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Which line is the best model for the data in the scatter plot? A scatterplot with an x axis labeled, variable A from twenty to s

Firlakuza [10]

The line which is the best model for the data in a scatter plot is: D. A scatterplot with an x axis labeled, variable A from twenty to sixty in increments of ten and the y axis labeled, variable B from twenty to forty in increments of five with a negative trend line through the middle of eight points.

<h3>What is a scatter plot?</h3>

A scatter plot can be defined as a graph which uses Cartesian coordinates (x-axis and y-axis) to display the values of two (2) variables, with the resulting points revealing any correlation between the data.

<h3>What is a trend line?</h3>

A trend line refers to an statistical (analytical) tool that is used in conjunction with a scatter plot, so as to determine whether or not there's any correlation between a data.

For example, a scatterplot with an x-axis with its points ranging from twenty (20) to sixty (60) in increments of ten (10) while the y-axis has its points ranging from twenty (20) to forty (40) in increments of five (5) with a negative trend line through the middle of eight (8) points i.e above and below the trend line.

Read more on scatterplot here: brainly.com/question/6592115

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2 years ago

Test the claim below about the mean of the differences for a population of paired data at the level of significance α Assume the

Evgen [1.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is F

$t = 1.17$

$t_ {\alpha , df} =t_ {0.10 , 8} = 1.86$

Since the test statistics<u> is outside </u>the rejection region , <u>we fail to reject </u>the null hypothesis ,There<u> is no</u> statistically significant evidence to reject the claim

Step-by-step explanation:

From the question we are told that

The claim is $\mu = 0$

Hence

The null hypothesis is $H_o : \mu = 0$

The alternative is $H_a : \mu \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= d - \mu_d }{ \frac{s_d}{ \sqrt{n} } }$

=> $t = \frac{3.5 -0 }{ \frac{8.94}{ \sqrt{9} } }$

=> $t = 1.17$

Generally the degree of freedom is mathematically represented as

$df = n- 1$

$df = 9 - 1$

$df = 8$

From the student t-distribution table the critical value of $\alpha$ at a degree of freedom of 8 is

$t_ {\alpha , df} =t_ {0.10 , 8} = 1.86$

Since the $t_ {\alpha , df}$ is outside the rejection region , we fail to reject the null hypothesis ,There is no sufficient evidence to reject the claim