The TreeDropp’d Fruit Company wants to sell its apples overseas in attractive pink and yellow gift boxes. To design the boxes, the company needs to estimate the average diameter of their apples. A random sample of 50 apples has a mean of 4.1 inches. You assume a population standard deviatino of .4 inches.

A. What is the point estimate in this scenario? Is it an unbiased estimator?

B. What is the critical z-value for a 95% interval?

C. What is the margin of error for a 95% interval in this scenario? Show your work.

Answer:

a. 10,000 Hours

Step-by-step explanation:

MTBF (Mean Time Between Failures) can be calculated using the formula:

[tex]MTBF={(L-F)}*{n}[/tex] where

[tex]{MTBF=(200-100)}*{100}=10000[/tex]

Final answer:

The MTBF for a set of devices tested by the manufacturer, with failures evenly distributed from 100 to 200 hours, is 15,000 hours, which is calculated using the sum of an arithmetic series formula.

Explanation:

The student is asking about the mean time between failures (MTBF) for a set of devices. MTBF is a basic measure of reliability for repairable items and represents the average time expected between failures. To calculate MTBF, you divide the total operational time by the number of failures. In this case, the device manufacturer tests 100 devices, with the first failing after 100 hours and the last after 200 hours, which suggests a linear distribution of failures over time.

Assuming a linear and even distribution of failures from 100 to 200 hours for the 100 devices, you would calculate the total operational time before each device failed and then divide by the number of devices. The calculation would be the sum of an arithmetic series:

MTBF = (100 hours + 101 hours + ... + 200 hours) / 100 devices

We can use the formula for the sum of an arithmetic series:

S = n/2 * (a1 + an)

Where:

n is the number of terms in the series (here, 100 devices),

a1 is the first term in the series (100 hours),

an is the last term in the series (200 hours),

Now apply the formula:

MTBF = 100/2 * (100 + 200)

MTBF = 50 * 300

MTBF = 15,000 hours

So, the answer is b. 15,000 hours.

Answer:

The percentage of regional phone companies whose operating expenses were closer to the mean than the operating expenses of Company A were to the mean is 73%.

Step-by-step explanation:

The Company A's operating expenses were $27.00. This is $2.93 less than the regional mean.

[tex]\Delta E=29.93-27.00=2.93[/tex]

The companies whose operating expenses are closer to the mean are the ones that have expenses $2.93 below or above the mean.

The fraction of companies that are closer to the mean is equal to the proability of having expenses between those two limits:

[tex]z_1=(M-\mu)/\sigma=-2.93/2.65=-1.105\\\\z_2=+1.105[/tex]

[tex]P(|z|\leq1.105)=P(z\leq 1.105)-P(z<-1.105)=0.86542-0.13458=0.73[/tex]

To find the percentage of companies with operating expenses closer to the mean than Company A, calculate the Z score for Company A. Then find the percentile, and double it to account for both sides of the normal distribution.

To find out the percentage of regional phone companies that had operating expenses closer to the mean than the ones of Company A, we need to calculate the Z-score for Company A. The

Z-score

is a measure of how many standard deviations an element is from the mean. In this particular case, you can calculate the Z score using the formula:

Z = (X - μ) / σ

, where X is the value from the dataset (in this case, Company A's operating expenses), μ is the mean of the dataset, and σ is the standard deviation of the dataset.

So using the data from the question:

This will give you a percentage that can be understood as the percentage of regional phone companies which had operating expenses closer to the mean than Company A's were to the mean.

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To find the 35th percentile of LSAT scores, a z-score corresponding to the 35th percentile is needed, which can then be applied to the formula using the provided mean and standard deviation of LSAT scores.

To find the 35th percentile of LSAT scores, we need to use a normal distribution with the given mean and standard deviation. We use a z-table or a percentile calculator, looking to find the z-score that corresponds with the 35th percentile. Once the z-score is identified, we can use the mean and standard deviation of the LSAT scores to calculate the actual score corresponding to that percentile.

The process involves the following calculations:

Unfortunately, without the z-score or the use of applets, as suggested in the question, we cannot provide the exact LSAT score that corresponds to the 35th percentile.

Answer: [tex]x=\dfrac{25}{3}[/tex]

Step-by-step explanation:

The given equation : [tex]\log_2(3x+7)=5[/tex]

Using logarithmic property : [tex]\log_a N=M\to N=a^M[/tex]

The given equation will be equivalent to [tex](3x+7)=2^5[/tex]

[tex]\Rightarrow\ 3x+7=32[/tex]

Subtract 7 from both sides , we get

[tex]3x=25[/tex]

Divide both sides by 3 , we get

[tex]x=\dfrac{25}{3}[/tex]

Hence, the solution is [tex]x=\dfrac{25}{3}[/tex]

Answer:

[tex]t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507[/tex]  

[tex]p_v =P(t_{(15)}>1.507)=0.076[/tex]

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

[tex]\bar X_{CTS}=2.35[/tex] represent the mean for the sample CTS

[tex]\bar X_{N}=1.83[/tex] represent the mean for the sample Normal

[tex]s_{CTS}=0.88[/tex] represent the sample standard deviation for the sample of CTS

[tex]s_{N}=0.54[/tex] represent the sample standard deviation for the sample of Normal

[tex]n_{CTS}=10[/tex] sample size selected for the CTS

[tex]n_{N}=7[/tex] sample size selected for the Normal

[tex]\alpha=0.01[/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 mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{CTS} \leq \mu_{N}[/tex]

Alternative hypothesis:[tex]\mu_{CTS} > \mu_{N}[/tex]

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

[tex]t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}[/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".

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507[/tex]  

P-value

The first step is calculate the degrees of freedom, on this case:

[tex]df=n_{CTS}+n_{N}-2=10+7-2=15[/tex]

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

[tex]p_v =P(t_{(15)}>1.507)=0.076[/tex]

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Answer:

[tex]t=\frac{27.8-28.5}{\frac{4.1}{\sqrt{100}}}=-1.707[/tex]    

[tex]p_v =P(t_{99}<-1.707)=0.0455[/tex]

If we compare the p value with a significance level for example [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.

Step-by-step explanation:

Data given and notation    

[tex]\bar X=27.8[/tex] represent the mean for the account balances of a credit company

[tex]s=4.1[/tex] represent the population standard deviation for the sample    

[tex]n=1000[/tex] sample size    

[tex]\mu_o =28.5[/tex] represent the value that we want to test  

[tex]\alpha=0.1[/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 determine if the mean for the person consume is more than 28.5 gallons, the system of hypothesis would be:    

Null hypothesis:[tex]\mu \geq 28.5[/tex]    

Alternative hypothesis:[tex]\mu < 28.5[/tex]    

We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:    

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic    

We can replace in formula (1) the info given like this:    

[tex]t=\frac{27.8-28.5}{\frac{4.1}{\sqrt{100}}}=-1.707[/tex]    

Calculate the P-value    

First we need to calculate the degrees of freedom given by:

[tex]df=n-1=100-1=99[/tex]

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

[tex]p_v =P(t_{99}<-1.707)=0.0455[/tex]

In Excel we can use the following formula to find the p value "=T.DIST(-1.707,99)"  

Conclusion    

If we compare the p value with a significance level for example [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, so there is not enough evidence to conclude that the mean for the consumption is less than 28.5 gallons at 0.1 of significance, so we can reject the claim that person consume more than 28.5 gallons.

Answer:  0.084

Step-by-step explanation:

Formula to find standard error of [tex]\hat{p}[/tex] for finding confidence interval for p:

[tex]SE=\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where [tex]\hat{p}[/tex] = sample proportion and n= sample size.

Let p be the population proportion of students named math as their favorite class.

As per given , we have

n= 32

[tex]\hat{p}=\dfrac{21}{32}=0.65625[/tex]

Substitute these values in the formula, we get

[tex]SE=\sqrt{\dfrac{0.65625(1-0.65625)}{32}}\\\\=\sqrt{0.00705}\\\\=0.0839642781187\approx0.084[/tex]

∴ The correct value for the standard error of [tex]\hat{p}[/tex] in this case = 0.084

Final answer:

The standard error of pˆ in this case is 0.0417.

Explanation:

The correct value for the standard error of pˆ in this case is 0.0417.

To calculate the standard error of pˆ, you can use the formula: SE = √((pˆ * q) / n), where pˆ is the sample proportion, q is 1 - pˆ (the proportion of non-favorites), and n is the sample size.

In this case, pˆ = 21/32 = 0.6563, q = 1 - 0.6563 = 0.3437, and n = 32. Plugging these values into the formula gives SE = √((0.6563 * 0.3437) / 32) = 0.0417.

Answer:

The objective function in terms of the x-coordinate is [tex]d=\sqrt{82x^2+72x+16}[/tex].

The point closest to the origin is [tex](-\frac{18}{41},\frac{2\sqrt{82}}{41})[/tex].

Step-by-step explanation:

The formula for the distance from point (x, y) to the origin is

[tex]d=\sqrt{x^{2}+y^{2} }[/tex]

So, in our case, [tex]y=9x+4[/tex] and the distance is

[tex]d=\sqrt{x^{2}+(9x+4)^{2}  }\\\\d=\sqrt{x^2+81x^2+72x+16} \\\\d=\sqrt{82x^2+72x+16}[/tex]

This is the objective function.

Next, we need to find the derivative of the function

[tex]d=\sqrt{82x^2+72x+16}\\\\\frac{d}{dx}d= \frac{d}{dx}\sqrt{82x^2+72x+16}\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\f=\sqrt{u},\:\:u=82x^2+72x+16\\\\\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(82x^2+72x+16\right)\\\\\frac{1}{2\sqrt{u}}\left(164x+72\right)\\\\\mathrm{Substitute\:back}\:u=82x^2+72x+16\\\\\frac{1}{2\sqrt{82x^2+72x+16}}\left(164x+72\right)\\\\\frac{d}{dx}d=\frac{2\left(41x+18\right)}{\sqrt{82x^2+72x+16}}[/tex]

Now, we set the derivative function equal to zero to find the critical points

[tex]\frac{2\left(41x+18\right)}{\sqrt{82x^2+72x+16}}=0[/tex]

[tex]\frac{f\left(x\right)}{g\left(x\right)}=0\quad \Rightarrow \quad f\left(x\right)=0\\\\2\left(41x+18\right)=0\\\\\frac{2\left(41x+18\right)}{2}=\frac{0}{2}\\\\41x+18=0\\\\x=-\frac{18}{41}[/tex]

We need to check that the value that we found is a minimum point for this we analyze the intervals of increase or decrease (First derivative test).

We can use a sign chart. In a sign chart, we pick a test value at each interval that is bounded by the critical points and check the derivative's sign on that value.

This is the sign chart for our function:

[tex]\left\begin{array}{ccc}\mathrm{Interval}&\mathrm{Test \:x-value}&f'(x)\\(-\infty,-\frac{18}{41} )&-1&-9.02\\(-\frac{18}{41},\infty )&1&9.05\end{array}\right\\[/tex]

d(x) decreases before [tex]x=-\frac{18}{41}[/tex], increases after it, and is defined at [tex]x=-\frac{18}{41}[/tex]. So d(x) has a relative minimum point at [tex]x=-\frac{18}{41}[/tex].

The point closest to the origin is

[tex]d=y=\sqrt{82(-\frac{18}{41})^2+72(-\frac{18}{41})+16}=\frac{2\sqrt{82}}{41}[/tex]

[tex](-\frac{18}{41},\frac{2\sqrt{82}}{41})[/tex]

The objective function in this problem is the x-coordinate of the closest point on the line y = 9x + 4 to the origin. To find this point, we need to find the intersection of the line and the perpendicular line passing through the origin.

The objective function in this case is the distance between the closest point on the line y = 9x + 4 and the origin. To find this point, we need to find the intersection of the line and the perpendicular line passing through the origin. The x-coordinate of this closest point will be our objective function.

The slope of the given line is 9, the negative reciprocal of which is -1/9. This means the perpendicular line will have a slope of -1/9 as well. Since the perpendicular line passes through the origin, its equation is given by y = -1/9x.

To find the intersection point, we can set the equations of the two lines equal to each other: 9x + 4 = -1/9x. Solving this equation, we find x = -4/81. Therefore, the objective function in terms of the x-coordinate is -4/81.

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Final answer:

The 99% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars, based on a sample of 1418 buyers where 354 prefer foreign cars, is between 0.2205 and 0.2787.

Explanation:

To construct a 99% confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars, we use the sample proportion and the Z-distribution since the sample size is large. Given that in a sample of 1418 new car buyers, 354 preferred foreign cars, we calculate the sample proportion (p-hat) as follows:

p-hat = 354 / 1418 ≈ 0.2496

The formula for a confidence interval is p-hat ± Z*(sqrt((p-hat*(1-p-hat))/n)), where Z* is the Z-score corresponding to the confidence level, and n is the sample size. For a 99% confidence interval, Z* is approximately 2.576.

Using the formula, the standard error (SE) for the proportion is calculated as:

SE = sqrt((0.2496*(1-0.2496))/1418) ≈ 0.0113

The margin of error (ME) is:

ME = Z* * SE ≈ 2.576 * 0.0113 ≈ 0.0291

Now, we can construct the 99% confidence interval:

99% CI = p-hat ± ME = 0.2496 ± 0.0291 = (0.2205, 0.2787)

Therefore, we are 99% confident that the true proportion of new car buyers who prefer foreign cars over domestic cars is between 0.2205 and 0.2787.

Answer:

1) n=114

2) n=59

3) On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by [tex]\alpha=1-0.80=0.2[/tex] and [tex]\alpha/2 =0.1[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.28, z_{1-\alpha/2}=1.28[/tex]

Part 1

The margin of error for the proportion interval is given by this formula:  

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]    (a)  

And on this case we have that [tex]ME =\pm 0.06[/tex] or 6% points, and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

Since we don't have a prior estimate of [tex]\het p[/tex] we can use 0.5 as a good estimate, replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.06}{1.28})^2}=113.77[/tex]  

And rounded up we have that n=114

Part 2

On this case we have a prior estimate for the population proportion and is [tex]\hat p =0.85[/tex] so replacing the values into equation (b) we got:

[tex]n=\frac{0.85(1-0.85)}{(\frac{0.06}{1.28})^2}=58.027[/tex]

And rounded up we have that n=59

Part 3

On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

Answer:

±1, ±2, ±7, ±14

Step-by-step explanation:

Answer:68.3 degrees

Step-by-step explanation:

The diagram of the triangle ABC is shown in the attached photo. We would determine the length of side AB. It is equal to a. We would apply the cosine rule which is expressed as follows

c^2 = a^2 + b^2 - 2abCos C

Looking at the triangle,

b = 75 miles

a = 80 miles.

Angle ACB = 180 - 42 = 138 degrees. Therefore

c^2 = 80^2 + 75^2 - 2 × 80 × 75Cos 138

c^2 = 6400 + 5625 - 12000Cos 138

c^2 = 6400 + 5625 - 12000 × -0.7431

c^2 = 12025 + 8917.2

c = √20942.2 = 144.7

To determine A, we will apply sine rule

a/SinA = b/SinB = c/SinC. Therefore,

80/SinA = 144.7/Sin 138

80Sin 138 = 144.7 SinA

SinA = 53.528/144.7 = 0.3699

A = 21.7 degrees

Therefore, theta = 90 - 21.7

= 68.3 degees

Answer:

[tex]\large\boxed{x=\pm\dfrac{3\pi}{20}+\dfrac{2n\pi}{5}\ \vee\ x=\pm\dfrac{\pi}{20}+\dfrac{2n\pi}{5}}[/tex]

Step-by-step explanation:

[tex][tex]\cos^2(5x)=\sin^2(5x)\qquad\text{substitute}\ t=5x\\\\\cos^2t=\sin^2t\qquad\text{use}\ \sin^2\theta+\cos^2\theta=1\to\sin^2\theta=1-\cos^2\theta\\\\\cos^2t=1-\cos^2t\qquad\text{add}\ \cos^2t\ \text{to both sides}\\\\2\cos^2t=1\qquad\text{divide both sides by 2}\\\\\cos^2t=\dfrac{1}{2}\Rightarrow \cos t=\pm\sqrt{\dfrac{1}{2}}\\\\\cos t=\pm\dfrac{\sqrt2}{2}\\\\\cos t=-\dfrac{\sqrt2}{2}\Rightarrow t=\pm\dfrac{3\pi}{4}+2n\pi\\\\\cos t=\dfrac{\sqrt2}{2}\Rightarrow t=\pm\dfrac{\pi}{4}+2n\pi[/tex]

[tex]t=5x\\\\5x=\pm\dfrac{3\pi}{4}+2n\pi\ \vee\ 5x=\pm\dfrac{\pi}{4}+2n\pi\qquad\text{divide both sides by 5}\\\\x=\pm\dfrac{3\pi}{20}+\dfrac{2n\pi}{5}\ \vee\ x=\pm\dfrac{\pi}{20}+\dfrac{2n\pi}{5}[/tex]

The trigonometric equation 2cos^2(x) + 2 = 4 has solutions x = 0 and x = π in the interval [0, 2π). To include all possible solutions, we express them as x = 2nπ and x = π + 2nπ with n as any integer.

To solve the trigonometric equation 2cos^2(x) + 2 = 4, we start by simplifying the equation:

This equation has two solutions in the interval [0, 2π): x = 0 and x = π (which are 0 and 3.1416 when rounded to four decimal places). To express the infinite set of solutions that occur at every period of cos(x), we add 2nπ to each solution, where n is any integer, giving the final solutions as x = 2nπ and x = π + 2nπ.

the complete Question is given below:

Use trigonometric identities and algebraic methods, as necessary, to solve the following trigonometric equation. Please identify all possible solutions by including all answers in [0, 2π)

and indicating the remaining answers by using n to represent any integer. Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution."

2cos^2 (x) + 2 = 4

Enter your answer in radians, as an exact answer when possible. Multiple answers should be separated by commas.

The probability density function (pdf) is;

f(x)=[tex]\frac{1}{b - a}[/tex] for a ≤ x ≤ b.

while the mean is given as [tex]\frac{a + b}{2}[/tex]

And the standard deviation given as [tex]\sqrt{\frac{(b - a)^{2} }{12} }[/tex]

Answer: -None of these choices.

PDF = 1/(b-a)

Step-by-step explanation:

The probability density function f(x) of a random variable X that has a uniform distribution between a and b is given by:

PDF = 1/(b-a) for X€[a,b]

otherwise zero.

Answer:

To prove

∠ABC = ∠PQR

Using euclidean distance, Length of each side can be found as

[tex]AB=\sqrt{65} ,BC=\sqrt{45} ,CA=\sqrt{50} \\PQ=\sqrt{65} ,QR=\sqrt{45} ,RP=\sqrt{50}[/tex]

As can be seen

AB ≅ PQ

BC ≅ QR

CA ≅ RP

As all the sides of ΔABC are equal and congruent to ΔPQR, this Proves that measure of all angles inside both triangles must be equal.

Answer:

The 95% confidence interval is given by: (0.226, 0.240)  

On this case we can't conclude that we have a significant reduction on the reaction time since the upper bounf of the interval is higher than the value of 0.239.

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 =0.233[/tex] represent the sample mean for the sample  

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

s=0.021 represent the sample standard deviation  

n=41 represent the sample size  

2) Calculate the confidence interval

Since the sample size is large enough n>30 but we don't know the population deviation. 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)  

First we need to calculate the degrees of freedom given by:

[tex]df=n-1=41-1=40[/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(0.025,40)".And we see that [tex]t_{\alpha/2}=2.02[/tex]

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

[tex]0.233-2.02\frac{0.021}{\sqrt{41}}=0.226[/tex]    

[tex]0.233+2.02\frac{0.021}{\sqrt{41}}=0.240[/tex]

The 95% confidence interval is given by: (0.226, 0.240)  

On this case we can't conclude that we have a significant reduction on the reaction time since the upper bounf of the interval is higher than the value of 0.239.

Answer:

B is greater than A.

Step-by-step explanation:

We have been given that 40% of A is equal to $300 and 30% of B is equal to $400. We are asked to compare both quantities.

Let us find A and B using our given information.

[tex]\frac{40}{100}*A=300[/tex]

[tex]0.40A=300[/tex]

[tex]\frac{0.40A}{0.40}=\frac{300}{0.40}[/tex]

[tex]A=750[/tex]

Similarly, we will find B.

[tex]\frac{30}{100}*B=400[/tex]

[tex]0.30*B=400[/tex]

[tex]\frac{0.30B}{0.30}=\frac{400}{0.30}[/tex]

[tex]B=1333.3333[/tex]

We know that smaller percent of big number is greater than bigger percent of a small number.

Therefore, B is greater than A.

Answer:

Step-by-step explanation:

40% Of A = 300

30% of B = 400

40a / 100 = 300, 40a = 3  divide both sides by 3 ,

then a = 13.33 approximately 13

30b = 400 = 400, 30b = 4 divide both sides by 30, then b =7.5

comparing both results where a = 13 , b = 7.5

a is greater than b with difference of 5.5

In probability, if two events are mutually exclusive (cannot occur at the same time), the probability of either event occurring is calculated by adding the probabilities of each event separately.

The concept being referred to in your question is related to probability within the study of mathematics. When two events are mutually exclusive, it means they cannot occur at the same time. For example, when rolling a die, the events of throwing a '6' and a '3' are mutually exclusive; you cannot throw both on a single toss.

In reference to your question, the probability that one event or the other occurs, given that they are mutually exclusive, is calculated by adding the probabilities of each individual event. So if the probability of Event A is P(A) and the probability of Event B is P(B), then the probability that either A or B will occur (P(A U B)) equals P(A) + P(B).

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Answer:

177,827.941

Step-by-step explanation:

The magnitude of an earthquake is given by:

[tex]R=log(\frac{l}{l_0})[/tex]

The intensity of the Nephi earthquake is:

[tex]3.0=log(\frac{l_N}{l_0})\\10^3 =\frac{l}{l_0} \\l_N=1,000*l_0[/tex]

The intensity of the San Francisco earthquake is

[tex]8.25=log(\frac{l_S}{l_0})\\10^{8.25} =\frac{l_S}{l_0} \\l_S=177,827,941*l_0[/tex]

The intensity ratio is:

[tex]r= \frac{l_S}{l_N}=\frac{177,827,941*l_0}{1,000*l_0}\\r= 177,827.941[/tex]

The San Francisco earthquake was 177,827.941 times as intensive as the Nephi earthquake.

When comparing the intensity of the San Francisco earthquake (magnitude 8.25) and the Nephi earthquake (magnitude 3.0) on the Richter Scale, we find that the San Francisco earthquake was approximately 177,827.94 times more intensive than the Nephi earthquake.

To compare the intensity of two earthquakes using the Richter Scale, we need to apply the formula:
R = log10 I/Io, where R is the Richter magnitude, I is the intensity of the earthquake, and Io is a reference intensity.

Now, we know that the Richter scale is a base-10 logarithmic scale. This means that each whole number increase on the scale represents a tenfold increase in measured amplitude and roughly 31.6 times more energy release.

The earthquakes in question have magnitudes of 3.0 and 8.25. So, the intensity of the San Francisco earthquake compared to the Nephi earthquake is 10^(8.25-3) = 10^5.25.

Therefore, the San Francisco earthquake was approximately 177,827.94 times more intensive as the Nephi earthquake.

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Answer:

Since p > alpha, we accept H0, there is no evidence to prove that p1 is greater than p2

Step-by-step explanation:

Set up hypotheses as

[tex]H_0: p1 = p2\\H_a: p1>p2[/tex]

(Right tailed t test)

Alpha = 0.05

Sample               I                II                Total

X                        117             141              248

N                        249           312            561

p                        0.4700     0.4519        0.4420

p difference = 0.0181    

Std dev = [tex]\sqrt{p(1-p)(\frac{1}{n_1} +\frac{1}{n_2} )}[/tex]

=0.0427

z statistic = 0.424

p value = 0.33724

Since p > alpha, we accept H0, there is no evidence to prove that p1 is greater than p2

Final answer:

To test whether p1 is greater than p2 using hypothesis testing, one needs to state the null and alternative hypotheses, calculate the test statistic and p-value, and then decide whether to reject or accept the null hypothesis based on the level of significance and the p-value. Calculating the test statistic and p-value requires specific formulas and is not provided here.

Explanation:

Conducting a Hypothesis Test for Two Population Proportions

A student would like assistance in conducting a hypothesis test for two population proportions. The steps to perform such a test include:

State the null and alternative hypotheses: The null hypothesis (H0) is that there is no difference between the two population proportions (p1 - p2 = 0), while the alternative hypothesis (Ha) posits that there is a difference (p1 > p2).

The random variable (P') represents the difference between the two sample proportions.

Calculate the test statistic: Using the provided sample sizes and successful outcomes, the test statistic is calculated using a formula based on the Z-distribution.

Calculate the p-value: The p-value is determined from the test statistic, indicating the probability of observing such a result if the null hypothesis were true.

At the 5 percent level of significance, compare the p-value to the alpha value of 0.05 to make a decision. If the p-value is less than alpha, reject the null hypothesis.

The Type I error would occur if the null hypothesis is incorrectly rejected when it is actually true.

The Type II error would occur if the null hypothesis is not rejected when it is actually false.

The test statistic and p-value cannot be calculated from the information provided without the appropriate formulas or statistical tools.

Decision Making and Errors

If the alpha level is greater than the p-value, the null hypothesis should be rejected, indicating there is evidence to suggest p1 is greater than p2.

Failure to reject the null hypothesis when it is false constitutes a Type II error.

The particular solution to the differential equation is:

[tex]y_p(t)= 16/25tcos(5t) - 8/5tsin(5t)[/tex]

Here, we have,

To find a particular solution to the given differential equation

y′′+25y=−40sin(5t),

we'll assume a particular solution of the form:

[tex]y_p(t)=Asin(5t)+Bcos(5t)[/tex]

where A and B are constants to be determined.

Now, let's find the first and second derivatives of [tex]y_p(t)[/tex] :

[tex]y_p'(t)=5Acos(5t)-5Bsin(5t)\\y_p''(t)=-25Asin(5t)-25Bcos(5t)[/tex]

Now, substitute these derivatives back into the original differential equation:

y′′+25y=(−25Asin(5t)−25Bcos(5t))+25(Asin(5t)+Bcos(5t))

Now, equate the coefficient of sin(5t) and cos(5t) on both sides of the equation:

For sin(5t):

−25A+25A=0⟹0=0

For cos(5t):

−25B+25B=−40⟹0=−40

The above equations show that there is no solution for A and B that satisfy the original equation.

This means that our initial assumption for the particular solution is not appropriate for this case.

To find a particular solution that works, we'll make another assumption. Since the right-hand side of the differential equation is −40sin(5t), we'll try a particular solution in the form:

[tex]y_p(t)=Atcos(5t)+Btsin(5t)[/tex]

where A and B are constants to be determined.

Now, let's find the first and second derivatives of this new

[tex]y_p'(t)=Acos(5t)-5Atsin(5t)+Bsin(5t)+5Btcos(5t)[/tex]

[tex]y_p''(t)=-10Asin(5t)-25Atcos(5t)+25Btsin(5t)-10Bcos(5t)[/tex]

Now, substitute these derivatives back into the original differential equation:

[tex]y''+25y=(-10Asin(5t)-25Atcos(5t)+25Btsin(5t -10Bcos(5t))+25(Atcos(5t)+Btsin(5t))[/tex]

Now, equate the coefficient of sin(5t) and cos(5t) on both sides of the equation:

For sin(5t):

25Bt=−40⟹B=− 8/5

For cos(5t):

−25At−10B=0

​⟹A= 16/25

​So, the particular solution to the differential equation is:

[tex]y_p(t)= 16/25tcos(5t) - 8/5tsin(5t)[/tex]

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To find a particular solution to the differential equation y″+25y=−40sin⁡(5t), assume a particular solution in the form y(t) = Asin(5t) + Bcos(5t) and solve for the coefficients A and B.

To find a particular solution to the given differential equation y″+25y=−40sin⁡(5t), we can assume a particular solution in the form of y(t) = Asin(5t) + Bcos(5t). Differentiating this equation twice and substituting it back into the original differential equation, we can solve for the coefficients A and B. In this case, the particular solution is y(t) = -8sin(5t) - 0.64cos(5t).

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Answer:

We can conclude that the drying time in hours for type A is significantly higher than the drying time for type B. And the margin above it's between 4.90 and 17.5 hours at 2% of significance.

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

[tex]\bar X_1 =76.3[/tex] represent the sample mean 1

[tex]\bar X_2 =65.1[/tex] represent the sample mean 2

n1=11 represent the sample 1 size  

n2=9 represent the sample 2 size  

[tex]s_1 =4.5[/tex] sample standard deviation for sample 1

[tex]s_2 =5.1[/tex] sample standard deviation for sample 2

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

Confidence interval

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

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

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

[tex]\bar X_1 -\bar X_2 =76.3-65.1=11.2[/tex]

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

[tex]df=n_1 +n_2 -1=11+9-2=18[/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 tabel to find the critical value. The excel command would be: "=-T.INV(0.01,18)".And we see that [tex]t_{\alpha/2}=\pm 2.55[/tex]  

The standard error is given by the following formula:

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

And replacing we have:

[tex]SE=\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}=2.175[/tex]

Confidence interval

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

[tex]11.2-2.55\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}=5.65[/tex]  

[tex]11.2+2.55\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}=16.75[/tex]  

So on this case the 98% confidence interval would be given by [tex]5.65 \leq \mu_1 -\mu_2 \leq 16.75[/tex]  

But let's assume that the confidence interval given is true 4.90 hrs < μ1 - μ2 < 17.50 hrs

What does the confidence interval suggest about the population means?

We can conclude that the drying time in hours for type A is significantly higher than the drying time for type B. And the margin above it's between 4.90 and 17.5 hours at 2% of significance.

Answer:

Step-by-step explanation:

The ages for Academy Award winning best actors in order from smallest to largest are

18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77

The total number of terms, n is 32

31℅ of 32 = 31/100 × 32 = 0.31 × 32 = 9.92. It is approximately 10

Counting from left to right, the 10th term is 31. This means that the 31st percentile is 31

Answer

b. Type II error

Step-by-step explanation:

The hypothesis was formulated as  

the solution can be seen in the attached document

Answer:

[tex]z=\frac{0.247-0.15}{\sqrt{0.191(1-0.191)(\frac{1}{300}+\frac{1}{400})}}=3.23[/tex]    

[tex]p_v =P(Z>3.23)=0.000619[/tex]  

Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the proportion of Italians who prefer white champagne at​ weddings it's significantly higher than the proportion of Americans.  

Step-by-step explanation:

1) Data given and notation  

[tex]X_{I}=74[/tex] represent the number of Italians that preferred white​ champagne

[tex]X_{A}=60[/tex] represent the number of Americans that preferred white​ champagne

[tex]n_{I}=300[/tex] sample of Italians selected  

[tex]n_{A}=400[/tex] sample of Americans selected  

[tex]p_{I}=\frac{74}{300}=0.247[/tex] represent the proportion of Italians that preferred white​ champagne

[tex]p_{A}=\frac{60}{400}=0.15[/tex] represent the proportion of Americans that preferred white​ champagne

z would represent the statistic (variable of interest)  

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

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if a higher proportion of Italians than Americans prefer white champagne at​ weddings, the system of hypothesis would be:  

Null hypothesis:[tex]p_{I} - p_{A} \leq 0[/tex]  

Alternative hypothesis:[tex]p_{I} - p_{A} > 0[/tex]  

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

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

Where [tex]\hat p=\frac{X_{I}+X_{A}}{n_{I}+n_{A}}=\frac{74+60}{300+400}=0.191[/tex]  

3) Calculate the statistic  

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

[tex]z=\frac{0.247-0.15}{\sqrt{0.191(1-0.191)(\frac{1}{300}+\frac{1}{400})}}=3.23[/tex]    

4) Statistical decision

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

[tex]p_v =P(Z>3.23)=0.000619[/tex]  

Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the proportion of Italians who prefer white champagne at​ weddings it's significantly higher than the proportion of Americans.  

To determine whether a higher proportion of Italians than Americans prefer white champagne at weddings, we need to conduct a hypothesis test. First, state the null and alternative hypotheses. Then, calculate the test statistic and compare it to the critical value at a significance level of 0.05.

To determine whether a higher proportion of Italians than Americans prefer white champagne at weddings, we need to conduct a hypothesis test. First, we need to state the null hypothesis (H0) and the alternative hypothesis (Ha). In this case, H0: p1 = p2 (the proportion of Italians who prefer white champagne is equal to the proportion of Americans who prefer white champagne) and Ha: p1 > p2 (the proportion of Italians who prefer white champagne is greater than the proportion of Americans who prefer white champagne).

Next, we calculate the test statistic using the formula: z = (p1 - p2) / √(p*(1-p)*((1/n1) + (1/n2))), where p = (x1 + x2) / (n1 + n2), x1 and x2 are the number of Italians and Americans who prefer white champagne respectively, and n1 and n2 are the sample sizes of Italians and Americans respectively.

We then compare the test statistic to the critical value at a significance level of 0.05. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that a higher proportion of Italians than Americans prefer white champagne at weddings.

To find f(a+h), substitute a+h into the function f(x). To find the difference quotient, subtract f(a) from f(a+h) and divide by h. To find the instantaneous rate of change of f when a = 77, substitute a = 77 into f(a).

To find f(a+h), we substitute a+h into the function f(x).
f(a+h) = 4(a+h)^2 - 3(a+h) - 1
Expanding and simplifying:
f(a+h) = 4(a^2 + 2ah + h^2) - 3a - 3h - 1
f(a+h) = 4a^2 + 8ah + 4h^2 - 3a - 3h - 1

To find f(a), we substitute a into the function f(x).
f(a) = 4a^2 - 3a - 1

To find the difference quotient, we subtract f(a) from f(a+h) and divide by h:
(f(a+h) - f(a))/h = [(4a^2 + 8ah + 4h^2 - 3a - 3h - 1) - (4a^2 - 3a - 1)]/h
Expanding and simplifying:
(f(a+h) - f(a))/h = (8ah + 4h^2 - 3h)/h
(f(a+h) - f(a))/h = 8a + 4h - 3

To find the instantaneous rate of change of f when a = 77, we substitute a = 77 into f(a) and simplify:
f(77) = 4(77)^2 - 3(77) - 1
f(77) = 4(5929) - 231 - 1
f(77) = 23716 - 231 - 1
f(77) = 23484

Answer: Houston residences

Step-by-step explanation:

Here , The objective of the researcher is to determine the average number of vehicles are registered to a typical Houston residence.

Clearly , the population is this situation is "Houston residences" having vehicles.

Note : 250 randomly selected residences are defining the sample of the entire population of Houston residences which is a subset of population.

Hence, the correct answer is Houston residences.

Answer:

C. No. In warm weather, more children will go outside and play

Step-by-step explanation:

The correct answer is option C: No. In warm weather, more children will go outside and play.

Correlation means that two variables are related, but it does not necessarily mean that one causes the other. In this case, there is a strong correlation between temperature and the number of skinned knees on playgrounds, but it does not tell us that warm weather causes children to trip.

Option A is incorrect because warm weather does not directly cause children to go outside and play. It may be a factor, but it is not the sole reason. Option B is incorrect because warm weather does not cause fewer children to trip and suffer skinned knees. In fact, the correlation suggests that more children are likely to be outside playing in warm weather, which could potentially increase the number of skinned knees. Option D is incorrect because warm weather does not directly cause more children to trip and suffer skinned knees.

The correlation suggests that the increase in skinned knees is likely due to more children being outside and playing, rather than the warm weather itself. To summarize, the strong correlation between temperature and the number of skinned knees on playgrounds indicates that in warm weather, more children will go outside and play. However, it does not tell us that warm weather causes children to trip.

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Answer:

-any value in an interval or collection of intervals

Step-by-step explanation:

Discrete random variables are only integers in the interval.

Continuous random variables are all the values(integers, fractional, negative, positive) in an interval, or a collection of intervals.

The correct answer is:

-any value in an interval or collection of intervals

A continuous random variable can assume any value in an interval or collection of intervals. Unlike discrete random variables, which can only take integer values, continuous variables can take a range of values, like measurements such as height.

A continuous random variable is a type of variable in statistics that can assume any value within an interval or collection of intervals. This is different from a discrete random variable which can only take on a finite or countable number of values (often integer values).

For example, if we measure the height of a person, it could be any value within the feasible range, say between 0 meter and 3 meters. This is a continuous variable because there are infinite possibilities between 0 and 3, including measurements like 1.73 meters or 2.45 meters etc. It's not limited to integer values (like 1 m, 2 m, etc.), nor only fractional values, nor just positive integers within an interval.

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Answer:

a) [tex]P(5.8<X<8.2)=P(\frac{5.8-7}{1.2}<Z<\frac{8.2-7}{1.2})=P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.841-0.159=0.683[/tex]

b) [tex]P(5.8<\bar X <8.2) = P(Z<3)-P(Z<-3)=0.999-0.0014=0.9973[/tex]

c) For part a we are just finding the probability that an individual baby would have a weight between 5.8 and 8.2. So we can't compare the result of part a with the result for part b.

For part b we are finding the probability that the mean of 9 babies (from random sampling) would be between 5.8 and 8.2, so on this case we have a distribution with a different deviation depending on the sample size. And for this reason we have different values

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  The letter [tex]\phi(b)[/tex] is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: [tex]\phi(b)=P(z<b)[/tex]

Let X the random variable that represent the weights of​ full-term newborn babies of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(7,1.2)[/tex]

a. What is the probability that one newborn baby will have a weight within 1.2 pounds of the meanlong dashthat ​is, between 5.8 and 8.2 ​pounds, or within one standard deviation of the​ mean?

We are interested on this probability

[tex]P(5.8<X<8.2)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

[tex]P(5.8<X<8.2)=P(\frac{5.8-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{8.2-\mu}{\sigma})[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(5.8<X<8.2)=P(\frac{5.8-7}{1.2}<Z<\frac{8.2-7}{1.2})=P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.841-0.159=0.683[/tex]

b. What is the probability that the average of nine ​babies' weights will be within 1.2 pounds of the​ mean; will be between 5.8 and 8.2 ​pounds?

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

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

On this case  [tex]\bar X \sim N(7,\frac{1.2}{\sqrt{9}})[/tex]

The z score on this case is given by this formula:

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

And if we replace the values that we have we got:

[tex]z_1=\frac{5.8-7}{\frac{1.2}{\sqrt{9}}}=-3[/tex]

[tex]z_2=\frac{8.2-7}{\frac{1.2}{\sqrt{9}}}=3[/tex]

For this case we can use a table or excel to find the probability required:

[tex]P(5.8<\bar X <8.2) = P(Z<3)-P(Z<-3)=0.999-0.0014=0.9973[/tex]

c. Explain the difference between​ (a) and​ (b).

For part a we are just finding the probability that an individual baby would have a weight between 5.8 and 8.2. So we can't compare the result of part a with the result for part b.

For part b we are finding the probability that the mean of 9 babies (from random sampling) would be between 5.8 and 8.2, so on this case we have a distribution with a different deviation depending on the sample size. And for this reason we have different values