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 < 0.01 Therefore we reject 0H , and conclude that there is sufficient evidence at 1% level of significance that the students from the school spend more time reading magazines as compared to the national average. No assumption needed. This is because the sample size is large and thus by Central Limit Theorem, X follows a normal distribution. (iii) Test 00 :H = Against 01 :H  The null hypothesis is rejected at 1% level of significance  04.8231 2.32634 1.2221 65 − − 0 5.1421  0 5.14 (to 3 s.f.) 

River Valley High School, Mathematics Department 2023 Hypothesis Testing T4 4(i) Unbiased estimate of population mean, 10012 166.87 167 (3 s.f.)60x = = = Unbiased estimate of population variance, 2 6259 106. 1008 s.f.)60 1 6 (3s =− == (ii) Let X denote the breaking strength of the rope (in kN) 0 1 To test H 169.7 Against H 6 . : : 1 9 7   =  Conduct 1-tail test at 2% significance level. Under H0, since 60 is large,n= by Central Limit Theorem, 106.08 ~ N 169.7, 60X   approximately. Using a Z-test, -value 0.0167 (3 s.f.)p = Since -value 0.02,p  we reject 0H and conclude that there is sufficient evidence at 2% significance level that the mean breaking strength is less than 169.7 kN / that the manufacturer’s claim is invalid / to reject the manufacturer’s claim (iii) It is not necessary to have any assumptions about the population. Since the sample size n is large, the sample mean X can be approximated to a normal distribution by Central Limit Theorem. (iv) Let Y denote the breaking strength of the rope using the new weaving process (in kN) 1 00 0 To test H Against : :H   =  Conduct 2-tail test at 2% significance level. Under H0, since 50 is large,n= by Central Limit Theorem, 0 90.2 ~ N , 50 5Y   approximately z 0 -2.3263 0.01 2.3263 0.01

River Valley High School, Mathematics Department 2023 Hypothesis Testing T5 Given 171,y = and in order not to reject 0H, 0 0 0 1712.3263 2.3263 90.25 50 3.1254 171 3.1254 167.87 174.13    −−   −  −      set of values of 0 is: 00{ :168 174}   5(i) ( ) ( ) ( ) ( ) 2 150 35 50 150 35 7465 7465 1493 149.350 10 1 167167 3.40816 3.41(3 sf)49 49 x x x x s − =− − =− = = = = = = =     (ii) 01: 150 against : 150 Test at 5% significance level, 2.68 0.00367 0.05 HH Z p = =− = Reject 0H and conclude that there is sufficient evidence at 5% significance level, the content of the drinks in the bottle is less than what the packaging claimed to be / the complaints are valid. (iii) 5% significance level means there is a 5% chance that we say that the complaints are valid when the contents are actually not less than 150ml. (iv)(a) If the sample was not randomly chosen, the conclusion made can be unreliable. For example, the sample may have been the first 50 bottles that are produced in the same batch and the mean quantity could have changed as production continued. (iv)(b) The conclusion is unaffected whether the distribution is normal or not as this is a large sample. By Central Limit Theorem, the mean quantity in the bottled drink is approximately normally distributed. (v) ( )( ) 01: 150 against : 150 0.00367 2 0.00734 0.05 HH p = = =  Reject 0H at 5% significance level that there is sufficient evidence that, on average, the bottled soft drink does not contain 150ml of drinks 6(ai) Unbiased estimate of population mean,

River Valley High School, Mathematics Department 2023 Hypothesis Testing T6 __ 970.02 10.77890 xx n= = = Unbiased estimate of population variance, ( ) ( ) ( ) 2 22 2 1 1 970.021 1132689 90 9.78792 9.79 3 s.f. xsx nn  =− −   =−  ==  (aii) Let X months be the r.v. denoting the walking age of a toddler and  months be the population mean walking age. To test 0H : 11.2= against 1H : 11.2  at 5% level of significance Since n is large, under 0H , approximately by Central Limit Theorem, Test statistic: ( ) __ 11.2 ~ 0,1 9.7879 90 XZN −= Using G.C, p-value = 0.20067= 0.201 (3s.f) Since p-value = 0.201 > 0.05, do not reject 0H and conclude that there is insufficient evidence to conclude at 5% level of significance that the toddlers’ walking ability is not equal to 11.2 months. (bi) To test 0H : 11.2 = against 1H : 11.2 (bii) At 5% level of significance Under H0, ( )12 0,110.3 20 XZN −= We do not reject H0 if z > -1.64485

River Valley High School, Mathematics Department 2023 Hypothesis Testing T7 11.2 1.6448510.3 20 10.311.2 1.64485 20 7.4117 k k k − − − −   Set of values of k =  : 7.41 (3 s.f.)kk 7 Let X = weight of students and μ = mean weight Under , Test statistics (exact) At 5%, we reject if z ≤ −1.9600 or z ≥ 1.9600. Since we do not reject 8(i) Unbiased estimate for population mean is 475 4.75100x == Unbiased estimate