Question

1. A random sample of subjects who tested positive for influenza during the 2014-15 flu season was obtained. There are four different influenza type: A/H1N1, A/H3N3, A-unsub, and B. Each subject was classified by age group and influenza type. Determine whether there is evidence at the 5% significance level that age group and influenza type are associated. Fully justify! Include hypotheses, test used, test statistic, p-value, decision and conclusion. You should put the frequencies into a matrix. Also report the Observed and Expected Matrices.

	Influenza type
age group	A/H1N1	A/H3N2	A unsub	B	Total
Younger than 5	150	999	1628	495	3272
5-19	96	748	1163	495	2502
20-44	319	1207	1617	512	3655
45-64	200	1206	1946	476	3828
65+	220	3299	5316	997	9832
	985	7459	11670	2975	23089

Answer 1

Solution:-

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

H₀: Age group and influenza type are independent.
H_a: Age group and influenza type are not independent.

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) = (5 - 1) * (4 - 1)
D.F = 12
E_r,c = (n_r * n_c) / n


Χ² = 517.65

where DF is the degrees of freedom.

The P-value is the probability that a chi-square statistic having 12 degrees of freedom is more extreme than 517.65.

We use the Chi-Square Distribution Calculator to find P(Χ² > 517.65) = 0.000

Interpret results. Since the P-value (0.000) is less than the significance level (0.05), we cannot accept the null hypothesis. Thus, we conclude that there is a relationship between age group and influenza type.