There are 39 employees in a particular division of a company. Their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000. The largest number on the list is $100,000. By accident, this number is changed to $1,000,000.
a)What is the value of the mean after the change? Write your answer in units of $1000.b)What is the value of the median after the change? Write your answer in units of $1000.c)What is the value of the standard deviation after the change? Write your answer in units of $1000.

Answer:

  31.747 GW

Step-by-step explanation:

The total cell capacity (C) manufactured in the time period is the integral of the rate of manufacture. So, we have ...

  [tex]C=\displaystyle\int\limits^3_0 {3.7e^{0.61t}} \, dt=\frac{3.7}{0.61}(e^{0.61\cdot 3}-1)\approx 31.747 \quad\text{GW}[/tex]

About 31.747 GW of generating capacity was manufactured in that time interval.

Answer:

80 ft x 40 ft

Step-by-step explanation:

Let 'L' be the length of the longer side and 'W' be the length of the shorter side (or the width).

The equations that compose the linear system are:

[tex]L+2W=160\\L=2W[/tex]

Solving the system:

[tex]L+L=160\\L=80\\W=\frac{L}{2}\\W=40[/tex]

The garden is a rectangle with dimensions 80 ft x 40 ft.

The dimensions of the garden are approximately 53.33 feet in width and 106.66 feet in length according to the given linear equations and conditions.

Given that the fencing of 160 feet encloses a rectangular garden on three sides and one side of the rectangle is parallel with the barn, we can derive the linear equations based on these parameters. Let's denote the width of the rectangle as x and the length of the rectangle as 2x, as stated the length is twice the width.

The perimeter of this garden will equal the amount of fencing, which is 160 feet. Because only three sides of the rectangle are enclosed by the fence, the equation for the perimeter becomes 2x + x = 160. Simplifying this, we get 3x = 160. Solving for x, we divide both sides by 3, hence, x = 53.33 ft approximately. The width of the garden is thus around 53.33 feet and the length (being twice the width) would be 2x = 2*53.33 = 106.66 ft.

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Answer:

a) Null hypothesis: [tex]p \geq 0.368[/tex]

Alternative hypothesis: [tex]p<0.368[/tex]

b) A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually is not true.

c) A type of error II for this case would be FAIL to reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually the alternative hypothesis is the true.

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".  

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

Type II error, also known as a "false negative" is the error of not rejecting a null  hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.

Part a

On this case we want to test if the proportion of students who enroll in her institution have a lower completion rate (0.398), so the system of hypothesis would be:

Null hypothesis: [tex]p \geq 0.368[/tex]

Alternative hypothesis: [tex]p<0.368[/tex]

Part b

A type of error I for this case would be reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually is not true.

Part c

A type of error II for this case would be FAIL to reject the null hypothesis that the population proportion is greater or equal than 0.368 when actually the alternative hypothesis is the true.

The null hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is the same as the national proportion. Making a Type I error would mean wrongly rejecting the null hypothesis, while making a Type II error would mean failing to reject the null hypothesis when it is actually false.

(a) The null hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is the same as the national proportion of 0.398. The alternative hypothesis would state that the proportion of students who earn a bachelor's degree within six years at the college is lower than 0.398.

(b) Making a Type I error in this context would mean rejecting the null hypothesis when it is actually true. This would imply that the college has a lower completion rate when in fact it does not.

(c) Making a Type II error in this context would mean failing to reject the null hypothesis when it is actually false. This would imply that the college has the same completion rate as the national proportion when it actually has a lower completion rate.

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Answer: [tex]0.12< \sigma<0.42[/tex]

Step-by-step explanation:

Confidence interval for standard deviation is given by :-

[tex]\sqrt{\dfrac{s^2(n-1)}{\chi^2_{\alpha/2}}}< \sigma<\sqrt{\dfrac{s^2(n-1)}{\chi^2_{1-\alpha/2}}}[/tex]

Given : Confidence level : [tex]1-\alpha=0.95[/tex]

⇒[tex]\alpha=0.05[/tex]

Sample size : n= 7

Degree of freedom = 6    (df= n-1)

sample standard deviation : s= 0.19

Critical values by using chi-square distribution table :

[tex]\chi^2_{\alpha/2, df}}=\chi^2_{0.025, 6}}=14.4494\\\\\chi^2_{1-\alpha/2, df}}=\chi^2_{0.975, 6}}=1.2373[/tex]

Confidence interval for standard deviation of the weights of the packages prepared by the machine is given by :-

[tex]\sqrt{\dfrac{ 0.19^2(6)}{14.4494}}< \sigma<\sqrt{\dfrac{ 0.19^2(6)}{1.2373}}[/tex]

[tex]\Rightarrow0.12243< \sigma<0.418400[/tex]

[tex]\approx0.12< \sigma<0.42[/tex]

Hence, the 95% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine.  :

[tex]0.12< \sigma<0.42[/tex]

Answer:

  b. y-y1 = m(x-x1)

Step-by-step explanation:

It's a matter of definition. There are perhaps a dozen useful forms of equations for a line. Each has its own name (and use). Here are some of them.

Adding y1 to the point-slope form puts it in an alternate form that is useful for getting to slope-intercept form faster: y = m(x -x1) +y1. I use this when asked to write the equation of a line with given slope through a point, with the result in slope-intercept form.

The equation of the line, in point-slope form, is given by:

[tex]y - y_1 = m(x - x_1)[/tex]

Option b.

-----------------------------------------

The equation of a line, in point-slope form, is given by:

[tex]y - y_1 = m(x - x_1)[/tex]

In which

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Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

[tex]T(t)=T_a+(T_0-T_a)e^{-kt}[/tex]

where [tex]T_a[/tex] is the ambient temperature, [tex]T_0[/tex] is the initial temperature, [tex]t[/tex] is the time and [tex]k[/tex] is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say [tex]T_0[/tex]

We know that the ambient temperature is [tex]T_a=65[/tex], so

[tex]T(t)=65+(T_0-65)e^{-kt}[/tex]

We also know that when [tex]t=5 \:min[/tex] the temperature is [tex]T(5)=35[/tex] and when [tex]t=10 \:min[/tex] the temperature is [tex]T(10)=50[/tex] which gives:

[tex]T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}[/tex]

[tex]T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}[/tex]

Rearranging,

[tex]35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}[/tex]

[tex]50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}[/tex]

If we divide these two equations we get

[tex]\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}[/tex]

[tex]\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}[/tex]

Now, that we know the value of [tex]k[/tex] we can use it to find the initial temperature of the beam,

[tex]35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5[/tex]

so the beam started out at 5 °F.

Using Newton's Law of Cooling, the initial temperature of the metal beam is estimated to be 5°F.

Estimating the Initial Temperature of the Beam Using Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its current temperature and the ambient temperature of its surroundings. Mathematically, it can be expressed as:

where:

Given:

Step-by-Step Solution:

Therefore, solving for k: k = ln(1/2) / 5 ≈ -0.1386

Finally, To = 65 - 60 = 5°F

Answer:

a) -0.842

b) $0

c) 11,664 cases

Step-by-step explanation:

a)

The price elasticity of demand E at a given point [tex]\large (p_1,q_1)[/tex] is defined as

[tex]\large E=\frac{p_1}{q_1}.\frac{\text{d}q}{\text{d}p}(p_1)[/tex]

and in this case, it would measure the possible response of tissues demand due to small changes in its price when the price is at [tex]\large p_1[/tex]

When the price is set at $32 the demand is

[tex]\large q=(108-32)^2=5,776 [/tex]

cases of tissues, so

[tex]\large (p_1,q_1)=(32,5776)\Rightarrow \frac{p_1}{q_1}=\frac{32}{5776}=0.00554[/tex]

Also, we have

[tex]\large \frac{\text{d}q}{\text{d}p}=-2(108-p)\Rightarrow \frac{\text{d}q}{\text{d}p}(32)=-2(108-32)=-152[/tex]

hence

[tex]\large E=\frac{p_1}{q_1}.\frac{\text{d}q}{\text{d}p}(p_1)=0.00554(-152)=-0.842[/tex]

That would mean the demand is going down about 0.842% per 1% increase in price at that price level.

b)

When the price is $108 the demand is 0, so the price should always be less than $108.

On the other hand, the parabola

[tex]\large q=(108-32)^2=5,776 [/tex] is strictly decreasing between 0 and 108, that means the maximum demand would be when the price is 0.

c)

When the price is 0 the demand is  

[tex]\large (108)^2=11,664[/tex] cases

The elasticity for p = 32 is E = -0.84.

The maximum revenue is obtained when p = 32, and the demand for that price is q = 5,184.

It is given by:

[tex]E = \frac{p}{q} *\frac{dq}{dp} (p)[/tex]

if p = $32, then:

q(32) = (108 − 32)^2 = 5,776

and:

dq/dp = -2*(108 - p)

Evaluating that in p = 32 we get:

-2*(108 - 32)  = -152

Then the elasticity is:

E = (32/5,776)*-152 = -0.84.

The revenue is equal to the demand times the price, so we get:

R = p*q = p*(108 - p)^2 = p*(p^2 - 216p + 11,664)

To maximize this we need to find the zeros of the derivation, we have:

R' = 3p^2 - 2*216*p + 11,664

The zeros of that equation are given by:

[tex]p = \frac{432 \pm \sqrt{(-432)^2 - 4*3*11,664} }{2*3} \\\\p = (432 \pm 216)/6[/tex]

Notice that we need to use the smaller of these values, so the demand never becomes zero, then we use:

p = (432 - 216)/6 = 36

(the other root gives p = 108, so that is a minimum).

c) For this price, the number of cases demanded is:

q = (108 - 36)^2 = 5,184

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Answer: f(-) = - 12

Step-by-step explanation:

The function is expressed as

f(x)= -x^2+10x-3

To determine f(-1), we will substitute

x = - 1 into the given function. It becomes

f(-1) = (- 1)^2 + 10(-1) - 3

f(-1) = 1 - 10 - 3

f(-1) = - 12

Answer:

36

Step-by-step explanation:

multiply the width & the height

Answer:

Mean = 71.4 years

Standard Deviation = 20.64 years

Step-by-step explanation:

We are given the following data set:

72, 68, 81, 93, 56, 19, 78, 94, 83, 84, 77, 69, 85, 97, 75, 71, 86, 47, 66, 27

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{1428}{20} = 71.4[/tex]

Sum of squares of differences =

0.36 + 11.56 + 92.16 + 466.56 + 237.16 + 2745.76 + 43.56 + 510.76 + 134.56 + 158.76 + 31.36 + 5.76, 184.96 + 655.36 + 12.96 + 0.16 + 213.16 + 595.36 + 29.16 + 1971.36 = 8100.8

[tex]S.D = \sqrt{\displaystyle\frac{8100.8}{19}} = 20.64[/tex]

Mean = 71.4 years

Standard Deviation = 20.64 years

To maximize and minimize the quantity given by p^2q^250 with the constraint p+q=10, express q in terms of p, derive a single-variable function, optimize using differentiation, find critical points, and evaluate at the endpoints of the domain [0, 10]. The difference between the maximum and minimum values would be the answer.

The student asks how to maximize and minimize the quantity p2q250, given the constraint that p + q = 10 and that p and q are nonnegative. To find the maximum and minimum values of this expression, we can use the concept of optimization in algebra. First, express q in terms of p, using the given equation p + q = 10. We get q = 10 - p. Substitute this into the original expression to obtain a single-variable function f(p) = p2(10 - p)250. Now, we can differentiate this function with respect to p and find its critical points.

To find the difference between the maximum and minimum values, evaluate f(p) at the critical points and at the endpoints of the domain (since p and q have to be nonnegative, the domain is [0, 10]). The required difference is the greatest value of f(p) minus the smallest value of f(p), which is the solution to the student's problem. This can be found by tedious algebra and the final answer would be expressed as a fraction.

Answer:

Step-by-step explanation:

(a) The average number of times a customer carries out banking transactions per week is 13 give or take 5 or so.

(b) The 90% confidence interval for the average number of banking operations per week is given by:

(Assuming a normal distribution of the number of banking operations)

CI=\overline{X}\pm 1.645\times \sigma/\sqrt{n}

CI=13\pm 1.645\times 5/\sqrt{350}

CI=13\pm 0.439645

CI=(12.560355, 13.439645)

The 99% confidence interval for the average number of banking operations per week is given by:

(Assuming a normal distribution of the number of banking operations)

CI=\overline{X}\pm 2.576\times \sigma/\sqrt{n}

CI=13\pm 2.576\times 5/\sqrt{350}

CI=13\pm 0.688465

CI=(12.311535, 13.688465)

The difference in the margin of error is 0.688465-0.439645=0.24882

The difference in the margin of error will make the confidence interval wider in the second case(99% confidence interval) as compared to the first case(90% confidence interval).

(c) Since, both our confident interval contains the value 10 hence we have sufficient evidence to conclude that the apparent difference in banking habits between the nation (It is given that the national survey suggest that on an average 10 transactions are performed per week by a single person) and the customer of the bank is not significant or is not real and is just due to the random errors/variations in sampling.

(d) A 95% confidence interval gives a range of values for the (A) Population Average which are plausible according to the observed data.

(The confidence interval always gives the range of possible values for the population values)

(e) The sample standard deviation measures how far (A) number of bank operations is from sample average.

The standard deviation for the sample average measures how far (B) Average number of bank operations is from the population average for typical (D) samples.

Answer: 7

Step-by-step explanation: Linear equations can be written in the form y = mx + b where the multiplier, m, represents the slope of the line and the base, b, represents the y-intercept of the line.

The linear equation y = -6x + 7 has a y-intercept equal to 7.

Answer:

(a) B. G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.

(b) Every function of the form [tex]4x^2+C[/tex] is an antiderivative of 8x

Step-by-step explanation:

A function F is an antiderivative of the function f if

[tex]F'(x)=f(x)[/tex]

for all x in the domain of f.

(a) If [tex]f(x) = 8x[/tex], then [tex]G(x)=4x^2[/tex] is an antiderivative of f because

[tex]G'(x)=8x=f(x)[/tex]

Therefore, G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.

Let F be an antiderivative of f. Then, for each constant C, the function F(x) + C is also an antiderivative of f.

(b) Because

[tex]\frac{d}{dx}(4x^2)=8x[/tex]

then [tex]G(x)=4x^2[/tex] is an antiderivative of [tex]f(x) = 8x[/tex]. Therefore, every antiderivative of 8x is of the form [tex]4x^2+C[/tex] for some constant C, and every function of the form [tex]4x^2+C[/tex] is an antiderivative of 8x.

Final answer:

G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x. The general antiderivative of f(x) = 8x is F(x) = 4x^2 + C.

Explanation:

To determine if G(x) is an antiderivative of f(x), we must consider the derivative of G(x), which is G'(x). If G'(x) is the same function as f(x), then by definition G is an antiderivative of f. Starting with G(x) = 4x2, the derivative is G'(x) = 8x, which is exactly the function f(x). Thus, the correct choice is B: G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.

Now, to find all antiderivatives of the function f(x) = 8x, we integrate f(x) to get the general antiderivative, which includes a constant of integration, C. The antiderivative of 8x is F(x) = 4x2 + C, representing all antiderivatives of f(x).

Answer:

Step-by-step explanation:

Hello!

You have X~Bi (n;ρ)

Where:

n=10

ρ= 0.10

For all asked probabilities you need to use a Binomial distribution table. Remember this table has the information of the cummulative probabilities P(X≤x).

a. f(0) ⇒ P(X=0) = 0.3487

b. f(2) ⇒ P(X=2) ⇒ P(X≤2) - P(X≤1) = 0.9298 - 0.7361 = 0.1937

c. P(X≤2) = 0.9298

d. P(X ≥ 1) = 1 - P(X ≤ 1) = 1 - 0.7361 = 0.2639

e. E(X)= nρ = 10*0.10 = 1

f. V(X)= nρ(1-ρ) = 10*0.1*0.9 = 0.9

σ= √V(X) = √0.9 = 0.9487

I hope it helps!

The binomial experiment depicts that f(0) will be 0.3487.

From the information given, it can be noted that:

n = 10

p = 0.10

q = 1 - 0.10 = 0.90

f(0) ⇒ P(X=0) = 0.3487

f(2) = P(X=2) ⇒ P(X≤2) - P(X≤1)

= 0.9298 - 0.7361

= 0.1937

P(X≤2) = 0.9298

P(X ≥ 1) = 1 - P(X ≤ 1)

= 1 - 0.7361 = 0.2639

E(X) = nρ

= 10 × 0.10 = 1

V(X) = nρ(1-ρ)

= 10 × 0.1 × 0.9

= 0.9

σ = √V(X) = √0.9

= 0.9487

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Answer:

can you provide the list?

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

  |x| < √7

Step-by-step explanation:

The product of the roots of the given quadratic equation is 2/a, so the other root (the one not given) is 2/(7a). It will be negative, since "a" is negative.

The roots of ax^2 +bx +2 = 0 are the values of x^2 that satisfy ax^4 +bx^2 +2 = 0. That is, roots of the latter equation will be the square root of the roots of the former equation.

We know that two of the zeros of the quartic are ±√7, and the other two are complex, as they are the square roots of a negative number. So, the graph of the quartic opens downward (because a < 0), and has real zeros at x=±√7.

The solution to the inequality must be ...

  -√7 < x < √7

_____

The graph shows an example of the quadratic (green) and quartic (black). The ripple in the quartic changes amplitude with different values of "a", but the locations of the zeros do not change.

Answer:

92 meters

Step-by-step explanation:

This is a problem that can be solved by the concept of "similar triangles", where we have two right angle triangles that share also another common angle" the angle that the rays of the sun form with them (see attached image).

The smaller triangle is formed by the limiting rays of the sun, the pole, and its shadow: it has a height of 3.3 m (the length of the pole) and a base of 1.69 m (the pole's shadow).

The larger triangle is formed by the rays of the sun, the building and its shadow: it has a base of 47.25 m and an unknown height that we named as "x" (our unknown).

The ratio of the bases of such similar triangles must be in the same proportion as the ratio of their heights, so we can create a simple  equation that equals such ratios, and then solve for the unknown "x":

[tex]\frac{H}{h} =\frac{B}{b}\\\frac{x}{3.3} =\frac{47.25}{1.69}\\x=\frac{47.25\,*\,3.3}{1.69} \\x=92.26331\, m[/tex]

which we can round to the nearest meter as: 92 m

Answer:

Less.

Step-by-step explanation:

Most tables are rectangle so hope it helps!

The P(69 ≤ X ≤ 71) when n = 16 is approximately 0.9312.

Here's the calculation for P(69 ≤ X ≤ 71) when n = 16:

1. Find the standard error (SE):

SE = σ / √n = 2.2 GPa / √16 = 0.55 GPa

2. Standardize the values:

Z1 = (69 - 70) / 0.55 = -1.82

Z2 = (71 - 70) / 0.55 = 1.82

3. Use a standard normal table or calculator to find the probabilities:

P(Z ≤ 1.82) = 0.9656

P(Z ≤ -1.82) = 0.0344

4. Calculate the probability of the interval:

P(69 ≤ X ≤ 71) = P(-1.82 ≤ Z ≤ 1.82) = P(Z ≤ 1.82) - P(Z ≤ -1.82)

= 0.9656 - 0.0344 = 0.9312

Therefore, P(69 ≤ X ≤ 71) when n = 16 is approximately 0.9312.

Answer:

0.5

Step-by-step explanation:

Since Ashok stays for exactly 15 minutes after his arrival, they will meet if he arrives in the 10 minutes between 9:10 and 9:20 independently of the time Melina arrives.

Since there are 20 minutes from 9:00 and 9:20 and Ashok arrives at a random time, the distribution of his arrival is uniform, so the probability that he arrives between 9:10 and 9:20 is

10/20 = 0.5

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].

Step-by-step explanation:

1) Data given and notation

[tex]s^2 =16[/tex] represent the sample variance

s=4 represent the sample standard deviation

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

n=20 the sample size

Confidence=98% or 0.98

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 mean or variance 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.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

[tex]df=n-1=20-1=19[/tex]

Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=36.191[/tex]

[tex]\chi^2_{1- \alpha/2}=7.633[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}[/tex]

[tex] 8.400 \leq \sigma^2 \leq 39.827[/tex]

So the 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].

The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points

Standard deviation = √variance = √16 = 4

The z score of  98% confidence interval is 2.326

The margin of error (E) is:

[tex]E = Z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } =2.326*\frac{4}{\sqrt{20} } =2.08[/tex]

The confidence interval = mean ± E = 72 ± 2.08 = (69.92, 74.08)

The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points

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Answer:

4,200 liters

Step-by-step explanation:

The flow rate is given by:

[tex]r(t) = 200 - 4t[/tex]

Integrating the flow rate expression from t=0 to t=30 minutes, yields the total volume that flows out of the tank during that period:

[tex]\int\limits^{30}_0 {r(t)} = \int\limits^{30}_0 {(200 - 4t} )\, dt \\V=(200t - 2t^2)|_0^{30}\\V= (200*30 -2*30^2)-(200*0 -2*0^2)\\V=4,200\ liters[/tex]

4,200 liters of water flow from the tank during the first 30 minutes.

Answer:

t= 32.327

Would be a significant difference in the average amount of saturated fat in solid and liquid fats.

Step-by-step explanation:

1) Data given and notation

Stick:[26.2,25.6,25.5,26.1,26.5,26.7]

Liquid:[16.3,16.4,16.3,16.7,16.6,17.7]

[tex]\bar X_{stick}[/tex] represent the mean for the sample Stick

[tex]\bar X_{Liquid}[/tex] represent the mean for the sample Liquid

[tex]s_{stick}[/tex] represent the sample standard deviation for the sample Stick

[tex]s_{Liquid}[/tex] represent the sample standard deviation for the sample Liquid

[tex]n_{stick}=6[/tex] sample size for the group Stick

[tex]n_{Liquid}=6[/tex] sample size for the group Liquid

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null Hypothesis :[tex]\mu_{stick}=\mu_{Liquid}[/tex]

Alternative Hypothesis :[tex]\mu_{stick} \neq \mu_{Liquid}[/tex]

If we analyze the size for the samples both are < 30 so for this case we can apply a t test to compare means, and the statistic formula is:

[tex]t=\frac{\bar X_{stick}-\bar X_{Liquid}}{\sqrt{\frac{s^2_{stick}}{n_{stick}}+\frac{s^2_{Liquid}}{n_{Liquid}}}}[/tex] (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

In order to calculate the mean and the sample deviation we can use the following formulas:

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

3) Calculate the statistic

First we need to calculate the mean and deviation for each sample, after apply the formulas (2) and (3) we got the following results:

[tex]\bar X_{stick}=26.10[/tex] [tex]s_{stick}=0.477[/tex]

[tex]\bar X_{Liquid}=16.67[/tex] [tex]s_{Liquid}=0.532[/tex]

And with this we can replace in formula (1) like this:

[tex]t=\frac{26.10-16.67}{\sqrt{\frac{0.477^2}{6}+\frac{0.532^2}{6}}}}=32.327[/tex]  

4) Statistical decision

For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

[tex]df=n_{stick}+n_{liquid}-2=6+6-2=10[/tex]

Since is a bilateral test the p value would be:

[tex]p_v =2*P(t_{(10)}>32.327)=1.8x10^{-11}[/tex]

So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average amount of saturated fat in solid and liquid fats.

Answer:

Amount of mercury in population is in between 0.2841 , 1.1531

Step-by-step explanation:

Step by step explanation is given in the attachments

Answer:

Step-by-step explanation:

Hello!

You have some p-values and different decisions that can be made using these p-values.

Remember the decision rule for the p-value method.

If p-value > α, you don't reject the null hypothesis.

If p-value ≤ α, you reject the null hypothesis.

With this in mind, I've arranged the decisions with the given p-values.

a. do not reject H₀ at 10% ⇒ p-value: 0.3

b. do reject H₀ at 10% but not at 5% p-value: ⇒ 0.07

c. do reject H₀ at 5% but not at 1% ⇒ p-value: 0.02

d. do reject H₀ at 1% ⇒ p-value: 0.006

I hope it helps!

Answer:

Null hypothesis:[tex]p_{SB} \geq p_{NSB}[/tex]  

Alternative hypothesis:[tex]p_{SB} < \mu_{NSB}[/tex]  

The p value is a very low value and using the significance given [tex]\alpha[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of people death using seat belts is significant lower than the proportion of deaths no using seat belts .  

Step-by-step explanation:

1) Data given and notation  

[tex]X_{NSB}=30[/tex] represent the number of people killed not using seat belts

[tex]X_{SB}=20[/tex] represent the number of people killed using seat belts

[tex]n_{NSB}=2902[/tex] sample of people not wearing seat belts

[tex]n_{SB}=7866[/tex] sample of people wearing seat belts

[tex]p_{NSB}=\frac{30}{2902}=0.0103[/tex] represent the proportion of people killed not using seat belts

[tex]p_{SB}=\frac{20}{7866}=0.00254[/tex] represent the proportion of people killed using seat belts

z would represent the statistic (variable of interest)  

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that seat belts are effective in reducing fatalities (If using seat belts reduce the proportion of deaths we need to see that the proportion of death using seat belts is lower than not using seat belts) , the system of hypothesis would be:  

Null hypothesis:[tex]p_{SB} \geq p_{NSB}[/tex]  

Alternative hypothesis:[tex]p_{SB} < \mu_{NSB}[/tex]  

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{SB}-p_{NSB}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{SB}}+\frac{1}{n_{NSB}})}}[/tex]   (1)

Where [tex]\hat p=\frac{X_{SB}+X_{NSB}}{n_{SB}+n_{NSB}}=\frac{20+30}{7866+2902}=0.00464[/tex]

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.00254-0.0103}{\sqrt{0.00464(1-0.00464)(\frac{1}{7866}+\frac{1}{2902})}}=-5.26[/tex]  

4) Statistical decision

Using the significance level provided [tex]\alpha=0.05[/tex], the next step would be calculate the p value for this test.  

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

[tex]p_v =P(Z<-5.26)=7.2x10^{-8}[/tex]  

So the p value is a very low value and using the significance given [tex]\alpha[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of people death using seat belts is significant lower than the proportion of deaths no using seat belts .  

To find the probability that the average tax paid on the sample forms is greater than $1980, standardize the sample mean using the Central Limit Theorem. To find the probability that more than 60 of the sampled forms have a tax greater than $3200, use a normal distribution approximation.

To solve this problem, we can use the Central Limit Theorem.

a. To find the probability that the average tax paid on the sample forms is greater than $1980, we need to standardize the sample mean. We calculate the z-score by subtracting the population mean from the sample mean and dividing by the population standard deviation divided by the square root of the sample size. Then, we can use a z-table or a calculator to find the probability.

b. To find the probability that more than 60 of the sampled forms have a tax greater than $3200, we can use a normal distribution approximation. We can calculate the z-score for 60 forms by subtracting the population mean from 60 and dividing by the square root of the population mean multiplied by (1 - the population mean) divided by the sample size. Then, we can use a z-table or a calculator to find the probability.

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Answer:

We conclude that the lifetime of tires is less than 30,000 miles.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30,000 miles

Sample mean, [tex]\bar{x}[/tex] = 29,400 miles

Sample size, n = 54

Alpha, α = 0.05

Population standard deviation, σ =1200 miles

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 30000\text{ miles}\\H_A: \mu < 30000\text{ miles}[/tex]

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

Formula:

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

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{29400 - 30000}{\frac{1200}{\sqrt{54}} } = -3.6742[/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 the lifetime of tires is less than 30,000 miles.