Question

2. (a) Dalam sampel rawak 200 orang lelaki, 54 menyatakan bahawa mereka menggunakan kad kredit dan, dalam sampel 160 orang wanita, 56 daripadanya menggunakan kad kredit. Pada aras keertian 0.01, adakah terdapat perbezaan yang bererti di antara kadar lelaki berbanding wanita yang menggunakan kad kredit? In a random sample of 200 men, 54 indicated they are using credit cards and, in sample of 160 women, 56 indicated of using credit cards. At the 0.01 level of significance, is there a significant difference botween the proportions of men versus the proportion of women who use credit cards? 8 markah/marks] (b) Sebuah kilang pemproses ingin membandingkan bahan pengawet yang manakah yang lebih berkesan dalam melambatkan kerosakan. Suatu sampel 185 potongan daging segar diawet menggunakan Pengawet A dan 15 potongan daging segar yang lain diawet menggunakan Pengawet B. Min dan sisihan piawai bagi kedua-dua sampel diberikan di bawah: A processor wants to compare the preservatives for their effects on retarding spolage. A sample of 15 cuts of fresh meat are treated with preservative A and 15 with preservative B Moan and standard deviation for both samples are given below: t APreservative A Pengawet B/Preservative B R,-1064hours X96.5 hours S-10.3 hours Sa-13.4 hours i Dengan menggunakan aras keertian 0.05, adakah bukti yang menunjukkan perbezaan dalam min masa mula berlakunya kerosakan bagi kedua-dua Using the 0.05 level of significance, is there evidence of a dference in the mean time until spoilage begins for the two preservatives? 7 markah/marks) 3/8 ESEE 2201 i. Apakah andaian yang perlu untuk melakukan ujian tersebut? What assumption is necessary to perforrm this test? iii. Dapatkan selang keyakinan 90% untuk min perbezaan masa pemprosesan di antara dua sistem tersebut. Find a 90% confidence interval for the mean difference in the mean time unti spoilage begins for the two preservatives. [3 markahimarks] Total 20 markah/marks)

answer QS 2,thank,

Answer 1

2.a.

$X_{1}\;and\;X_{2}$ be variables for men and women respectively.

be population proportion for men and women using credit cards

Let pi and pz

This is a two tailed binomial hypothesis test for two proportions difference

Null hypo. Ho) . P1 = p, VS Alternat i Ue Hypo( Hı) : Pıtp2

Test Statistics : $|\frac{(\hat{p_{1}}-\hat{p_{2}})-(p_{1}-p_{2})}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}|\sim Z_{\alpha\%/2}$

n2
$\hat{p}=\frac{\hat{p_{1}}n_{1}+\hat{p_{2}}n_{2}}{n_{1}+n{2}}$

r!= 54 n1= 200 .r2= 56 n2= 160

p1= 0.27 0.35 =ב p= 0.306

Test Statistic =

0.27 0.35)-0 to) 0.306( 1-0.306) (2 K

Test Statistic = 1.6367

Decision criteria : Reject null hypothesis if T.S. >  $Z_{\alpha\%/2 }$ at $\alpha \%$ level of significance.

$\alpha =1\%$

$Z_{\alpha\%/2 }$ = $Z_{0.005}=2.5758$

T.S < $Z_{\alpha\%/2 }$

Conclusion; We do not reject null hypothesis and concludethat proportion of men and women using credit cards is same at 1% level of significance.

2.b.i

Let be variable for treating with preservative A and B respectively.

Ai and B2

This is an independent two-tailed t-test for checking difference between twopopulation means.

We use sample S.D. because population S.D. is unknown

$Null\;hypo(H_{0}):\mu_{a}=\mu_{b}\;\;VS\;\;Alternative\;Hypo(H_{1}):\mu_{a}\neq \mu_{b}$

Test Statistics :$\frac{(\bar{a}-\bar{b})-(\mu_{a}-\mu_{b})}{S_{p}\sqrt{(1/n_{1}+1/n_{2})}}\sim t_{n_{a}+n_{b}-2,\alpha\%/2}$

$\bar{a}=106.4\;\;\bar{b}=96.5\;\;\mu_{a}-\mu_{b}=0$  $n_{a}=n_{b}=15$

$S_{p}=\sqrt{\frac{S_{a}^{2}n_{a}+S_{b}^{2}n_{b}}{n_{a}+n_{b}}}$  $=\sqrt{\frac{10.3^{2}*15+13.4^{2}*15}{30}}$= 11.823

Test Stat: = $\frac{106.4-96.5}{11.823\sqrt{2/15}}$ = 2.2932

Decision criteria : Reject null hypothesis if T.S. >  
natn,-2.0%/2
at $\alpha \%$ level of significance.

$\alpha =5\%$

natn,-2.0%/2
= $t_{28,0.025}=$ 2.048

T.S >

natn,-2.0%/2

Conclusion: We reject null hypothesis and conclude that there is difference between the mean time until spoilage of the 2 preservatives at 5% level of significance.

ii. The assumptions to conduct the above test are:

• The samples should be random and independent
• the samples should come from a population following a normal distribution
• The samples shall have same population variance

iii.

90% Confidence interval for differences in mean

$((\bar{a}-\bar{b})-t_{\alpha\%/2,n_{a}+n_{b}-2}*S_{p}\sqrt{1/n_{a}+1/n_{b}}, (\bar{a}-\bar{b})+t_{\alpha\%/2,n_{a}+n_{b}-2}*S_{p}\sqrt{1/n_{a}+1/n_{b}})$

% = 10

$t_{5\%,28}=1.701$

$((106.3-96.5)-1.701*11.823\sqrt{2/15}, (106.3-96.5)-1.701*11.823\sqrt{2/15})$

Final Ans: (2.557, 17.24)