Question

11. Essential Wear is a company that sends 5 items of apparel every 3 months. They specialize in women's clothes and the boxes consist of undergarments, accessories, and loungewear. The subscriber pays a flat fee but they do not know what they will be getting in each box. The company claims that 20% of the goods sent are underwear, and 10% are bras, 18% are socks, 36% are loungewear, and 16% are accessories. Suppose a suspicious customer tally's the totals for the last 3 years and finds that the company sent 18 pairs of underwear, 4 bras, 16 pairs of socks, 12 loungewear pieces and 10 accessories. Test the claim that the actual proportions of content differ from Essential Wear's claim. Use a 0.05 level of significance.

Answer 1

Solve using Excel:

B D 1 Apparels Observed Frequency (O) Expected Frequency ('E) (0,-E)? / 2 Underwear 18 12.00 3.00 3 Bras 4 6.00 0.67 4 Socks 16 10.80 2.50 5 Loungewear 12 21.60 4.27 6 Accessories 10 9.60 0.02 7 Total 60 Test statistic 10.45 8 9 Apparels Observed Frequency (0) Expected Frequency ('E) (0,-E)?/E 10 Underwear 18 =0.2*60 =(B2-C2)^2/C2 11 Bras 41=0.1*60 =(B3-C3)^2/03 12 Socks 16=0.18*60 =(B4-C4)^2/04 13 Loungewear 121=0.36*60 =(B5-C5)^2/C5 14 Accessories 101=0.16*60 =(B6-C6)^2/C6 15 Total I=SUM(B2:36) Test statistic =SUM(D2:06) 16

Apparels	Observed Frequency (O)	Expected Frequency ('E)	(Oi - Ei)2 / Ei
Underwear	18	12.00	3.00
Bras	4	6.00	0.67
Socks	16	10.80	2.50
Loungewear	12	21.60	4.27
Accessories	10	9.60	0.02
Total	60	Test statistic	10.45
Apparels	Observed Frequency (O)	Expected Frequency ('E)	(Oi - Ei)2 / Ei
Underwear	18	=0.2*60	=(B2-C2)^2/C2
Bras	4	=0.1*60	=(B3-C3)^2/C3
Socks	16	=0.18*60	=(B4-C4)^2/C4
Loungewear	12	=0.36*60	=(B5-C5)^2/C5
Accessories	10	=0.16*60	=(B6-C6)^2/C6
Total	=SUM(B2:B6)	Test statistic	=SUM(D2:D6)

H₀: Actual proportions of content do not differ Essential Wear's claim

H₁: Actual proportions of content differ Essential Wear's claim

Test statistic = 10.45

Degrees of freedom: df = k-1 = 5-1 = 4, Level of significance = 0.05

Critical value (Using Excel function CHISQ.INV.RT(probability, df)) = CHISQ.INV.RT(0.05,4) = 9.49

Since test statistic is greater than critical value, we reject the null hypothesis.

So, actual proportions of content differ Essential Wear's claim.