The accompanying TI-83/84 Plus calculator display results from thehypothesis test described in Exercise 1. Assume that the hypothesis test requirements are allsatisfied. Identify the test statistic and the P-value (expressed in standard form and rounded tothree decimal places), and then state the conclusion about the null hypothesis.

The hypothesis test is conducted for the data in exercise 1. It is assumed that all requirements of the test are satisfied and the results for the test are given in the accompanying image.

The hypothesis for independence of ear preference and handedness is tested using chi-square test when the assumptions are satisfied.

From the provided result, the test statistic can be observed as\[{\chi ^2} = 64.5172694\].

Thus, the chi- square test statistic is 64.517.

P-value is the probability of getting the value as extreme as test statistic.

Thus, the associated P-value can be observed as 0.000.

Assume the level of significance is 0.05.

Since the P-value (0.000) is less than the level of significance (0.05), the null hypothesis is rejected at 0.05 level of significance.

Therefore, the decision is to reject the null hypothesis.

There is not enough evidence at 0.05 level of significance to conclude that the handedness and cell phone ear preference are independent.