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1 year ago

What is the 95% confidence interval of E(Y) at X = 4?

Mathematics

1 answer:

diamong [38]1 year ago

4 0

Based on the calculations, the 95% confidence interval of E(Y) is equal to 52.3% ± 3.26%.

<h3>How to construct a 95% confidence interval?</h3>

Mathematically, a confidence interval of 95% is given by;

α = 1 - 0.95

α = 0.05.

α/2 = 0.05/2 = 0.025.

Next, we would determine the standard deviation of the mean:

Standard deviation of mean = 50/√800

Standard deviation of mean = 1.77.

From the z-table, the z-score of a 95% confidence interval is equal to 1.96.

Confidence interval (0.05, 50, 800) = 1.77 × 1.96

Confidence interval (0.05, 50, 800) = 3.46.

Mathematically, the confidence interval for a mean is given by:

Mean ± (t-critical × (standard deviation/√(sample size)))

Confidence interval = 52.3% ± 3.26%.

Read more on confidence interval here: brainly.com/question/25779324

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Complete Question:

A research team studied Y, the percentage of voters in favor of a candidate. The random variable Y had standard deviation o = 50%. A random sample of 800 voters was selected, and their average was y-bar(800) = 52.3%. What is the 95% confidence interval for E(Y)?

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If T: (x, y) → (x + 6, y + 4), then T-1: (x,y) → _____.

alexgriva [62]

T is a linear transformation from R²→R² with basis {(1,0),(0,1)}
T: (x,y)→(x+6,y+4)

A function from one vector space to another that preserves the underlying (linear) structure of each vector space is called a linear transformation.

Then the vector (1,0) goes to (1+6,4)=(7,4)=7(1,0)+4(0,1)

and the vector (0,1) goes to (6,1+4)=(6,5)=6(1,0)+5(0,1)

So, the matrix of the transformation is

$\left[\begin{array}{ccc}7&6\\4&5\end{array}\right]$

The inverse of the matrix is

$\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]$

So, the Inverse Transformation is given by

$T^{-1}(x,y)=\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]\left[\begin{array}{ccc}x\\y\end{array}\right] =(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$

So, no option is correct. And the answer is

$T^{-1}(x,y)=(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$

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5 0

1 year ago

Is the following growth or decay? f(x)= 3(0.4)^x

Anna11 [10]

The graph is a decay

7 0

2 years ago

8/7 = 12/k solve for k

Serjik [45]

Answer:

k = 10.5

Step-by-step explanation:

8/7 = 12/k

Using cross products

8k = 7*12

8k = 84

Divide each side by 8

8k/8 = 84/8

k = 10.5

7 0

2 years ago

Read 2 more answers

Pleaseee helppp with thissss asapppp

Natasha2012 [34]

As we know that the standard equation of circle is ${\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}}$ , where r is the radius of circle and centre at (h,k) 

Now , as the circle passes through (2,9) so it must satisfy the above equation after putting the values of h and k respectively

${:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}$

${:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}$

${:\implies \quad \sf 9+16=r^{2}}$

After raising ½ power to both sides , we will get r = +5 , -5 , but as radius can never be -ve . So r = +5 

Now , putting values in our standard equation ;

${:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}$

${:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}$

This is the required equation of Circle

Refer to the attachment as well !

8 0

1 year ago

Help and explain how you got it. no trolling or report!

Galina-37 [17]

Answer:

I am 100% confident in my calculations. I'm not 100% sure about interpreting. Sorry I can't be of more help.

Part A:

Median: 4.5

IQR: 3.5

Mode: 6

I think the median/IQR is a better measure of center and variability because the range of data is so big (10 - 0 = 10), and the data skews to the right with the 6s.

Part B:

I would say no. It is the one that occurs most frequently, but there are many more times she rides less than 6, which is why the median is 4.5

Step-by-step explanation:

Mean: 4.714

MAD: 2.43

Median: 4.5

IQR: 3.5

Mode: 6

the number set in order:

0, 2, 2, 2.5, 2.5, 3, 4, 5, 6, 6, 6, 8, 9, 10

The median (middle number) is two numbers: 4 and 5, so the median is 4.5

The first quartile/Q1 (middle of the bottom half of data) is 2.5, the third quartile/Q3 (middle of the top half of data) is 6. So the IQR (interquartile range) is 6 - 2.5 = 3.5

The mean is the sum of the numbers divided by the quantity of the numbers, so 66/14 = 4.714

the mean absolute deviation is found by calculating the mean, and then finding the absolute value of the difference between each number of the set and the mean...then adding those together and finding the mean of that set. Labor intensive, but that's 34/14 = 2.42857143

4 0

2 years ago