Question

listed below are numbers of internet users per 100 people and numbers of scientific award winners per 10 million people for different countries. construct a scatterplot, find the value of the linear correlation coefficient r, and find the p - value of r. determine whether there is sufficient evidence to support a claim of linear correlation between the two variables. use a significance level of α = 0.05.
internet users | 79.4 79.7 56.9 67.2 76.8 38.6
award winners | 5.5 9 3.3 1.6 11 0.1

determine the null and alternative hypotheses.
h₀: ρ ▼
h₁: ρ ▼
(type integers or decimals. do not round.)
the test statistic, is t = □
(round to two decimal places as needed.)
the p - value is □
(round to three decimal places as needed.)
because the p - value of the linear correlation coefficient is ▼ the significance level, there ▼ sufficient evidence to support the claim that there is a linear

Explanation:

Step1: Define Hypotheses

For testing linear correlation, the null hypothesis \( H_0 \) is that the population correlation coefficient \(
ho = 0 \) (no linear correlation), and the alternative hypothesis \( H_1 \) is that \(
ho
eq 0 \) (there is linear correlation). So \( H_0:
ho = 0 \), \( H_1:
ho
eq 0 \).

Step2: Calculate Sample Correlation (r)

First, list the data:

• \( x \) (Internet Users): 79.4, 79.7, 56.9, 67.2, 76.8, 38.6
• \( y \) (Award Winners): 5.5, 9, 3.3, 1.6, 11, 0.1

Calculate \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^2 \), \( \sum y^2 \):

• \( \sum x = 79.4 + 79.7 + 56.9 + 67.2 + 76.8 + 38.6 = 398.6 \)
• \( \sum y = 5.5 + 9 + 3.3 + 1.6 + 11 + 0.1 = 30.5 \)
• \( \sum xy = (79.4 \times 5.5) + (79.7 \times 9) + (56.9 \times 3.3) + (67.2 \times 1.6) + (76.8 \times 11) + (38.6 \times 0.1) \)

\( = 436.7 + 717.3 + 187.77 + 107.52 + 844.8 + 3.86 = 2298.95 \)

• \( \sum x^2 = 79.4^2 + 79.7^2 + 56.9^2 + 67.2^2 + 76.8^2 + 38.6^2 \)

\( = 6304.36 + 6352.09 + 3237.61 + 4515.84 + 5898.24 + 1489.96 = 27798.1 \)

• \( \sum y^2 = 5.5^2 + 9^2 + 3.3^2 + 1.6^2 + 11^2 + 0.1^2 \)

\( = 30.25 + 81 + 10.89 + 2.56 + 121 + 0.01 = 245.67 \)

The formula for \( r \) is:

$$ r = \frac{n\sum xy - \sum x \sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} $$

where \( n = 6 \).

Substitute values:

• Numerator: \( 6 \times 2298.95 - 398.6 \times 30.5 = 13793.7 - 12157.3 = 1636.4 \)
• Denominator part 1: \( 6 \times 27798.1 - (398.6)^2 = 166788.6 - 159281.96 = 7506.64 \)
• Denominator part 2: \( 6 \times 245.67 - (30.5)^2 = 1474.02 - 930.25 = 543.77 \)
• Denominator: \( \sqrt{7506.64 \times 543.77} \approx \sqrt{4082000} \approx 2020.396 \) (approximate calculation, more accurately: \( 7506.64 \times 543.77 \approx 7506.64 \times 543.77 \approx 4,082,000 \) (exact: \( 7506.64\times543.77 = 7506.64\times(500 + 43.77)= 3,753,320 + 328,600 = 4,081,920 \))
• \( r = \frac{1636.4}{2020.4} \approx 0.81 \) (more accurately, using calculator: \( r \approx 0.8097 \))

Step3: Calculate Test Statistic (t)

The formula for the test statistic \( t \) is:

$$ t = \frac{r\sqrt{n - 2}}{\sqrt{1 - r^2}} $$

Substitute \( r \approx 0.8097 \), \( n = 6 \):

$$ t = \frac{0.8097\sqrt{6 - 2}}{\sqrt{1 - (0.8097)^2}} = \frac{0.8097 \times 2}{\sqrt{1 - 0.6556}} = \frac{1.6194}{\sqrt{0.3444}} \approx \frac{1.6194}{0.5868} \approx 2.76 $$

Step4: Calculate P - value

The degrees of freedom \( df = n - 2 = 6 - 2 = 4 \). The test is two - tailed (since \( H_1:
ho
eq 0 \)). Using a t - distribution table or calculator, the P - value for \( t = 2.76 \) with \( df = 4 \) is the probability that \( |t| > 2.76 \). From t - table, for \( df = 4 \), \( t_{0.025}=2.776 \), \( t_{0.05}=2.132 \). Since \( 2.132 < 2.76 < 2.776 \), the P - value is between \( 0.02 < P - value < 0.05 \). Using a calculator (t - test for correlation), the exact P - value is approximately \( 0.045 \) (more accurately, using statistical software: \( P - value \approx 0.0447 \)).

Step5: Decision Rule

Compare the P - value with \( \alpha = 0.05 \). If \( P - value \leq \alpha \), we reject \( H_0 \); otherwise, we fail to reject \( H_0 \). Since \( P - value \approx 0.045 < 0.05 \), we reject \( H_0 \).

Answer:

• Null hypothesis \( H_0:

ho = 0 \)

• Alternative hypothesis \( H_1:

ho
eq 0 \)

• Test statistic \( t \approx 2.76 \) (rounded to two decimal places)
• P - value \( \approx 0.045 \) (rounded to three decimal places)
• Because the P - value of the linear correlation coefficient is less than the significance level, there is sufficient evidence to support the claim that there is a linear correlation between the two variables.