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Katen [24]
2 years ago
10

26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics

course. Then they are given a post-test once the professor has concluded lecturing on the material. Pre-test and post-test scores for 4 students in an elementary statistics class are given below. Can it be concluded that the test scores have improved? The standard deviation is 6.02. Use α = .05.
Pre-test 65 70 75 84
Post-test 80 82 85 85
Difference 15 12 10 1

a. Hypotheses:




b. Test Statistic:






c. Critical Value(s):





d. Decision:






e. Conclusion:
Mathematics
1 answer:
hjlf2 years ago
7 0

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

\mu _d = \mu _{post} - \mu _{pre}

Then the null hypotheses and alternative hypotheses will be

H₀: \mu _d = 0

Hₐ: \mu _d > 0

Then the test statistic will be

\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5

Then

\rm S_d = 6.02

The test statistic value is given by

\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16

Since this is a right-tailed test, so the critical value is given by

\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that \mu _d > 0 that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

#SPJ1

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ZanzabumX [31]

Answer:

The slope is -1

Y-intercept is (0, -6)

Equation is y = -1x - 6 or y = -x - 6 either is right

Step-by-step explanation:

We already know that the slope is negative because the line is pointed this way \ pointed up to the left.

You have to do rise/run to find the slope.

The rise is 6 and the run is 6

6/6 = 1 since the line is negative the slope is -1x or -x

The y intercept is -6

Use the point (-2, -4)

Plug in

-4 = -(-2) + b

-4 = 2 + b (subtract 2 on both sides)

-6 = b

y = -x - 6

Hope this helps ya!!

4 0
2 years ago
Sarah ran 3/4 mile on Monday, 3/4 mile on Tuesday, and 1/4 mile on Wednesday. How many miles did Sue run in the 3 days
SCORPION-xisa [38]

Answer:

1 3/4 miles

Step-by-step explanation:

add 3/4 +3/4+1/4 to get your answer

6 0
3 years ago
When evaluating (6 x + 9) squared minus 5 for x = 1, the given value of 1 is substituted for what part of the expression?
STALIN [3.7K]

Answer:

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Step-by-step explanation:

i got it right on edge :)

4 0
3 years ago
Read 2 more answers
What is the distance between the lines y = 2x and y = 2x+5? Round your answer to the nearest tenth.
Ilia_Sergeevich [38]

The distance between both lines is 2.24 units

Step-by-step explanation:

There is no straight method to find the distance between two lines. The distance can be found out by finding a point on one line and then finding the distance of that point from the other line. The y-coordinate of point is obtained by putting any value of x in equation . The x and y combined give us the point.

Given

y=2x\\Putting\ x=1\\y=2(1)\\y=2\\So, the\ point\ on\ y=2x\ is\ (1,2)

We have to find the distance of this point from y=2x+5

y=2x+5\\Subtracting\ y\ from\ both\ sides\\y-y=2x+5-y\\2x-y+5=0

The formula for finding distance of a point (x,y) from a line is:

d=\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}

A=2

B=-1

C=5

Putting the values in the formula

d=\frac{|(2)(1)+(-1)(2)+5|}{\sqrt{(2)^2+(-1)^2}}\\d=\frac{|2-2+5|}{\sqrt{4+1}}\\d=\frac{|5|}{\sqrt{5}}\\d=\frac{5}{\sqrt{5}}\\d=2.236\ units

Rounding off will give: 2.24 units

The distance between both lines is 2.24 units

Keywords: Equations of lines

Learn more about distance in lines at:

  • brainly.com/question/7449065
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#LearnwithBrainly

7 0
3 years ago
Which of the following equations is of the parabola whose vertex is at (2, 3), axis of symmetry parallel to the y-axis and p = 4
lora16 [44]

Answer:

option A.) y-3 = 1/16 (x-2)^2

Step-by-step explanation:

we know that

If the axis of symmetry is parallel to the y-axis, then we have a vertical parabola

The equation of a vertical parabola is equal to

4p(y-k)=(x-h)^{2}

where

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In this problem we have

(h,k)=(2,3)

p=4

substitute

4(4)(y-3)=4(4)(x-2)^{2}

16(y-3)=(x-2)^{2}

(y-3)=(1/16)(x-2)^{2}

8 0
3 years ago
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