Question

QUESTIONS 3, 9, 12, 13 and 14

10:384 . osu.instructure.com Scores on the anxiety scale are causally related to feelings of comfort in a social gathering Scores on the anxiety scale are significantly related to feelings of comfort in a social gathering Question 3 0.5 pts After analyzing a large database, researchers found a statistically significant relationship between consumption of breakfast cereal per day and daily spending on pet care. If this relationship is only due to random chance it would be considered mixed correlation weak relationship spurious correlation partial correlation D Question 4 0.5 pts If all the points on a scatterplot fall on a single line.

Answer 1

3.

Spurious correlation is nothing but a strong relationship between two variables, often due to coincidence (the relationship does not make any sense, in real world) or due to a confounding variable (- an unseen factor that might be correlated with both variables and thus bring a synchronous change in the two study variables).

Here, in the given problem, Consumption of breakfast cereal per day and Daily spending on pet care hardly sounds related, but the correlation coefficient between them is found to be statistically significant. This can be best regarded as Spurious correlation.

9. Given: r = 0.35

By definition of pearson's correlation coefficient,

• r explains the strength and direction of this linear relationship.
• It ranges from -1 to 1, negative and positive values indicating a negative and positive linear relationship respectively.
• Values close to unity, depicts a strong linear relationship and those close to zero implies weak or no linear relationship.
• According to Cohen, an absolute value of r = 0.1, 0.3 and 0.5 implies the presence of Weak, Medium and Large correlation respectively.

Here , R =0.35. Hence, this may be interpreted as: The strength of the correlation between the two variables is Medium.

10.

To test: H₀: $\rho$ = 0 Vs H_a: $\rho \neq$ 0 at 5% level

Given r = 0.372, p-value = 0.073

Since the p-value corresponding to the correlation coefficient r = 0.372 : p-value =  0.073 > 0.05 is not significant at 5% level, we fail to reject H₀. Hence, the correct option would be:

Fail to reject H₀ since, the p-value is > 0.05.

12.

To test: H₀: $\rho$ = 0 Vs H_a: $\rho \neq$ 0 at 5% level

Computing the correlation coefficient r:

For variables X and Y,

Υ Ξ ηΣ ΧΥ - (ΣΧ)(ΣΥ) [ΗΣΧ2 - (ΣΧ)2][nΣΥ2 – (ΣΥ2]

Using excel,

Font G 1 Allg SUM Xfx =CORREL(A2:A6,B2:36 A B с D E F. Life expectancy People / TV 53.5 315 Correlation =CORREL(A2:A6, B2:B6 61 96 CORREL(array1, array2) 70 90 2 3 4 5 75.5 1 6 78.5 4 7 SR x 10 8 9

We get r = -0.92

The significance of this correlation can be tested by comparing r to its critical value for n - 2 df:

1-tailed 0.05 0.005 0.025 2-tailed Degrees of Freedom = N-2 N 1 2 0.1 0.988 0.900 0.805 0.729 0.669 0.621 0.05 0.997 0.950 0.878 0.811 3 0.01 0.999 0.990 0.959 0.917 0.875 0.834 4 5 0.754 6 0.707

r_3,0.05 = 0.878

13.Since  r = -0.92 for n = 5, with critical value r_5-2=3,0.05 = 0.878, the rejection / critical region being $|r|\geq r_{critical}$ , we find that  |r| = 0.92 > 0.878 lies in the rejection region, we may reject the null hypothesis at 5% level. We may conclude that the correlation is statistically significant at 5% level.

The correct option would be:

Reject null since r statistic lies in the critical region.

14.

From the scatterplot, although we find the data points to be widely scattered and strong clusters could be identified, we still can make out that the data points reveal a slight downward trend. We may guess that the variables Qualtiy of Life and Negativity are negatively correlated (weak linear negative relationship).