Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 27 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was four years with a standard deviation of 1.9 years. Suppose that it is somehow known that the population standard deviation is 1.4. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal.
Since both ơ and sx are given, which should be used?
O σ
O sx
In one to two complete sentences, explain why?

Answer:

1044 sq. inches

Step-by-step explanation:

The frame adds 2 inches on each side of the window, so the width becomes 29 in. and the length becomes 36 in.

Area = width × length

So the total area of the window and the frame is:

A = 29 × 36 = 1044 in²

Answer:

Initial velocity    192.666 ft/sec

a) 1.41 sec

Step-by-step explanation:

Equations for

V(f)  = V₀ ± at

V(f)²  =( V₀)² ± 2 a*d

d =   V₀*t  ±  a*t²/2

We are going to use a(t)  =  g  = 32 ft/sec²

a) V₀   =  ??         a = -g      

Then   as  d = 580 ft       and  V(f)  =  0

we have

V(f)²  =  ( V₀)²  - 2*32*580      ⇒  ( V₀)²   = 64*580  (ft²/sec²)

( V₀)²   =  37120          V₀  = 192,666 ft/sec      

a)  t  = ??   in this case V₀ = 0

d  =    V₀*t + gt²/2       ⇒    32  =   ( 32 t²)  /2

t²  = 2  

t =   1.41 sec

Answer:

Step-by-step explanation:

The equation for the number of can tabs y that she has collected is

y = 4x + 35

where x is the number of weeks. To plot the graph, values for different number of weeks are inputted to get the corresponding y values.

Comparing the equation with the slope intercept form,

y = mx + c,

m represents slope.

So slope of the equation is 4. It represents the rate at which the total number of can tabs that she has collected is increasing with respect to the increase in the number of weeks. So the total number increases by 4 each week

The y intercept is the point at x = 0

It is equal to 35. It represents the number of can tabs that she had initially collected before she started collecting 4 can tabs weekly. At this point, her weekly collection is 0

Answer: 10

Step-by-step explanation:

We know that the formula to find the sample size is given by :-

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

, where [tex]\sigma[/tex] = population standard deviation.

[tex]z_{\alpha/2}[/tex] = Two -tailed z-value for [tex]{\alpha[/tex] (significance level)

E= margin of error.

Given : Confidence level : C =99%=0.99

i.e. [tex]1-\alpha=0.99[/tex]

⇒Significance level :[tex]\alpha=1-0.99=0.01[/tex]

By using z-value table ,Two -tailed z-value for [tex]\alpha=0.01 [/tex]:

[tex]z_{\alpha/2}=2.576[/tex]

E= 2 minutes

[tex]\sigma=\text{2.4 minutes}[/tex]

The required sample size will be :-

[tex]n= (\dfrac{2.576\cdot 2.4}{2})^2\\\\= (3.0912)^2\\\\=9.55551744\approx10[/tex]

Hence, the required sample size = 10

To estimate the mean time college students spend watching online videos each day within 2 minutes of the population mean at a 99% confidence level, at least 10 students should be sampled assuming a population standard deviation of 2.4 minutes.

To determine the required sample size, we use the formula for the sample size of a mean:

n = (z*σ/E)^2

Where n is the sample size, z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and E is the maximum error allowed (margin of error).

For a 99% confidence interval, the z-score (from the standard normal distribution) is approximately 2.576. The population standard deviation σ is given as 2.4 minutes, and the margin of error E is 2 minutes. Using these values:

n = (2.576*2.4/2)^2

n ≈ (6.1824/2)^2

n ≈ (3.0912)^2

n ≈ 9.554

Since the sample size must be a whole number, we would round up to ensure the sample size is large enough to achieve the desired margin of error, which means at least 10 students should be surveyed to construct the 99% confidence interval for the mean time spent watching online videos.

It is important to note that the larger the sample size, the more accurate the estimate of the population mean will be.

Answer:

a)  13/52    and the second  12/51

b) Two solutions:

b.1 if we did not picked up an eight in the first two cards   4/50

b.2 there is an eight i the two previous    3/59

c ) (48/52)*(47/51)*(46/50)

Step-by-step explanation:

Condition: Cards are taken out without replacement

a) Probability of first card is heart

There are 52 cards and 4 suits with the same probability , so you can compute this probability in two ways

we have  13 heart cards and 52 cards  then probability of one heart card is  13/52   = 0.25

or you have 4 suits, to pick up one specific suit the probability is 1/4 = 0,25

Now we have a deck of 51 card with 12 hearts, the probability of take one heart is : 12/51

b) There are 4 eight (one for each suit )   P =  4/50 if neither the first nor the second card was an eight of heart, if in a) previous we had an eight, then this probability change to 3/50

c) The probability of the first card different from an ace is 48/52 , the probability of the second one different of an ace is 47/51 and for the thirsd card is 46/50. The probability of none of the three cards is an ace is

(48/52)*(47/51)*(46/50)

Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

[tex]\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy[/tex]

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

[tex] \int\limits^1_0\int\limits^1_0 {2xy} \, dxdy[/tex]

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

[tex]\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2[/tex]

We conclude that the line integral is 1/2

To evaluate the line integral using Green's Theorem, we need to find the curl of the vector field and the area enclosed by the square. The line integral of the vector field along the square is equal to the double integral of the curl over the region enclosed by the square. Using this method, we can find the value of the line integral to be 1/3.

To evaluate the line integral using Green's Theorem, we first need to find the curl of the vector field. In this case, the vector field is F(x, y) = 7y^2x i + 8x^2y j. Taking the partial derivatives of its components with respect to x and y, we get curl(F) = (8x^2 - 14xy^2) k.

Next, we need to find the area enclosed by the square C, which is 1 unit^2. Using Green's Theorem, the line integral of F along C is equal to the double integral of curl(F) over the region D enclosed by C. Integrating curl(F) with respect to y, we get -7xy^2 + 6x^2y. Integrating this with respect to x over the given limits, we find the value of the line integral to be 1/3.

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Answer:

a)[tex]H_0 :\mu = 4\\ H_1 : \mu \neq 4[/tex] , n = 15 , X=3.4 , S=1.5 , α = .05

Formula : [tex]t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{3.4-4}{\frac{1.5}{\sqrt{15}}}[/tex]

[tex]t =-1.549[/tex]

p- value = 0.607(using calculator)

α = .05

p- value > α

So, we failed to reject null hypothesis

b)[tex]H_0 :\mu = 21\\ H_1 : \mu < 21[/tex] , n =75 , X=20.12 , S=2.1 , α = .10

Formula : [tex]t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{20.12-21}{\frac{2.1}{\sqrt{75}}}[/tex]

[tex]t =-3.6290[/tex]

p- value = 0.000412(using calculator)

α = .1

p- value< α

So, we reject null hypothesis

(c) [tex]H_0 :\mu = 10\\ H_1 : \mu \neq 10[/tex], n = 36, p-value = 0.061.

Assume α = .05

p-value = 0.061.

p- value > α

So, we failed to reject null hypothesis

The decision to reject the null hypothesis in each scenario depends on comparing calculated test statistics (or given p-value) to critical values associated with the significance level. In (a) and (c), we may not reject H0, and in (b) we would reject H0 if our test statistic is larger than the critical value.

In hypothesis testing, we compare the test statistic, often a t-value, to the critical value determined by our chosen significance level, α. If the test statistic is greater, we reject the null hypothesis (H0).

(a) In the first scenario, we must calculate the test statistic: t = (X - µ) / (S/√n) = (3.4 - 4) / (1.5/√15); if this absolute value is less than our critical value associated with α = .05 and degrees of freedom = 14, we do not reject H0.

(b) For the second, the test statistic would be (20.12 - 21) / (2.1/√75); compare its value to the critical value associated with α = .10 and degrees of freedom = 74. If it is larger, we reject H0 due to the lower tail test in this scenario.

(c) In the third scenario, the p-value is given. Since our p-value = 0.061 is larger than our α = .05, we would not reject H0.

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Answer:

57,600,000

Step-by-step explanation:

Final answer:

In simple regression analysis, a positive correlation coefficient means that the slope of the regression line must also be positive, as both the dependent and independent variables increase or decrease together.

Explanation:

In simple regression analysis, the correlation coefficient's sign (positive or negative) informs us about the direction of the relationship between the two variables.

A positive correlation coefficient indicates that the variables move in the same direction: as one increases, so does the other, and as one decreases, so does the other. This characteristic of correlation directs us to the correct answer to the student's question.

Considering the options provided:

The slope of the regression line is directly determined by the correlation coefficient in the context of simple linear regression. Given that the correlation coefficient is positive, the slope of the regression line (which represents the change in the dependent variable for a unit change in the independent variable) must also be positive.

Therefore, option A is correct.

Answer:

0.082

Step-by-step explanation:

Number of attempts = n = 100

Since there are only two outcomes and in-dependent of each other, the probability of missing a shot  = 1 - Probability of making each shot

p = 1 - 0.95 = 0.05

Possion Ratio (λ) = np where n is the number of events and p is the probability of the shot missing

λ = 100 x 0.05 = 5

Define X such that X = Number of misses and X ≅ Poisson (λ = 5)

P [X ≤ 2] = P [X = 0] + P [X = 1] + P [X = 2]

P [X ≤ 2] = e⁻⁵ + e⁻⁵ x 5 + e⁻⁵ x 5²/2!

P [X ≤ 2] = e⁻⁵ [1 + 5 + 5²/2!]

P [X ≤ 2] = e⁻⁵ x 12.25 = 0.082

The required probability that there are at most 2 misses in the first 100 attempts is 0.082

[tex]\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k[/tex]

is conservative if there is a scalar function [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\vec F[/tex]. This would require

[tex]\dfrac{\partial f}{\partial x}=y[/tex]

[tex]\dfrac{\partial f}{\partial y}=x[/tex]

[tex]\dfrac{\partial f}{\partial z}=3[/tex]

(or perhaps the last partial derivative should be 4 to match up with the integral?)

From these equations we find

[tex]f(x,y,z)=xy+g(y,z)[/tex]

[tex]\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]

[tex]f(x,y,z)=xy+h(z)[/tex]

[tex]\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C[/tex]

[tex]f(x,y,z)=xy+3z+C[/tex]

so [tex]\vec F[/tex] is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:

[tex]\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r[/tex]

[tex]=f(5,7,-2)-f(1,2,3)=\boxed{18}[/tex]

Answer:

absolute max= (4.243,18)

absolute min =(-1,-5.916)

absolute max=(pi/6, 2.598)

absolute min = (pi/2,0)

Step-by-step explanation:

a) [tex]f(t) = t\sqrt{36-t^2} \\[/tex]

To find max and minima in the given interval let us take log and differentiate

[tex]log f(t) = log t + 0.5 log (36-t^2)\\Y(t) = log t + 0.5 log (36-t^2)[/tex]

It is sufficient to find max or min of Y

[tex]y'(t) = \frac{1}{t} -\frac{t}{36-t^2} \\\\y'=0 gives\\36-t^2 -t^2 =0\\t^2 =18\\t = 4.243,-4.243[/tex]

In the given interval only 4.243 lies

And we find this is maximum hence maximum at  (4.243,18)

Minimum value is only when x = -1 i.e. -5.916

b) [tex]f(t) = 2cost +sin 2t\\f'(t) = -2sint +2cos2t\\f"(t) = -2cost-4sin2t\\[/tex]

Equate I derivative to 0

-2sint +1-2sin^2 t=0

sint = 1/2 only satisfies I quadrant.

So when t = pi/6 we have maximum

Minimum is absolute mini in the interval i.e. (pi/2,0)

Answer:

  3 : 5

Step-by-step explanation:

vowels: a o o (3 of them)

consonants: b l l n s (5 of them)

The ratio of vowels to consonants is 3 : 5.

Answer:

option (c) n = 201

Step-by-step explanation:

Data provided in the question:

Standard deviation, s = 5.5 ounce

Confidence level = 99%

Length of confidence interval = 2 ounces

Therefore,

margin of error, E = (Length of confidence interval ) ÷ 2

= 2 ÷ 2

= 1 ounce

Now,

E = [tex]\frac{zs}{\sqrt n}[/tex]

here,

z = 2.58 for 99% confidence interval

n = sample size

thus,

1 = [tex]\frac{2.58\times5.5}{\sqrt n}[/tex]

or

n = (2.58 × 5.5)²

or

n = 201.3561 ≈ 201

Hence,

option (c) n = 201

Answer:

f(2)=8

Step-by-step explanation:

f(2)=(-5)*f(1)-7=

(-5)*(-3)-7=8

f(2)=8

The probability of a pug approaching first in a park of 25 dogs is 5/25 or 1/5. The average grade of a larger class is typically a more reliable representation of the group's performance than that of a smaller one.

For the first question, the probability of a pug being the first dog to approach you in a park with 25 dogs, 5 of them being pugs, is indeed 5/25, or 1/5. This fraction represents the ratio of the desired outcome (a pug approaching you first) to the total number of possible outcomes (any of the 25 dogs in the park could potentially be the first to approach).

For the second part of the question - determining the more reliable average grade between two classes - the class with 50 students is the better choice. This conclusion is based on statistical reasoning. A larger sample size (in this case, the larger class) generally provides a more reliable average than a smaller sample size.

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Answer:

Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929

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{340119}{10} = 34011.9[/tex]

Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

[tex]S.D = \sqrt{\frac{2711418821}{9}} = 17357.09[/tex]

Confidence interval:

[tex]\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621[/tex]

[tex]34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)[/tex]

6n - 3(2n-5)

mutiply the bracket by -3

(-3)(2n)= -6n

(-3)(-5)= 15

6n-6n+15

0+15

answer:

15

Answer:

± 0.0736

Step-by-step explanation:

Data provided in the question:

randomly chosen graduates of California medical schools last year intended to specialize in family practice, p = [tex]\frac{48}{120}[/tex] = 0.4

Confidence level = 90%

sample size, n = 120

Now,

For 90% confidence level , z-value = 1.645

Width of the confidence interval = ± Margin of error

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

= ± [tex]1.645\times\sqrt\frac{0.4\times(1-0.4)}{120}[/tex]

= ± 0.07356 ≈ ± 0.0736

Hence,

The correct answer is option  ± 0.0736

The correct option is a. [tex]\±0.0736.[/tex] The width of a [tex]90\%[/tex] confidence interval for the proportion that plan to specialize in family practice

To find the width of a confidence interval for a proportion, we can use the formula:

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

where:

[tex]\( Z \)[/tex] is the Z-score corresponding to the desired confidence level,

[tex]\( p \)[/tex] is the sample proportion,

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

Given:

[tex]Sample\ proportion \( p = \frac{48}{120} = \frac{2}{5} = 0.4 \)[/tex],

[tex]Sample\ size \( n = 120 \)[/tex]

[tex]Confidence \ level = 90\%[/tex], which corresponds to a Z-score of approximately [tex]1.645.[/tex]

Substitute the values into the formula:

[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.4 \times 0.6}{120}} \][/tex]

[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.24}{120}} \][/tex]

[tex]\[ \text{Width} = 1.645 \times \sqrt{0.002} \][/tex]

[tex]\[ \text{Width} = 1.645 \times 0.0447 \][/tex]

[tex]\[ \text{Width} = 0.0736 \][/tex]

Answer: The value of test statistic is 0.696.

Step-by-step explanation:

Since we have given that

[tex]n_1=1399\\\\x_1=411\\\\p_1=\dfrac{x_1}{n_1}=\dfrac{411}{1399}=0.294[/tex]

Similarly,

[tex]n_2=759, x_2=211\\\\p_2=\dfrac{x_2}{n_2}=\dfrac{211}{759}=0.278[/tex]

At 0.01 level of significance.

Hypothesis are :

[tex]H_0:P_1=P_2=0.5\\\\H_a:P_1\neq P_2[/tex]

So, the test statistic value would be

[tex]z=\dfrac{(p_1-p_2)-(P_1-P_2)}{\sqrt{\dfrac{P_1Q_1}{n_1}+\dfrac{P_2Q_2}{n_2}}}\\\\z=\dfrac{0.294-0.278}{\sqrt{0.5\times 0.5(\dfrac{1}{1399}+\dfrac{1}{759})}}\\\\z=\dfrac{0.016}{0.023}=0.696[/tex]

Hence, the value of test statistic is 0.696.

Answer:

D. Null Hypothesis:μ=25 Alt Hypothesis:μ < 25

Step-by-step explanation:

Previous concepts and notation

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

[tex]\bar X= 23.9[/tex] represent the sample mean

[tex]s=3[/tex] represent the sample standard deviation

n=25 represent the sample selected

Solution

The claim that we want to test is that: "You believe that the mean BMI for college students is less than 25". So this statement needs to be on the Alternative hypothesis ([tex]\mu <25[/tex]) and the Null hypothesis would be the complement ([tex]\mu =25[/tex]) or ([tex]\mu \geq 25[/tex]).

So the best option on this case is:

D. Null Hypothesis:μ=25 Alt Hypothesis:μ < 25

Answer:

y = 2x^2

Step-by-step explanation:

The the coordinates (2,8) are expressed in terms of (x,y). Substitute the x with 2 and y with 8 and plug them into the formulas. The one that has the two sides equal to each other is the function that has (2,8) on its line.

Final answer:

To test if there has been a significant decrease in the average price of homes sold in the U.S., we will conduct a hypothesis test using the p-value approach. We will define the null and alternative hypotheses, calculate the Z-test statistic, compare the p-value to the significance level, and make a conclusion based on the results.

Explanation:

To test if there has been a significant decrease in the average price of homes sold in the U.S., we will conduct a hypothesis test using the p-value approach. Let's define our hypotheses:

Null hypothesis (H0): The average price of homes sold in the U.S. is not significantly different from $220,000.

Alternative hypothesis (Ha): The average price of homes sold in the U.S. is significantly less than $220,000.

Next, we need to calculate the test statistic. Since the population standard deviation is known, we can use a Z-test. The formula for the Z-test statistic is:

Z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))

In this case, the sample mean is $210,000, the population mean is $220,000, the population standard deviation is $36,000, and the sample size is 81. Plugging these values into the formula:

Z = (210,000 - 220,000) / (36,000 / sqrt(81))

Calculating this gives us a Z-statistic of -1.6875.

The final step is to compare the p-value of the test statistic to the significance level. Since the significance level is 1%, the critical value is 0.01. We can use a Z-table or calculator to find the p-value corresponding to a Z-statistic of -1.6875. The p-value is approximately 0.0466.

Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. This means that there is sufficient evidence to conclude that there has been a significant decrease in the average price of homes sold in the U.S.

Answer:

a) By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.

[tex]X \sim N(\mu, \sigma=60)[/tex]

And we are interested on the distribution for the sample mean [tex]\bar X[/tex], we know that distribution for the sample mean is given by:

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

And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for [tex]\bar X[/tex] is normal.

b) [tex]270.283\leq \mu \leq 279.717[/tex]

c) We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)

Step-by-step explanation:

Previous concepts

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

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

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

[tex]\bar X=275[/tex] represent the sample mean for the sample  

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

[tex]\sigma=60[/tex] represent the population standard deviation

n=1077 represent the sample size  

Part a

By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.

[tex]X \sim N(\mu, \sigma=60)[/tex]

And we are interested on the distribution for the sample mean [tex]\bar X[/tex], we know that distribution for the sample mean is given by:

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

And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for [tex]\bar X[/tex] is normal.

Part b

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

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

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]

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

[tex]275-2.58\frac{60}{\sqrt{1077}}=270.283[/tex]    

[tex]275+2.58\frac{60}{\sqrt{1077}}=279.717[/tex]

So on this case the 99% confidence interval would be given by (270.283;3279.717)    

[tex]270.283\leq \mu \leq 279.717[/tex]

Part c

We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)

Final answer:

A 99% confidence interval for the mean quantitative scores for young women, based on the NAEP data, is found to be between 270.27 and 279.73. We can state with 99% confidence that the true mean quantitative score for young women falls within this range.

Explanation:

We are asked to find a 99% confidence interval for the mean quantitative scores for young women, based on the National Assessment of Educational Progress (NAEP) sample data.

a) Check that the normality assumptions are met.

The normality assumption is met because individual NAEP scores are given to have a Normal distribution. Additionally, with a large sample size (more than 30), the Central Limit Theorem ensures the sampling distribution of the mean is approximately normal, regardless of the distribution of the population from which the sample is drawn.

b) What is the 99% confidence interval for the mean quantitative scores for young women?

Using the given data, the sample mean, \\bar{x}\, is 275 and the population standard deviation, \(\sigma\), is 60. Since the population standard deviation is given and the sample size is large (1077), we will use the z-distribution to find the 99% confidence interval.

The z-score corresponding to a 99% confidence level is approximately 2.576. Therefore, the margin of error (ME) can be calculated as:

ME = z * [tex](\(\sigma/\sqrt{n}\))[/tex]

ME = 2.576 * [tex](60/\sqrt{1077})[/tex] \approx 4.73

The 99% confidence interval is:

Lower limit =[tex]\(\bar{x}\)[/tex] - ME = 275 - 4.73 = 270.27

Upper limit =[tex]\(\bar{x}\)[/tex] + ME = 275 + 4.73 = 279.73

The 99% confidence interval for the mean quantitative scores for young women is 270.27 ≤ μ ≤ 279.73.

c) Interpret the confidence interval obtained in previous question

The interpretation of this confidence interval is that we are 99% confident that the true mean quantitative score for young women falls between 270.27 and 279.73.

x is replaced with Y, so it is a reflection across the line y = x

Answer:

y=x

Step-by-step explanation:

Answer:

hi

Step-by-step explanation:

Answer:

[tex]\mu_{\hat{p}}=0.19[/tex]

[tex]\sigma_{\hat{p}}=0.0785[/tex]

Step-by-step explanation:

We know that the mean and the standard error of the sampling distribution of the sample proportions will be :-

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

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

, where p=population proportion and n= sample size.

Given : The proportion of students at a college who have GPA higher than 3.5 is 19%.

i.e. p= 19%=0.19

The for sample size n= 25

The mean and the standard error of the sampling distribution of the sample proportions will be :-

[tex]\mu_{\hat{p}}=0.19[/tex]

[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{0.19(1-0.19)}{25}}\\\\=\sqrt{0.006156}=0.0784601809837\approx0.0785[/tex]

Hence , the mean and the standard error of the sampling distribution of the sample proportions :

[tex]\mu_{\hat{p}}=0.19[/tex]

[tex]\sigma_{\hat{p}}=0.0785[/tex]

Answer:

Step-by-step explanation:

Descartes' rule of signs tells you this function, with its signs {- - - + +}, having one sign change, will have one positive real root.

When odd-degree terms have their signs changed, the signs {- + - - +} have three changes, so there will be 1 or 3 negative real roots.

The constant term (10) tells us the y-intercept is positive. The sum of coefficients is -11 -5 -9 +12 +10 = -3, so f(1) < 0 and there is a root between 0 and 1.

When odd-degree coefficients change sign, the sum becomes -11 +5 -9 -12+10 = -17, so there is a root between -1 and 0.

__

Synthetic division by (x+1) gives a quotient of -11x^3 +6x^2 -15x +27 -17/(x+1), which has alternating signs, indicating -1 is a lower bound on real roots.

Real roots are located in the intervals [-1, 0] and [0, 1].

_____

The remaining roots are complex. All roots are irrational.

The attached graph confirms that roots are in the intervals listed here. Newton's method iteration is used to refine these to calculator precision. Dividing them from f(x) gives a quadratic with irrational coefficients and complex roots.

Answer: D) 0.438 < p < 0.505

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where n= sample size

[tex]\hat{p}[/tex] = Sample proportion.

z* = critical value.

Given : A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature.

i.e. n= 865

[tex]\hat{p}=\dfrac{408}{865}\approx0.4717[/tex]

Two-tailed critical avlue for 95% confidence interval : z* = 1.96

Then, the 95 % confidence interval for the true proportion of all voters in the state who favor approval will be :-

[tex]0.4717\pm (1.96)\sqrt{\dfrac{0.4717(1-0.4717)}{865}}\\\\\approx0.4717\pm 0.03327\\\\=(0.4717-0.03327,\ 0.4717+0.03327)=(0.43843,\ 0.50497)\approx(0.438,\ 0.505)[/tex]

Thus, the required 95% confidence interval : (0.438, 0.505)

Hence, the correct answer is D) 0.438 < p < 0.505