In the Country A legal​ system, a defendant is presumed innocent until proven guilty. Consider a null​ hypothesis, Upper H 0​, that the defendant is​ innocent, and an alternative​ hypothesis, Upper H 1​, that the defendant is guilty. A jury has two possible​ decisions: Convict the defendant​ (i.e., reject the null​ hypothesis) or do not convict the defendant​ (i.e., do not reject the null​ hypothesis). Explain the meaning of the risks of committing either a Type I or Type II error in this example.

(A) A Type I error would be incorrectly convicting the defendant when he is guilty. A Type II error would be incorrectly failing to convict the defendant when he is innocent.
(B) A Type I error would be incorrectly convicting the defendant when he is innocent. A Type II error would be incorrectly failing to convict the defendant when he is guilty.
(C) A Type I error would be incorrectly failing to convict the defendant when he is guilty. A Type II error would be incorrectly convicting the defendant when he is innocent.
(D) A Type I error would be incorrectly failing to convict the defendant when he is innocent. A Type II error would be incorrectly convicting the defendant when he is guilty.

Answer: 31

Step-by-step explanation:

2^x=2 000 000 000

log2^x=log2 000 000 000

xlog2 = log 2 000 000 000

x= log (2000 000 000)/log 2

x= 30.897352854

round to 31

gotchu bro

Answer:

B) 4.07

Step-by-step explanation:

First we need to calculate the mean of all the data, which is the same as the mean of the means of each grade of gasoline:

Regular    BelowRegular   Premium   SuperPremium

39.31             36.69                38.99             40.04

39.87            40.00                40.02             39.89

39.87            41.01                  39.99             39.93

X1⁻=39.68    X2⁻= 39.23       X3⁻= 39.66    X4⁻=  39.95

Xgrand⁻ = (39.68+39.23+39.66+39.95)/4 = 39.63

Next we need to calculate the sum of squares within the group (SSW) and the sum of squares between the groups (SSB), and the respective degrees of freedom):

SSW = [ (39.31-39.68)² + (39.87-39.68)² + (39.87-39.68)² ] + [ (36.69-39.23)² + (40.00-39.23)² + (41.01-39.23)² ] + [ (38.99-39.66)² + (40.02-39.66)² + (39.99-39.66)² ] + [ (40.04-39.95)² + (39.89-39.95)² + (39.93-39.95)² ] = [0.2091] + [10.2129] + [0.6874] + [0.0121] = 11.12

SSW =  11.12

Degrees of freedom in this case is calculated by m(n-1), with m being the number of grades of gasoline (4) and n being the number of trial results for each one (3), so we would have 4(3-1) = 8 degrees of freedom

SSB = [ (39.68-39.63)² + (39.68-39.63)² + (39.68-39.63)²] + [ (39.23-39.63)² + (39.23-39.63)² + (39.23-39.63)² ] + [ (39.66-39.63)² + (39.66-39.63)² + (39.66-39.63)² ] + [ (39.95-39.63)² + (39.95-39.63)² +(39.95-39.63)² ] = [0.0075] + [0.48] + [0.0027] + [0.3072] = 0.7974

SSB =  0.80

For this case, the degrees of freedom are m-1, so we would have 4-1 = 3 degrees of freedom

Now we can establish the hypothesis for the test:

H0: μ1 = μ2 = μ3 = μ4

The null hypothesis states that the means of miles per gallon for each fuel are the same, indicating that the drade of gasoline does not make a difference, therefore our alternative hypothesis will be:

H1: the grade of gasoline does makes a difference

We will use the F statistic to test the hypothesis, which is calculated like follows:

F - statistic = (SSB/m-1) / (SSW/m(n-1)) = (0.80/3) / (11.12/8) = 0.19

We know that the level of significance we are using is α = 0.05, so to find the critical value F we need to look at some table of critical values for the F distribution for the 0.05 significance level (like the attached image). Then we just need to look fot the value that is located in the intersection between the degrees of freedom we have in the numerator (horizontal) and the denominator (vertical) of the statistic (3 and 8). That critical value is:

Fc = 4.07

Final answer:

The critical value of F used to test the hypothesis that the miles per gallon for each fuel is the same is 3.49.

Explanation:

To test the hypothesis that the miles per gallon for each fuel is the same, we can use an ANOVA test. The critical value of F at the 0.05 level can be found using the F-distribution table or by using statistical software. Since we have three trial runs for each grade of gasoline, we have a total of 12 observations. At a significance level of 0.05 and with 3 degrees of freedom for the numerator and 8 degrees of freedom for the denominator, the critical value of F is 3.49.

Answer:

The area of the searched region is [tex]A= a+b+ \frac{2n}{n+1}- \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1}-2[/tex]

Step-by-step explanation:

If you want to find the area of a region bounded by functions f(x) and G(x) between two limits (a,b), you have to do a double integral. you must first know which of the functions is greater than the other for the entire domain.

In this case, for 0<x<1, f(x)<g(x)

while for 1<x, g(x)<f(x).

Therefore if our domain is all real numbers superior to 0 (where the limit 0<a<1 and 1<b), we have to do 2 integrals:

A=A(a<x<1)+A(1<x<b)

[tex]A(a<x<1)=\int\limits^1_a {\int\limits^{x^{\frac{1}{n}} }_{x} } {} \, dy } \, dx = \int\limits^1_a {x^{\frac{1}{n} } -x \, dx = a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1}[/tex]

[tex]A(1<x<b)=\int\limits^b_1 {\int\limits^{x}_{x^{\frac{1}{n} } } {} \, dy } \, dx = \int\limits^b_1 {x-x^{\frac{1}{n} } \, dx =b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1}[/tex]

[tex]A=a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1} +  b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1} = a+b+ \frac{2n}{n+1}  - \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1}   -2[/tex]

Answer:

Check the explanation below

Step-by-step explanation:

Hello!

To make a pooled variance t-test you have to make the following assumptions:

The study variables X₁ and X₂ must be independent.

Both variables should have a normal distribution, X₁~N(μ₁; σ₁²) and X₂~N(μ₂; σ₂²)

The population variances should be equal but unknown, σ₁² = σ₂² = ?.

You have the information of two samples:

Sample 1

n₁=8

sample mean X[bar]₁= 42

sample standard deviation S₁=4

Sample 2

n₂=15

sample mean X[bar]₂= 34

sample standard deviation S₂= 5

For the hypothesis:

H₀: μ₁ = μ₂

H₁: μ₁ > μ₂

The statistic is:

t=  (X[bar]₁ - X[bar]₂) - (μ₁ - μ₂) ~[tex]t_{n_1 + n_2 - 2}[/tex]

Sa[tex]\sqrt{\frac{1}{n_1} + \frac{1}{n_2} }[/tex]

Sa²= [tex]\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}[/tex]

Sa²= 22

Sa= 4.69

[tex]t_{H0}[/tex]= 3.8962 ≅ 3.9

The critical region is one-tailed, for example for α: 0.05

[tex]t_{n_1 + n_2 - 2; 1 - \alpha } = t_{21; 0.95} = 1.721[/tex]

Since [tex]t_{H0}[/tex] > 1.721, then the decision is to reject the null hypothesis.

I hope it helps!

Answer:

70 is answer

Step-by-step explanation:

Given that a function in x is

[tex]f(x) = 5x^2[/tex]

we have to find f'(7)

we know by derivative rule derivative of a function is

[tex]f'(x) = lim_({h-->0}) \frac{f(x+h)-f(x)}{h}[/tex]

For finding out at 7 we replace x by 7

[tex]f'(7) = lim_({h-->0}) \frac{f(7+h)-f(7)}{h}[/tex]

=[tex]lim\frac{5(7+h)^2-5*7^2}{h} \\= lim \frac{10h*7+h^2}{h} \\= 70+h = 70[/tex]

So f'(7) = 70

answer is 70

Answer:

f'(7)=70

Step-by-step explanation:

We have the definition of the derivative as:

[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]

Now we have a function [tex]f(x)=5x^2[/tex] and we want to approximate the first derivative around x=7, that is [tex]f'(7)[/tex].

We can replace this in the first formula as:

[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}= \lim_{h \to 0} \dfrac{5(x+h)^2-5x^2}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{5(x^2+2xh+h^2-x^2)}{h}\\\\f'(x)=\lim_{h \to 0}\dfrac{5(2xh+h^2)}{h}\\\\f'(x)=\lim_{h \to 0}5(2x+h)\\\\f'(x)=10x+lim_{h \to 0}h=10x+0=10x[/tex]

Then, the value for f'(7) is:

[tex]f'(7)=10\cdot 7=70[/tex]

Answer:

A) [tex]P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}[/tex]

B) P(t→∞)=2000

C) [tex]P=\frac{K}{e}=\frac{carrying capacity}{e}[/tex]

Step-by-step explanation:

Given differential eq is

                      [tex]\frac{dP}{dt}=c ln (\frac{K}{P})P[/tex] --- (1)

Eq is separable

                     [tex]\frac{1}{ln (\frac{K}{P})P}dP=cdt[/tex] --- (2)

                     [tex]let \\u = ln\frac{K}{P}\\du= \frac{1}{\frac{K}{P}}(\frac{-K}{P^{2}}).dP\\du=\frac{-1}{P}.dP\\dP=-P.du[/tex]

substituting in  (2)

[tex]-\frac{du}{u}=dt[/tex]

Integrating both sides

[tex]\int {-\frac{1}{u}} \, du=\int{c}\,dt\\-ln|u|=ct +B\\ln|u|=-ct -B\\[/tex]

Back substituting value of u

[tex]ln |ln\frac{K}{P}|=-ct-B\\|ln\frac{K}{P}|=e^{-ct-B}\\ln|\frac{K}{P}|=be^{-ct}\\[/tex]---(3)

at t =0

[tex]ln|\frac{K}{P}|=be^{-ct}\\b=ln|\frac{K}{P}|\\b=ln\frac{2000}{500}\\b=ln|4|[/tex]

from (3)

[tex]ln|\frac{K}{P}|=be^{-ct}\\\frac{K}{P}=e^{ln4e^{-ct}}\\P(t)=\frac{K}{e^{ln4e^{-ct}}}[/tex]

[tex]P(t)=\frac{2000}{e^{ln4e^{-0.1t}}}[/tex]

B) [tex]\lim{t \to \infty}[/tex]

[tex]P( {t \to \infty} )=\frac{2000}{e^{ln4e^{-0.1\infty}}}\\e^{-0.1\infty}=0\\\implies P( {t \to \infty} )=\frac{2000}{e^{0}}}\\\\P(\infty)=2000\\[/tex]

which is the carrying capacity.

C) To find the fastest growth rate we have to maximize [tex]\frac{dP}{dt}[/tex]

From given differential eq

[tex]\frac{dP}{dt}=cln|\frac{K}{P}|P[/tex]

so function to maximize is

[tex]f(P)=cln|\frac{K}{P}|P[/tex]

[tex]f'(P)=cln|\frac{K}{P}|+c\frac{1}{\frac{K}{P}}\frac{-K}{P^{2}}.P[/tex]

[tex]f'(P)=c[ln|\frac{K}{P}|-1][/tex]

To maximize find f'(P)=0

[tex]c[ln|\frac{K}{P}|-1]=0[/tex]

[tex]ln|\frac{K}{P}|=1[/tex]

[tex]\frac{K}{P}=e[/tex]

[tex]P=\frac{K}{e}=\frac{carrying capacity}{e}[/tex]

The Gompertz function models population growth considering the carrying capacity. solve the Gompertz differential equation for c=0.1, K=2000, P0=500. The carrying capacity (K) gives the limiting value of the population size (2000). The time at which the population growth is fastest can be calculated by taking the derivative of the population function, setting it to zero and solving for P.

The Gompertz function is a model of growth that was developed to model population growth, considering the factor of carrying capacity. Instead of compound exponential growth, it models the growth as slowing down as it reaches the limit of the carrying capacity. In your particular case, solving the differential equation for the variables provided (c=0.1, K=2000, P0=500) would require the integration techniques and use of logarithmic functions.

(a): The full solution for the Gompertz function involves advanced mathematics, you should expect some intricate function in the form of P(t) = ... The particulars depend on the specifics of the integration process.

(b): The limiting value of the population size as t→infinity (limt→[infinity]P(t) will be K, which in this instance equals 2000. This is due to the concept of carrying capacity. Beyond this value, the environment/conditions can no longer support additional growth.

(c): Finding the time at which population growth is fastest involves setting the derivative of the population function to zero and solving for P. The solution P=... can be calculated using standard techniques of calculus.

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Answer:

a) Discrete

b) continuous

Step-by-step explanation:

Given that there are two random variables

1) The number of points scored during a basketball game

2) The weight of a

T bone steak.

The number of points scored during a basketball game

-- Can take values as 0,1,2.....

This can be counted i.e. we can have a one to one correspondence with natural numbers.

So this is a discrete variable

The weight of a T-bone steak

-- This can take any value in decimal or fraction.  This can be between an interval comprising all values over the interval.  Hence we cannot set one to one correspondence with set of natural numbers.

So continuous variable.

Answer:

Reject at 5%, accept at 1% the null hypothesis

Step-by-step explanation:

Set up hypotheses as

[tex]H_0: \bar x = 60\\H_a: \bar x < 60[/tex]

(Left tailed test)

Population std dev = 6

Sample std error = [tex]\frac{6}{\sqrt{49} } \\=0.8555[/tex]

Mean difference = -1.5

Since sigma is known we can use Z test

Z = mean diff/std error = -1.7533

p value = 0.039

a) Since p value <0.05 we reject H0.  There is evidence  that the true mean is different from 60 wpm

b) Yes, because his sample would have been biased since he may want to prove his belief so slow or inefficient persons he would have selected in the sample.

c) For 99% confidence interval critical value = 2.58

Confidence interval for population mean = 58.5±2.58*std error

=(56.2928, 60.7072)

Since this contains 60, the hypothesized mean, we accept null hypothesis.

we do not have evidence to reject the claim that the average graduate can type 60 wpm at 1% level of significance.

The best predicted value of 'y' when 'x' is 2, using the linear regression equation ŷ = 2 + 3x, is 8. However, the correlation coefficient of 0.003 indicates this prediction may not be accurate due to the weak linear relationship between the variables.

The question is about predicting a value using a given linear regression equation. Given the regression equation ŷ = 2 + 3x, to predict 'y' when x = 2, we just replace 'x' with '2' in the regression equation. The equation becomes ŷ = 2 + 3*2 = 2 + 6 = 8. Therefore, the best predicted value of 'y' when 'x' is 2 is 8.

Note that the provided correlation coefficient (r) of 0.003 indicates a very weak linear relationship between the variables, hence this prediction might not be very reliable.

We use the regression line equation to make the prediction, this line of best fit has been calculated using the data provided. These predictions are most reliable when there is a strong correlation between the variables used.

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Answer: [tex]\frac{\sqrt{6}-\sqrt{2}}{2} [/tex]

Step-by-step explanation:

We apply the formula [tex]\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y) [/tex].

Note that  [tex]\cos(\frac{41}{12}\pi)=\cos((\frac{36}{12}+\frac{7}{12})\pi)=\cos(3\pi + \frac{7}{12})\pi) [/tex]. Take  [tex]x=3\pi[/tex] and [tex]y=\frac{7}{12}\pi[/tex] in the formula above to get

[tex]\cos(\frac{41}{12}\pi)=\cos(3\pi)\cos(\frac{7}{12}\pi)-\sin(3\pi)\sin(\frac{7}{12}\pi)=(-1)\cdot \cos(\frac{7}{12}\pi)-0\cdot\sin(\frac{7}{12}\pi)=-\cos(\frac{7}{12}\pi)[/tex]

Then the value of this expression is [tex]-\cos(\frac{7}{12}\pi) [/tex]

We can use the cosine addition formula again to simplify further. Decompose the fraction in the argument as:

[tex]\cos(\frac{7}{12}\pi)=\cos((\frac{3}{12}+\frac{4}{12})\pi)=\cos((\frac{1}{4}\pi + \frac{1}{3})\pi) [/tex]

Applying the formula with [tex]x=\frac{1}{4}\pi[/tex] and [tex]y=\frac{1}{3}\pi[/tex] we obtain

[tex]\cos(\frac{7}{12}\pi)=\cos(\frac{1}{4}\pi)\cos(\frac{1}{3}\pi)-\sin(\frac{1}{4}\pi)\sin(\frac{1}{3}\pi)=\frac{\sqrt{2}}{2}\cdot\frac{1}{2} -\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{2}-\sqrt{6}}{2} [/tex]

We conclude that this expression has the value [tex]-\frac{\sqrt{2}-\sqrt{6}}{2}=\frac{\sqrt{6}-\sqrt{2}}{2} [/tex]

Answer:

Fraction

Step-by-step explanation:

Fraction is a symbol that represents a part of a whole. It consists of a numerator and a denominator. The numerator is the number above the fraction bar (also known as "Vinculum), while the denominator is the number below the fraction bar. The denominator is the total number of equal parts in a whole.

Examples of Fraction: [tex]\frac{1}{2}[/tex], [tex]\frac{2}{7}[/tex] and [tex]\frac{5}{8}[/tex].

In the first example, [tex]\frac{1}{2}[/tex], 1 is the numerator, while 2, is the denominator.

Additional Information

When the numerator is smaller than the denominator, the fraction is called "proper fraction." On the contrary, when the numerator is bigger than the denominator, the fraction is called "improper fraction."

$ 0.65 more dollars is the price of a box Sophia's original brand of cereal than the price of a box of the super-saving cereal

Amount saved each month with the change in cereal brand if he buys 5 cereal boxes each month is $ 3.25

Solution:

Given that Sophia buys a certain brand of cereal that costs $5 per box

The equation shows the price, y, as a function of the number of boxes, x, for the new brand:

y = 4.35x

Part A:

New brand, y = 4.35x where y is the price and x is the number of boxes

Original brand, y = 5x since given that cereal that costs $5 per box

If Sophia old cereal preference was $5, and the equation shows that the new cereal preference is $4.35, if I subtract the amount of the new one from the old,

we get , 5 - 4.35 = 0.65

Therefore, $ 0.65 more dollars is the price of a box Sophia's original brand of cereal than the price of a box of the super-saving cereal

Part B:

Given that if he buys 5 cereal boxes, let us calculate price for old and new brand

New brand, y = 4.35x

New brand, y = 4.35(5) = 21.75

Original brand, y = 5x = 5(5) = 25

Amount saved = $ 25 - $ 21.75 = $ 3.25

Thus amount saved each month with the change in cereal brand if he buys 5 cereal boxes each month is $ 3.25

Answer:

It is high likely that the filling operation is capale of meeting design specifications.

Step-by-step explanation:

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 = 355, \sigma = 2[/tex]

Is the filling operation capable of meeting the design specifications?

It will be capable if it is highly likely that the specifications will be met. A probability is said to be high likely when it is of at least 95%.

In this case, the probability of containing between 350 and 360 ml of liquid is the pvalue of Z when X = 360 subtracted by the pvalue of Z when X = 350.

X = 360

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

[tex]Z = \frac{360 - 355}{2}[/tex]

[tex]Z = 2.5[/tex]

[tex]Z = 2.5[/tex] has a pvalue of 0.9938.

X = 350

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

[tex]Z = \frac{350 - 355}{2}[/tex]

[tex]Z = -2.5[/tex]

[tex]Z = -2.5[/tex] has a pvalue of 0.0062.

This means that there is a 0.9938 - 0.0062 = 0.9876 = 98.76% probability that the filling operation is capable of meeting the design specifications. It is high likely that the filling operation is capale of meeting design specifications.

Answer:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]

Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:

[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]

[tex]t=\frac{19 -22)-(0)}{4.095\sqrt{\frac{1}{8}+\frac{1}{7}}}=-1.416[/tex]

Step-by-step explanation:

Data given

American: 21,17,17,20,25,16,20,16 (Sample 1)

French: 24,18,20,28,18,29,17 (Sample 2)

When we have two independent samples from two normal distributions with equal variances we are assuming that  

[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]

And the statistic is given by this formula:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]

Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:

[tex]S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]

This last one is an unbiased estimator of the common variance [tex]\sigma^2[/tex]

The system of hypothesis on this case are:

Null hypothesis: [tex]\mu_1 = \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]

Or equivalently:

Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]

Alternative hypothesis: [tex]\mu_1 -\mu_2 \neq 0[/tex]

Our notation on this case :

[tex]n_1 =8[/tex] represent the sample size for group 1

[tex]n_2 =7[/tex] represent the sample size for group 2

[tex]\bar X_1 =19[/tex] represent the sample mean for the group 1

[tex]\bar X_2 =22[/tex] represent the sample mean for the group 2

[tex]s_1=3.117[/tex] represent the sample standard deviation for group 1

[tex]s_2=5.0[/tex] represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

[tex]S^2_p =\frac{(8-1)(3.117)^2 +(7 -1)(5.0)^2}{8 +7 -2}=16.770[/tex]

And the deviation would be just the square root of the variance:

[tex]S_p=4.095[/tex]

And now we can calculate the statistic:

[tex]t=\frac{19 -22)-(0)}{4.095\sqrt{\frac{1}{8}+\frac{1}{7}}}=-1.416[/tex]

Now we can calculate the degrees of freedom given by:

[tex]df=8+7-2=13[/tex]

And now we can calculate the p value using the altenative hypothesis:

[tex]p_v =2*P(t_{13}<-1.416) =0.1803[/tex]

So with the p value obtained and using the significance level assumed [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance we don't have significant differences between the two means.  

Answer:

The margin of error for this study is 2.5 pounds.

Step-by-step explanation:

The margin of error is the subtraction of the mean by the lower end of the confidence interval, and this must be equal to the subtraction of the upper end to the mean.

In this problem, we have that:

M - 13 = 18 - M

2M = 31

M = 15.5

The mean is 15.5 pounds.

So the margin of error for this study is 15.5 - 13 = 18 - 15.5 = 2.5 pounds.

Answer:

We conclude that the  bag filling machine works correctly at the 444.0 gram setting.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 444.0 gram

Sample mean, [tex]\bar{x}[/tex] = 443.0 grams

Sample size, n = 40

Alpha, α = 0.02

Population standard deviation, σ = 23.0 grams

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 444.0\text{ grams}\\H_A: \mu < 444.0\text{ grams}[/tex]

We use one-tailed(left) 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{443 - 444}{\frac{23}{\sqrt{40}} } =-0.274[/tex]

Now, [tex]z_{critical} \text{ at 0.02 level of significance } = -2.054[/tex]

Since,  

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

We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that the  bag filling machine works correctly at the 444.0 gram setting.

The decision rule for the hypothesis test to check machine calibration involves finding the z-value for a sample mean of 443.0 grams and comparing it with the critical z-value corresponding to a 0.02 significance level for a left-tailed test. If the z-value is less than the critical value, the null hypothesis is rejected.

To determine the decision rule for testing whether the bag filling machine is correctly set at 444.0 grams when we suspect underfilling, we need to perform a hypothesis test using the z-test since the standard deviation is known. Given the sample size (n = 40), sample mean (μ = 443.0 grams), population mean (μ0 = 444.0 grams), population standard deviation (σ = 23.0 grams), and a level of significance of 0.02, we can formulate the null hypothesis (H0: μ = μ0) and the alternative hypothesis (H1: μ < μ0).

The decision rule involves comparing the computed z-value to the critical value from the standard normal distribution at the 0.02 level of significance for a left-tailed test. If the computed z-value is less than the critical z-value, we reject the null hypothesis and accept that the machine is underfilling. The critical z-value is found using the z-table, which corresponds to a cumulative probability of 0.02.

Answer:

Answer: 0.6022

Consider the following calculation

Step-by-step explanation:

Let F shows the event that guardian father is biological father. So

P(F) = 0.75

By the complement rule,

P(F') = 1 - P(F) =1 - 0.75 = 0.25

Let B shows the event that child has blood type B. So we have

P(B|F) = 0.50, P(B' |F') = 0.0091

By the complement rule we have

P(B|F') = 1 - P(B' |F') = 0.9909

The probability that the guarding father is the child’s biological father given that child have blood type B is

P(BFPF) P(F|B) = PRI P(BF)P(F) + P(BF)P(F) 0.50 -0.75 0.50 -0.75 +0.9909 - 0.25

0.375 0.622725 = 0.6022

Answer: 0.6022

Answer:

Step-by-step explanation:

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

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 = 9.6, \sigma = 0.5[/tex].

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when [tex]X = 10[/tex].

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

[tex]Z = \frac{10 - 9.6}{0.5}[/tex]

[tex]Z = 0.8[/tex]

[tex]Z = 0.8[/tex] has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

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

[tex]Z = \frac{8 - 9.6}{0.5}[/tex]

[tex]Z = -3.2[/tex]

[tex]Z = -3.2[/tex] has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have [tex]n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1[/tex].

This probability is 1 subtracted by the pvalue of Z when [tex]X = 9.8[/tex]. So:

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

[tex]Z = \frac{9.8 - 9.6}{0.1}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Answer:

Step-by-step explanation:

When a value is decreasing at a rate proportional to that value, it can be modeled by the formula

  a = a0·e^(-kt)

where k is the constant of proportionality.

Alternatively, we can write the exponential function describing the pool volume* as ...

  a = 15000·(138/150)^(t/6) = 15000·((138/150)^(1/6))^t

Comparing these, we see that ...

  e^(-kt) = (138/150)^(t/6)

or ...

  k = -ln(138/150)/6 ≈ 0.0138969

__

So, when 14000 gallons remain, the rate of decrease is ...

  14000·0.0138969 ≈ 194.6 . . . gallons per minute

When 5000 gallons remain, the rate of decrease is ...

  5000·0.0138969 ≈ 69.5 . . . gallons per minute

_____

* The generic form of this is ...

  (initial value) · (multiplier per interval)^(number of intervals)

Here, the multiplier over a 6-minute period is 13800/15000 = 138/150, and the number of 6-minute intervals is t/6 when t is in minutes.

_____

Effectively, we make use of the fact that for ...

  a = a0·e^(-kt)

the derivative is ...

  da/dt = -k(a0·e^(-kt)) = -k·a

That is, k is the constant of proportionality mentioned in the first sentence of the problem statement.

Answer:

a) Two tailed test

Null hypothesis:[tex]p=0.217[/tex]  

Alternative hypothesis:[tex]p \neq 0.217[/tex]  

b) [tex]p_v =2*P(Z>2.17)=0.03[/tex]  

c) If we compare the p value obtained and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We Fail to reject the null hypothesis H0

Step-by-step explanation:

Data given and notation

n represent the random sample taken

X represent the outcomes desired in the sample

[tex]\hat p[/tex] estimated proportion of interest

[tex]p_o[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is 0.217 or no:  

a. Identify the hypothesis test as being​ two-tailed, left-tailed, or​ right-tailed.

Two tailed test

Null hypothesis:[tex]p=0.217[/tex]  

Alternative hypothesis:[tex]p \neq 0.217[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

For this case the calculated value is given z =2.17  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

b. Find the​ P-value

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

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(Z>2.17)=0.03[/tex]  

c. Using a significance level of alphaαequals=0.01, should we reject Upper H 0 or should we fail to reject Upper H 0​?

If we compare the p value obtained and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We Fail to reject the null hypothesis H0

Answer:

C. 0.0207

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they vote for Carl, or they vote for Nathan. So we use the binomial probability distribution.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/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.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]\mu = 100*0.4 = 40[/tex]

[tex]\sigma = \sqrt{100*0.4*0.6} = 4.9[/tex]

What is the likelihood that the poll will show a majority in favor of Carl?

This is 1 subtracted by the pvalue of Z when X = 50. So

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

[tex]Z = \frac{50 - 40}{4.9}[/tex]

[tex]Z = 2.04[/tex]

[tex]Z = 2.04[/tex] has a pvalue of 0.9793. So the answer is 1-0.9793 = 0.0207.

Answer:

Option B.

Step-by-step explanation:

The given data set is

34, 35, 41, 28, 26, 29, 32, 36, 38, 40

We need to find the mean deviation of the given data.

Number of observations, n = 10

Mean of the data is

[tex]Mean=\dfrac{\sum x}{n}[/tex]

[tex]Mean=\dfrac{34+35+41+28+26+29+32+36+38+40}{10}[/tex]

[tex]Mean=\dfrac{339}{10}[/tex]

[tex]Mean=33.9[/tex]

Formula for mean deviation is

[tex]\text{Mean deviation}=\dfrac{\sum |x-mean|}{n}[/tex]

[tex]\sum |x-mean|=|34-33.9|+|35-33.9|+|41-33.9|+|28-33.9|+|26-33.9|+|29-33.9|+ |32-33.9|+|36-33.9|+|38-33.9|+|40-33.9|=41.2[/tex]

[tex]\text{Mean deviation}=\dfrac{41.2}{10}[/tex]

[tex]\text{Mean deviation}=4.12[/tex]

The mean deviation of the ratings is 4.12.

Therefore, the correct option is B.

Answer:

b. 4.12

Step-by-step explanation:

We have been given that 10 experts rated a newly developed chocolate chip cookie on a scale of 1 to 50. Their ratings were:

34, 35, 41, 28, 26, 29, 32, 36, 38, and 40.

First of all, we will find the mean of the ratings.

[tex]\text{Mean of ratings}=\frac{34+35+41+28+26+29+32+36+38+40}{10}[/tex]

[tex]\text{Mean of ratings}=\frac{339}{10}[/tex]

[tex]\text{Mean of ratings}=33.9[/tex]

Let us find absolute deviation of each point from mean.

[tex]|34-33.9|=0.1[/tex]

[tex]|35-33.9|=1.1[/tex]

[tex]|41-33.9|=7.1[/tex]

[tex]|28-33.9|=5.9[/tex]

[tex]|26-33.9|=7.9[/tex]

[tex]|29-33.9|=4.9[/tex]

[tex]|32-33.9|=1.9[/tex]

[tex]|36-33.9|=2.1[/tex]

[tex]|38-33.9|=4.1[/tex]

[tex]|40-33.9|=6.1[/tex]

Now we will use mean deviation formula.

[tex]\text{Absolute mean deviation}=\frac{\Sigma |x-\mu|}{N}[/tex], where,

[tex]\mu=\text{Mean}[/tex] and N = Number of data points.

[tex]MD=\frac{0.1+1.1+7.1+5.9+7.9+4.9+1.9+2.1+4.1+6.1}{10}[/tex]

[tex]MD=\frac{41.2}{10}[/tex]

[tex]MD=4.12[/tex]

Therefore, the mean deviation of the ratings is 4.12 and option 'b' is the correct choice.

Answer:

e. 1.28

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.8}{2} = 0.1[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex]. This is our critical value.

So it is z with a pvalue of [tex]1-0.1 = 0.9[/tex], so [tex]z = 1.28[/tex]

The correct answer is:

e. 1.28

Answer:

Becomes less able to repair duplication errors

Step-by-step explanation:

This is premised on the fact that aging has been connected with the deterioration of DNA maintenance and repair machinery, which tends to lose its ability to replicate new cell as a person age with time.

Answer:

[tex]\bar X \sim N(\mu=0.948, \sigma_{\bar X}=\frac{0.006}{\sqrt{35}}=0.00101)[/tex]

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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".

Let X the random variable who represents the bio/total carbon ratio. We know from the problem that the distribution for the parameters for the random variable X are:

[tex]\mu = 0.948[/tex]

[tex]\sigma=0.006[/tex]

We select a sample of n=35 nails. That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

So on this case :

[tex]\bar X \sim N(\mu=0.948, \sigma_{\bar X}=\frac{0.006}{\sqrt{35}}=0.00101)[/tex]

Answer:

They are not uniformly distributed.

The expected frequency of each group is 50

Step-by-step explanation:

In probability distributions, uniform distribution refers to a probability distribution for which all of the values that a random variable can take on occur with equal probability.

In other words, for n number of events, the probability of occurrence 1,2,3,4......n is 1/n

There are 3 possible occurrence in the question above

1. Yes

2. No

3. No Opinion.

For the above events to have a uniform distribution, then they must have a probability of ⅓ each.

The expected frequency of each would then be ⅓ of n where n = 150

⅓ of 150 = 50

Answer:

Null hypothesis:[tex]p\leq 0.79[/tex]  

Alternative hypothesis:[tex]p > 0.79[/tex]  

[tex]z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786[/tex]  

[tex]p_v =P(z>2.786)=0.00267[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.  

Step-by-step explanation:

1) Data given and notation

n=750 represent the random sample taken

X=314 represent the respondents that use social media in their job search.

[tex]\hat p=\frac{314}{370}=0.849[/tex] estimated proportion of respondents that use social media in their job search

[tex]p_o=0.79[/tex] is the value that we want to test

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

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is hgiher than 0.79.:  

Null hypothesis:[tex]p\leq 0.79[/tex]  

Alternative hypothesis:[tex]p > 0.79[/tex]  

When we conduct a proportion test we need to use the z statisticc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.849 -0.79}{\sqrt{\frac{0.79(1-0.79)}{370}}}=2.786[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

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

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

[tex]p_v =P(z>2.786)=0.00267[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.79.  

The context of a restaurant owner examining dessert orders involves several statistical concepts: Data is the list of Yes or No answers, the Sample is the 65 restaurant patrons, Population is all restaurant customers, the Statistic is the proportion of the sample who ordered dessert, the Parameter is the proportion of all customers who order dessert, and the Variable is whether an individual customer ordered dessert.

The question involves the concepts of statistics in Mathematics, particularly focusing on how different terminologies are used. Here is the matching:

https://brainly.com/question/33544588

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