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liubo4ka [24]
4 years ago
6

Write an equation in slope -intercept form for a line that passes through (-2,7) and (14,21)

Mathematics
1 answer:
Georgia [21]4 years ago
7 0
(-2,7) and (14,21)

Find the slope by using the formula m =  \frac{y2 - y1}{x2 - x1}

Plug in the numbers. m = \frac{21 - 7}{14 - - 2} 
 \\  m= \frac{14}{16}  =  \frac{7}{8} 


Write it in point slope form first

y - y1 = m(x - x1)
y - 7 = 7/8(x - -2)
y - 7 = 7/8x + 1.75
   + 7            +7

y = 7/8x + 8.75

Our final equation would be: <span>y = 7/8x + 8.75</span>

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Using the t-distribution, it is found that since the p-value of the test is 0.72 > 0.05, there is not significant difference between the values of each group.

At the null hypothesis, we <u>test if the groups have the same values</u>, that is, the subtraction of their means is 0:

H_0: \mu_2 - \mu_1 = 0

At the alternative hypothesis, we <u>test if the groups have different values</u>, that is, the subtraction of their means is different of 0:

H_1: \mu_2 - \mu_1 \neq 0

First, we have to find the mean and the standard deviation for both samples, hence, using a calculator:

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The standard errors are:

s_1 = \frac{\sigma_1}{\sqrt{n_1}} = \frac{6.2507}{\sqrt{8}} = 2.21

s_2 = \frac{\sigma_2}{\sqrt{n_2}} = \frac{4.2678}{\sqrt{8}} = 1.5089

For the distribution of differences, we have that:

\overline{x} = \mu_2 - \mu_1 = 19.75 - 18.75 = 1

s = \sqrt{s_1^2 + s_2^2} = \sqrt{2.21^2 + 1.5089^2} = 2.676

We have the <u>standard deviation for the samples</u>, hence, the t-distribution is used. The test statistic is given by:

t = \frac{\overline{x} - \mu}{s}

In which \mu is the value tested at the null hypothesis, which is 0 for this problem. Hence, the value of the test statistic is:

t = \frac{\overline{x} - \mu}{s}

t = \frac{1 - 0}{2.676}

t = 0.37

To find the p-value, we have a <u>two-tailed test</u>, as we test if the means are different from a value, with <u>t = 0.37</u> and 8 + 8 - 2 = <u>14 df</u>.

Using a t-distribution calculator, this p-value is of 0.72.

Since the p-value of the test is 0.72 > 0.05, there is not significant difference between the values of each group.

A similar problem is given at brainly.com/question/15682365

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