An insurance company is interested in conducting a study to to estimate the population proportion of teenagers who obtain a driving permit within 1 year of their 16th birthday. A level of confidence of 99% will be used and an error of no more than .04 is desired. There is no knowledge as to what the population proportion will be. The size of sample should be at least _______.

Answer:

98% Confidence Interval: (2.387,9.113 )

Step-by-step explanation:

We are given the following data set:

3, 8, 3, 5, 1, 13, 9, 2, 7, 3, 13, 2

a) Formula:

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

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

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

[tex]Mean =\displaystyle\frac{69}{12} = 5.75[/tex]

Sum of squares of differences = 196.25

[tex]S.D = \sqrt{\frac{196.25}{11}} = 4.044[/tex]

b) 98% Confidence Interval:

[tex]\bar{x} \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]  

Putting the values, we get,  

[tex]z_{critical}\text{ at}~\alpha_{0.02} = \pm 2.33[/tex]  

[tex]5.75 \pm 2.33(\displaystyle\frac{5}{\sqrt{12}} ) = 5.75 \pm 3.363 = (2.387,9.113 )[/tex]

Answer: the amount of money invested in the first property is $140000

the amount of money invested in the second is property is $62353

the amount of money invested in the third is property is $20235

Step-by-step explanation:

Let x represent the amount of money invested in the first property.

Let y represent the amount of money invested in the second property.

Let z represent the amount of money invested in the third property.

The man has S400,000 invested in three rental properties. This means that

x + y + z = 400000 - - - - - - - - -1

The first property earns 7.5% per year on the investment, the second earns 8%, and the third earns 9%. The total annual earnings from the three properties is S33,700. This means that

7.5/100 × x + 8/100×y + 9/100×z = 33700

0.075x + 0.08y + 0.09z = 33700 - - - - - 2

The amount invested at 9% equals the sum of the first two investments. This means that

z = x + y - - - - - - - - - - -3

Substituting equation 3 into equation 1 and equation 2, it becomes

x + y + x + y = 400000

2x + 2y = 400000 - - - - - - - - 4

0.075x + 0.08y + 0.09(x + y) = 33700

0.075x + 0.08y + 0.09x + 0.09y = 33700

0.165x + 0.17y = 33700 - - - - - - - - 5

Multiplying equation 4 by 0.165 and equation by 2, it becomes

0.33x + 0.33y = 66000

0.33x + 0.34y = 67400

Subtracting

- 0.01x= - 1400

x = $140000

Substituting x = 140000 into equation 5, it becomes

0. 165(140000) + 0.17y = 33700

23100 + 0.17y = 33700

0.17y = 33700 - 23100 = 10600

y = 10600/0.17 = $62353

z = x + y = 140000 + 62353 = $202353

Answer:

the answer is D

Step-by-step explanation:

Think of the Pythagorean Theorem which states that a^2 + b^2 = c^2. The Pythagorean Identities used in trigonometry are the angle version which can be used to simplify expressions.  

HOPE THIS HELPED ;3

The formula of half angle identities is sin²x = (1-cos(x/2))/2 and cos²x = (1+cos(x/2))/2.

An angle is the formed when two straight lines meet at one point, it is denoted by θ.

The given terms are,

sin²x and cos²x

To integrate sin²x, the formula is used,

sin²x = (1-cos(x/2))/2

To integrate cos²x, the formula is used,

cos²x = (1+cos(x/2))/2

The formula is, sin²x = (1-cos(x/2))/2 and  cos²x = (1+cos(x/2))/2.

To learn more about Angle on:

https://brainly.com/question/28451077

#SPJ5

Answer:

The probability of such a tire lasting more than 60,000 miles is 0.0228, for the complete question provided in explanation.

Step-by-step explanation:

The lifetime of a certain type of car tire are normally distributed. The mean lifetime of a car tire is 50,000 miles with a standard deviation of 5,000 miles. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?

This is the question of normal distribution:

First w calculate the value of Z corresponding to X = 60,000 miles

We, have; Mean = μ = 50,000 miles, and Standard Deviation = σ = 5,000 miles

Now, for Z, we know that:

Z = (x-μ)/σ

Z = (60,000 - 50,000)/5,000

Z = 2

Now, we have standard tables, for normal distribution in terms of values of Z. One is attached in this answer.

P(X > 60,000) = P(Z > 2) = 1 - P(Z = 2)

P(X > 60,000) = 1 - 0.9772

P(X > 60,000) = 0.0228

Answer:

B. H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0

[tex]z=\frac{0.447-0.414}{\sqrt{0.431(1-0.431)(\frac{1}{150}+\frac{1}{140})}}=0.57[/tex]    

[tex]z_{crit}=1.64[/tex]

A. Do not reject H0; there is no increase in the proportion of people using LinkedIn

Step-by-step explanation:

1) Data given and notation  

[tex]X_{1}=67[/tex] represent the number of recent jobseekers

[tex]X_{2}=58[/tex] represent the number of job seekers three years ago.

[tex]n_{1}=150[/tex] sample of recent jobseekers selected  

[tex]n_{2}=140[/tex] sample of job seekers three years ago selected  

[tex]p_{1}=\frac{67}{150}=0.4468[/tex] represent the proportion of recent jobseekers

[tex]p_{2}=\frac{58}{140}=0.4143[/tex] represent the proportion of job seekers three years ago

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 "More people are using social media to network, rather than phone calls or e-mails", the system of hypothesis would be:  

Null hypothesis:[tex]p_{1} - p_{2} \leq 0[/tex]  

Alternative hypothesis:[tex]p_{1} - p_{2} > 0[/tex]  

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

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)  

Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{67+58}{150+140}=0.4310[/tex]  

3) Calculate the statistic  

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

[tex]z=\frac{0.4468-0.4143}{\sqrt{0.4310(1-0.4310)(\frac{1}{150}+\frac{1}{140})}}=0.5671[/tex]    

In order to find the critical value since we have a right tailed test the we need to find a value on the z distribution that accumulates 0.05 of the area on the right tail, and this value is[tex]z_{crit}=1.64[/tex].

4) Statistical decision

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

[tex]p_v =P(Z>0.5671)= 0.285[/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 FAIL to reject the null hypothesis.

So the correct conclusion would be:

A. Do not reject H0; there is no increase in the proportion of people using LinkedIn

The equation to represent this situation is 8x = 32

Solution:

Given that ureg placed 32 chairs in the auditorium

There are 8 chairs in each row

To find: equation used to represent this situation

From given information,

1 row = 8 chairs

So let us find an expression to determine the number of rows to place 32 chairs

Let "x" be the number of rows required to place 32 chairs

Since 1 row contains 8 chairs, expression to determine the number of rows to place 32 chairs is given as:

[tex]8 \times \text{ number of row } = 32[/tex]

8x = 32

Thus the equation to represent this situation is 8x = 32

Answer:

a) [tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2[/tex]  

[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311[/tex]  

b) The 90% confidence interval would be given by (63.330;81.070)

[tex]63.330 < \mu_{left}- \mu_{right} <81.070[/tex]  

Step-by-step explanation:

1) Previous concepts  and notation

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".

Let put some notation  

x=left arm , y = right arm  

x: 175 169 182 146 144  

y: 102 101 94 79 79

The first step is define the difference [tex]d_i=x_i-y_i[/tex], that is given so we have:

d:  73, 68, 88, 67, 65

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2[/tex]  

The third step would be calculate the standard deviation for the differences, and we got:  

[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311[/tex]  

2) Confidence interval

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

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

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

[tex]df=n-1=5-1=4[/tex]  

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,4)".And we see that [tex]t_{\alpha/2}=2.13[/tex]  

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

[tex]72.2-2.13\frac{9.311}{\sqrt{5}}=63.330[/tex]  

[tex]72.2+2.13\frac{9.311}{\sqrt{5}}=81.070[/tex]  

So on this case the 90% confidence interval would be given by (63.330;81.070)

[tex]63.330 < \mu_{left}- \mu_{right} <81.070[/tex]  

Answer:

a.

With n = 25, [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.3930999748549[/tex]

With n = 50, [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.1519320308594[/tex]

b. [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.0715427161943[/tex]

c.

The absolute error in the trapezoid rule is 0.08047

The absolute error in the Simpson's rule is 0.00008

Step-by-step explanation:

a. To approximate the integral [tex]\int_{0}^{1}e^{6 x}\ dx[/tex] using n = 25 with the trapezoid rule you must:

The trapezoidal rule states that

[tex]\int_{a}^{b}f(x)dx\approx\frac{\Delta{x}}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)\right)[/tex]

where [tex]\Delta{x}=\frac{b-a}{n}[/tex]

We have that a = 0, b = 1, n = 25.

Therefore,

[tex]\Delta{x}=\frac{1-0}{25}=\frac{1}{25}[/tex]

We need to divide the interval [0,1] into n = 25 sub-intervals of length [tex]\Delta{x}=\frac{1}{25}[/tex], with the following endpoints:

[tex]a=0, \frac{1}{25}, \frac{2}{25},...,\frac{23}{25}, \frac{24}{25}, 1=b[/tex]

Now, we just evaluate the function at these endpoints:

[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]

[tex]2f\left(x_{1}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]

[tex]2f\left(x_{2}\right)=2f\left(\frac{2}{25}\right)=2 e^{\frac{12}{25}}=3.23214880438579[/tex]

...

[tex]2f\left(x_{24}\right)=2f\left(\frac{24}{25}\right)=2 e^{\frac{144}{25}}=634.696657835701[/tex]

[tex]f\left(x_{25}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]

Applying the trapezoid rule formula we get

[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{50}(1+2.54249830064281+3.23214880438579+...+634.696657835701+403.428793492735)\approx 67.3930999748549[/tex]

We have that a = 0, b = 1, n = 50.

Therefore,

[tex]\Delta{x}=\frac{1-0}{50}=\frac{1}{50}[/tex]

We need to divide the interval [0,1] into n = 50 sub-intervals of length [tex]\Delta{x}=\frac{1}{50}[/tex], with the following endpoints:

[tex]a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b[/tex]

Now, we just evaluate the function at these endpoints:

[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]

[tex]2f\left(x_{1}\right)=2f\left(\frac{1}{50}\right)=2 e^{\frac{3}{25}}=2.25499370315875[/tex]

[tex]2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]

...

[tex]2f\left(x_{49}\right)=2f\left(\frac{49}{50}\right)=2 e^{\frac{147}{25}}=715.618483417705[/tex]

[tex]f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]

Applying the trapezoid rule formula we get

[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{100}(1+2.25499370315875+2.54249830064281+...+715.618483417705+403.428793492735) \approx 67.1519320308594[/tex]

b. To approximate the integral [tex]\int_{0}^{1}e^{6 x}\ dx[/tex] using 2n with the Simpson's rule you must:

The Simpson's rule states that

[tex]\int_{a}^{b}f(x)dx\approx \\\frac{\Delta{x}}{3}\left(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right)[/tex]

where [tex]\Delta{x}=\frac{b-a}{n}[/tex]

We have that a = 0, b = 1, n = 50

Therefore,

[tex]\Delta{x}=\frac{1-0}{50}=\frac{1}{50}[/tex]

We need to divide the interval [0,1] into n = 50 sub-intervals of length [tex]\Delta{x}=\frac{1}{50}[/tex], with the following endpoints:

[tex]a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b[/tex]

Now, we just evaluate the function at these endpoints:

[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]

[tex]4f\left(x_{1}\right)=4f\left(\frac{1}{50}\right)=4 e^{\frac{3}{25}}=4.5099874063175[/tex]

[tex]2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]

...

[tex]4f\left(x_{49}\right)=4f\left(\frac{49}{50}\right)=4 e^{\frac{147}{25}}=1431.23696683541[/tex]

[tex]f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]

Applying the Simpson's rule formula we get

[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{150}(1+4.5099874063175+2.54249830064281+...+1431.23696683541+403.428793492735) \approx 67.0715427161943[/tex]

c. If B is our estimate of some quantity having an actual value of A, then the absolute error is given by [tex]|A-B|[/tex]

The absolute error in the trapezoid rule is

The calculated value is

[tex]\int _0^1e^{6\:x}\:dx=\frac{e^6-1}{6} \approx 67.0714655821225[/tex]

and our estimate is 67.1519320308594

Thus, the absolute error is given by

[tex]|67.0714655821225-67.1519320308594|=0.08047[/tex]

The absolute error in the Simpson's rule is

[tex]|67.0714655821225-67.0715427161943|=0.00008[/tex]

Answer:

Step-by-step explanation:

Given that many couples believe that it is getting too expensive to host an "average" wedding in the United States.

Population mean =24066

Sample mean =  23224

Sample size = 40

Sample std dev = 2903

Since sample std dev is known, we use t critical value.

df =39

Sample mean follows a normal distribution with mean = 23224, and std dev = [tex]\frac{s}{\sqrt{n} } \\=\frac{2903}{\sqrt{40} } \\=459.005[/tex]

t critical value = 2.023

Margin of error = 2.023*459.005

Confidence interval[tex](22295.43, 24152.57)[/tex]

Final answer:

The question involves calculating a 95% confidence interval for the average cost of weddings in the U.S., resulting in a range of $22,296 to $24,152, based on recent sample data.

Explanation:

The question pertains to the formation of a confidence interval for the average cost of weddings in the U.S. based on a sample. The provided data asserts that the average cost of a wedding in 2009 was $24,066, while a more recent sample reveals an average of $23,224 with a standard deviation of $2,903. Calculating a 95% confidence interval results in a range of $22,296 to $24,152. This implies that we can be 95% confident that the true average cost of weddings in the whole population falls within this interval.

Answer:the number of dimes is 22

the number of quarters is 11

Step-by-step explanation:

A dime is worth 10 cents. Converting to dollars, it becomes 10/100 = $0.1

A quarter is worth 25 cents. Converting to dollars, it becomes 25/100 = $0.25

Let x represent the number of dimes.

Let y represent the number of quarters.

Daniele has 33 quarters and dimes in her piggy bank. This means that

x + y = 33

The piggy bank contains a total of $4.95. This means that

0.1x + 0.25y = 4.95 - - - - - - - - -1

Substituting x = 33 - y into equation 1, it becomes

0.1(33 - y) + 0.25y = 4.95

3.3 - 0.1y + 0.25y = 4.95

- 0.1y + 0.25y = 4.95 - 3.3

0.15y = 1.65

y = 1.65/0.15 = 11

Substituting y = 11 into x = 33 - y, it becomes

x= 33 - 11 = 22

Final answer:

To find the number of dimes and quarters in Daniele's piggy bank, a system of equations can be solved by substitution method.

Explanation:

System of Equations:

Let x be the number of dimes and y be the number of quarters.

We have the equations: 0.10x + 0.25y = 4.95 and x + y = 33.

Solving by substitution, we first solve x + y = 33 for x to get x = 33 - y.

Substitute x = 33 - y into 0.10x + 0.25y = 4.95 and solve to find the values of x and y.

The solution is x = 15, y = 18, so there are 15 dimes and 18 quarters.

Answer:

[tex]t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036[/tex]  

[tex]p_v =P(t_{12}>3.036) =0.0051[/tex]  

So with the p value obtained 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 mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).  

Step-by-step explanation:

The system of hypothesis on this case are:  

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

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

Or equivalently:  

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

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

Our notation on this case :  

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

[tex]n_2 =6[/tex] represent the sample size for group 2  (Edne)

We can calculate the sampel means and deviations with the following formulas:

[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

[tex]\bar X_1 =10[/tex] represent the sample mean for the group 1  

[tex]\bar X_2 =7[/tex] represent the sample mean for the group 2  

[tex]s_1=2.268[/tex] represent the sample standard deviation for group 1  

[tex]s_2=1.414[/tex] represent the sample standard deviation for group 2  

If we see the alternative hypothesis we see that we are conducting a right tailed test.

On this case since the significance assumed is 0.05 and we are conducting a bilateral test we have one critica value, and we need on the right tail of the distribution [tex]\alpha/2 = 0.05[/tex] of the area.  

The distribution on this case since we don't know the population deviation for both samples is the t distribution with [tex]df=8+6 -2=12[/tex] degrees of freedom.

We can use the following excel code in order to find the critical value:

"=T.INV(1-0.05,12)"

And the rejection zone is: (1.78,infinity)

The statistic is given by this formula:  

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{S^2_2}{n_2}}}[/tex]  

And now we can calculate the statistic:  

[tex]t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036[/tex]  

The degrees of freedom are given by:  

[tex]df=8+6-2=12[/tex]

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

[tex]p_v =P(t_{12}>3.036) =0.0051[/tex]  

So with the p value obtained 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 mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).  

Answer:

Step-by-step explanation:

Final answer:

The decision rule for the hypothesis test at a 0.10 significance level involves rejecting the null hypothesis if the test statistic is greater than the critical value. The calculated Z-value is approximately 1.8233, and the corresponding p-value will determine our decision to reject or not reject the null hypothesis.

Explanation:

Decision Rule and Test Statistic

For a hypothesis test comparing two population means with known standard deviations, we use a Z-test for two samples. The decision rule at a significance level of 0.10 involves rejecting the null hypothesis (H0: μ1 = μ2) if the absolute value of the test statistic exceeds the critical value from the standard normal distribution.

Calculation of Test Statistic

The test statistic (Z) is calculated using the formula: Z = (101.0 - 99.3)/ √[ (4.6²/40) + (4.0²/47) ]

Plugging in the numbers: Z = (1.7)/ √[ (21.16/40) + (16/47) ] Z = (1.7)/ √[ 0.529 + 0.3404255 ] Z = (1.7)/√[0.8694255] Z = (1.7)/0.932447 Z ≈ 1.8233

Decision Based on P-value

The p-value associated with this Z-value can be found using standard normal distribution tables or software. If the p-value is less than 0.10, we reject the null hypothesis. Otherwise, we fail to reject it.

Answer:

Consider the following explanation

Step-by-step explanation:

Completely Randomized Design :-

A completely randomized design is probably the simplest experimental design, in terms of data analysis and convenience. With this design, subjects are randomly assigned to treatments.

This completely randomized design relies on randomization to control for the effects of extraneous variables. The experimenter assumes that, on average, extraneous factors will affect treatment conditions equally; so any significant differences between conditions can fairly be attributed to the independent variable.

Double Blind Design :-

In an experiment, if subjects in the control group know that they are receiving a placebo, the placebo effect will be reduced or eliminated; and the placebo will not serve its intended control purpose.

Blinding is the practice of not telling subjects whether they are receiving a placebo. In this way, subjects in the control and treatment groups experience the placebo effect equally. Often, knowledge of which groups receive placebos is also kept from analysts who evaluate the experiment. This practice is called double blinding. It prevents the analysts from "spilling the beans" to subjects through subtle cues; and it assures that their evaluation is not tainted by awareness of actual treatment conditions.

Matched Pairs Design :-

A matched pairs design is a special case of a randomized block design. It can be used when the experiment has only two treatment conditions; and subjects can be grouped into pairs, based on some blocking variable. Then, within each pair, subjects are randomly assigned to different treatments.

Randomized Block Design :-

With a randomized block design, the experimenter divides subjects into subgroups called blocks, such that the variability within blocks is less than the variability between blocks. Then, subjects within each block are randomly assigned to treatment conditions. Compared to a completely randomized design, this design reduces variability within treatment conditions and potential confounding, producing a better estimate of treatment effects.

The experimental design that is given in the problem, is an example of a Randomized Block Design. Here, the different type of soils can be considered different blocks. In these blocks, the variability among within the blocks is minimum and between the blocks, it is maximum. Also, the treatments ( fertilizers ) are assigned randomly to the plts in each block (different soil type ).

Answer:

Step-by-step explanation:

Given that a new drug test needs to be evaluated. The probability of a random person taking drugs is 4%.

The drug test tested positive for a person not taking drugs 2% of the time. The drug test tested negative for a person taking drugs 1% of the time.

                          Tested positive     Tested negative       Total

Having drug               0.98                        0.02                  1.00

not hav drug               0.01                        0.99                   1.00

a)            the probability that a random person tests negative

= Prob ( really negative and test negative) + Prob (positive and test negative)

[tex]= 0.96*0.99+0.04*0.02=    0.9504+0.0008=0.9512[/tex]

b) the probability that a person who tests negative is without any drug problems

= [tex]\frac{0.9504}{0.9504+0.0008} \\=0.9992[/tex]

Answer:

b) 36

Step-by-step explanation:

We can use combinations to solve this problem.

The binomial coefficient [tex]\binom{n}{k}=\frac{n!}{k!(n-k)!}[/tex] counts the number of ways of choose k elements from a set of n elements.

The product rule from combinatorics says that if there are N ways of doing something and M ways of doing another thing, the number of ways of doing both things is equal to NM.

First, we choose the blue balls. The urn contains 4 blue balls and we select 2 so there are [tex]N=\binom{4}{2}=6[/tex] ways of doing this. Similarly, we choose the 5 orange balls from the set of 6 in the urn, which can be done in [tex]M=\binom{6}{5}=6[/tex] ways. By the product rule, there are MN=6(6)=36 ways of selecting all the balls.

The number of ways 2 blue balls and 5 orange balls can be selected from the urn is 36 number of ways: Option C is correct

Combination has to do with selection.

If r number is selected from n number, this is expressed using the formula:

[tex]nC_r=\frac{n!}{(n-r)!r!}\\[/tex]

If 2 blue balls are selected from 4 blue balls, this is expressed as:

[tex]4C_2=\frac{4!}{(4-2)!2!}\\4C_2=\frac{4\times 3 \times 2!!}{2!2!}\\4C_2=\frac{12}{2} = 6 ways[/tex]

Similarly, if 5 orange balls are selected from 5 orange balls, this is expressed as:

[tex]6C_5=\frac{6!}{(6-5)!5!}\\6C_5=\frac{6\times 5!}{1!5!}\\6C_5=\frac{6}{1} = 6 ways[/tex]

The number of ways 2 blue balls and 5 orange balls can be selected from the urn is 36 number of ways

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Answer:

Option A) 0.0127  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 16 weeks

Standard Deviation, σ = 2 weeks

Sample size = 20

We are given that the distribution of time taken to find a job is a bell shaped distribution that is a normal distribution.

Formula:  

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

P(20 young workers average less than 15 weeks)  

P(x < 15)  

[tex]P( x < 15) = P( z < \displaystyle\frac{15-16}{\frac{2}{\sqrt{20}}}) = P(z < -2.236)[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(x < 15) =0.0127= 1.27\%[/tex]

Thus, 0.0127 is the probability that 20 young workers average less than 15 weeks to find a job.

The probability of a group of 20 workers finding jobs in less than 15 weeks, given the specific mean and standard deviation, was calculated using z-scores and the normal distribution. The result was approximately 0.0571, but this was an interpretation that wasn't presented as an answer choice.

In this question, we're asked to calculate the probability that a group of 20 young workers on average find a job in less than 15 weeks, given that the mean is 16 weeks and standard deviation is 2 weeks. This is a problem of statistics, specifically involving the

normal distribution

and

z-scores

. The z-score is calculated using the formula Z = (X - μ) / (σ/√n), where X is the random variable (15 weeks in this case), μ is the population mean (16 weeks), σ is the standard deviation (2 weeks) and n is the sample size (20 workers). This gives us Z = (15 - 16) / (2/√20), which is approximately -1.58. Looking up this z-score in a standard normal distribution table gives us a probability of 0.0571, which isn't an option provided in the question. However, we may have made an error in calculation or interpretation. It's important to thoroughly understand and practice these statistical concepts.

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Answer:

a) P=0.84

b) Mean=0.33

Standard deviation=0.356

Step-by-step explanation:

The probabilty that the measurement error in a randomly selected instance us less than 0.6 µs is P=0.84.

The mean of a Beta(α = 1, β = 2) is

[tex]\mu=\frac{\alpha}{\alpha+\beta}=\frac{1}{1+2}=0.33[/tex]

The standard deviation of a Beta(α = 1, β = 2) is

[tex]\sigma=\sqrt{\frac{\alpha\beta}{(\alpha+\beta)^2*(\alpha+\beta+1)}}\\\\\\\sigma= \sqrt{\frac{1*2}{(1+2)^2*(1+2+1)}}=\sqrt{\frac{2}{(2)^2*(4)}}=\sqrt{\frac{2}{16} } =\sqrt{0.125}= 0.356[/tex]

Given a Beta Distribution with α = 1 and β = 2, probability calculations typically rely on integration of the probability density function, often executed with statistical software. The mean and standard deviation can be mathematically derived with the given α and β, giving us mean= 0.33 and standard deviation= 0.236.

The question falls under the domain of probability distribution, specifically the Beta Distribution. The Beta distribution is a family of continuous probability distributions defined on the interval [0, 1] parameterized by two positive shape parameters, denoted by α and β. For a Beta distribution, the probability density function is given by f(x; α, β). When α = 1 and β = 2, the beta distribution becomes a decreasing linear function.

To find the probability that the measurement error is less than 0.6, we need to integrate the probability density function from 0 to 0.6. However, it's important to note that directly performing these integrations and calculations is a task typically performed by statistical software.

Concerning mean and standard deviation of the Beta Distribution, the mean (µ) is given by α / (α + β), and the variance (σ²) is given by (αβ) / [(α+β)²*(α+β+1)]. Here, given α=1 and β=2, we find µ=1/(1+2) = 0.33 and σ² = (1*2)/(3²*4)=0.0556, and taking square root of variance we get standard deviation σ = 0.236.

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Answer:

a.the goodness of fit for the estimated multiple regression equation increases.

Step-by-step explanation:

As the value of the multiple coefficient of determination increases,

a. the goodness of fit for the estimated multiple regression equation increases.

As we know that the coefficient of determination measures the variability of response variable with the help of regressor. As we know that if the value of the coefficient of determination increases strength of fit also increases.  

Answer:

The required formula for y, the number of cars sold is: [tex]y = 0.008x +200[/tex].

Step-by-step explanation:

Consider the provided information.

Let x represents the amount of money spend on advertising .

y is number of cars sold,

For each additional 5000 dollars spent, they sold 40 more cars.

[tex]Slope = m = \frac{Rise}{Run}=\frac{40}{5000}=0.008[/tex]

The y-intercept form is: [tex]y = mx + b[/tex]

Substitute x=60,000, y=680 and m=0.008 in the above equation.

[tex]680= 0.008(60,000)+b[/tex]

[tex]b=680-480[/tex]

[tex]b=200[/tex]

Thus, the required formula for y, the number of cars sold is: [tex]y = 0.008x +200[/tex].

To create a formula for the number of cars sold (y) as a function of advertising expenditure (x, in thousands of dollars), we use the data points provided to establish a linear equation: y = 8x + 200, where x represents the advertising spending and y the number of cars sold.

To find a formula for y, the number of cars sold, given the linear relationship between advertising expenditure (x, in thousands of dollars) and sales, we first establish the given data points:

An additional $5,000 in advertising (or an increase of 5 in x since it's measured in thousands) results in selling 40 more cars.

We can express this relationship with the linear equation y = mx + b, where m is the slope (change in y over change in x) and b is the y-intercept (the number of cars sold when no money is spent on advertising).

Let's calculate the slope, m:

For every increase of 5 in x, y increases by 40, so m = 40/5 = 8.

We can then use this slope and one of the data points to find the y-intercept, b:

680 = 8(60) + b, so b = 680 - (8*60) = 680 - 480 = 200.

The linear equation representing the relationship between advertising spend and cars sold is:

y = 8x + 200

Answer: 95% of the time the maximum monthly temperature is between 25.28 degrees Celsius and 34.52 degrees Celsius.

Step-by-step explanation:

The Range Rule of Thumb tells that the range is approximately four times the standard deviation.

95% of the data lies within 2 standard deviations from the mean .

Maximum usual value = Mean +2 (Standard deviation )

Minimum usual value = Mean - 2 (Standard deviation)

Given : Mean =  29.9 degrees Celsius

Standard deviation = 2.31 degrees Celsius

Then, according to the Range Rule of Thumb , we have

Maximum usual value = 29.9 +2 (2.31) = 34.52 degrees Celsius

Minimum usual value = 29.9 - 2 (2.31) = 25.28 degrees Celsius

i.e. 95% of the time the maximum monthly temperature is between 25.28 degrees Celsius and 34.52 degrees Celsius.

Using the empirical rule for normal distribution, 95% of the time, the maximum monthly temperature is between 25.3 and 34.5 degrees Celsius, given the mean is 29.9°C and the standard deviation is 2.31°C.

The question asks for the range of temperatures within which 95% of the maximum monthly temperatures fall, given that the average maximum monthly temperature is 29.9 degrees Celsius with a standard deviation of 2.31 degrees, assuming a normal distribution. To find this, we can apply the empirical rule that states approximately 95% of the data in a normal distribution falls within two standard deviations of the mean. Therefore:

Thus, 95% of the time, the maximum monthly temperature is between 25.3 and 34.5 degrees Celsius.

Answer:

17%

Step-by-step explanation:

If 83% is removed than

total is always 100%

that's why

so the new sound level will sound

100%-83%=17% remains.

hence 17 % is correct answer .

Answer:

about 9 passengers will not keep theire reservation

Step-by-step explanation:

given that an airline finds that about 8 percent of the time, a person who makes an advance reservation does not keep the reservation

Capacity of the airline per trip = 105 only

But company schedules = 113 people with advance reservations

Since 8% do not turn up

out of 113 passengers who reserved in advance we can say

that 8% of 113 passengers will not keep their reservation

Based on this information, number  of the scheduled people will not keep their reservation

= [tex]8%of 113\\= \frac{8*113}{100} \\=9.04[/tex]

Since persons cannot be in decimals we can expect about 9 passengers will not keep theire reservation

Based on the given information, about 9 people out of the scheduled 113 will not keep their reservation.

The airline finds that 8 percent of the time, a person with an advance reservation does not keep it. This means that 8% of the scheduled passengers will not keep their reservations.

Percentage of passengers not keeping reservation = 8% = 0.08

Number of scheduled passengers = 113

Number of passengers not keeping reservation = 0.08 * 113 = 9

Therefore, about 9 of the scheduled people will not keep their reservation.

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Answer:

a) The mean of the sampling distribution of the sample means is 92.

b) The variance of the sampling distribution of the sample mean is 3.24.

c) The standard error of the sampling distribution of the sample mean is 1.8.

d) 28.77% probability that the sample mean ex-ceeds 93.0 units.

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 = 92, \sigma = 3.6[/tex].

a. Find the mean of the sampling distribution of thesample means.

The mean is the same as the mean of the population. So the mean of the sampling distribution of the sample means is 92.

b. Find the variance of the sampling distribution ofthe sample mean.

The standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{4}} = 1.8[/tex]

The variance is [tex]s^{2} = 1.8^{2} = 3.24[/tex]

c. Find the standard error of the sampling distribution of the sample mean.

This is the same as the standard deviation of the sample. So the standard error of the sampling distribution of the sample mean is 1.8.

d. What is the probability that the sample mean ex-ceeds 93.0 units?

This is 1 subtracted by the pvalue of Z when X = 93. So

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

[tex]Z = \frac{93 - 92}{1.8}[/tex]

[tex]Z = 0.56[/tex]

[tex]Z = 0.56[/tex] has a pvalue of 0.7123.

So there is a 1-0.7123 = 0.2877 = 28.77% probability that the sample mean ex-ceeds 93.0 units.

The mean of the sampling distribution of the sample means is 92.0. The variance is 0.81. The standard error is 0.9.

a. The mean of the sampling distribution of the sample means can be calculated by finding the mean of the original population, which is 92.0.

b. The variance of the sampling distribution of the sample mean can be calculated using the formula (standard deviation/original population size)^2. In this case, it is (3.6/4)^2 = 0.81.

c. The standard error of the sampling distribution of the sample mean can be calculated by taking the square root of the variance. So, in this case, it is √0.81 = 0.9.

d. To find the probability that the sample mean exceeds 93.0 units, you can calculate the z-score using the formula (sample mean - population mean)/standard error. Then, you can find the probability of the z-score using a standard normal distribution table or calculator.

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Answer:

Policeman is correct.

Step-by-step explanation:

Consider the provided information.

The tires left three skid marks of 69 ft, 70 ft, and 74 ft.

The road had a drag factor of 0.95. Your brakes were operating at 98% efficiency.

Thus, drag factor (f) = 0.95 and brakes efficiency (n) = 0.98

Compute the skid distance by finding the mean of length of 3 skids.

[tex]D=\frac{69+70+74}{3}\\D=\frac{213}{3}\\D=71[/tex]

Therefore, the skid distance is 71 ft.

Now find the speed of car using skid speed formula:

[tex]S=\sqrt{30Dfn}[/tex]

Where, S is the speed of car, D is the skid distance, f is the drag factor and n is the beak efficiency.

Substitute the respective values in above formula.

[tex]S=\sqrt{30\times 71\times0.95 \times0.98}[/tex]

[tex]S=\sqrt{1983.03}[/tex]

[tex]S\approx 44.5312[/tex]

Your speed was 44.5312 mi/h which is greater than 40 mi/hr.  So policeman is correct.

Answer:

a) The 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

b) Since the confidence interval NOT contains the value 0.06 we  have anough evidence to reject the claim at 5% 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]p[/tex] represent the real population proportion of interest  

[tex]\hat p =0.09[/tex] represent the estimated proportion for the sample  

n=800 is the sample size required  

[tex]z[/tex] represent the critical value for the margin of error  

Confidence =0.95 or 95%

Part a

The population proportion have the following distribution  

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

The confidence interval would be given by this formula  

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

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.96[/tex]  

The margin of error is given by :

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

[tex]Me=1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0198[/tex]

And replacing into the confidence interval formula we got:  

[tex]0.09 - 1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0702[/tex]  

[tex]0.09 + 1.96 \sqrt{\frac{0.108(1-0.09)}{800}}=0.1098[/tex]  

And the 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

Part b

Since the confidence interval NOT contains the value 0.06 we  have anough evidence to reject the claim at 5% of significance.

The 95% confidence interval for the true percentage of men taking on second jobs is approximately 7.02% to 10.98%. The pundit's claim of 6% is not plausible.

To construct a 95% confidence interval for the true percentage of men taking on second jobs, we'll use the sample proportion (\( \hat{p} \)) and the margin of error formula. Then, we'll use this interval to test the pundit's claim.

Given:

- Sample size ( n ) = 800

- Sample proportion [tex](\( \hat{p} \))[/tex] = 9% = 0.09

a) Constructing a 95% confidence interval:

The margin of error ( E ) for a 95% confidence interval can be calculated using the formula:

[tex]\[ E = Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \][/tex]

Where:

-  Z  is the Z-score corresponding to the desired confidence level (95%)

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

-  n  is the sample size

We can find the Z-score corresponding to a 95% confidence level, which is approximately 1.96.

Now, let's calculate the margin of error and construct the confidence interval.

b) Testing the pundit's claim:

We'll compare the pundit's claim (6%) with the confidence interval constructed in part (a). If the pundit's claim falls within the confidence interval, it is plausible. Otherwise, it is not plausible.

Let's proceed with the calculations.

a) Constructing a 95% confidence interval:

First, let's calculate the margin of error ( E ):

[tex]\[ E = 1.96 \times \sqrt{\frac{0.09 \times (1 - 0.09)}{800}} \][/tex]

[tex]\[ E \approx 1.96 \times \sqrt{\frac{0.09 \times 0.91}{800}} \][/tex]

[tex]\[ E \approx 1.96 \times \sqrt{\frac{0.0819}{800}} \][/tex]

[tex]\[ E \approx 1.96 \times \sqrt{0.000102375} \][/tex]

[tex]\[ E \approx 1.96 \times 0.010118 \][/tex]

[tex]\[ E \approx 0.0198 \][/tex]

Now, let's construct the confidence interval:

Lower bound: [tex]\( \hat{p} - E = 0.09 - 0.0198 = 0.0702 \)[/tex]

Upper bound: [tex]\( \hat{p} + E = 0.09 + 0.0198 = 0.1098 \)[/tex]

So, the 95% confidence interval for the true percentage of men taking on second jobs is approximately [tex]\( (7.02\%, 10.98\%) \).[/tex]

b) Testing the pundit's claim:

The pundit's claim is 6%, which falls below the lower bound of the confidence interval (7.02%). Therefore, it is not plausible given the poll data.

Answer:

0.0561,0.0673,0.999997

Step-by-step explanation:

Given that a sign on  the pumps at a gas station encourages customers to have their oil checked, and claims that one out of 8 cars needs to have oil added.

X- no of cars that need oil in a sample of 8 cars is binomial with p = 1/8

since each car is independent of the other and there are only two outcomes.

a) Exactly three out of the next eight cars needs oil.

Probability = [tex]P(X=3) =0.0561[/tex]

b) At least three out of the next eight cars needs oil.

Probability = [tex]P(X\geq 3)\\= 0.0673[/tex]

c) At most 6 out of the next 8 cars needs oil.

Probability =[tex]P(X\leq 6)\\=0.999997[/tex]

Answer:

The statement which is true is:

B. This is a double blind study

Step-by-step explanation:

Answer:

C. Observational Study

Step-by-step explanation:

There arent any treatments being manipulated so it can't be an experiment. The researcher is aware of the experiment being performed so its not double-blind. The only option it could be is C, because the researcher is observing the people eating and isn't manipulating their eating habits in any way.

Answer:

a) The 90% confidence interval would be given by (73.56;79.84)  

b) The 90% confidence interval for the deviation would be [tex] 6.44 \leq \sigma \leq 11.116[/tex].

c) 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 reject the null hypothesis, so we can conclude that the mean is significantly higher than 70 at 1% of significance.  So the claim of the teacher makes sense.

d) Since is a one-side upper test the p value would given by:  

[tex]p_v =P(t_{19}>3.69)=0.00078[/tex]  

Step-by-step explanation:

Data given: 82 94 66 87 68 85 68 84 70 83 65 70 83 71 82 72 73 81 76 74

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".  

We can calculate the sample mean and deviation with the following formulas:

[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

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

And we got the following results:

[tex]\bar X=76.7[/tex] represent the sample mean  

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

[tex]s=8.112[/tex] represent the population standard deviation  

n=20 represent the sample size  

90% confidence interval  

Part a

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)  

The degrees of freedom are given by:

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

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,19)".And we see that [tex]t_{\alpha/2}=1.73[/tex]  

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

[tex]76.7-1.73\frac{8.112}{\sqrt{20}}=73.56[/tex]  

[tex]76.7+1.73\frac{8.112}{\sqrt{20}}=79.84[/tex]  

So on this case the 90% confidence interval would be given by (73.56;79.84)  

Part b

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

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

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

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

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

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,19)" "=CHISQ.INV(0.95,19)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=30.143[/tex]

[tex]\chi^2_{1- \alpha/2}=10.117[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(19)(8.112^2)}{30.143} \leq \sigma^2 \leq \frac{(19)(8.112^2)}{10.117}[/tex]

[tex] 41.472 \leq \sigma^2 \leq 123.574[/tex]

So the 90% confidence interval for the deviation would be [tex] 6.44 \leq \sigma \leq 11.116[/tex].

Part c

Null hypothesis:[tex]\mu \leq 70[/tex]  

Alternative hypothesis:[tex]\mu > 70[/tex]  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

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{76.7-70}{\frac{8.112}{\sqrt{20}}}=3.69[/tex]  

Part d

P-value  

Since is a one-side upper test the p value would given by:  

[tex]p_v =P(t_{19}>3.69)=0.00078[/tex]  

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 reject the null hypothesis, so we can conclude that the mean is significantly higher than 70 at 1% of significance.  

Answer:

(a) 95% confidence interval for the population mean score on Long Island is (73, 80.40)

(b) 90% confidence interval for the population standard deviation of the scores on Long Island is (4.85, 10.97)

(c) The teacher's claim that the mean score on Long Island is higher than the mean for New York State is reasonable

(d) p-value is 0.01

Step-by-step explanation:

From the data values from Long Island,

Mean is 76.7 and standard deviation is 7.91

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

(a) mean = 76.7, sd = 7.91, n = 20, degree of freedom = n-1 = 20-1 = 19, t-value corresponding to 19 degrees of freedom and 95% confidence level is 2.093

Lower bound = 76.7 - (2.093×7.91)/√20 = 76.7 - 3.70 = 73

Upper bound = 76.7 + (2.093×7.91)/√20 = 76.7 + 3.70 = 80.40

95% CI is (73, 80.40)

(b) CI = sd + or - (t×sd)/√n

sd = 7.91, t-value corresponding to 19 degrees of freedom and 90% confidence level is 1.729

Lower bound = 7.91 - (1.729×7.91)/√20 = 7.91 - 3.06 = 4.85

Upper bound = 7.91 + (1.729×7.91)/√20 = 7.91 + 3.06 = 10.97

90% CI is (4.85, 10.97)

(c) Null hypothesis: The mean score on Long Island is 70

Alternate hypothesis: The mean score on Long Island is greater than 70

Z = (sample mean - population mean)/(sd/√n) = (70 - 76.7)/(7.91/√20) = -6.7/1.77 = -3.79

Using 0.01 significance level, the critical value is 2.326

Since -3.79 is less than 2.326, reject the null hypothesis. The teacher's claim is reasonable

(d) p-value = 1 - cumulative area of test statistic = 1 - 0.9900 = 0.01

Answer: 176.

Step-by-step explanation:

Formula to find the sample size is given by :-

[tex]n=(\dfrac{z^*\cdot \sigma}{E})^2[/tex]

, where [tex]\sigma[/tex] = Population standard deviation from prior study.

E = margin of error.

z* = Critical value.

As per given , we have

[tex]\sigma=\$675000[/tex]

E= $100,000

We take 95% confidence interval.

Critical value (Two tailed)=[tex]z^*=1.96[/tex]

The required sample size = [tex]n=(\dfrac{(1.96)\cdot 675000}{100000})^2[/tex]

[tex]n=(13.23)^2\\\\ n=175.03299\approx176[/tex] [Round to next integer]

Hence, the required sample size = 176.

Answer:

V = 8π/3

Step-by-step explanation:

Haven't done calc in years so I might be wrong.

Graph out all the lines needed so you can have a better look at it.

I set y=2x = y=2 to find where the intersect each others so I can have my boundaries for integration.

You goal is to find the area so you can integrate around that area. We're revolving around the x-axis so the area will be a circle.

V = ∫A(x)dx =  ∫(πr²)dr

Since we have two different radius, we subtract them from each others.

∫(πr₂² - πr₁²)dr

∫(π(2)² - π(2x)²)dr

∫(4π - 4πx²)dr

4π∫(1 - x²)dr

integrate from 0 to 1 since that's where our boundary is.

V = 4π∫(1 - x²)dr = 8π/3

To find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis, we can use the method of cylindrical shells and integrate the expression 2πx(2-2x) from 0 to 1.

To find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis, we can use the method of cylindrical shells. The region is bounded by the line y = 2x, the horizontal line y = 2, and the vertical line x = 0. First, we need to determine the limits of integration by finding the points where the curves intersect. Setting y = 2x and y = 2 equal to each other, we find x = 1. Next, we integrate the expression 2πx(2-2x) with respect to x from 0 to 1. Simplifying and evaluating the integral gives us the volume of the solid generated as 2π/3 cubic units.

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