An economist proposed that there might be some level of relationship between a p

Question

An economist proposed that there might be some level of relationship between a person's height and his or her annual income. A student of ECON1005, Introductory Statistics, decided to test this hypothesis using data on height and income of 185 persons. The people in the survey were categorized as follows: Category 1 Category 2 Category 3 Less than 5 feet tall 5 to 6 feet tall Over 6 feet tall Each person in the survey was classified as having one of the following annual salaries: Less than $20,000 $20,000 to less than $30,000 $30,000 to less than $40,000 $40,000 to less than $50,000 S50,000 and over The Exhibit overleaf was generated by MINITAB. Use the Exhibit to answer the question below Carefully define the null and alternative hypotheses of the x test underlying the generation of the above table. (a) (2 marks) (b) il in the missing values *", **", **, (c What is the conclusion of this test? Give reasons for your answer. (3 marks) (d) State one (I) limitation of the x test. (4 marks) (1 mark) Chi-Square Test for Association: Height, Annual Income E Total Categoryl 50 20 15.40 1.37 8.648 0.64 13.51 4.18 4.59 0.04 0.00 Category2 15 16.95 0.22 18 14.86 8.62 0.02 5.05 0.75 9.51 1.30 15 0.10 32 26 0.89 50 Category3 80 12 24.65 13.84 21.62 12.54 0.02 7.35 0.75 0.28 Total 57 29 17 185 Cell Contents: Count Expected count Contribution to Chi-square Chi-Square 10.47, DF P-Value Total 10 marks

Explanation / Answer

(a)

H0: The factors 'Category' and 'Height ranges' are independent
H1: The factors are dependent

(b)

Under each of the categories,

* = E14 = R1 x C4 / G = 50 x 29 / 185 = 7.84
** = (O23 - E23)2/E23 = (18 - 14.86)^2/14.86 = 0.66
*** = DF = (# of categories - 1)*(# of height ranges - 1) = (3 - 1)*(5 - 1) = 8
**** = p-value for test statistic 11.22 at a DF of 8 = =1-CHISQ.DIST(11.13,8,TRUE) = 0.19

Note that the test statistic is i,j(Oij - Eij)2/Eij = Sum of the 3rd rows = 10.47 + 0.66 = 11.13

(c)

The p-value is greater than 0.05, we fail to reject the null hypothesis with a confidence level of 95% and thus conclude that the two factors are independent of each other i.e. not associated.

(d)

Limitation - The Chi-square test statistic is very sensitive to small frequencies in the cells of the contingency table. Therefore, when the expected frequency in a cell is less than 5, chi-square test may lead to erroneous conclusions.

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