Mathematics, 06.03.2020 21:23 CobyHageman

I took a sample of the grade point averages for students in my class. For the 25 students, the standard deviation of grade points was 0.65 and the mean was 2.89. A 95% confidence interval for the average grade point average for all students in my class is: ( 2.62 , 3.16 ) ( 2.53 , 3.25 ) ( 2.64 , 3.14 )

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Answer from: Sophiiiiamm374

$2.89-2.06\frac{0.65}{\sqrt{25}}=2.62$

$2.89+2.06\frac{0.65}{\sqrt{25}}=3.16$

So on this case the 95% confidence interval would be given by (2.62;3.16)

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".

$\bar X= 2.89$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=0.65 represent the sample standard deviation

n=25 represent the sample size

Solution to the problem

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

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

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

df=n-1=25-1=24

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,24)".And we see that $t_{\alpha/2}=2.06$

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

$2.89-2.06\frac{0.65}{\sqrt{25}}=2.62$

$2.89+2.06\frac{0.65}{\sqrt{25}}=3.16$

So on this case the 95% confidence interval would be given by (2.62;3.16)

Answer from: Quest

whats the question?

step-by-step explanation:

Answer from: Quest

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.