Question

terences Mailings Review View x' | A. g. Δ. | 삐리캐 ■ 1 : . I 희·EB. | | T List Para..! 1 Normal No Spaci Heading 1 Heading 2 Paragraph Question One a) The table below shows the investment plans made by people of all ages. Type of investment Age 25 -30 36 -50 50 70 Common Fixed 30 95 15 46 37 52 Mixed 69 18 38 Does this data support the hypothesis that investment type is independent of the age of the person? The life time distribution for a sample of 1000 electronic components produced by Chege and Ouma Industries is shown in the following table: b) No. of electroniccomponents Life time (hours) Less than 2000 2000 and less than 2020 2020 and less than 2040 2040 and less than 2060 2060 and less than 2080 2080 and less than 2100 2100 and less than 2120 2120 and more 90 122 135 160 175 124 110 84 (0 Compute quartile coefficient of skewness for the given distribution and comment on the () Compute quartile coefficient of variation for the given distribution result. State clearly any assumptions made before starting the calculations. c) The length distribution of widgets manufactured by ABC Industries Ltd. is normally distributed with mean 100cm and variance of 64cm.If the firm produces 10, 000 such widgets per month, how many widgets produced months would be of length? () Less than 92 cm () More than 120cm (ini) Between 86 cm and 96cm tween 84 cm and 112 cm

Answer 1

Solution

The solution is based on Chi-square Test for Independence, also known as Contingency Chi-square.

Back-up Theory

Suppose we have a contingency table with r rows representing r levels/grades of one attribute and c columns representing c levels/grades of another attribute.

The Chi-square Test of Independence is designed to test if the two attributes are associated.

Hypotheses

Null H₀: The two attributes are independent Vs

Alternative H₁: The two attributes are not independent

Test Statistic

χ² = ∑(i = 1 to r, j = 1 to c){(Oij - Eij)²/Eij}, where Oij and Eij are respectively, the observed and the expected frequencies of the ijth cell of the contingency table.

Distribution

Under H₀, χ² ~ χ²_n, where n = {(r - 1)(s - 1)}

Critical Value

Given level of significance as α, critical value is the upper α% point of χ²_n.

p-value

P(χ²_n > χ²_cal)

Rejection Region

χ²_cal > χ²_crit or equivalently, p-value < α.

Decision

Since χ²_cal > χ²_crit, H₀is rejectedor equivalently, p-value < α, H₀is rejected.

Conclusion

There is not enough evidence to suggest that The two attributes, namely, degree of interest in political election and level of education are independent.

Now to work out the solution,

Row represents the age group and column represents the Type of Investment.

CONTINGENCY		(3 x 3)	r = 3	c = 3
Oij			DF =	4
	1	2	3	Oi.
1	10	40	17	67
2	20	35	12	67
3	21	15	30	66
O.j	51	90	59	200
Eij = (Oi. X O.j)/N
	1	2	3	Total
1	17.085	30.15	19.765	67
2	17.085	30.15	19.765	67
3	16.83	29.7	19.47	66
Total	51	90	59	200
Chi-square = (Oij - Eij)^2/Eij
	1	2	3	Total
1	2.938088	3.217993367	0.386806	6.542887
2	0.49735	0.780182421	3.050606	4.328138
3	1.033209	7.275757576	5.694961	14.00393
Total	4.468646	11.27393336	9.132374	24.87495

So, assuming 5% significance level,

χ2cal =	24.875	α =	0.05
χ2crit =	9.49	for DF =	4

Decision: Since χ²_cal> χ²_crit, null hypothesis is rejected.

Conclusion:

There is not sufficient evidence to suggest that type of investment depends on the age. Answer

DONE