The answer because I hate homework it's to hard<br>

Which of the following equations best represents the regression line for the data given in the

True [87]

The regression equation for the data given is y= -8.57 -2.31x

Step-by-step explanation:

The first step is to form a table as shown below;

x y xy x² y²

1 4 4 1 16

2 1 2 4 1

3 5 15 9 25

4 10 40 16 100

5 16 80 25 256

6 19 114 36 361

7 15 105 49 225

28 60 360 140 984 ------sum

A linear regression equation is in the form of y=A+Bx

where ;

x=independent variable

y=dependent variable

n=sample size/number of data points

A and B are constants that describe the y-intercept and the slope of the line

Calculating the constants;

A=(∑y)(∑x²) - (∑x)(∑xy) / n(∑x²) - (∑x)²

A=(60)(140) - (28)(360) / 7(140)-(28)²

A=8400 - 10080 /980-784

A= -1680/196

A= - 8.57

B= n(∑xy) - (∑x) (∑y) / n(∑x²) - (∑x)²

B= 7(360)-(28)(60) / 7(60) - (28)²

B=2520 - 1680 /420-784

B=840/-364

B= -2.31

y=A+Bx

y= -8.57 -2.31x

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Regression equation :brainly.com/question/12280902

Keywords : equations, regression line, data

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7 0

3 years ago

A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e

professor190 [17]

Answer:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

$n=15$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =16$ represent the sample variance

$\sigma^2_0 =25$ represent the value that we want to verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 25$

Alternative hypothesis: $\sigma^2$

The statistic is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And replacing we got:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

6 0

2 years ago

Someone help me<br>10-14x-7-7x=6

Burka [1]

Answer: your answer is going to be 1/7

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

25 POINTS! MATCHING PROBLEM!

Nataly_w [17]

1. f(x) = 1
2. f(x) = x
3. f(x) = 5 - x
4. f(x) = 3