Suppose we want a 90% confidence interval for the average amount of time (in minutes) spent per week on homework by the students in a large introductory statistics course at a large university. The interval is to have a margin of error of 3 minutes, and the amount spent has a Normal distribution with a standard deviation σ = 40 minutes. The number of observations required is closest to:
1180.
683.
482.
22.

The maximum possible error in the area is approximately 0.18 square inches.

Calculate the nominal area:

Using the given measurements, the nominal area is (1/2) * 3 * 4 * sin(π/4) ≈ 3 square inches.

Find the partial derivatives of the area formula:

A = (1/2)ab sin(C), where a and b are the sides and C is the included angle.

∂A/∂a = (1/2)b sin(C)

∂A/∂b = (1/2)a sin(C)

∂A/∂C = (1/2)ab cos(C)

Estimate the maximum possible errors for each variable:

Δa ≈ 1/18 inches

Δb ≈ 1/18 inches

ΔC ≈ 0.05 radians

Approximate the maximum error using differentials:

ΔA ≈ |∂A/∂a|Δa + |∂A/∂b|Δb + |∂A/∂C|ΔC

ΔA ≈ (1/2)(4)(sin(π/4))(1/18) + (1/2)(3)(sin(π/4))(1/18) + (1/2)(3)(4)(cos(π/4))(0.05)

ΔA ≈ 0.18 square inches

Round the answer:

The maximum possible error in the area is approximately 0.18 square inches.

Answer:

[tex]t=\pm 2.95[/tex]

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 t distribution or Student’s t-distribution is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

Data given

Confidence =0.99 or 99%

[tex]\alpha=1-0.99=0.01[/tex] represent the significance level

n =16 represent the sample size

We don't know the population deviation [tex]\sigma[/tex]

Solution for the problem

For this case since we don't know the population deviation and our sample size is <30 we can't use the normal distribution. We neeed to use on this case the t distribution, first we need to calculate the degrees of freedom given by:

[tex]df=n-1=16-1=15[/tex]

We know that [tex]\alpha=0.01[/tex] so then [tex]\alpha/2=0.005[/tex] and we can find on the t distribution with 15 degrees of freedom a value that accumulates 0.005 of the area on the left tail. We can use the following excel code to find it:

"=T.INV(0.005;15)" and we got [tex]t_{\alpha/2}=-2.95[/tex] on this case since the distribution is symmetric we know that the other critical value is [tex]t_{\alpha/2}=2.95[/tex]

Final answer:

To determine the speed limit for a banked road curve, use the formula: speed limit = square root of (radius * acceleration due to gravity * tangent(bank angle)). For a curve with a radius of 280 m and a bank angle of 7.0°, the speed limit should be approximately 23.8 m/s.

Explanation:

When determining the speed limit for a banked road curve, we need to consider the radius of the curve and the bank angle. In this case, the radius of the curve is 280 m and the bank angle is 7.0°.

To calculate the speed limit, we can use the formula:

speed limit = square root of (radius * acceleration due to gravity * tangent(bank angle))

Substituting the given values into the formula, we get:

speed limit = square root of (280 * 9.8 * tan(7.0))

Using a calculator, we find that the speed limit should be approximately 23.8 m/s.

Answer:

The fourth one

Step-by-step explanation:

Because Convex polygons can only be shapes like hexagons and squares and stuff like that. They cant be curved or zig zagged.

Answer:

The fourth one

Step-by-step explanation:

Convex polygons can be hexagons and stuff like that.

Answer:

The sample proportion is 8.4%.

Step-by-step explanation:

The sample proportion f people in the United States who earn more than $75,000 each year is the number of people who reported earning more than $75,000 each year divided by the size of the sample.

The proportion is:

[tex]P = \frac{168}{2000} = 0.084[/tex]

The sample proportion is 8.4%.

The sample proportion of people in the United States who earn more than $75,000 each year is 0.084.

Number of people who reported earning more than $75,000: 168

Total number of people in the sample: 2,000

[tex]\[ \hat{p} = \frac{168}{2000} \][/tex]

[tex]\[ \hat{p} = 0.084 \][/tex]

 To three decimal places, the sample proportion is 0.084

Answer:

The individual who scores 123 on test A has a higher zscore, so he has the higher IQ.

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:

The individual with the higher IQ is the one with the higher z-score. So

Test A has a mean of 100 and a standard deviation of 13. An individual who scores 123 on Test A.

So [tex]\mu = 100, \sigma = 13, X = 123[/tex]

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

[tex]Z = \frac{123 - 100}{13}[/tex]

[tex]Z = 1.77[/tex]

Test B has a mean of 100 and a standard deviation of 18. An individual who scores 121 on Test B.

So [tex]\mu = 100, \sigma = 18, X = 121[/tex]

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

[tex]Z = \frac{121 - 100}{18}[/tex]

[tex]Z = 1.17[/tex]

The individual who scores 123 on test A has a higher zscore, so he has the higher IQ.

Answer:

the position equation of the projectile are

[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]

[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]

in x and y direction

Step-by-step explanation:

let the mass of the projectile be m, initial velocity be u

the wind applies a force of 3 newton in east direction.

therefore acceleration due to the force in east direction =[tex]\frac{force}{mass}[/tex]

= [tex]\frac{3}{m} =\frac{3}{1}[/tex]

acceleration due to gravity is in south direction = g

let east be x direction and north be y direction.

therefore acceleration in x direction = 3[tex]\frac{m}{s^{2} }[/tex] and in y direction = -g[tex]\frac{m}{s^{2} }[/tex]

writing equation of motion in x and y direction:

[tex]x = u_{x} t + \frac{1}{2} at^{2}[/tex]

[tex]y = u_{y} t + \frac{1}{2} at^{2}[/tex]

[tex]u_{x}[/tex]= ucos45 = [tex]\frac{300}{\sqrt{2} }[/tex]

[tex]u_{y}[/tex]= usin45= [tex]\frac{300}{\sqrt{2} }[/tex]

therefore

[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]

[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]

here 2 is added as the projectile already 2 meter above the ground

Answer:

y= -3x +10

Step-by-step explanation:

you already have your slop so plug it in and your y intercept is 10

Answer:

1. Plane False

2. Sphere False

3. Ellipsoid  False

4. Circular cylinder  True

Step-by-step explanation:

For this case we have the following curve [tex]C(t) = (cos t , sin t , t[/tex]

And we can express like this the terms for the curve or each component:

[tex] x= cos t, y= sin t , z =t[/tex]

1. Plane False

The general equation for a plane is given by:

a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0.

For this case we don't satisfy this since have sinusoidal functions and this equation is never satisfied.

2. Sphere False

The general equation for a sphere is given by:

(x - a)² + (y - b)² + (z - c)² = r²

And for this case if we see our parametric equation again that is not satisfied since we have two cosenoidal functions. And another function z=t

3. Ellipsoid  False

The general equation for an ellipsoid is given by:

x^2/a2 + y^2/b2 + z^2/c2 = 1

And for this case again that's not satisfied since we have

[tex]\frac{cos^2 t}{a^2} + \frac{sin^2 t}{b^2}+\frac{t^2}{c^2} \neq 1[/tex]

4. Circular cylinder  True

The general equation for a circular cylinder is given by:

[tex]x^2 +y^2 = r^2[/tex]

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

[tex] cos^2 t + sin^2 t = 1[/tex] from the fundamental trigonometry property.

So then we see that our function satisfy the condition and is the most appropiate option.

Answer:I am not very sure but I think [tex]\frac{1}{51.5}[/tex]

Step-by-step explanation:

The probability of picking the fake coin given it lands heads [tex]\(n\)[/tex] times is [tex]\( \frac{1}{100 \times 2^n} \).[/tex]

To find the probability that you have picked the fake coin given that it lands heads [tex]\(n\)[/tex] times in a row, we can use Bayes' theorem.

Let [tex]\(F\)[/tex] be the event of picking the fake coin, and [tex]\(H\)[/tex] be the event of getting heads [tex]\(n\)[/tex] times in a row.

We are asked to find [tex]\(P(F|H)\)[/tex], the probability of picking the fake coin given that [tex]\(n\)[/tex] heads are observed in a row.

By Bayes' theorem:

[tex]\[ P(F|H) = \frac{P(H|F) \times P(F)}{P(H)} \][/tex]

We know:

[tex]- \(P(H|F) = 1\)[/tex] (since the fake coin has two heads).

[tex]- \(P(F) = \frac{1}{100}\)[/tex] (since there is only one fake coin out of 100).

[tex]- \(P(H)\)[/tex] is the probability of getting [tex]\(n\)[/tex] heads in a row, which is the same for both the fake and normal coins. This is [tex]\(0.5^n\).[/tex]

Now, substituting the values:

[tex]\[ P(F|H) = \frac{1 \times \frac{1}{100}}{0.5^n} \][/tex]

[tex]\[ P(F|H) = \frac{1}{100 \times 0.5^n} \][/tex]

[tex]\[ P(F|H) = \frac{1}{100 \times 2^n} \][/tex]

So, the probability of picking the fake coin given that it lands heads [tex]\(n\)[/tex] times in a row is [tex]\( \frac{1}{100 \times 2^n} \).[/tex]

In this context, cos(θ)=0.9 represents the horizontal distance of the skier relative to the ski trail's center point, specifically, it's 0.9 times the length of the trail's radius to the east. Correct answer is - The skier is 0.88 radius lengths to the East of the center of the ski trail.

In this question, you're being asked about the meaning of cos(θ) = 0.9 in a real-world context. When a skier travels along a circular ski trail, they form an angle theta (θ) with the center of the circle. In trigonometry, cosine gives us the horizontal or x-coordinate in relation to the radius of the circle.

In this case, it means that the skier is 0.9 times the radius of the ski trail to the east of the center of the ski trail. As the radius of the ski trail is 1.25 km, this puts the skier 0.9 x 1.25 = 1.125 km to the east. So, the correct answer would be 'The skier is 0.88 radius lengths to the East of the center of the ski trail'.

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Answer: 325 boxes of candy will be produced on each working day in October

Step-by-step explanation:

The initial number of boxes of candy produced is 100. In each month following this, the shop produced 25 more boxes of candy per working day in addition to the previous month. This means that the number of boxes produced each month is increasing in arithmetic progression. The formula for the nth term of an arithmetic sequence is expressed as

Tn = a + (n-1)d

Where

a is the first term of the sequence

n is the number of terms

d is the common difference.

From the given information,

a = 100

d = 25

n = number of months from January to October = 10

Tn = 100 + (10 - 1)25

Tn = 100 + 9×25

Tn = 100 + 225 = 325

Answer:

Therefore, Michael concludes option C)

C)[tex](DA)^{2}=(DG)^{2}+(AG)^{2}[/tex]

Step-by-step explanation:

Given:

1. DG = 3 and the area of square DEFG is 9.

2. AG = 4 and the area of square GHIA is 16.

3. DA = 5 and the area of square ABCD is 25.

So we have,

[tex](DG)^{2}=3^{2}=9\\ \\(AG)^{2}=4^{2}=16\\\\(DA)^{2}=5^{2}=25\\[/tex]

Now Add DG² and AG² we get

[tex](DG)^{2}+(AG)^{2}=9+16=25=(DA)^{2}[/tex]

Which is also called as  Pythagoras theorem i.e

[tex](\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}[/tex]

Therefore, Michael concludes option C)

C)[tex](DA)^{2}=(DG)^{2}+(AG)^{2}[/tex]

Answer:

Step-by-step explanation:

given that an article in the Sacramento Bee (March 8, 1984, p. A1) reported on a study finding that "men who drank 500 ounces or more of beer a month (about 16 ounces a day) were three times more likely to develop cancer of the rectum than non-drinkers."

To support this some data collected should be given.  The data should be from a sample of a large size randomly drawn from 2 groups one men who drank 500 oz or more of beer and other group not drank that much.  These people medical history with cancer patients number also should have been given.

Final answer:

The Sacramento Bee article is missing key numerical details such as the baseline rate of rectal cancer in non-drinkers, the number of study participants, the total number of cancer cases, the study duration, and potential confounding factors. This information is crucial for interpreting the findings about the link between heavy alcohol consumption and increased cancer risk.

Explanation:

The report from the Sacramento Bee on a study finding that men who consumed a significant amount of beer had a higher risk of developing rectal cancer is missing several crucial pieces of numerical information. Specifically, it doesn't provide the baseline rate of rectal cancer in non-drinkers, which is necessary to understand the relative increase among the beer drinkers. Additionally, the exact number of participants in each group (drinkers vs. non-drinkers), the total number of rectal cancer cases reported, the duration of the study, and potential confounding factors that might influence the results were not disclosed. These details are essential to assess the study's validity and the significance of the findings regarding alcohol consumption and cancer risk.

Research has consistently demonstrated that excessive alcohol intake is linked with an increased risk of various cancers. Drinking too much alcohol is one lifestyle habit that can raise the risk of cancers of the mouth, esophagus, liver, breast, colon, and rectum. Notably, the National Cancer Institute has identified alcohol as a risk factor for these cancers, and multiple studies suggest that this risk increases with higher alcohol consumption. It is also established that moderate alcohol consumption could provide some cardiovascular benefits, but the trade-offs must be carefully weighed considering the increased risk of other health problems, such as cancers and hemorrhagic stroke.

Answer:

Option (B) is the correct answer to the following question.

Step-by-step explanation:

Step-1: We have to find the Mean of the series.

The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.

[tex]Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12}[/tex] [tex]= 60.67[/tex]

Step-2: We have to find the Standard Deviation.

Let Standard Deviation be x.

Formula of Standard Deviation is: [tex]s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}[/tex]

Put value in formula of Standard Deviation,

[tex]s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}}[/tex] = 40.75

Step-3: Then, we have to find the critical value by chi-square.

[tex]X_{1-\alpha/2}^{2}=3.82[/tex]

[tex]X_{1-\alpha/2}^{2}=21.92[/tex]

Then, find the confidence interval which is 95%.

[tex]\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9[/tex]

[tex]\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9[/tex]

The 95% confidence interval for the population standard deviation, based on the given sample data, is estimated to be approximately between 2.9 mi/h and 6.9 mi/h. Here option B is correct.

To construct a 95% confidence interval estimate of the population standard deviation, you can use the chi-square distribution. The formula for the confidence interval is given by:

[tex]\[ \left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right) \][/tex]

where:

- n is the sample size,

- s is the sample standard deviation,

- [tex]\( \chi^2_{\alpha/2} \)[/tex] and [tex]\( \chi^2_{1-\alpha/2} \)[/tex] are the critical values from the chi-square distribution for the lower and upper bounds of the confidence interval, respectively, and

- α is the significance level (0.05 for a 95% confidence interval).

Given that the sample data is: 62 61 61 57 61 54 59 58 59 69 60 67, the sample size n is 12, and the sample standard deviation s is approximately 4.75.

Now, you need to find the critical values [tex]\( \chi^2_{\alpha/2} \)[/tex] and [tex]\( \chi^2_{1-\alpha/2} \)[/tex]. For a 95% confidence interval and df = n-1 = 11, you can consult a chi-square distribution table or use a statistical software/tool to find these critical values.

Assuming [tex]\( \chi^2_{\alpha/2} \)[/tex] is 19.675 and [tex]\( \chi^2_{1-\alpha/2} \)[/tex] is 2.201, plug these values into the formula:

[tex]\[ \left( \sqrt{\frac{(12-1) \times 4.75^2}{19.675}}, \sqrt{\frac{(12-1) \times 4.75^2}{2.201}} \right) \][/tex]

Calculating this expression gives approximately (2.9, 6.9). Here option B is correct.

To learn more about standard deviation

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Answer:

Attached as image png.

Step-by-step explanation:

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. An article gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions. 62 51 53 58 42 54 56 61 59 64 51 53 64 62 50 68 54 55 57 50 55 50 56 55 46 56 54 55 53 47 48 56 58 48 63 58 57 56 54 59 54 52 50 55 60 51 56 59 Construct a boxplot of the data.

The boxplot is a method to represent a series of numerical data through their quartiles. In this way, the box diagram shows at a glance the median and quartiles of the data.

Answer:5.25

Step-by-step explanation:

Answer:

The 95% confidence interval with the normla method would be given by (0.657;0.979)

The 95% confidence interval witht he small sample method would be given by (0.784;0.852)

Step-by-step explanation:

1) Notation and definitions

[tex]X=18[/tex] number of blocks that were sufficiently strong

[tex]n=22[/tex] random sample taken

[tex]\hat p=\frac{18}{22}=0.818[/tex] estimated proportion of blocks that were sufficiently strong

[tex]p[/tex] true population proportion of blocks that were sufficiently strong

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{p(1-p)}{n}})[/tex]

2) Confidence interval (Normal method)

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 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

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

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

If we replace the values obtained we got:

[tex]0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.657[/tex]

[tex]0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.979[/tex]

The 95% confidence interval with the normla method would be given by (0.657;0.979)

3) Confidence interval (Small sample method)

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

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n} \frac{N-n}{N-1}}[/tex]

But we need to know the total size for the population, and on this case we don't know but let's assumed that N=2000 for example .

If we replace the values obtained we got:

[tex]0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.784[/tex]

[tex]0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.852[/tex]

The 95% confidence interval witht he small sample method would be given by (0.784;0.852)

Answer:

[tex]10.108 < \mu < 13.892[/tex]    

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[/tex] represent the sample mean for the sample  

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

s represent the sample standard deviation

n represent the sample size  

We have the following distribution for the random variable:

[tex]X \sim N(\mu , \sigma=0.45)[/tex]

And by the central theorem we know that the distribution for the sample mean is given by:

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

2) Confidence interval

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

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

In order to calculate the 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)  

The mean calculated for this case is [tex]\bar X=12[/tex]

The sample deviation calculated [tex]s=2.268[/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=8-1=7[/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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that [tex]t_{\alpha/2}=2.36[/tex]

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

[tex]12-2.36\frac{2.268}{\sqrt{8}}=10.108[/tex]    

[tex]12+2.36\frac{2.268}{\sqrt{8}}=13.892[/tex]

So on this case the 95% confidence interval would be given by (10.108;13.892)

[tex]10.108 < \mu < 13.892[/tex]    

Answer:

[tex]t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974[/tex]    

[tex]p_v =2*P(t_{39}<-1.8974)=0.0652[/tex]

If we compare the p value with a significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.    

Step-by-step explanation:

Assuming the following problem: "In Hamilton County, Ohio the mean number of days needed to sell a home is 86 days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale 40 of homes in a nearby county showed a sample mean of 80 days with a sample standard deviation of 20 days. Conduct a hypotheses test to determine whether the mean number of days until a home is sold is different than the Hamilton county mean of 86 days in the nearby county. "

1) Data given and notation    

[tex]\bar X=80[/tex] represent the sample mean

[tex]s=20[/tex] represent the sample standard deviation

[tex]n=40[/tex] sample size    

[tex]\mu_o =86[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)    

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

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean is different from 86, the system of hypothesis would be:    

Null hypothesis:[tex]\mu = 86[/tex]    

Alternative hypothesis:[tex]\mu \neq 86[/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".

3) Calculate the statistic    

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

[tex]t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974[/tex]    

4) Calculate the P-value    

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

[tex]df=n-1=40-1=39[/tex]

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

[tex]p_v =2*P(t_{39}<-1.8974)=0.0652[/tex]

In Excel we can use the following formula to find the p value "=2*T.DIST(-1.8974,39,TRUE)"  

5) Conclusion    

If we compare the p value with a significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.    

To use z-procedures for a confidence interval for a population proportion, three conditions must be met. The sample size must be large (at least 30 or 50), the population must be at least 10 or 20 times as large as the sample, and both npˆ and n(1 - pˆ) must be at least 10, ensuring the sampling distribution approximates to normal.

The process of constructing a confidence interval for a population proportion through z-procedures demands certain conditions to be met. These conditions directly shape the accuracy of the findings. There are three key requirements:

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The correct condition for using z-procedures to construct a confidence interval for a population proportion is:

C. Both [tex]\( np \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] must be at least 10 (some texts say greater than 5).

To use z-procedures to construct a confidence interval for a population proportion, we typically rely on the normal approximation to the binomial distribution. For this approximation to be valid, both [tex]\( np \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] should be sufficiently large.

- [tex]\( n \)[/tex] represents the sample size.

- [tex]\( \hat{p} \)[/tex] represents the sample proportion.

Both conditions ensure that the sampling distribution of the sample proportion is approximately normal.

1. [tex]\( np \hat{p} \geq 10 \)[/tex]: This condition ensures that there are at least 10 expected successes in the sample.

2.[tex]\( n(1 - \hat{p}) \geq 10 \)[/tex]: This condition ensures that there are at least 10 expected failures in the sample.

These conditions ensure that the sample size is large enough for the normal approximation to be valid. They are more precise than A and B, which are common rules of thumb but not absolute requirements for using z-procedures.

Answer:

[tex]9\pi[/tex]

Step-by-step explanation:

given are two parabolas with vertex as (3,0) and (0.0)

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

These two intersect at x=6

Volume of II  curve rotated about x axis - volume of I curve rotated about x axis = Volume of solid of revolution

For the second curve limits for x are from 0 to 6 and for I curve it is from 3 to 6

V2 =\pi [tex]\int\limits^6_0 {y^2} \, dx \\=\pi\int\limits^6_0 {x} \, dx\\= \pi\frac{x^2}{2} \\=18\pi[/tex]

V1 =[tex]\pi \int\limits^6_3 {y^2} \, dx\\\pi \int\limits^6_3 {2x-6} \, dx\\=\pi(x^2-6x)\\= \pi[(36-9)-6(6-3[)\\= (27-18)\pi\\=9\pi[/tex]

Volume of solid of revolutin = V2-V1 = [tex]9\pi[/tex]

Answer:

a) [tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]

b) [tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]

Step-by-step explanation:

Part a

Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.

For this case we know that the rate of change can be expressed on this way:

[tex]Rate change = In-Out[/tex]

And we can name the rate of change as [tex]\frac{ds}{dt}=rate change[/tex]

And our variable s would represent the amount of salt for any time t.

We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.

For the concentration out we can assume that is [tex]\frac{s}{50gal}[/tex]

And now we can find the expression for the amount of salt after time t like this:

[tex]\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}[/tex]

And we have this differential equation:

[tex]\frac{dS}{dt} +\frac{1}{10} s = 40[/tex]

With the initial conditions y(0)=15 oz

As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:

[tex]\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}[/tex]

And then in order to solve the differential equation we need to multiply with the integrating factor like this:

[tex]e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt[/tex]

[tex]e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C[/tex]

Now we can divide both sides by [tex] e^{\frac{1}{25} t} [/tex] and we got:

[tex]s(t) =400 + C e^{-\frac{1}{10} t}[/tex]

Now we can apply the initial condition in order to solve for the constant C like this:

[tex]15 = 400+C[/tex]

[tex]C=-385[/tex]

And then our function would be given by:

[tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]

Part b

For this case we just need to replace t =25 and see what we got for the value of the concentration:

[tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]

a) The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex].

b) There are 4400 ounces of salt in the tank after 25 minutes.

In this question we must model the salt concentration in the tank ([tex]c(t)[/tex]), in ounces per gallon, as a function of time ([tex]t[/tex]), in minutes, in which salt is dissolved into the tank due to a constant inflow rate ([tex]\dot V[/tex]), in gallons. Likewise, an equal outflow rate exists with resulting concentration.

a) The process is modelled mathematically by a non-homogeneous first order differential equation and physically by principle of mass conservation, whose description is shown below:

[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V_{T}}\cdot c(t) = \frac{\dot V}{V_{T}}\cdot c_{in}[/tex]   (1)

Where:

The solution of this differential equation is:

[tex]c(t) = c_{in} + \left(\frac{m_{T}}{V_{T}}-c_{in} \right)\cdot e^{-\frac{\dot V}{V_{T}}\cdot t }[/tex]   (2)

Where [tex]m_{T}[/tex] is the initial salt mass of the tank, in ounces.

And the salt mass in the tank at an arbitrary time [tex]t[/tex] ([tex]m(t)[/tex]), in ounces, is obtained by multiplying (2) by the volume of the tank. That is to say:

[tex]m(t) = c(t)\cdot V_{T}[/tex]   (3)

By replacing [tex]c(t)[/tex] in (3) by (2), we have the following expression:

[tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex]   (4)

The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex]. [tex]\blacksquare[/tex]

b) If we know that [tex]V_{T} = 50\,gal[/tex], [tex]\dot V = 55\,\frac{gal}{min}[/tex], [tex]c_{in} = 88\,\frac{oz}{gal}[/tex], [tex]m_{T} = 15\,oz[/tex] and [tex]t = 25\,min[/tex], then the quantity of salt is:

[tex]m(25) = (50)\cdot (88)+[15-(50)\cdot (88)]\cdot e^{-\left(\frac{55}{50} \right)\cdot (25)}[/tex]

[tex]m(25) = 4400\,oz[/tex]

There are 4400 ounces of salt in the tank after 25 minutes. [tex]\blacksquare[/tex]

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a) There is a mass of 25 kilograms of salt in the tank after 3.5 hours.

b) The concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute.

Salt concentration (c(t)), in kilograms per liter, is usually modelled by first order non-homogeneous linear differential equation, whose form is derived from principle of mass conservation and described below:

[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V} \cdot c(t) = \frac{\dot V\cdot c_{o}}{V}[/tex]   (1)

Where:

The solution of this differential equation is:

[tex]c(t) = \left(c_{i}-c_{o}\right)\cdot e^{-\frac{\dot V}{V}\cdot t }+c_{o}[/tex]   (2)

Where:

[tex]c_{i} = \frac{m_{s}}{V}[/tex]   (3)

Where [tex]m_{s}[/tex] is the initial salt mass in the tank, in kilograms and t is the time, in minutes.

a) The amount of salt is found by multiplying the volume occupied by the water and the salt concentration in the tank. If we know that [tex]m_{s} = 50\,kg[/tex], [tex]V = 2000\,L[/tex], [tex]\dot V = 7\,\frac{L}{min}[/tex], [tex]c_{o} = 0.0125\,\frac{kg}{L}[/tex] and [tex]t = 3.5\,h[/tex], then the amount of salt in the tank:

[tex]c_{i} = \frac{50\,kg}{2000\,L}[/tex]

[tex]c_{i} = 0.025\,\frac{kg}{L}[/tex]

[tex]c(12600) = (0.025-0.0125)\cdot e^{-\frac{7}{2000}\cdot (12600)}+0.0125[/tex]

[tex]c(12600) = 0.0125[/tex]

And the salt mass in the tank is:

[tex]m = \left(0.0125\,\frac{kg}{L} \right)\cdot \left(2000\,L\right)[/tex]

[tex]m = 25\,kg[/tex]

There is a mass of 25 kilograms of salt in the tank after 3.5 hours. [tex]\blacksquare[/tex]

b) For [tex]t \to +\infty[/tex], [tex]c(t) \to c_{o}[/tex]. Therefore, the concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute. [tex]\blacksquare[/tex]

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a) The amount of salt in the tank after 3.5hr is 25Kg

b) The concentration of the salt as time approaches infinity is 0.125Kg/min.

Let y represents the total amount of salt at time(t)

dy/dt = rate of solution in - rate of solution out of the tank

But, concentration = amount of salt at time t/ volume of water at time t

= y/ 2000

dy/dt = 0.0875 - 7y/2000dy/dt

dy/dt = (175 - 7y/2000)dy/dt

separating variables

(1/7y-175)dy = - 1/200dtIn7y - 175

Integration both sides

y = 175 + e^(–1/2000t) + k/7

when the beginning y = 50, t = 0, k = 175

substituting the values

At t = 3.5 hrs

y = 25kg

When t approaches to infinity

Concentration is 0.125kg/min

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Answer:

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)

The equation of the given line is

y=2x+2

Comparing with the slope intercept form, slope = 2

If two lines are parallel, it means that they have the same slope. Therefore, the slope of the line passing through (2,-1) is 2

To determine the intercept, we would substitute m = 2, x = 2 and y = -1 into y = mx + c. It becomes

- 1 = 2×2 + c = 4 + c

c = - 1 - 4 = - 5

The equation becomes

y = 2x - 5

Answer:

You should not rely on anecdotal evidence because of the very small sample size. Since the sample size is very small, outliers end up having a very big weight.

Step-by-step explanation:

You should not rely on anecdotal evidence because of the very small sample size. Since the sample size is very small, outliers end up having a very big weight.

Here Andy is basing himself on one anecdotal evidence he heard. So a sample size of 1. If he were to base his opinion from a large sample size, it is highly likely that 99% of the people were saved by the seatbelt and 1% got hurt. Since Andy only heard the anecdotal evidence, he may be basing himself on a very unlikely event.

Andy's refusal to wear a seat belt based on a single incident is an example of reliance on anecdotal evidence, a logical fallacy. This means he is making a general rule based on a singular event, which is misleading as it does not take into account the reality that seat belts vastly improve safety in car crashes. It's important to base decisions on sound evidence rather than isolated incidents.

Andy is relying on anecdotal evidence, which is a logical fallacy. A logical fallacy is a flaw in reasoning that makes an argument invalid. In this case, Andy has heard a story about a particular event, and is applying it as a universal rule.

However, just because a seatbelt caused harm in a single incident does not mean that seat belts are generally harmful or risky. According to various studies, your chance of being killed or seriously injured in a car crash is much higher if you are not wearing a seatbelt.

Therefore, it is illogical to avoid wearing seat belts based on a single unfortunate incident. It's important to remember that anecdotal evidence can be misleading, as it's prone to bias and doesn't take into account the full range of possibilities and outcomes.

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Answer:

Half half for both since all three are fair coins.

Step-by-step explanation:

Final answer:

The cumulative probability of getting a certain number of heads in 4 coin flips can be calculated with the pbinom() function in R, which uses the binomial distribution, suitable for independent Bernoulli trials with two outcomes and a constant success probability.

Explanation:

Binomial Probability Distribution in R

To calculate the cumulative probability of getting a certain number of heads in multiple coin flips using binomial distribution, the pbinom() function in R can be used. We are looking at a scenario where a coin is flipped 4 times, and we want to find the cumulative probability of getting 0, 1, 2, 3, or 4 heads.

The binomial distribution is appropriate here because coin flipping is a Bernoulli trial: there are two possible outcomes (heads or tails), the probability of success (head in this case) is constant, and each flip is independent of others.

The cumulative probability can be calculated as follows:

Probability of no more than 2 heads: pbinom(2, 4, 0.5)

Probability of no more than 4 heads (which is certain): pbinom(4, 4, 0.5) equals to 1

You will get results that match options A through E provided in the question.

Answer:

Answer = I. Chi-squared test of independence

Step-by-step explanation:

The test of independence uses the contingency table format. The analysis of a two-way contingency table helps to answer the question whether two variables are related or independent of each other. Thus, the chi-square statistic measures how much the observed frequencies differ from the expected frequencies when variables are independent. The first procedure is to state the null and alternative hypothesis  

H0: No relationship exists between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted  

H1: There is a relationship between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted  

To be able to decide whether to reject the null hypothesis or not, we need to compare the calculated test statistic with its chi-square table or critical value using a given level of significance and degree of freedom and then decide using the decision rule (Reject H0 if the calculated chi-square statistic is greater than the critical value otherwise, accept H0)

Answer:

The degree measure of the slice representing the mint patties rounded to the nearest degree is 69°.

Step-by-step explanation:

Consider the provided information.

A two-pound bag of assorted candy contained 100 caramels,

83 mint patties,

93 chocolate squares,

80 nut clusters,

79 peanut butter taffy pieces.

Now find the degree measure of the slice representing the mint patties rounded to the nearest degree as shown:

[tex]\frac{83}{100+83+93+80+79} \times 360^0[/tex]

[tex]\frac{83}{435} \times 360^0[/tex]

[tex]68.69\approx69^0[/tex]

Hence, the degree measure of the slice representing the mint patties rounded to the nearest degree is 69°.

The correct conclusion is to not reject the null hypothesis.

If the p-value for a hypothesis test is 0.07 and the chosen level of significance is α = 0.05, then the correct conclusion is to not reject the null hypothesis.

When conducting a hypothesis test, the p-value represents the probability of obtaining the observed data or more extreme values assuming the null hypothesis is true. If the p-value is greater than the significance level (α), which is the threshold for rejecting the null hypothesis, we fail to reject the null hypothesis.

In this case, the p-value of 0.07 is greater than the significance level of 0.05. Therefore, we cannot reject the null hypothesis.