NYJC 2026 FM TP - FM Stats 2 (Solutions)

2026 J2 H2 FM FM Statistics 2 – Hypothesis Testing – Solutions 1 (a) In a test, students were asked to interpret the phrase "95% confidence interval". A student wrote that "95% confidence interval for μ = (77, 83) means that there is a 95% chance that the true population mean falls between 77 and 83." Explain why his answer is incorrect and give a correct interpretation. [2] (b) Prior to an election contested by 2 candidates A and B, a news agency conducted an online survey amongst its subscribers. Of 21,356 respondents, 16,331 indicated that they will vote for A. Calculate the 95% confidence interval for the population proportion that will vote for A. Show your workings clearly. [3] (c) For a confidence interval for population proportion (a, b), 2 baM −= is defined as the margin of error. Before collecting any sample, it is a common industrial practice to have a target Mmax and calculate the minimum required sample size to achieve a confidence interval that is guaranteed to have maxMM . Find the minimum sample size so that Mmax = 0.03 for a 95% confidence interval. [4] Solution: 2017/HCI/MYE/II/3 (a) He is incorrect as μ either falls within (77, 83) or does not fall within (77, 83), so the probability is either 1 or 0. The correct interpretation is that if we take repeated samples, 95% of the confidence intervals constructed using this method will contain the actual value of μ. B1 B1 (b) Let X be the number of respondents who indicated that they will vote for A out of 21,356. ~ B(21,356, ) AXp . Since n is large, by Central Limit Theorem, (1ˆˆ ) ˆ ~ N(0,1) AA A pp n AZ pp − −= approximately. ˆ 0.764703 1633 2 56 1 1p== , n = 21356. 95% confidence interval for ( ) ˆ ˆ ˆ ˆ(1 ) (1 )ˆˆ 1.95996 1.9599 , 6 A A A A A nn p p p p p pp −− −= + ( )0.759,0.770= M1 M1 A1 (c) For max 0.03MM= ( )ˆˆ 11.960 0.03pp n −  We want to find the value of p such that (1 )pp − maximum, so that regardless the value of p, 0.03M  (i.e. if (1 )pp − takes a smaller value, the inequality is still satisfied). (1 )pp − is largest when 0.5p= . ( )0.5 1 0.51.960 0.03 n −  ( ) 2 1.960 0.5 1 0.50.03n − 1067.1n Minimum sample size n = 1068 M1 M1 M1 A1

2 An investigation was carried out into the effectiveness of two well -known drugs, A and B, in relieving pain for adult arthritis sufferers. Ten adult arthritis sufferers each volunteered to record the number of hours of relief from pain gained when taking drug A and when taking drug B. The adults agreed to take one of the drugs, A or B, on one day and the other drug on another day. The order of taking the drugs was randomly assigned. The recorded times, in hours, are given in the table. (a) Explain the purpose of randomly assigning the order of taking drug A or drug B. [2] (b) It was believed that drug B is more effective in relieving pain gained by arthritis sufferers than drug A. Determine if the data support this belief, at the 1% level of significance, using (i) the Wilcoxon matched-pairs signed rank test, [4] (ii) a t-test, given that the corresponding populations are normally distributed.[4] (iii) State, giving a reason, which of the two tests is more appropriate. [1] (c) Give a reason why your conclusions in part (b) might not apply to all adult arthritis sufferers. [1] The investigator claims that the average number of hours of relief from pain gained by arthritis sufferers when taking drug A is 4.5 hours. (d) Use a non-parametric test, at the 10% significance level, to test this claim against the alternative hypothesis that the average number of hours is less than 4.5 hours. [3] Adult 1 2 3 4 5 6 7 8 9 10 Drug A 2.0 3.6 2.6 2.6 7.2 3.4 6.5 2.5 6.1 8.5 Drug B 3.5 5.7 2.8 2.3 9.8 3.3 5.9 3.7 9.1 11.9 Solution: 2017/VJC/Prelim/II/10 [15 Marks] (a) This ensures that any influence of the order of taking drugs does not affect the outcome of the investigation. (b) Let AX and BX hours be the number of hours of relief from pain gained by arthritis sufferers taking drug A and drug B respectively. Let ABD X X=− be the paired difference. (i) Let m hours be the population median of D. 0H : 0m= 1H : 0m Level of significance: 1% ABd x x=− rank of d Signed rank 1.5− 6 6− 2.1− 7 7− 0.2− 2 2− 0.3 3 3 2.6− 8 8− 0.1 1 1 0.6 4 4 1.2− 5 5− 3.0− 9 9− 3.4− 10 10−

8, 47 min( , ) 8 PQ T P Q == == Rejection H0 if 5T  Since 5calT  , 0H is not rejected at 1% level of significance. Hence there is insufficient evidence to conclude that drug B is more effective in relieving pain gained by arthritis sufferers than drug A at 1% significance level. (b) Let A and B hours be the population mean number of hours of relief from pain gained by arthritis sufferers taking drug A and drug B respectively. Let D be the population mean of the paired differences. 213, 35.92, 10d d n=− = = 1.30dd n  = =− , ( ) 2 2 22 1 1.453731 d sd nn  = − =−   0H : 0 D = 1H : 0 D  Level of significance: 1% Test Statistic: Under 0H , (9) 10 tDT S = Rejection region: 0.05p value− Computation: 2.8279calt =− value 0.0098956p−= Conclusion: Since p-value = 0.009896 < 0.01 0H is rejected at 1% significance level. Hence, there is sufficient evidence that the data support the belief that drug B is more effective in relieving pain gained by arthritis sufferers than drug A at the 1% significance level. (iii) Given the populations are normally distributed, the t –test is more appropriate as it is a more powerful test because it takes into consideration the size of differences between the paired observations while the Wilcoxon matched-pairs signed rank test only considers the rank of the absolute difference between paired observations. (c) The conclusion in part (b) might not apply to all adult arthritis sufferers since it is based on a sample in which adults volunteered to take part in the experiment, the sample is not a random sample. (d) Let Am hours be the population median number of hours of relief from pain gained by arthritis sufferers taking drug A. A A 0 1 H : 4.5 H : 4.5 m m =  Level of significance: 10%

Test statistic: Let S be the number of plus signs from ( )A 4.5X − in a sample of size 10. Under 0H , ( ) 1B 10, 2S  Computation: From the sample, there are 4 observations greater than 4.5, 4s= p – value ( )P4S= 0.37695 0.377= Conclusion: Since p – value = 0.377 > 0.10, 0H is not rejected at 10% level of significance. Hence there is insufficient evidence that the average number of hours is less than 4.5 hours at the 10% significance level. 3 A To explore the effectiveness of a newly launched Mathematics remedial programme, Ms Chan randomly selected six JC2 students who were identified as underperforming in the subject. She implemented an assessment to these six students before the remedial programme. After each student had gone through 10 hours of remedial programme, they were given a second assessment. Their scores were recorded in the table below. (a) Carry out a suitable test, at the 2% level of significance, to determine whether there is any difference in the students’ scores, on average, after the remedial programme. State the assumptions necessary for your test. [6] (b) To ensure that the two assessments are set at comparable standards, M s Chan invited another Math tutor M s Soh to examine the two assessments. Ms Soh commented that the first assessment was too easy and recommended that each score for the first