5. Hypothesis Testing Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Hypothesis Testing T1 Hypothesis Testing 1 mass of a packet of cat foodX = ( ) 2~ N , 2.7 Test 0H : 375= Against 1H : 375  Two-tail at 5% significance level Test statistic: Under 0H , ( ) 22.7 100~ N 375, ~ N(0,1)XZ  Reject 0H if p-value < 0.05. Observed sample mean 375.31x = p-value = ( )2 P 375.31 0.251 0.05X  =   , 0H is not rejected. There is insufficient evidence at 5% significance level that the mean mass of the packets of cat food differs from 375 grams. Alternatively, At 5% significance level, the rejection region is 374.471 or 375.529XX . Since the observed sample mean 375.31x = does not fall in either rejection region, 0H is not rejected. Let mass of a packet of cat food packed by the new machineryY = ( ) 2~ N , 2.7 Test 0H : 375= Against 1H : 375  One-tail test at 5% significance level Under 0H , ( ) 22.7 100~ N 375, ~ N(0,1)YZ  To reject H0, zcalculate > zcritical 375 1.6448536262.7 100 375.4441105 y y −  Therefore, ( )375.5 1 d.p.y  1.644853626

River Valley High School, Mathematics Department 2023 Hypothesis Testing T2 2a(i) Unbiased estimate of population mean = 3072 5.12600y== Unbiased estimate of population variance = s2 21 307216688 1.601602671 1.60 (3 sf)599 600 = − = = (ii) Let  denote the population mean weight of the bags of multigrain rice. To test H0 :  = 5 Against H1 :  > 5 at 2% significance level Under H0 , 2 5 N(0,1) 600 yZ s −= where 2 1.601602671s = p-value = 0.0100995741 = 0.0101 (3 sf) < 2%. Reject H0. There is sufficient evidence at 2% level of significance to conclude that the machine dispenses more than 5 kg of rice. (iii) “Machine dispenses more than 5 kg of rice”, i.e. reject H0  p-value < 100  i.e.  % > 0.0100995741 The smallest level of significance to reject H0 is 1.01 %. b(i) Let  denote the population mean weight of the bags of rice. To test H0 :  = 5 Against H1 :   5 at 5% sig level Under H0 , 2 5 N(0,1) 0.153 55 XZ −= by Central Limit Theorem since n is large. Do not reject H0 so 51.95996 1.95996 0.153/ 55 x−−   giving 4.96 < x < 5.04 (3 sf) (ii) There is NO need for any assumption about the population distribution because Central Limit Theorem is applicable due to the large sample size, to have X follow normal distribution.

River Valley High School, Mathematics Department 2023 Hypothesis Testing T3 3(i) Unbiased estimate of the population mean ( 4.5) 4.565 21 4.565 4.823076923 4.82 (3 s.f.) x−=+ =+ =  Unbiased estimate of the population variance 21 21(85 )65 1 65=− − = 1.222115385  1.22 (3 s.f) (ii) Test 0 : 4.5H  = Against 1 : 4.5H  Level of significance: 1% = 0.01 Test statistic: 0 ~ N(0,1)XZ s n −= , where 4.8231x= , 1.22212s= and n = 65 Using G.C., p-value = 0.00923 <