Question

A. As a health policy expert, you are interested in knowing whether men and women in the United States differ in the frequency they visit doctors when sick. So you conduct a survey of 134 people and ask them how often they visit a doctor when they are sick.  The results from this sample are as follows:

	Frequency visit doctor
Sex	Always	Sometimes	Never	Total
Men	13	41	27	81
Women	24	19	10	53
Total	37	60	37	134

Perform the appropriate hypothesis test, at alpha = .01, to determine whether there is a relationship between sex and the frequency with which individuals in the United States visit doctors.  You must show your hand calculations for this problem. Describe your hypotheses, results, and conclusion in a brief paragraph.  If it is appropriate, you should also discuss the strength of the relationship based upon a suitable measure.

B . Show in a table or explain in words what the crosstabulation for your two variables from

Question above would look like if there were absolutely no association between the independent variable and the dependent variable.

Answer 1

Solution:-

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

H₀: Sex and the frequency with which individuals in the United States visit doctors are independent.
H_a: Sex and the frequency with which individuals in the United States visit doctors are not independent.

Formulate an analysis plan. For this analysis, the significance level is 0.01. 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) * (3 - 1)
D.F = 2
E_r,c = (n_r * n_c) / n

Χ² = 13.90

where DF is the degrees of freedom.

The P-value is the probability that a chi-square statistic having 2 degrees of freedom is more extreme than 13.90.

We use the Chi-Square Distribution Calculator to find P(Χ² > 13.90) = 0.001

Interpret results. Since the P-value (0.001) is less than the significance level (0.05), we cannot accept the null hypothesis.

Thus, we conclude that there is a relationship between Sex and the frequency with which individuals in the United States visit doctors.