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2 years ago

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Jaime finished analyzing a set of data with an explanatory variable x and a response variable y. He finds that the mean and stan

dard deviation for x are 5.43 and 1.12, respectively. The mean and standard deviation for y are 10.32 and 2.69, respectively. The correlation was found to be 0.893.
Select the correct slope and y-intercept for the least-squares line.

1 answer:

malfutka [58]2 years ago

7 0

Answer:

The slope is m=2.145.

The y-intercept is b=-1.33.

Step-by-step explanation:

We have this data:

- The mean and standard deviation for x are 5.43 and 1.12, respectively.

- The mean and standard deviation for y are 10.32 and 2.69, respectively.

- The correlation is 0.893.

We have to calculate the slope and the y-intercept of the least-squares line.

With the given data, we can calculate the slope m as:

$m=r\;\dfrac{s_y}{s_x}=0.893\;\dfrac{2.69}{1.12}=2.145$

Then, the y-intercept is calculated as:

$b=\bar y-m\cdot \bar x=10.32-2.145\cdot 5.43=10.32-11.65=-1.33$

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Answer:

(negative 7 x + 4)(negative 7 x minus 4)

Step-by-step explanation:

Consider two real numbers a and b. A difference of squares involving a and b is usually given as;

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The difference of the two squares above can be factored to yield;

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This implies that in order to have the difference of squares we must have a product involving the difference and the sum of the numbers.

The expression;

(negative 7 x + 4)(negative 7 x minus 4) can also be written as ( 4 - 7x) ( -4 - 7x)

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olga2289 [7]

Answer:

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Step-by-step explanation:

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For which discriminant is the graph possible<br> b2-4ac=0<br><br> b2-4ac=-1<br><br> b2-4ac=4

VARVARA [1.3K]

Answer:

The graph is possible for $b^2-4ac=4$

Step-by-step explanation:

we know that

The discriminant of a quadratic equation of the form

$ax^{2} +bx+c=0$

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$D=b^2-4ac$

If D=0 the quadratic equation has only one real solution

If D>0 the quadratic equation has two real solutions

If D<0 the quadratic equation has no real solution (complex solutions)

In this problem , looking at the graph, the quadratic equation has two real solutions (the solutions are the x-intercepts)

so

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The graph is possible for $b^2-4ac=4$

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An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They

arlik [135]

Answer:

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

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".

$\bar X_1 =2.0233$ represent the sample mean 1

$\bar X_2 =3.048$ represent the sample mean 2

n1=15 represent the sample 1 size

n2=15 represent the sample 2 size

$s_1 =4.893387$ population sample deviation for sample 1

$s_2 =5.12399$ population sample deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: $\mu_1 = \mu_2$

H1: $\mu_1 \neq \mu_2$

And we can do this using the confidence interval for the difference of means.

Solution to the problem

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

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

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 15+15-2=28$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that $t_{\alpha/2}=\pm 1.701$

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

$-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137$

$-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087$

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

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