In Exercises 5 and 6, use the scatterplot to find the value of the rank correlation coefficient\({r_s}\)and the critical values corresponding to a 0.05 significance level used to test the null hypothesis of\(\rho \)= 0. Determine whether there is a correlation.

Distance/Time Data for a Dropped Object

Short Answer

Expert verified

The value of\({r_s}\)is approximately equal to 1.

The critical values of Spearman rank correlation coefficient at the 0.05 level of significance are equal to -0.886 and 0.886.

There is enough evidence to conclude that there is a significant correlation between distance and time.

Step by step solution

01

Given information

A scatterplot is constructed between two variables, distance and time.

02

Interpretation of scatterplot

By observing the scatterplot, it can be seen that the points fall along a straight line.

This means that the value of the rank correlation coefficient between distance and time is approximately equal to 1.

In terms of notation, the rank correlation coefficient is:

\({r_s} \approx 1\)

03

Perform the hypothesis testing

The null hypothesis to test the significance of the correlation is as follows:

\({H_0}:\)There is no correlation between the variables distance and time,i.e.\({\rho _s} = 0\).

The alternative hypothesis is as follows:

\({H_1}:\)There is a correlation between the variables distance and time,i.e.\({\rho _s} \ne 0\).

The total number of points plotted in the scatterplot is6.

Therefore, the sample size (n) is 6.

The critical values of the rank correlation corresponding to n=6 at the 0.05 level of significance are -0.886 and 0.886.

Since the value of the rank correlation coefficient does not fall in the interval bounded by the two critical values, the null hypothesis is rejected.

Thus, there is enough evidence to conclude that there is a significant correlation between distance and time.

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