If the sample size is n = 75, what are the degrees of freedom for the appropriate chi-square distribution when testing for independence of two variables each with three categories? a. 4 b. 69 c. 74

Answer:

a) The 95% confidence interval would be given by (29.639;31.361 )

And the results are not very different from 28.9 hg< μ < 31.9 hg

b)The 95% confidence interval would be given by (28.939;31.861)

And the results are not very different from 28.9 hg< μ < 31.9 hg

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".

[tex]\bar X=30.5[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s=6.7 represent the sample standard deviation

n=235 represent the sample size  

2) First confidence interval

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

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

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=235-1=234[/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 table to find the critical value. The excel command would be: "=-T.INV(0.025,234)".And we see that [tex]t_{\alpha/2}=1.97[/tex]

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

[tex]30.5-1.97\frac{6.7}{\sqrt{235}}=29.639[/tex]    

[tex]30.5+1.97\frac{6.7}{\sqrt{235}}=31.361[/tex]    

So on this case the 95% confidence interval would be given by (29.639;31.361 )

And the results are not very different from 28.9 hg< μ < 31.9 hg

3) Second confidence interval

[tex]\bar X=30.4[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s=2.3 represent the sample standard deviation

n=12 represent the sample size  

[tex]df=n-1=12-1=11[/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 table to find the critical value. The excel command would be: "=-T.INV(0.025,11)".And we see that [tex]t_{\alpha/2}=2.20[/tex]

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

[tex]30.4-2.20\frac{2.3}{\sqrt{12}}=28.939[/tex]    

[tex]30.4+2.20\frac{2.3}{\sqrt{12}}=31.861[/tex]    

So on this case the 95% confidence interval would be given by (28.939;31.861)

And the results are not very different from 28.9 hg< μ < 31.9 hg

Answer:

Please read the answers below.

Step-by-step explanation:

Let's recall that in a square all its sides are equal length and all the four internal angles measure 90 °

5. If LN = 46, then we have:

OM = 46 (Same length than LN)

PN = LN/2 = 46/2 = 23

ON = √LN²/2 = √ 46²/2 = √ 2,116/2 = √ 1,058 = 32.53 (Rounding to two decimal places)

MN = ON = 32.53

6. m ∠EFG = 90°

m ∠GDH = ∠GDH/2 = 90/2 = 45°

m ∠FEG = ∠DEF/2 = 90/2 = 45°

m ∠DHG = 180 - (∠GDH + ∠DGH) = 180 - (45 + 45)= 180 - 90 = 90°

7. Solve for x

6x - 21 = ∠PQR/2

6x - 21 = 90/2

6x - 21 = 45

6x = 45 +21

6x = 66

x = 11

Answer:

110

Step-by-step explanation:

%20 increase = [tex]\frac{20}{100} 1300 = 260\\[/tex]

%50 decrease = [tex]\frac{50}{100}300 = 150[/tex]

%30 no change = [tex]\frac{30}{100}0 = 0[/tex]

Expected change = Increase - Decrease + No Change

Expected change = 260 - 150 + 0

Expected change = 110

Answer:

We conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30 minutes

Sample mean, [tex]\bar{x}[/tex] = 31.5 minutes

Sample size, n = 41

Alpha, α = 0.05

Sample standard deviation, s = 3.5 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 30\text{ minutes}\\H_A: \mu > 30\text{ minutes}[/tex]

We use one-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{31.5 - 30}{\frac{3.5}{\sqrt{41}} } = 2.744[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]

Since,  

[tex]z_{stat} > z_{critical}[/tex]

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.

Answer:

Step-by-step explanation:

1) 1) 97 – 39. It becomes

100 - 40 = 60

2)2) 812,344 – 187,675. It becomes

800000 - 200000 = 600000

3)321 + 79. It becomes

300 + 80 = 410

4) 315 x 821. It becomes

300 + 800 = 1100

5) 562 x 791. It becomes

600× 800 = 480000

6) 82 x 156. It becomes

80×200 = 16000

7) 711 x 884. It becomes

700×900 = 630000

8) 126 x 952. It becomes

100×1000 = 100000

9) 824 x 541. It becomes

800× 500 = 400000

10) 4027 x 78. It becomes

4000×80 = 320000

11) 796 x 123. It becomes

800×100 = 80000

12) 817 19. It becomes

800/20 = 40

13) 3615 72. It becomes

4000/70 = 57.1 = 60

14) 232 64. It becomes

200/60 = 33.3 = 30

15) 559 81. It becomes

600/80 = 7.5 = 8

16) 2986 222. It becomes

3000/200 = 15

17) 10275 232. It becomes

10000/200 = 50

18) 7428 286. It becomes

7000/300 = 23.3 = 20

19) 7143 369. It becomes

7000/400 = 17.5 = 18

20) 628,597 1525, it becomes

600000 - 2000 = 300

Answer:

1) 100 - 40 = 60

2) 800000 - 200000 = 600000

3) 300 + 80 = 410

4) 3000 + 800 = 1100

5) 600× 800 = 480000

6) 80×200 = 16000

7) 700×900 = 630000

8) 100×1000 = 100000

9) 800× 500 = 400000

10) 4000×80 = 320000

11) 800×100 = 80000

12) 800/20 = 40

13) 4000/70 = 57.1 = 60

14) 200/60 = 33.3 = 30

15) 600/80 = 7.5 = 8

16) 3000/200 = 15

17) 10000/200 = 50

18) 7000/300 = 23.3 = 20

19) 7000/400 = 17.5 = 18

20) 600000 / 2000 = 300

Step-by-step explanation:

1) 1) 97 – 39. It becomes

100 - 40 = 60

2)2) 812,344 – 187,675. It becomes

800000 - 200000 = 600000

3)321 + 79. It becomes

300 + 80 = 410

4) 315 x 821. It becomes

300 + 800 = 1100

5) 562 x 791. It becomes

600× 800 = 480000

6) 82 x 156. It becomes

80×200 = 16000

7) 711 x 884. It becomes

700×900 = 630000

8) 126 x 952. It becomes

100×1000 = 100000

9) 824 x 541. It becomes

800× 500 = 400000

10) 4027 x 78. It becomes

4000×80 = 320000

11) 796 x 123. It becomes

800×100 = 80000

12) 817 19. It becomes

800/20 = 40

13) 3615 72. It becomes

4000/70 = 57.1 = 60

14) 232 64. It becomes

200/60 = 33.3 = 30

15) 559 81. It becomes

600/80 = 7.5 = 8

16) 2986 222. It becomes

3000/200 = 15

17) 10275 232. It becomes

10000/200 = 50

18) 7428 286. It becomes

7000/300 = 23.3 = 20

19) 7143 369. It becomes

7000/400 = 17.5 = 18

20) 628,597 1525, it becomes

600000 - 2000 = 300

Answer:

a.  [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ \frac{9}{x^2}} \, dx[/tex]

b.  [tex]\displaystyle AL = 4.10322[/tex]

General Formulas and Concepts:

Calculus

Differentiation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Integration

Integration Property [Multiplied Constant]:                                                         [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:                                                       [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

U-Substitution

Arc Length Formula [Rectangular]:                                                                       [tex]\displaystyle AL = \int\limits^b_a {\sqrt{1+ [f'(x)]^2}} \, dx[/tex]

Step-by-step explanation:

Step 1: Define

Identify

y = 3ln(x)

Interval [2, 5]

Step 2: Find Arc Length

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Applications of Integration

Book: College Calculus 10e

The arc length of the curve [tex]\( y = 3 \ln x \)[/tex] over the interval [2, 5] is approximately 12.092 units, obtained through numerical methods such as Simpson's rule or a CAS like Wolfram Alpha.

a. Writing and Simplifying the Integral:

The arc length L of the curve y = f(x) over the interval [a, b] is given by the following definite integral:

L = [tex]\int_a^b \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} dx[/tex]

First, find the derivative dy/dx of the given function:

y' = [tex]\dfrac{d}{dx} (3 \ln x) = \dfrac{3}{x}[/tex]

Now, plug f(x) = 3 \ln x and [tex]f'(x) = \dfrac{3}{x}[/tex] into the arc length formula for the interval [2, 5]:

L = [tex]\int_2^5 \sqrt{1 + \left( \dfrac{3}{x} \right)^2} dx[/tex]

L = [tex]\int_2^5 \sqrt{1 + \dfrac{9}{x^2}} dx[/tex]

b. Evaluating or Approximating the Integral:

This integral cannot be solved analytically using standard techniques. Therefore, we can use numerical methods like Simpson's rule or a CAS (Computer Algebra System) to approximate the value.

Using Simpson's rule with 10 subintervals, we get an approximate arc length of:

L ≈ 12.092

Alternatively, using a CAS like Wolfram Alpha, we can obtain a more precise numerical value:

L ≈ 12.091971077

Therefore, the arc length of the curve is approximately 12.092 units.

Answer:

a) [0.278; 0.342]

b) No, the population the sample comes from may not be representative of college students, violating the Fundamental Rule for Using Data For Inference.

Step-by-step explanation:

Hello!

a)

The study variable is

X: Number of American adults that said that they do not get enough sleep each night.

The sample is n= 935 American Adults

Sample proportion 'p= 0.31

The margin of error of the 95% CI d= 0.032

The formula for a CI for the population proportion is:

['p ± [tex]Z_{1-\alpha /2}[/tex]*[tex]\sqrt{\frac{'p(1-'p)}{n} }[/tex]]

The basic structure of the CI formula is "Point estimator" ± "Margin of error"

So with the given data, you can calculate the CI.

[0.31 ± 0.032]

[0.278; 0.342]

So with a confidence level of 95%, you'd expect that the interval [0.278; 0.342] will contain the population proportion of the American adults that don't get enough sleep at night.

b)

The answer is no. The sample is of "American adults" there is no information on their education level, therefore it is possible that all of them have a college education or that none of the have. Since no further information about their education then you cannot be certain that this sample will be representative of the American College Students.

Then the correct option is B

I hope you have a SUPER day!

Answer:

a) Null hypothesis:[tex]\mu \geq 10[/tex]  

Alternative hypothesis:[tex]\mu < 10[/tex]  

b) [tex]p_v =P(t_{17}<-2.3)=0.017[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

c) [tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]

And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.

d) [tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

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

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

[tex]n=18[/tex] sample size  

[tex]\mu_o =10[/tex] represent the value that we want to test

[tex]\alpha[/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 true mean is at least 10 hours, the system of hypothesis would be:  

Part a

Null hypothesis:[tex]\mu \geq 10[/tex]  

Alternative hypothesis:[tex]\mu < 10[/tex]  

If we analyze the size for the sample is < 30 and 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)  

Part b

For this case we have t=-2.3 , [tex]\alpha=0.05[/tex]

First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]

Now since we are conducting a left tailed test the p value is given by:

[tex]p_v =P(t_{17}<-2.3)=0.017[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

Part c

For this case we have t=-1.8 , [tex]\alpha=0.01[/tex]

First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]

Now since we are conducting a left tailed test the p value is given by:

[tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]

And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.

Part d

For this case we have t=-3.6 , [tex]\alpha=0.05[/tex]

First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]

Now since we are conducting a left tailed test the p value is given by:

[tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

The student is learning about conducting a one-sample t-test in a hypothesis testing problem related to the lifetime of a pen. The null and alternative hypothesis are defined, and conclusions are drawn based on various test statistics and significance levels.

This situation relates to hypothesis testing in statistics, in particular the usage of a one-sample t-test.

The null hypothesis (H0) and alternative hypothesis (Ha) to be tested would be:

The t value represents the test statistic for conducting the hypothesis test, and the 'α' and 'u' are the significance levels.

With t = -23 and α = 0.05,  this falls within the rejection region, as the t value is less than the critical value for a one-tailed test at significance level α = 0.05. Hence, we reject the null hypothesis. There is evidence to suggest that the pen's lifetime is less than 10 hours.

Similarly, with t = -1.8 and u = 0.01, the null hypothesis would be retained, as this value is within the acceptance region.The conclusion would be that there is not sufficient evidence to suggest that the pen's lifetime is less than 10 hours.

Finally,  with t = -3.6, the null hypothesis is rejected, suggesting that the pen's lifetime is less than 10 hours. However, this is only applicable if the significance level is above the calculated p-value for t = -3.6.

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Answer:

We accept  H₀   with the information we have, we can say level of ozone is under the major limit

Step-by-step explanation:

Normal Distribution

population mean  =   μ₀   =  7.5   ppm

Sample size     n   =  16      df  =  n  -  1  df  =  15

Sample mean      =    μ   =  7.8  ppm

Sample standard deviation   =  s  =  0.8

We want to find out if ozono level, is above normal level that is bigger than 7.5

1.- Hypothesis Test

null hypothesis                          H₀         μ₀   =  7.5

alternative hypothesis             Hₐ          μ₀  >  7.5  

2.-Significance level    α  =  0.01   we will develop one tail-test (right)

then   for   df  =  15    and    α =  0,01   from t -student table we get

t(c)  = 2.624

3.-Compute  t(s)

t(s)  =  (  μ -  μ₀ ) / s /√n            ⇒   t(s)  = ( 7.8  -  7.5 )*4/0.8

t(s)  = 0.3*4/0.8

t(s)  = 1.5

4.-Compare   t(s)   and  t(c)

t(s)  <  t(c)       1.5  <  2.64

Then  t(s) is inside the acceptance region. We accept  H₀

Answer:

Step-by-step explanation:

To answer this question, First, we define a direction in space, which defines

the North Pole of each planet.

We also assume that the direction is not at 90° perpendicular to the axis of any two planet.

Assume we now define an order on the set of planet by saying that planet A ≥ B, when removing all the other planet from space, the north pole B, is visible from A.

If we refer to the direction of the planet in the diagram, we discover that,

1, A ≥ A

2, If A ≥ B and B≥ A,

Then A = B

3, If A ≥ B, and B ≥ C , then A ≥ C

4, Either A ≥ B or B ≥ A

It should also be noted that the array of order above has a unique maximal element  (M). This is the only  planet whose North Pole is not visible from another.

If we now consider a sphere of the same radius as the planets. Remove

from it all North Poles defined by directions that are perpendicular

to the axis of two of the planets. This is a set of area zero.

For every other point on this sphere, there exists a direction in

space that makes it the North Pole, and for that direction, there

exists a unique North Pole on one of the planets which is not visible

from the others. Hence the total area of invisible points is equal

to the area of this sphere, which in turn is the area of one of the  planets.

To test the hypothesis that the average debt load of graduating students with a Bachelor's degree is equal to $17,000, a hypothesis test needs to be conducted using the given sample data. The p-value needs to be calculated and compared to the significance level α to determine the conclusion. Without the p-value, we cannot make a definitive conclusion.

To test the hypothesis that the average debt load of graduating students with a Bachelor's degree is equal to $17,000, the Department of Education would conduct a hypothesis test using the given sample data. The null hypothesis, denoted as H0, states that the average debt load is equal to $17,000, while the alternative hypothesis, denoted as Ha, states that the average debt load is not equal to $17,000.

Based on the given information, the sample mean debt load is $18,200, the population standard deviation is $4,200, and the sample size is 34. Using these values, we can calculate the p-value, which represents the probability of obtaining a sample mean as extreme as $18,200 or more extreme, assuming the null hypothesis is true. The p-value is the probability of observing a sample mean of $18,200 or more extreme, given that the average debt load is $17,000.

To determine the conclusion of the hypothesis test, we compare the p-value to the significance level α. In this case, the given α is 0.05. If the p-value is less than α, we reject the null hypothesis and conclude that the average debt load is not equal to $17,000. If the p-value is greater than or equal to α, we fail to reject the null hypothesis and cannot conclude that the average debt load is not equal to $17,000.

In this scenario, the p-value is not given, so we cannot make a definitive conclusion. We would need to calculate the p-value based on the given information to determine which statement is true.

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The correct option is A.  [tex]\text{Because the p-value is greater than}[/tex] [tex]\alpha,[/tex] [tex]\text{ we fail to reject the null hypothesis and conclude that the average debt load is equal to }[/tex] [tex]\$17,000.[/tex]

To determine the correct statement based on the hypothesis test conducted:

Given data:

Population standard deviation[tex](\( \sigma \)): \$4,200[/tex]

Sample size [tex](\( n \)): 34[/tex]

Sample mean [tex](\( \bar{x} \)): \$18,200[/tex]

Hypothesized population mean [tex](\( \mu_0 \))[/tex]: [tex]\$17,000.[/tex]

Significance level [tex](\alpha): 0.05[/tex]

Hypotheses

We are testing whether the average debt load [tex](\( \mu \))[/tex] of graduating students with a Bachelor's degree is equal to [tex]\$17,000.[/tex]

Null hypothesis [tex](\( H_0 \)): \( \mu = 17,000 \)[/tex]

Alternative hypothesis [tex](\( H_1 \)): \( \mu \neq 17,000 \)[/tex]

This is a two-tailed test because [tex]\( H_1 \)[/tex] states that [tex]\( \mu \)[/tex] is not equal to [tex]\$17,000.[/tex]

Test Statistic

Since the population standard deviation [tex](\( \sigma \))[/tex] is known and the sample size [tex](\( n \))[/tex] is greater than [tex]30[/tex], we use a z-test.

The test statistic [tex]\( z \)[/tex] is calculated as:

[tex]\[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \][/tex]

Calculate [tex]\( z \)[/tex]

[tex]\[ z = \frac{18,200 - 17,000}{\frac{4,200}{\sqrt{34}}} \][/tex]

[tex]\[ z = \frac{1,200}{\frac{4,200}{5.83}} \][/tex]

[tex]\[ z = \frac{1,200 \times 5.83}{4,200} \][/tex]

[tex]\[ z = 1.66 \][/tex]

P-Value Calculation

The p-value is the probability of observing a sample mean at least as extreme as [tex]18,200[/tex] if the null hypothesis [tex](\( \mu = 17,000 \))[/tex] is true. Since this is a two-tailed test, we consider both tails of the normal distribution.

From the standard normal distribution table:

The area to the right of [tex]\( z = 1.66 \)[/tex] (since [tex]\( z \)[/tex] is positive) is approximately [tex]0.0485.[/tex]

Therefore, the p-value for the two-tailed test is approximately [tex]\( 2 \times 0.0485 = 0.097 \).[/tex]

Conclusion

Compare the p-value [tex](0.097)[/tex] with the significance level [tex]\alpha = 0.05)\( \text{p-value} (0.097) > \alpha (0.05) \)[/tex]

Since the p-value is greater than the significance level ([tex]\alpha[/tex]), we fail to reject the null hypothesis [tex]\( H_0 \)[/tex]. This means we do not have enough evidence to conclude that the average debt load is different from [tex]\$17,000.[/tex]

The 1.125 in the exponential function V(t) = 30,000(1.125)^t represents the growth factor of the investment, indicating an annual growth of 12.5%.

In the exponential function V(t) = 30,000(1.125)^t, the 1.125 represents the growth factor of the investment. This means each year, the value of the investment grows by 12.5%. It's the rate at which the initial value of the investment, 30,000 in this case, increases on an annual basis. Therefore, the exponential function shows us how the value of the investment changes over time with this constant growth factored in.

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Answer:

Slope: 3

Y intercept: y = 75

Step-by-step explanation:

The slope of a linear function is the coefficient in front of x. In this case, 3 is being multiplied by x, indicating the slope is 3.

The y-intercept is the y value where the graph of the function crosses the y-axis. In other words, it's the y value when x = 0. To find the y-intercept, simply substitute 0 for x:

[tex]y = 3(0) + 75 \\ y = 75[/tex]

We find that the y-intercept is y = 75.

In the linear function y=3x+75, the slope 3 represents the rate of cost per mile to tow a car ($3/mile), and the y-intercept 75 is the basic fee for the towing service. So, the total cost is a fixed fee of $75 plus $3 per mile towed.

In the linear function y = 3x + 75, the slope, represented by '3', is the cost in dollars per mile to tow a car. This means for each mile the car is towed, the cost increases by $3.

The y-intercept, represented by '75', is the basic fee for the service, or the cost in dollars that one must pay even if the car isn't towed any miles.

So, to interpret the equation function: the total cost (y) of having a car towed is composed of a fixed fee of $75 plus an additional $3 for each mile (x) the car is towed.

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Answer:

On demand

Step-by-step explanation:

On demand (fund Value Streams, not projects).

Lean Budgets is a collection of methods that significantly reduce overhead by financing and encouraging Value Streams instead of projects while preserving financial and user-fit governance. This is done by comprehensive work program assessment, active management of Epic finances, and complex budget changes.

Hence, the recommended frequency for updating lean budget distribution on demand that is fund value streams.

Lean budget distribution should be updated at least once per quarter, or as often as required by the specific circumstances of your organization. The goal of a lean budget is to optimize resource use and reduce waste to deliver more value to customers.

The recommended frequency for updating lean budget distribution is not fixed and largely depends on the specific needs, context, and circumstances of your organization. However, a common practice is to review and adjust the lean budget allocation at least once per quarter. This allows you to take into account any changes in strategy, project progress, and business environment. It's important to highlight that these revisions should be strategic and informed by data and not just random shifts in resource allocation. Remember, the main goal of a lean budget is to minimize waste and optimize resource utilization to bring more value to customers.

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Answer:

The correct option is B) 0.0228

Step-by-step explanation:

Consider the provided information.

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.

Thus, the value of n = 100, [tex]\bar x=3.1[/tex] and σ = 0.5

We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

The null and alternative hypotheses are,

[tex]H_0:\mu\leq3[/tex] and [tex]H_a:\mu>3[/tex]

Compute the test statistic [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z=\frac{3.1-3}{\frac{0.5}{\sqrt{100}}}[/tex]

[tex]z=\frac{0.1}{0.05}[/tex]

[tex]z=2[/tex]

By using the table.

P value = P(Z>2) = 0.0228

Hence, the correct option is B) 0.0228

Answer:

The standard error increases and the width of the interval increases

Step-by-step explanation:

The variance is equal to the square of the standard deviation. An increase in variance will increase the standard deviation.

The width of the confidence interval can be defined as:

[tex]W=UL-LL=2z\sigma/\sqrt{n}[/tex]

The width is proportional to the standard deviation, so when the variance increases, the standard deviation and also the width of the interval increases.

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:

H0: [tex]\sigma \geq 4.1[/tex]

H1: [tex]\sigma <4.1[/tex]

[tex]\Chi^2_{crit} =10.851[/tex]

[tex] t=(21-1) [\frac{2.5}{4.1}]^2 =7.436[/tex]

[tex]p_v = P(\Chi^2_{20}<7.436)=0.0050[/tex]

True, since our p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. And we can conclude that the population standard dviation is less than 4.1 (Or the variance is less than 16.81)

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

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

n = 21 sample size

[tex]s^2 =6.25[/tex] represent the sample variance

[tex]s=2.5[/tex] represent the sample deviation

[tex]\sigma^2_0 =16.81[/tex] represent the target variance

[tex]\sigma_o =4.1[/tex] the value that we want to test

[tex]p_v [/tex] represent the p value for the test

t represent the statistic

[tex]\alpha=0.05[/tex] significance level

State the null and alternative hypothesis

On this case we want to check if the population variance is lower (or if the deviation is lower than 4.1) than 16.81, so the system of hypothesis are:

H0: [tex]\sigma \geq 4.1[/tex]

H1: [tex]\sigma <4.1[/tex]

In order to check the hypothesis we need to calculate th statistic given by the following formula:

[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a left tailed test the critical zone it's on the left tail of the distribution. On this case we need a quantile on the chi square distribution with 20 degrees of freedom that accumulates 0.05 of the area on the left tail and 0.95 on the right tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.05,20)". And our critical value would be [tex]\Chi^2_{crit} =10.851[/tex]

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

[tex] t=(21-1) [\frac{2.5}{4.1}]^2 =7.436[/tex]

What is the approximate p-value of the test?

For this case since we have a left tailed test the p value is given by:

[tex]p_v = P(\Chi^2_{20}<7.436)=0.0050[/tex]

Based on your answer for the p value, answer True or False as to whether the null hypothesis will be rejected.

True, since our p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. And we can conclude that the population standard dviation is less than 4.1 (Or the variance is less than 16.81)

Answer:

92.79% probability that the mean clock life would be greater than 15.1 years.

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].

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 16, \sigma = 4, n = 42, s = \frac{4}{\sqrt{42}} = 0.6172[/tex]

What is the probability that the mean clock life would be greater than 15.1 years?

This probability is 1 subtracted by the pvalue of Z when [tex]X = 15.1[/tex]. So:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{15.1 - 16}{0.6172}[/tex]

[tex]Z = -1.46[/tex]

[tex]Z = -1.46[/tex] has a pvalue of 0.0721.

So there is a 1-0.0721 = 0.9279 = 92.79% probability that the mean clock life would be greater than 15.1 years.

Final answer:

Using the Central Limit Theorem and the given standard deviation and mean, the standard error was calculated, and then the z-score was determined. After finding the z-score, the area to the right of it on the standard normal distribution curve gives the probability that the mean clock life is greater than 15.1 years, which is 0.9274.

Explanation:

To find the probability that the mean clock life of a sample of 42 wall clocks is greater than 15.1 years, we can use the Central Limit Theorem which tells us that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough. Since we are dealing with a sample of 42, which is generally considered sufficiently large, the sampling distribution of the sample mean is normally distributed. Given the population mean (μ) is 16 years, the standard deviation (σ) is 4 years, the sample size (n) is 42, and we want to find the probability of the sample mean (μ_x) > 15.1 years, we first need to find the standard error (SE) which is σ/sqrt(n). Then we can find the z-score which is (15.1 - μ) / SE. Lastly, we use the z-score to find the probability from z-tables or a calculator with normal distribution functions.

The standard error (SE) is:
SE = 4 / sqrt(42) ≈ 0.6172
The z-score (z) is:
z = (15.1 - 16) / 0.6172 ≈ -1.4573

To find the probability of the sample mean being greater than 15.1 years, we want the area to the right of the z-score on the standard normal distribution curve. Looking up the z-score of -1.4573 on the z-table or using a normal distribution calculator gives us the probability to the left of the z-score. We subtract this value from 1 to find the probability to the right:
P(X > 15.1) = 1 - P(Z < -1.4573) ≈ 1 - 0.0726 = 0.9274

Therefore, the probability that the mean clock life is greater than 15.1 years is 0.9274 when rounded to four decimal places.

Answer:

Step-by-step explanation:

Given

mean [tex]\mu =20\ ounces[/tex]

standard deviation [tex]\sigma =1\ ounce[/tex]

The no of sample boxes weigh Every morning is 25

Average weight is 1 % more than average

i.e. [tex]20+20\times 0.1=20.2 ounce[/tex]

The company re-calibrates the machine

[tex]P\left ( \bar{x}> 20.2\right )=P\left ( z-\frac{20.2-20}{\frac{1}{\sqrt{25}}}\right )[/tex]

[tex]=P\left ( z> 1\right )=1-P\left ( z\leq 1\right )[/tex]

[tex]=1-0.8413[/tex]

[tex]=0.1587[/tex]

Therefore the Probability that the average weight of box is more than 20.2 ounce is 0.1587

No of days the machine is expected to re-calibrate is [tex]100\times 0.1587=16\ days[/tex]

Answer:

The equation of the line using function notation is

[tex]y=f(x)=-\frac{1}{4}x[/tex]

Step-by-step explanation:

Given points are (-20,5) and (-36, 9)

Now to find the equation of the line passes through these points

Let [tex](x_{1},y_{1})[/tex] and  [tex](x_{2},y_{2})[/tex] be the two given points  (-20,5) and (-36, 9) respectively.

To find slope

[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]m=\frac{9-5}{-36-(-20)}[/tex]

[tex]m=\frac{4}{-36+20}[/tex]

[tex]m=\frac{4}{-16}[/tex]

[tex]m=-\frac{1}{4}[/tex]

Therefore  [tex]m=-\frac{1}{4}[/tex]

The equation of the line is of thr form y=mx+c

The point (-20,5) passes through the above line and  [tex]m=-\frac{1}{4}[/tex]

[tex]5=-\frac{1}{4}(-20)+c[/tex]

[tex]5=5+c[/tex]

[tex]c=0[/tex]

[tex]y=-\frac{1}{4}x+0[/tex]

Therefore  [tex]y=-\frac{1}{4}x[/tex]

Therefore the equation of the line using function notation is

[tex]y=f(x)=-\frac{1}{4}x[/tex]

Answer:

The tangent line equations are:

- The one that has the smaller slope [tex]\bold{y =3}[/tex]

- The one that has the larger slope [tex]\bold{y = \cfrac 14 x - \cfrac{15}{4}}[/tex]

Step-by-step explanation:

In order to find the tangent lines we need to find first the first derivative since the first derivative evaluated at a point give us the slope of the tangent line.

Working with implicit differentiation.

In order to find the derivative we can work with implicit differentiation and we get

[tex]2x+18y \cfrac{dy}{dx}= 0[/tex]

Notice that only when we are finding the derivative of an expression that has y we multiply by dy/dx due chain rule.

Solving for the first derivative we get

[tex]18y\cfrac{dy}{dx}=-2x[/tex]

[tex]\cfrac{dy}{dx}=-\cfrac{x}{9y}[/tex]

So the equation of the first derivative at the point (x,y) give us the first equation for the slope.

[tex]m=-\cfrac{x}{9y}[/tex]

Slope of a line using definition.

We can also find the slope of the line that will pass the point (x,y) trough  the point (27,3) using the definition:

[tex]m = \cfrac{y_2-y_1}{x_2-x_1}[/tex]

Replacing the point we get

[tex]m=\cfrac{y-3}{x-27}[/tex]

That is another equation for the slope.

Finding the value of the slopes.

We can now set both slope equations equal to each other to find the point (x,y) where they intersect and the value of the slopes.

[tex]-\cfrac{x}{9y}= \cfrac{y-3}{x-27}[/tex]

We can work with cross multiplication to get

[tex]-x(x-27)=9y(y-3)[/tex]

And we can distribute and simplify

[tex]-x^2+27x=9y^2-27y[/tex]

Moving everything to the right side

[tex]0=x^2-27x+9y^2-27y\\\\0=x^2+9y^2-27x-27y[/tex]

At this point we can use the ellipse equation so we can replace [tex]x^2+9y^2[/tex] with 81, that will give us

[tex]0=81-27x-27y[/tex]

Then we can divide by 27 and solve for y.

[tex]0=3-x-y\\y= 3-x[/tex]

At this point we can replace on the ellipse equation.

[tex]x^2+9(3-x)^2=81[/tex]

And we can distribute and simplify.

[tex]x^2+9(9-6x+x^2)=81\\x^2+81-54x+9x^2=81\\x^2-54x+9x^2=0\\10x^2-54x=0\\[/tex]

So then we can factor and solve for x.

[tex]2x(5x-27)=0[/tex]

Setting each factor and solve for x we get

[tex]x= 0, \cfrac{27}{5}[/tex]

At x = 0, we can find the y value.

[tex]y= 3-x\\y= 3-0\\y= 3[/tex]

And the slope

[tex]m = -\cfrac{x}{9y}\\m = -\cfrac{0}{9y}\\m = 0[/tex]

So the line equation is

[tex]y-3=0(x-0)\\y=3[/tex]

For the second point x = 27/5 we have:

[tex]y= 3-\cfrac{27}{5}\\y=-\cfrac{12}{5}[/tex]

So slope is:

[tex]m = -\cfrac{\cfrac{27}{5}}{9\left(-\cfrac{12}{5}\right)}[/tex]

[tex]m = \cfrac 14[/tex]

So the line equation is

[tex]y-\left(-\cfrac{12}{5}\right)= \cfrac 14 \left(x -\cfrac{27}{5}\right)\\y = \cfrac 14 x - \cfrac{15}{4}[/tex]

Thus the equations of tangent lines are:

[tex]\bold{y =3}[/tex]

and

[tex]\bold{y = \cfrac 14 x - \cfrac{15}{4}}[/tex]

To find the equations of the tangent lines to the ellipse that passes through a specific point, differentiate the equation of the ellipse to get a formula for the slope at any point on the ellipse. Set this equal to the slope of the line from the point on the ellipse to the specific point, and solve. The resultant solutions give the points of tangency, and the lines through these points and the specific point are the desired tangent lines.

The subject of the question is about finding the equations of the tangent lines to the ellipse x^2 + 9y^2 = 81 that pass through a specific point (27, 3). We first differentiate the equation of the ellipse to get a general formula for the slope of the tangent line at any point on the ellipse. The differentiation of x^2 + 9y^2 = 81 with respect to x gives 2x + 18yy' = 0, so y' = -x/ (9y). Now, we put the slopes y' equal to the slopes of line from (x, y) to (27, 3), and solve the resulting system of equations. The solutions give the points of tangency, and the lines through these points and (27, 3) are the desired tangent lines. The exact computations involve solving a quadratic equation and it'll give two slopes for the tangent line that passes through the point (27,3).

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Answer:

i don't know sorry hope this helps

Step-by-step explanation:

Answer:

Part (a): 25 people became ill with the flu when the epidemic​ began.

Part (b): 1373 people were ill by the end of the fourth​ week.

Part (c): The limiting size of the population that becomes​ ill is 107,000.

Step-by-step explanation:

Consider the provided logistic growth function

[tex]f (t )=\frac{ 107,000}{1 +4200e^{-t}}[/tex]

Part (a) How many people became ill with the flu when the epidemic​ began?

Substitute t = 0 in above function.

[tex]f (t )=\frac{ 107,000}{1 +4200e^{-0}}[/tex]

[tex]f (t )=\frac{ 107,000}{4201}[/tex]

[tex]f (t )=25.47[/tex]

Hence, 25 people became ill with the flu when the epidemic​ began.

Part (b) How many people were ill by the end of the fourth​ week?

Substitute t = 4 in above function.

[tex]f (t )=\frac{ 107,000}{1 +4200e^{-4}}[/tex]

[tex]f (t )=1373.10[/tex]

Hence, 1373 people were ill by the end of the fourth​ week.

Part (c) What is the limiting size of the population that becomes​ ill?

Substitute t = ∞ in above function.

[tex]f (t )=\frac{ 107,000}{1 +4200e^{-\infty}}[/tex]

[tex]f (t )=\frac{ 107,000}{1 +0}[/tex]

[tex]f (t )=107,000[/tex]

Hence, the limiting size of the population that becomes​ ill is 107,000.

Using the logistic growth function, the number of sick people at the start of the epidemic and at the end of the fourth week can be found by substituting the appropriate values for time into the function. The limiting size of the potentially ill population is the maximum value of the function, which is 107,000.

The logistic growth function provided describes the number of people, represented by f(t), who have become ill with influenza, t weeks after its initial outbreak. The function is given as f(t) = 107000/(1 + 4200 * e-t).

a. At the beginning of the epidemic (i.e., when t=0), the number of ill people, f(0), can be calculated by substituting 0 for t in the equation, resulting in 107,000/(1+4200) people.

b. By the end of the fourth week (i.e., when t=4), the number of ill people, f(4), can be calculated by substituting 4 for t in the equation.

c. The limiting size of the population that becomes ill is represented by the maximum value the growth function can take, which in this case is 107,000 (as seen by the numerator of the function). This means that illness will eventually spread to reach a potential maximum of 107,000 people in this specific community.

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A financial analyst wanted to estimate the mean annual return on mutual funds. A random sample of 60 funds' returns shows an average rate of 12%. If the population standard deviation is assumed to be 4%, the 95% confidence interval estimate for the annual return on all mutual funds is

A. 0.037773 to 0.202227

B. 3.7773% to 20.2227%

C. 59.98786% to 61.01214%

D. 51.7773% to 68.2227%

E. 10.988% to 13.012%

Answer: E. 10.988% to 13.012%

Step-by-step explanation:

Given;

Mean x= 12%

Standard deviation r = 4%

Number of samples tested n = 60

Confidence interval is 95%

Z' = t(0.025)= 1.96

Confidence interval = x +/- Z'(r/√n)

= 12% +/- 1.96(4%/√60)

= 12% +/- 0.01214%

Confidence interval= (10.988% to 13.012%)

Answer:

[tex]\frac{dV}{dt}=525 cm^{3}/min[/tex]

Step-by-step explanation:

The volume of a sphere is:

[tex]V=\frac{4}{3}\pi r^{3}[/tex] (1)

We know that:

If we take the derivative of (1) with respect of time t, we ca n find the rate of increase of the volume.

[tex]\frac{dV}{dt}=\frac{4}{3}\pi 3r^{2}\frac{dr}{dt}=4\pi r^{2}\frac{dr}{dt}[/tex] (2)

We also know that the volume is 40 cm³, then using the (1) we can get the radius at this value.

Solving (1) for r, we have:

[tex]r=\left(\frac{3\cdot V}{4\pi}\right)^{1/3}=\left(\frac{3\cdot 400}{4\pi}\right)^{1/3}=4.57 cm[/tex]

Finally dV/dt will be:

[tex]\frac{dV}{dt}=4\pi (4.57)^{2}\cdot 2=525 cm^{3}/min[/tex]

I hope it helps you!

Answer:

See the explanation

Step-by-step explanation:

(a)  

H0: Consultant with more experience has the higher population mean service rating.

H1: Consultant with more experience doesn't have the higher population mean service rating.

(b)  

t = 1.9923 (see the attached image)

(c)

The degrees of freedom for the test statistic,

df = 16

The P-value of the one tailed t- test with 16 degrees of freedom is,

P−value = tdist(X,df,tails)

P-value = tdist(1.9923,16,1)

P-value = 0.032

(d)​

Since, P-value 0.032 is less than the significance level 0.05, there is an enough evidence to reject the null hypothesis.

Hence, there is a sufficient evidence to conclude that Consultant with more experience doesn't have the higher population mean service rating.

Hope this helps!

Answer:

[tex]E(X)= n\frac{M}{N}=10 \frac{5}{25}=2[/tex]

[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1[/tex]

Step-by-step explanation:

Previous concepts

The hypergeometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

[tex]P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}[/tex]  

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes  

The expected value and variance for this distribution are given by:

[tex]E(X)= n\frac{M}{N}[/tex]

[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}[/tex]

What is the distribution of X?  

For this case the random variable X follows a hypergometric distribution.

Compute the values for E(X) and Var(X)  

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

[tex]E(X)= n\frac{M}{N}=10 \frac{5}{25}=2[/tex]

[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1[/tex]

What is the probability that none of the animals in the second sample are tagged?  

So for this case we want this probability:

[tex]P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565[/tex]  

What is the probability that all of the animals in the second sample are tagged?  

So for this case we want this probability:

[tex]P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474[/tex]  

Answer:

The equation of the line with slope is 3  and point (2,17) is [tex]y=3x+11[/tex]

Step-by-step explanation:

Given slope is 3 ie., m=3 and point (2,17)

To find the equation of the line with slope and a point:

Using slope point formula

[tex]y-y_{1}=m(x-x_{1})[/tex]

Let the point (2,17) be [tex](x_{1},y_{1})[/tex]

Substitute the values of m=3 and point (2,17)

[tex]y-y_{1}=m(x-x_{1})[/tex]

[tex]y-17=3(x-2)[/tex]

[tex]y-17=3x-6[/tex]

[tex]y=3x-6+17[/tex]

[tex]y=3x+11[/tex]

Therefore the equation of the line is [tex]y=3x+11[/tex]

Answer:

A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

Test statistic(Z) = 2.31

P-value = 0.0104

Step-by-step explanation:

Step 1

Null hypothesis: The average damaging temperature of nine iPods is 5°F

Alternate hypothesis: The average damaging temperature of nine iPods differs from 5°F

Step 2

Mean=5°F, Sd=3, df=n-1=9-1=8

The t-value corresponding to 8 degrees of freedom and 95% confidence level (5% significance level) is 2.306

Confidence Interval(CI) = (mean + or - t×sd/√n)

CI = (5 + 2.306×3/√8) = 5 + 6.918/2.828= 5+2.45=7.45°F

CI = (5 - 2.306×3/√8) = 5-2.45 = 2.55°F

Z = (sample mean - population mean)/(sd÷√n) = (5-2.55)/(3÷√8) = 2.45/1.061 = 2.31

Step 3

Using the standard distribution table, the cumulative area to the left of Z = 2.31 is 0.9896

P-value = 1 - 0.9896 = 0.0104

Step 4

Conclusion: A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F