Can you please help me thanks

Answer:

67 degrees

Step-by-step explanation:

(a) Using a calculator with mean and standard deviation keys, we get:
x₁ = 257.19, S₁ = 12.0794, X2 = 660832.61, s₂ = 13.5992

(b) To find the 99% confidence interval for μ₁-μ₂, we can use the formula:

(x₁ - x₂) ± tα/2 * sqrt(S₁²/n₁ + S₂²/n₂)

where tα/2 is the critical value from the t-distribution with degrees of freedom equal to (n₁ - 1) + (n₂ - 1) = 38 and α/2 = 0.005 (since we want a 99% confidence interval). Using a t-table or calculator, we find tα/2 = 2.704.

Substituting the values, we get:

(257.19 - 204.26) ± 2.704 * sqrt(12.0794²/21 + 13.5992²/19)

= 52.93 ± 8.8529

So the 99% confidence interval for μ₁-μ₂ is (44.1, 61.76).

(c) The confidence interval means that we are 99% confident that the true population mean weight of pro football players is between 44.1 and 61.76 pounds more than the true population mean weight of pro basketball players. Since the interval contains only positive numbers, we can say that professional football players have a higher mean weight than professional basketball players at the 99% level of confidence.

(d) The Student's t-distribution was used because both ₁ and ₂ are unknown and the sample sizes are small (less than 30).
(a) After calculating the mean and standard deviation for both sets of data, we get the following results:

Football players (x₁):
Mean (x₁) = 258.4286
Standard Deviation (s₁) = 11.7043

Basketball players (x₂):
Mean (x₂) = 208.0526
Standard Deviation (s₂) = 12.7779

(b) To find a 99% confidence interval for μ₁ - μ₂, we can use the formula for the confidence interval of the difference between two means:

CI = (x₁ - x₂) ± t * √[(s₁²/n₁) + (s₂²/n₂)]

Using the t-distribution with the appropriate degrees of freedom (determined by the sample sizes, n₁ and n₂) and a 99% confidence level, we find the t-value, which is approximately 2.963.

CI = (258.4286 - 208.0526) ± 2.963 * √[(11.7043²/21) + (12.7779²/19)]
CI = 50.376 ± 2.963 * √[(162.0714/21) + (163.2774/19)]
CI = 50.376 ± 2.963 * 4.6571
CI = 50.376 ± 13.7903

The 99% confidence interval is:
Lower limit: 36.6
Upper limit: 64.1

(c) The confidence interval consists of only positive numbers. This means that, at the 99% level of confidence, professional football players have a higher population mean weight than professional basketball players.

(d) The Student's t-distribution was used because s₁ and s₂ are unknown.

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Answer:

The first option will be your answer

Step-by-step explanation:

To create a vector of the first and last element in R, you can use the `c()` function to concatenate the two values into a vector. You can access the first element of a vector by using the index 1 and the last element by using the index `length(vector_name)`.

Here is an example code:

```
# create a vector
my_vector <- c(3, 7, 9, 12, 4)

# create a vector of the first and last element
first_last_vector <- c(my_vector[1], my_vector[length(my_vector)])

# print the vector
print(first_last_vector)
```

The output will be: `3 4`, which is the first and last element of the `my_vector` concatenated into a new vector.
Hi! To create a vector containing the first and last elements of an existing vector in R, you can use the following code:

```R
original_vector <- c(2, 4, 6, 8, 10)
new_vector <- original_vector[c(1, length(original_vector))]
```

In this example, `original_vector` contains the values 2, 4, 6, 8, and 10. The `new_vector` is created by selecting the first (1) and last (length of the original vector) elements from `original_vector`. The result will be a new vector containing the values 2 and 10.

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The lower bound of the confidence interval is 0.197 and the upper bound is 0.283. Therefore, we can say with 90% confidence that the true population proportion of people with kids is between 0.197 and 0.283. Both values are rounded to three decimal places.

To construct a 90% confidence interval for the true population proportion of people with kids, we need to use the sample proportion and sample size. The sample proportion is the number of people with kids divided by the total number of people sampled, which is 144/600 = 0.24. The sample size is 600.

Next, we need to calculate the standard error, which is the square root of (sample proportion x (1 - sample proportion) / sample size). Plugging in the values, we get:

SE = √(0.24 x 0.76 / 600) = 0.026

To find the margin of error, we multiply the standard error by the z-score corresponding to a 90% confidence level, which is 1.645. Therefore, the margin of error is:

ME = 1.645 x 0.026 = 0.043

Finally, we can construct the confidence interval by adding and subtracting the margin of error to the sample proportion:

p ± ME = 0.24 ± 0.043

The lower bound of the confidence interval is 0.197 and the upper bound is 0.283. Therefore, we can say with 90% confidence that the true population proportion of people with kids is between 0.197 and 0.283. Both values are rounded to three decimal places.

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Let A={A_α ∶α∈Δ} be a family of sets, Δ≠∅ and let B be an arbitrary set, The statement is false.

To see why, consider the following example:

Let A = {{1}, {2}, {3}} be a family of sets and let B = {4}.

Then, A ∪ {B} = {{1}, {2}, {3}, {4}}, which is a family of sets with one additional set (B) compared to A.

However, this does not necessarily mean that A ∪ {B} is a family of sets, because B could be unrelated to the sets in A.

In this case, the elements of A are all singletons (sets with one element), while B has one element as well. Therefore, A ∪ {B} does not satisfy the definition of a family of sets, which requires that all elements are sets.

So, we have shown that the statement is false by providing a counter example.

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Answer:

78

Step-by-step explanation:

Curvature at t=1 is approximately 0.0867.

To find the derivative of r(t), we simply take the derivative of each component:

r'(t) = ⟨4, -5, -16t⟩

To find the second derivative, we take the derivative of each component of r'(t):

r''(t) = ⟨0, 0, -16⟩

To find the curvature at t=1, we use the formula:

κ(t) = ||r'(t) x r''(t)|| / ||r'(t)||³

Plugging in t=1 and using the values we found above, we get:

κ(1) = ||⟨4, -5, -16⟩ x ⟨0, 0, -16⟩|| / ||⟨4, -5, -16⟩||³

Simplifying, we get:

κ(1) = ||⟨320, 64, 0⟩|| / (4³ + (-5)³ + (-16)³)³/²
κ(1) = ||⟨320, 64, 0⟩|| / 1477.74
κ(1) = 128 / 1477.74

Therefore, the curvature at t=1 is approximately 0.0867.

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Answer: D

Step-by-step explanation:

The inverse Laplace transform are:

a. f(t) = (1/2)e^49/7 * uc(t-7) - (1/2)e^-49/7 * uc(t+7)

b. f(t) = (1/2)e^4 * uc(t-4) - (1/2)e^6 * uc(t-6)

a. F(s) = (7e^-7s)/(s^2-49)

We notice that the denominator of F(s) can be factored as (s-7)(s+7). We can use partial fraction decomposition to write F(s) in the form:

To find the values of A and B, we can multiply both sides by (s-7)(s+7) and then substitute s=7 and s=-7:

When we substitute s=7, we get:

Similarly, when we substitute s=-7, we get:

Now, we can write F(s) as:

To take the inverse Laplace transform, we can use the formula:

where uc(t) is the Heaviside step function.

Thus, we have:

b. F(s) = ((s-5)e^-s)/(s^2-10s+24)

The denominator of F(s) can be factored as (s-4)(s-6). We can use partial fraction decomposition to write F(s) in the form:

To find the values of A and B, we can multiply both sides by (s-4)(s-6) and then substitute s=4 and s=6:

When we substitute s=4, we get:

Similarly, when we substitute s=6, we get:

Now, we can write F(s) as:

To take the inverse Laplace transform, we can use the formula:

Thus, we have:

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Answer:

-27.89%, the portfolio loses about 28% of its value every year on average.

Step-by-step explanation:

The annual growth-rate is -52.90%

The annual growth rate can be calculated using the formula:

i = (1 - (final value / initial value)^(1/n)) x 100%

Where "final value" is the current value of the portfolio, "initial value" is the original value of the portfolio, "n" is the number of years, and "i" is the annual growth rate.

In this case, the portfolio drops to one-tenth of its former value, which means the final value is 1/10 or 0.1 times the initial value. The time period is 6 years. So we have:

i = (1 - (0.1)^(1/6)) x 100%
i = (1 - 0.471) x 100%
i = 52.9%

Therefore, the annual growth rate is -52.9% (since the portfolio has decreased in value). Rounded to two decimal places, the answer is -52.90%.

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[tex]\qquad \textit{power of a complex number} \\\\\ [\quad r[\cos(\theta)+i\sin(\theta)]\quad ]^n\implies r^n[\cos(n\cdot \theta)+i\sin(n\cdot \theta)] \\\\[-0.35em] ~\dotfill\\\\ z=4\left[ \cos\left( \frac{\pi }{2} \right)+i\sin\left( \frac{\pi }{2} \right)\right] \\\\\\ z^3=4^3\left[ \cos\left( 3\cdot \frac{\pi }{2} \right)+i\sin\left( 3\cdot \frac{\pi }{2} \right)\right]\implies z^3=64\left[ \cos\left( \frac{3\pi }{2} \right)+i\sin\left( \frac{3\pi }{2} \right)\right][/tex]

[tex] \blue{\boxed{\sf z^3 = 64[cos(\dfrac{3\pi}{2}) + isin(\dfrac{3\pi}{2}) ]}} [/tex]

[tex] \\ [/tex]

We are given a complex number, z, in its trigonometric form.

To find the trigonometric form of z³, we will apply De Moivre's Theorem.

[tex] \\ [/tex]

[tex] \Large{\boxed{\boxed{\sf [cos(\theta) + isin(\theta)]^n = cos(n\theta) + isin(n\theta)}}} [/tex]

[tex] \\ \\ [/tex]

[tex] \sf z^3 = \Bigg(4[cos( \dfrac{\pi}{2}) + isin( \dfrac{\pi}{2})]\Bigg) ^{3} \\ \\ \Longleftrightarrow \sf z^3 = 4^3 \times [cos( \dfrac{\pi}{2}) + isin( \dfrac{\pi}{2})]^3 \\ \\ \Longleftrightarrow \sf z^3 = 64[cos(\dfrac{\pi}{2}) + isin(\dfrac{\pi}{2})]^3 [/tex]

Let's apply the theorem with our values:

[tex] \: \star \: \theta = \dfrac{\pi}{2} \\ \\ \star \: \sf n = 3 [/tex]

[tex] \\ [/tex]

[tex] \sf z^3 = 64[cos(3 \times \dfrac{\pi}{2}) + isin(3 \times \dfrac{\pi}{2})] \\ \\ \Longleftrightarrow \boxed{\sf z^3 = 64[cos(\dfrac{3\pi}{2}) + isin(\dfrac{3\pi}{2})]} [/tex]

[tex] \\ \\ [/tex]

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Answer:

(1 , 4)

Step-by-step explanation:

x=1

y=4

The term most likely to be considered an ordinal variable among the given options is:
c. Education level

How do you know if a variable is ordinal?

A purely nominal variable is one that simply allows you to assign categories but you cannot clearly order the categories. If the variable has a clear ordering, then that variable would be an ordinal variable.

Your answer: Education level (c) is most likely to be considered an ordinal variable because it involves a ranking or order, such as high school diploma, bachelor's degree, master's degree, etc., where each level has a specific order but the differences between levels are not numerically equal.

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(a) The median tomato sales is 25 kg, (b) The minimum tomato sales is 3 kg, (c) The interquartile range can be found by subtracting the first quartile from the third quartile: Q3-Q1 = 40-11 = 29 kg.

(d) (i) To find the number of days tomato sales were between 42 kg and 50 kg, we look at the box-and-whisker diagram and count the number of days within that range. It looks like there are no days within that range, so the answer is 0. (ii) To find the number of days tomato sales were between 26 kg and 55 kg, we look at the box-and-whisker diagram and count the number of days within that range. It looks like there are 50 days within that range.

(e) To determine if the day with only 8 kg of tomato sales is an outlier, we need to calculate the lower and upper bounds for outliers. The lower bound is Q1 - 1.5(IQR) and the upper bound is Q3 + 1.5(IQR). Using the values we found earlier, the lower bound is 11 - 1.5(29) = -28.5 kg and the upper bound is 40 + 1.5(29) = 83.5 kg. Since 8 kg is outside of this range, it would be considered an outlier.

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The function value at x = 9/10 is approximately 0.406 when rounded to three decimal places.

To evaluate the function h(x) = e−x at x = 9/10, we substitute 9/10 in place of x:

h(9/10) = e−(9/10)

Using a calculator or mathematical software, we can approximate this value to three decimal places:

h(9/10) ≈ 0.406

Therefore, the rounded result of evaluating the function at x = 9/10 is 0.406.
To evaluate the function h(x) = e^(-x) at the indicated value of x = 9/10, substitute the value of x into the function and round the result to three decimal places.

h(9/10) = e^{(-9/10)}

Using a calculator or mathematical software, we get:

h(9/10) ≈ 0.406

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As a result, the following quadratic function matches this table:

y = (1/2)x² - (3/2)x - 1 A has a positive value, while B has a negative value.

A polynomial function of degree two is a quadratic function. Where a, b, and c are constants, it has the form f(x) = ax² + bx + c. A quadratic function's graph is a parabola that slopes upward if a > 0 and downward if a 0. The vertical line x = -b/2a serves as the axis of symmetry, and the parabola's apex is located at (-b/2a, f(-b/2a)).

A quadratic function's input values are represented by x in your table and its output values by y. This quadratic function's answer is y = (1/2)x2 - (3/2)x - 1. Setting y = 0 and using the quadratic formula to solve for x will get the roots of this quadratic function: x = (-b √(b² - 4ac)) / (2a). replacement of the values of

You illustrated a quadratic function in the table. We can use the method of finite differences to locate the quadratic equation that best fits this table.

The initial deviations are -1, 1, 3, 5, and 7. 2, 2, 2, 2 make up the second difference. We can infer the function's quadratic nature from the fact that the second differences are constant. The quadratic function's standard form formula is: y = ax²+ bx + c.

Any point on the graph can be used to calculate a. Let's employ (0,-1). Adding x=0 and y=-1 to the equation results in:

y = ax²+ bx + c.

-1 = a(0)² + b(0) + c -1 = c

So c = -1.

We must now locate a and b. To generate two equations and find the values of a and b, we can use two points.

. Using (2, 0) and (4,3). These values are substituted into the equation to produce:

0 = a(2)² + b(2) - 1 3 = a(4)² + b(4) - 1

These equations are simplified to give:

4a + 2b = 1 16a + 4b = 4

Calculating a and b results in:

a = 1/2 b = -3/2

As a result, the following quadratic equation matches this table:

y = (1/2)x² - (3/2)x - 1

A has a positive value, while B has a negative value.

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Answer:

C

Step-by-step explanation:

We can solve for r using numerical methods or trial and error. One possible way is to use Excel's goal seek function, which gives us a discount rate of approximately 16.5%. Therefore, at a discount rate of 16.5%, the NPV of the project is zero, and we would be indifferent between accepting the project and rejecting it.

Using the formula for NPV, we have:

[tex]NPV = -Cost + (CF1 / (1+r)^1) + (CF2 / (1+r)^2) + ... + (CFn / (1+r)^n)[/tex]

where CF is the annual cash flow, r is the required rate of return, and n is the number of years.

Plugging in the given values, we get:

At a required return of 9%:

[tex]NPV = -9100 + (2200 / (1+0.09)^1) + (2200 / (1+0.09)^2) + ... + (2200 / (1+0.09)^9)\\NPV = -9100 + 1704.13 + 1562.30 + ... + 653.89\\NPV = $404.19[/tex]

At a required return of 25%:

NPV = -9100 + (2200 / (1+0.25)^1) + (2200 / (1+0.25)^2) + ... + (2200 / (1+0.25)^9)

[tex]NPV = -9100 + 1548.28 + 1179.08 + ... + 160.69\\NPV = -$1,489.91[/tex]

To find the discount rate at which we would be indifferent between accepting the project and rejecting it, we can use the NPV formula and set it equal to zero:

[tex]0 = -9100 + (2200 / (1+r)^1) + (2200 / (1+r)^2) + ... + (2200 / (1+r)^9)[/tex]

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In the first 20 minutes, 96 words are memorized

To find the total number of words memorized in the first 20 minutes, you need to integrate the memory rate function M'(t) from t = 0 to t = 20.

M(t) = ∫(-0.009t^2 + 0.6t) dt

Integrating, we get:

M(t) = -0.003t^3 + 0.3t^2 + C

To find the total number of words memorized in the first 20 minutes, evaluate M(t) at t = 20 and subtract M(t) at t = 0:

M(20) - M(0) = (-0.003 * 20^3 + 0.3 * 20^2) - (0)
M(20) = -0.003 * 8000 + 0.3 * 400
M(20) = -24 + 120
M(20) = 96

In the first 20 minutes, 96 words are memorized.

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7a = 7. So, a = 1. Consequently, the $1 apple price is the suggested cost option. (A).

A linear equation is an algebraic equation of the form y=mx+b, where m is the slope and b is the y-intercept, and only a constant and a first-order (linear) term are included. The variables in the preceding equation are y and x, and it is occasionally referred to as a "linear equation of two variables."

Let a represent the cost of an apple and b represent the cost of a banana. Using the information provided, we can then create two equations:

2a + 5b = 17 (equation 1)

3a + 4b = 15 (equation 2)

By removing b, we can find a solution for a. When we multiply equations 1 and 2 by 4 and by 5, we obtain:

8a + 20b = 68

15a + 20b = 75

Equation 1 minus equation 2 results in:

7a = 7

So, a = 1.

Consequently, the $1 apple price is the suggested option. (A).

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Complete question is:

Two apples and five bananas cost $17, while three apples and four bananas cost $15. What is the price of an apple?

(A) $ 1

(B) $ 1.50

(C) $ 2

(D) $ 2.50

(E) $ 3

The Jazz and opera types of music were chosen by 40% of the people surveyed.

A percentage is a figure or ratio that reflects a portion of one hundred.

Given: total number of respondents = 55

As a result, 40% of all respondents equals 40% of 55 = [tex]\frac{40}{100} * 55[/tex] = 22.

County and Opera are chosen by 15 + 10 = 25 respondents, i.e. less than 40%.

Jazz and opera are chosen by = 12 + 10 = 22 respondents or 40%.

Jazz, Opera,  and Rock  are chosen by = 12 + 10 + 18 = 40

Country, Jazz and Rock = 15 + 12 + 18 = 45

Therefore Jazz and opera are chosen by 40% of people.

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Complete question:

maribel surveyed 55 people to find out their favorite types of music. The result is shown in the bar graph

The projection matrices and as we have verified that [tex]P = A(A^TA)^{-1}A^T.[/tex]

Let's start by defining the matrices we will use for this problem. The matrix A is given by A = [1 1 0 1 0 0], which means that A is a 3x6 matrix with three rows and six columns. The vector b is given by b = [2 3 4], which is a 3x1 matrix with three rows and one column.

To project b onto the column space of A, we need to find a vector p that is in the column space of A and is as close as possible to b. We can do this by solving the equation [tex]A^T Ax = A^T b[/tex], where [tex]A^T[/tex] is the transpose of matrix A. This equation is known as the normal equation of the least-squares problem, and it gives us the vector p that is the projection of b onto the column space of A.

We can also find the error vector e = b - p, which is the difference between b and its projection onto the column space of A. This error vector is perpendicular to the column space of A, which means that it lies in the null space. To verify this, we can take the dot product of e with each column of A, which should be zero for each column.

To compute the projection matrix P, we can use the formula

[tex]P = A(A^TA)^{-1}A^T.[/tex]

This matrix projects any vector onto the column space of A.

We can also verify that P² = P and P = [tex]P^T[/tex], which means that P is an idempotent matrix and a symmetric matrix.

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Answer: 257

Step-by-step explanation: Theres no remainder, and u couldve used a calculator

Answer:

257

Step-by-step explanation:

2056 : 8 = 257

For initial value problem x^2 y''-xy'+y=0, y(1)=3, y'(1)=-1, its general solution is y=c1x+c2x lnx, (0, infinity), the solution for the initial value problem is y(x) = 3x - 4x ln(x) for x in (0, infinity).

To find the solution for the initial value problem with the given general solution and initial conditions, follow these steps:

1. Write down the general solution: y(x) = c1x + c2x ln(x), where x is in (0, infinity).

2. Apply the initial conditions: y(1) = 3 and y'(1) = -1.

3. To apply the first initial condition, replace x with 1 in the general solution:
y(1) = c1(1) + c2(1) ln(1) = 3.
Since ln(1) = 0, the equation becomes:
c1 = 3.

4. To apply the second initial condition, first find the derivative of the general solution with respect to x:
y'(x) = c1 + c2(1 + ln(x)).

5. Replace x with 1 and y'(1) with -1 in the derivative equation:
-1 = c1 + c2(1 + ln(1)).
Substitute the value of c1 found in step 3:
-1 = 3 + c2(1 + 0).
Solve for c2:
c2 = -4.

6. Now that we have the values of c1 and c2, substitute them back into the general solution:
y(x) = 3x - 4x ln(x), where x is in (0, infinity).

So, the solution for the initial value problem is y(x) = 3x - 4x ln(x) for x in (0, infinity).

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The smallest length of fencing required to enclose the rectangular area of 12000 square feet and divide it in half with a fence down the middle parallel to one side is[tex]$380\sqrt{2}$[/tex] feet.

Let the length of the rectangular area be [tex]$l$[/tex] and the width be [tex]$w$[/tex]. Then, we have the equation:

[tex]$lw = 12000$[/tex]

We want to divide the region in half with a fence down the middle parallel to one side, so we will need two equal rectangular sections with area:

[tex]$(l/2)w = 6000$[/tex]

The total length of fencing required will be the perimeter of the rectangular region, plus the length of the fence down the middle. Therefore, we have:

[tex]$\text{Total length of fencing} = 2l + 2w + l = 3l + 2w$[/tex]

Substituting[tex]$w = 12000/l$[/tex], we get:

[tex]$\text{Total length of fencing} = 3l + 24000/l$[/tex]

To find the - of fencing required, we can take the derivative of this expression with respect to [tex]$l$[/tex], set it equal to zero, and solve for [tex]l$:[/tex]

[tex]$\frac{d}{dl} (3l + 24000/l) = 3 - \frac{24000}{l^2} = 0$[/tex]

Solving for [tex]$l$[/tex], we get:

[tex]l = \sqrt{8000} = 40\sqrt{2}$ feet[/tex]

Substituting this value of [tex]$l$[/tex] back into the expression for the total length of fencing, we get:

[tex]$\text{Total length of fencing} = 3(40\sqrt{2}) + 2(12000/40\sqrt{2}) = 80\sqrt{2} + 300\sqrt{2} = 380\sqrt{2}$ feet.[/tex]

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Answer:

Step-by-step explanation:

I think it would be the first one.(Sorry if I'm wrong!)

Random sampling is preferred over non-random sampling processes because it promotes sample representativeness, group equivalency for experiments, and reduces the likelihood of sample bias.

Random sampling does not necessarily promote external validity.

As the external validity depends on various other factors such as the sample size, sampling frame, and the research design.

Random sampling allows every member of the population to have an equal chance of being selected.

Which helps to ensure that the sample is representative of the population, and therefore promotes group equivalency for experiments.

Random sampling can also promote group equivalency for experiments.

As it helps to ensure that the groups being compared are similar in terms of their composition and characteristics.

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The critical χ2 value you are looking for is 16.151, which corresponds to option C.

To find the critical X2-value to test the claim σ2 < 5.6 with n=28 and α=0.10, we need to use the Chi-square distribution table. The degrees of freedom for this test is n-1 = 28-1 = 27.

The critical X2-value for a one-tailed test with α=0.10 and 27 degrees of freedom is 16.151 (option C).

To perform the test, we calculate the test statistic as:

X2 = (n-1) * s^2 / σ^2

where s is the sample standard deviation and σ is the population standard deviation.

If X2 < critical value, we reject the null hypothesis and accept the claim. Otherwise, we fail to reject the null hypothesis.

In this case, we have:

X2 = (28-1) * s^2 / 5.6

We don't have the sample standard deviation s or the population standard deviation σ, so we can't calculate X2 directly.

However, we can use the critical X2-value and the given significance level to find a confidence interval for the population standard deviation σ.

The confidence interval is given by:

s^2 / X2 < σ^2 < s^2 / χ^2(α/2, n-1)

where χ^2(α/2, n-1) is the Chi-square distribution value for a two-tailed test with significance level α/2 and degrees of freedom n-1.

Using the values given in the problem, we get:

s^2 / 16.151 < σ^2 < s^2 / χ^2(0.05, 27)

We don't know the value of s^2, but we can use the sample size and the given confidence level to find a confidence interval for s^2.

The confidence interval for s^2 is given by:

(n-1) * s^2 / χ^2(α/2, n-1) < σ^2 < (n-1) * s^2 / χ^2(1-α/2, n-1)

where χ^2(1-α/2, n-1) is the Chi-square distribution value for a two-tailed test with significance level 1-α/2 and degrees of freedom n-1.

Using the values given in the problem, we get:

27 * s^2 / χ^2(0.005, 27) < σ^2 < 27 * s^2 / χ^2(0.995, 27)

We can use a statistical software or a Chi-square distribution table to find the values of χ^2(0.005, 27) and χ^2(0.995, 27).

Assuming that s^2 is a reasonable estimate of σ^2, we can use the confidence interval for s^2 to estimate the confidence interval for σ^2.

For example, if we find that:

27 * s^2 / χ^2(0.005, 27) = 3.45

27 * s^2 / χ^2(0.995, 27) = 10.66

Then we can say with 90% confidence that:

3.45 < σ^2 < 10.66

This interval does not contain the value 5.6, so we can reject the claim that σ2 < 5.6 at the 0.10 significance level.

Thus,the  critical χ2 value you are looking for is 16.151, which corresponds to option C.

To learn more about critical value visit :

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Answer:

7

Step-by-step explanation:

(16/4+90/9)x2

(4+10)x2

14/2

7