[tex]\[\sum x = 31\][/tex]

[tex]\[\sum y = 804\][/tex]

[tex]\[\sum xy = 2439\][/tex]

[tex]\[\sum x^2 = 121.25\][/tex]

[tex]\[b1 = \frac{(10 \times 2439) - (31 \times 804)}{(10 \times 121.25) - 31^2} \][/tex]

[tex]\[b1 = \frac{24390 - 25404}{1212.5 - 961}\][/tex]

[tex]\[b1 = \frac{-1014}{251.5}\][/tex]

[tex]\[b1 \approx -4.028 \ (rounded to three decimal places)\][/tex]

b. Find the estimated y-intercept (bo):

[tex]\[bo = \frac{804 - (-4.028 \times 31)}{10}\][/tex]

[tex]\[bo = \frac{804 + 125.188}{10}\][/tex]

[tex]\[bo \approx \frac{929.188}{10}\][/tex]

[tex]\[bo \approx 92.919 \ (rounded to three decimal places)\][/tex]

c. Find the estimated value of y when x=5:

[tex]\[y = bo + b1x\][/tex]

[tex]\[y = 92.919 + (-4.028 \times 5)\][/tex]

[tex]\[y = 92.919 - 20.14\][/tex]

[tex]\[y \approx 72.779 \ (rounded to three decimal places)\][/tex]

d. Find the error prediction when x=2:

First, find the predicted y using the regression line:

[tex]\[y = bo + b1x\][/tex]

[tex]\[y = 92.919 + (-4.028 \times 2)\][/tex]

[tex]\[y = 92.919 - 8.056\][/tex]

[tex]\[y \approx 84.863 \ (rounded to three decimal places)\][/tex]

Now, find the error prediction:

[tex]\[Error = |Observed\, y - Predicted\, y|\][/tex]

[tex]\[Error = |79 - 84.863|\][/tex]

[tex]\[Error \approx 5.863 \ (rounded to three decimal places)\][/tex]

If all the students at the university spent 2.2 hours per day studying, with a standard deviation of 2 hours, and we find the probability of observing a sample mean of 1.8 hours studying has an extremely low probability, we say that the observed time is statistically significant.

Statistical significance does not imply practical or substantive significance but indicates strong evidence against the null hypothesis.

The number of students randomly sampled at the studied university = 200

The average time spent by the sampled students per day on social media = 2.3 hours

The average time spent by the sampled students per day studying = 1.8 hours

Population mean on studying = 2.2 hours

Assumed level of statistical significance = 5%

Thus, if the probability of observing a sample mean of 1.8 hours of studying is extremely low given the population mean of 2.2 hours, we would say that the observed average study time of 1.8 hours is statistically significantly different from the population mean. This suggests that the difference is unlikely to have occurred by chance alone.

The correct answer is:

b. statistically significant.

A sample mean of 1.8 hours studying is significantly different from the population mean of 2.2 hours, indicating an important deviation.

Certainly! In statistical terms, "statistically significant" means that an observed result is unlikely to have occurred by chance alone.

In this scenario:

- The population mean (average time spent on studying) is 2.2 hours per day.

- The sample mean (average time spent on studying) is 1.8 hours per day.

- The standard deviation of the population is 2 hours.

To determine whether the observed sample mean of 1.8 hours studying is statistically significant, we can use hypothesis testing or calculate the z-score and find the corresponding p-value.

A low probability associated with the observed sample mean suggests that it's unlikely to occur under the assumption that the population mean is 2.2 hours. This indicates that the difference between the sample mean and the population mean is not likely due to random chance, but rather reflects a true difference in the population.

Therefore, we conclude that the observed time spent on studying (1.8 hours) is statistically significant, as it deviates significantly from the expected population mean of 2.2 hours.

The complete question is here:

Researchers doing a study comparing time spent on social media and time spent on studying randomly sampled 200 students at a major university. They found that students in the sample spent an average of 2.3 hours per day on social media and an average of 1.8 hours per day on studying. If all the students at the university in fact spent 2.2 hours per day on studying, with a standard deviation of 2 hours, and we find the probability of observing a sample mean of 1.8 hours studying has an extremely low probability, we say that observed time is:

a. statistically unlikely.

b. statistically significant.

c. statistically wrong.

d. statistically rare.

Answer:

Step-by-step explanation:

Given that the housing market has recovered slowly from the economic crisis of 2008.​ Recently, in one large​ community, realtors randomly sampled 38 bids from potential buyers to estimate the average loss in home value.

s = sample std deviation = 3000

Sample mean = 9379

Sample size n = 38

df = 37

Std error of sample mean = [tex]\frac{s}{\sqrt{n} } \\=486.66[/tex]

confidence interval 95% = Mean ± t critical * std error

=Mean ±1.687*486.66 = Mean ±821.003

=(8557.997, 10200.003)

a) If std deviation changes to 9000 instead of 3000, margin of error becomes 3 times

Hence 2463.008

b) The more the std deviation the more the width of confidence interval.

Candy can arrange in 1, 2, 3, 5, 6, 10, 15 ways

2, 3, 4, 6, 8, 12 number of people can sit at each table

Solution:

Candy brought 30 cookies to a party. List all the ways she can arrange them on a plate in equal rows.

Given that Candy brought 30 cookies to party

She has to arrange the ways she can arrange them on plate in equal rows

We have to list the factors of 30

factors of 30 = 1, 2, 3, 5, 6, 10, 15

So she can arrange in 8 ways

Possible arrangement are 1, 2, 3, 5, 6, 10, 15 rows

David is setting up tables for 24 people at the party. The same number of people will sit at each table, and no one will sit alone. How many people can sit at each table? List all possibilities.

Fcators of 24 = 1, 2, 3, 4, 6, 8, 12

Here we can eliminate 1 since given that no one will sit alone

Thus 2, 3, 4, 6, 8, 12 number of people can sit at each table

Answer:

[tex]z=\frac{13-20}{4}=-1.75[/tex]

Assuming:

H0: [tex]\mu \geq 20[/tex]

H1: [tex]\mu <20[/tex]

[tex]p_v = P(Z<-1.75) = 0.0401[/tex]

Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest (number of correct answers in the test), on this case we now that:

[tex]X \sim Binom(n=100, p=0.2)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

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

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

We need to check the conditions in order to use the normal approximation.

[tex]np=100*0.2=20 \geq 10[/tex]

[tex]n(1-p)=20*(1-0.2)=16 \geq 10[/tex]

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

[tex]E(X)=np=100*0.2=20[/tex]

[tex]\sigma=\sqrt{np(1-p)}=\sqrt{100*0.2(1-0.2)}=4[/tex]

So we can approximate the random variable X like this:

[tex]X\sim N(\mu =20, \sigma=4)[/tex]

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".  The letter [tex]\phi(b)[/tex] is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: [tex]\phi(b)=P(z<b)[/tex]

The z score is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we replace we got:

[tex]z=\frac{13-20}{4}=-1.75[/tex]

Let's assume that we conduct the following test:

H0: [tex]\mu \geq 20[/tex]

H1: [tex]\mu <20[/tex]

We want to check is the score for the student is significantly less than the expected value using random guessing.

So on this case since we have the statistic we can calculate the p value on this way:

[tex]p_v = P(Z<-1.75) = 0.0401[/tex]

Answer:

a) p=0.39, where p the parameter of interest represent the true proportion of adults that would erase all their personal information online if they could

b) Null hypothesis:[tex]p = 0.39[/tex]

Alternative hypothesis:[tex]p \neq 0.39[/tex]

Step-by-step explanation:

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".

On this case the claim that they want to test is: "The true proportion of adults that would erase all their personal information online if they could is 0.39 or 39%". So we want to check if the population proportion is different from 0.39 or 0.39%, so this needs to be on the alternative hypothesis and on the null hypothesis we need to have the complement of the alternative hypothesis.

Part a. Express the original claim in symbolic form. Let the parameter represent the adults that would erase their personal information.

p=0.39, where p the parameter of interest represent the true proportion of adults that would erase all their personal information online if they could

Part b. Identify the null and alternative hypotheses.

Null hypothesis:[tex]p = 0.39[/tex]

And for the alternative hypothesis we have    

Alternative hypothesis:[tex]p \neq 0.39[/tex]  

Answer:

c.  There is no difference in the length between male and female babies

Step-by-step explanation:

When a confidence interval contains a zero, then there is no statistically significant different between the two group means because this means that there is a zero in the difference between the groups and a zero in the difference means there is no difference.

Answer: The correct answer is C.

Step-by-step explanation: The y-intercept is the point on the graph at which the line crosses over the y-axis. In this case, your y-intercept is 1, because the line crosses over the y-axis at 1. The slope formula is rise/run, or y2 -y1/x2-x1. However since you were not supplies with any coordinates, you'll have to go with rise (up/down) over run (left/right). The slope of the line falls one cube down ward, making the slope negative 1. The slope  "runs" to the right 4 cubes, making the slope -1 (falls 1 cube)/4 (moves to the left/right 4 cubes).

For a sample of size 81 with a sample variance of 625, testing the hypothesis H₀: σ² = 500 against the alternative Hₐ: σ² ≠ 500, the chi-square test statistic is 100.

The correct answer is option C.

To find the test statistic for this hypothesis test, we can use the chi-square test statistic formula:

Chi-square = ((n - 1) * s^2) / σ₀^2

where:

n is the sample size (81),

s^2 is the sample variance (625),

σ₀^2 is the hypothesized population variance under the null hypothesis (500).

Calculations:

Chi-square = ((81 - 1) * 625) / 500 = (80 * 625) / 500 = 100

c. 100

Interpretation:

The calculated chi-square test statistic is 100.

In conclusion, the correct test statistic for this problem is 100, and the correct option from the given choices is c. 100.

Final answer:

The test statistic for the given hypothesis test of a single variance where n = 81, s² = 625, and σ² under the null hypothesis is 500 is calculated using the chi-square statistic formula and equals 100.

Explanation:

The student's question pertains to finding the test statistic for a hypothesis test of a single variance. To find the test statistic in this scenario, the chi-square statistic is used, which is calculated using the formula:

X² = (n - 1)*s² / σ²₀

In this case, n = 81, s² = 625, and the null hypothesis ℓ₀ states that σ² = 500. Plugging in these values:

X² = (81 - 1)*625 / 500 = 80*1.25 = 100

Therefore, the test statistic for this problem equals 100, which corresponds to option c.

Answer:

[tex]t=\frac{75.5-69}{\frac{7.007}{\sqrt{6}}}=2.272[/tex]  

[tex]p_v =2*P(t_{(5)}>2.272)=0.072[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean for the Co concentracion it's not significant different from 69.

Step-by-step explanation:

Data given and notation  

Data: 77, 82, 72, 68, 69, 85

The mean and sample deviation can be calculated from the following formulas:

[tex]\bar X =\frac{\sum_{i=1}^n x_i}{n}[/tex]

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}[/tex]

[tex]\bar X=75.5[/tex] represent the sample mean  

[tex]s=7.007[/tex] represent the sample standard deviation  

[tex]n=6[/tex] sample size  

[tex]\mu_o =69[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/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 population mean is different from 69, the system of hypothesis are :  

Null hypothesis:[tex]\mu = 69[/tex]  

Alternative hypothesis:[tex]\mu \neq 69[/tex]  

Since we don't know the population deviation, 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{75.5-69}{\frac{7.007}{\sqrt{6}}}=2.272[/tex]  

P-value  

We need to calculate the degrees of freedom first given by:  

[tex]df=n-1=6-1=5[/tex]  

Since is a two tailed test the p value would given by:  

[tex]p_v =2*P(t_{(5)}>2.272)=0.072[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean for the Co concentracion it's not significant different from 69.