The problem asks us to test for a statistically significant relationship between the variables using the data from problem 7. Since the data from problem 7 is not provided, I'll assume a generic dataset for illustration. Let's assume we have the following data relating two variables, x and y: x = [1, 2, 3, 4, 5] y = [2, 4, 5, 4, 5] We will perform a linear regression and a hypothesis test to check for significance.

1. Problem Description

The problem asks us to test for a statistically significant relationship between the variables using the data from problem

7. Since the data from problem 7 is not provided, I'll assume a generic dataset for illustration. Let's assume we have the following data relating two variables, x and y:

x = [1, 2, 3, 4, 5]

y = [2, 4, 5, 4, 5]

We will perform a linear regression and a hypothesis test to check for significance.

2. Solution Steps

Step 1: State the null and alternative hypotheses.

Null hypothesis ($H_0$): There is no statistically significant relationship between x and y. The slope of the regression line is zero.

Alternative hypothesis ($H_1$): There is a statistically significant relationship between x and y. The slope of the regression line is not zero.

Step 2: Calculate the linear regression line. The formula for the slope (b) and y-intercept (a) are:

$b = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}$

$a = \frac{\sum y - b(\sum x)}{n}$

First, calculate the necessary sums:

$\sum x = 1 + 2 + 3 + 4 + 5 = 15$

$\sum y = 2 + 4 + 5 + 4 + 5 = 20$

$\sum xy = (1*2) + (2*4) + (3*5) + (4*4) + (5*5) = 2 + 8 + 15 + 16 + 25 = 66$

$\sum x^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55$

n = 5 (number of data points)

Now calculate b:

$b = \frac{5(66) - (15)(20)}{5(55) - (15)^2} = \frac{330 - 300}{275 - 225} = \frac{30}{50} = 0.6$

Now calculate a:

$a = \frac{20 - 0.6(15)}{5} = \frac{20 - 9}{5} = \frac{11}{5} = 2.2$

So, the regression line is $y = 2.2 + 0.6x$

Step 3: Calculate the standard error of the slope ($SE_b$).

First, we need to calculate the sum of squares error (SSE):

$SSE = \sum (y_i - \hat{y_i})^2$, where $\hat{y_i} = a + bx_i$

$\hat{y_1} = 2.2 + 0.6(1) = 2.8$

$\hat{y_2} = 2.2 + 0.6(2) = 3.4$

$\hat{y_3} = 2.2 + 0.6(3) = 4.0$

$\hat{y_4} = 2.2 + 0.6(4) = 4.6$

$\hat{y_5} = 2.2 + 0.6(5) = 5.2$

$SSE = (2-2.8)^2 + (4-3.4)^2 + (5-4)^2 + (4-4.6)^2 + (5-5.2)^2 = (-0.8)^2 + (0.6)^2 + (1)^2 + (-0.6)^2 + (-0.2)^2 = 0.64 + 0.36 + 1 + 0.36 + 0.04 = 2.4$

Now calculate the standard error of the estimate ($s_e$):

$s_e = \sqrt{\frac{SSE}{n-2}} = \sqrt{\frac{2.4}{5-2}} = \sqrt{\frac{2.4}{3}} = \sqrt{0.8} \approx 0.8944$

Calculate the standard error of the slope ($SE_b$):

$SE_b = \frac{s_e}{\sqrt{\sum (x_i - \bar{x})^2}}$

First, we need to find $\bar{x}$:

$\bar{x} = \frac{\sum x}{n} = \frac{15}{5} = 3$

Then, $\sum (x_i - \bar{x})^2 = (1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2 = (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 = 4 + 1 + 0 + 1 + 4 = 10$

$SE_b = \frac{0.8944}{\sqrt{10}} \approx \frac{0.8944}{3.1623} \approx 0.2828$

Step 4: Calculate the t-statistic.

$t = \frac{b - 0}{SE_b} = \frac{0.6}{0.2828} \approx 2.121$

Step 5: Determine the p-value.

Degrees of freedom = n - 2 = 5 - 2 =

3. Using a t-table or calculator with df = 3, the p-value for t = 2.121 for a two-tailed test is approximately 0.

1

2

7.

Step 6: Make a decision.

Let's assume a significance level of $\alpha = 0.05$.

Since the p-value (0.127) > $\alpha$ (0.05), we fail to reject the null hypothesis.

3. Final Answer

There is no statistically significant relationship between x and y based on this sample data and the assumed significance level.