Question

Question 1:

A researcher is interested in knowing if there are differences between the incidence of accidents

among pilots of different experience levels. They were able to collect 62 responses from a random number of pilots. Among them, 17 new pilots were in accidents, 26 were not;; 13 experienced pilots were in accidents, 6 were not.

• Construct a contingency table.

• Run the Chi-­Square test for independence using StatCrunch.

• To access, click STAT>Tables>Contingency> With summary

• State your findings and their significance. Do not forget to give summary statistics in

APA format.

Question 2:

A researcher has found that pilots taking an over the counter antihistamine have poor reaction

times. They performed an experiment on 22 pilots: 11 took the medication and 11 did not. Their

reaction times to a command by Air Traffic Control was considered to be adequate or

inadequate. Among those taking the medication, 8 were rated inadequate and 3 adequate.

Among those not-­taking the medication, 10 were rated adequate and only 1 inadequate.

• Construct a contingency table.

• Run the Chi-­Square test for independence using StatCrunch.

• To access, click STAT>Tables>Contingency> With summary

• State your findings and their significance. Do not forget to give summary statistics in

APA format.

• What is the problem with this research study? Is the Chi-­Square test appropriate? Why

or why not? What could the researcher do to potentially rectify the problem

Answer 1

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

H₀: The null hypothesis states that the proportion of accidents by experienced drivers is identical to the proportion of accidents by non-experienced drivers.

Alternative hypothesis: At least one of the null hypothesis statements is false.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for independence.

Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = (r - 1) * (c - 1) = (2 - 1) * (2 - 1)
D.F = 1
E_r,c = (n_r * n_c) / n

$\chi ^{2}=\sum \left [\frac{(O_{r,c}-E_{r,c})^{2}}{E_{r,c}} \right ]$

Χ² = 4.40

where DF is the degrees of freedom.

The P-value is the probability that a chi-square statistic having 1 degrees of freedom is more extreme than 4.40.

We use the Chi-Square Distribution Calculator to find P(Χ² > 4.40) = 0.036.

Interpret results. Since the P-value (0.036) is less than the significance level (0.05), we cannot accept the null hypothesis.