2019 H2 Prelim (Hypothesis Testing) Qns with Sol

Unbiased estimate of the population variance =   2 2 2000 59 59 t t s     = 33.89830508 = 33.90 (iv) Let  denote the population mean speed of the vehicles traveling along the stretch of the road. To test H 0 : μ = 90.0 Against H 1 : μ < 90.0 Conduct a one- tail test at 5% level of significance, i.e., α = 0.05 Under H 0 , since n = 60 and is sufficiently large, by Central Limit Theorem, 33.89830508 ~ 90.0, 60 T N       approximately. Using GC, p-value = 0.0481545117 Since p-value < 0.05, we reject H 0. There is sufficient evidence at 5% level of significance to conclude that that the mean speed along the 3km stretch of road has been reduced. (v) If the null hypothesis is not rejected, z calc must lie outside the critical region. Critical Region: 1.644853626 z   Test Statistics, 90.0 33.89830508 T Z n   ~ N(0, 1)  z calc = 89 90.0 33.89830508 n  > – 1.644853626 1.644853626 33.89830508 n    9.576708062 n  n < 91.713 Since n is an integer, the largest possible value n can take is 91.

(ii) Using the existing sample, the school carried out a test at 1% significance level and found that there was sufficient evidence to reject the trainer’s claim. Find the set of values that n can take, stating any necessary assumption(s) needed to carry out the test. [4] DHS JC2 Prelim 9758/2019/02/Q9 (Solutions) Qn Suggested Solution 9 (a) An unbiased estimate for the population variance :   2 2 16 4 1 1 n n s n n     minutes 2 (b) (i) Let  be the population mean time taken for a 17-year-old student to complete a 5 km run. To test at 10 % significance level, 0 1 : 30.0 min : 30.0 min H H     For 40, n  2 9 16(40) 6 39 40 3 s   Test Statistic: Under 0 H , 640 39 ~ N 30.0, 40 T         approximately by Central Limit Theorem since n is large p -value =   2P 28.9 0.0859 0.10 T    , we reject 0 H and conclude that there is sufficient evidence at the 10 % significance level that the population mean time taken has changed. (ii) The p -value is the probability of obtaining a sample mean at least as extreme as the given sample, assuming that the population mean time taken has not changed from 30.0 min. OR The p -value is the smallest significance level to conclude that the population mean time has changed from 30.0 min. (iii) Since the sample size of 40 is large, by Central Limit Theorem, T follows a normal distribution approximately. Thus no assumptions are needed. (c) (i) New population mean timing = 0.95 30 28.5   min To test at 5 % significance level, 0 1 : 28.5 min : 28.5 min H H    

(iii) Since p -value is 0.055322, 6   (iv) If we tested H 1 : 1053   , Either -value 0.027661 0.05 p   or -value 1.9163 1.645 z     So we may reject H 0 and conclude at 5% level of significance that the population mean amount of sodium had decreased. (v) To test H 0 : 1053   against H 1 : 1053   at 5% level of significance Since 50 n  is large, by Central Limit Theorem, under H 0 , 2 6.0 4 ~ N 10 , 0 53 X       approximately To reject H 0 , p -value < 0.05 1051.4 (to 1 d.p.) y   It is not necessary to assume anything about the population distribution, as sample size (= 40) is large enough, so the Central Limit Theorem says the sample mean amount of sodium approximately follows a normal distribution. HCI JC2 Prelim 9758/2019/02/Q9 A company purchased a machine to pack shower gel into its bottles. The expected mean volume of shower gel in a bottle is 950 ml. (a) The floor supervisor believes that the machine is packing less amount of shower gel than expected. A random sample of 80 bottles is taken and the data is as follows: Volume of shower gel in a bottle (correct to nearest ml) 948 949 950 951 952 953 955

JPJC JC2 Prelim 9758/2019/02/Q7 A company sells peanut butter in jars. Each jar is labelled as containing m grams of peanut butter on average. A consumer group suspects that the average mass of peanut butter in a jar is overstated. To test this suspicion, the consumer group checks a random sample of 50 jars and the mass of peanut butter per jar, x grams, are summarized by ( 390) 120 x    and 2 ( 390) 3100 x    . The consumer group uses the above data to carry out a test at the 2% level of significance. The result leads the consumer group to conclude that the company has overstated the average mass of peanut butter in a jar. (i) Explain why the consumer group is able to carry out a hypothesis test without knowing anything about the distribution of the mass of the peanut butter of the jars. [1] (ii) Find the least possible value of m , to the nearest gram that leads to the result of the hypothesis test as stated above. [7] JPJC JC2 Prelim 9758/2019/02/Q7 (Solutions) 7 (i) Central Limit Theorem states that sample means will follow a normal distribution approximately when sample size is big enough. (ii) ( 390) 120 x    and 2 ( 390) 3100 x    120 390 392.4 50 x    2 2 1 120 3100 57.3877551 57.388 50 1 50 s            0 H : m   vs 1 H : m   Since n = 50 is large, by Central Limit Theorem, 57.388 N( , ) 50 X m approximately Level of significance: 2% Critical region: Reject 0 H when 2.0537 z   Consider 392.4 2.0537 57.388 50 m    394.6 m  (5 s.f.) The least possible value of m is 395grams. MI PU3 Prelim 9758/2019/02/Q9 A chemist is conducting experiments to analyse the amount of active ingredient A in a particular type of health supplement tablets. Each tablet is said to contain an average amount of 50 mg of ˗ 2.0537

  : 37 n n   NJC JC2 Prelim 9758/2019/02/Q11 In view of the rise in the number of accidents involving Personal Mobility Devices (PMDs) in the past year, regulations on PMDs are being tightened to step up safety for pedestrians on footpaths. Since February 2019, the speed limit for riding a PMD on footpaths has been lowered to 10 km/h. In June 2019, a concerned citizen believes that the mean speed of PMD riders still exceeds the new speed limit. To test his belief, a team of enforcement officers capture the speeds, x km/h, of a random sample of 250 PMD riders at five hotspots. The findings are summarised by     2 10 38.9, 10 390.31 x x       . Assume that the speed of a PMD rider follows a normal distribution. (i) Calculate unbiased estimates of the population mean and variance of the speed of PMD riders on footpaths. [2] (ii) Test at the 4% level of significance whether the belief should be accepted. [4] (iii) Explain, in the context of the question, the meaning of ‘at the 4% level of significance’. [1] (iv) Without any further test, explain if the conclusion in part (ii) will change if instead we are testing that the mean speed of PMD riders on footpaths is not 10 km/h. [1] Three months later, to investigate if the mean speed of PMD riders on footpaths differs from the regulated speed limit, another large random sample is collected at the same hotspots. It is found that the mean speed captured is 9.5 km/h, with standard deviation 3.2 km/h. (v) Find the smallest sample size that will provide significant evidence at the 5% level of significance that the mean speed of PMD riders on footpaths differs from the regulated speed limit. [4] NJC JC2 Prelim 9758/2019/02/Q11 (Solutions) 11 Suggested Solution (i) Unbiased estimate of the population mean, 38.9 10 10.1556 10.2 250 x     Unbiased estimate of the population variance,

(v) Assume that the standard deviation has changed, test at the 4% level of significance whether the sample suggests that the process is not in control. [3] NYJC JC2 Prelim 9758/2019/02/Q10 (Solutions) 10(i) Let X be the thickness of the coating on a randomly chosen computer device. Let  be the mean thickness of the coating of a computer device. Assume that the standard deviation of the coating of a computer device remains unchanged. To test : 0 1 : 100 : 100 H H     Level of Significance: 5% Under 0 H , since sample size 50 n  is large, by Central Limit Theorem, 100 ~ (0,1) approx. 50 10 / X N Z   Reject 0 H if 0.05 p value   . Calculations: 103.4 x  0.0162 p value   Conclusion: Since 0.05 p value   , we reject 0 H and conclude that there is significant evidence at 5% level of significance that the process is not in control. 10(ii) Reject 0 H is | | 1.960 calc z  For 0 H to be rejected, 100 1.95996 50 10 10 100 1.95996 or 100 1.95996 50 50 97.228 or 102 / 2 0 .77 1 x x x x x                       Thus the required range of values of x is 0 97.2 or 102.8 x x    . 10(iii) 4164 104.1 40 y   ( 100) 4164 4000 164 y     

We reject 0 H , and conclude that there is sufficient evidence, at the 5% significance level, to support the administrator’s claim that the average consultation time at private clinics is less than 15 minutes. RVHS JC2 Prelim 9758/2019/02/Q10 Engineers claimed that a newly developed engine is efficient. The number of hours x the engine ran on one litre of fuel was measured on 70 occasions. The results are summarized by   46 27 x     ,   2 46 30939. x    (i) Calculate unbiased estimates for the population mean and variance of the number of hours the engine runs on one litre of fuel. [2] (ii) Determine the greatest mean number of hours the engineers should claim so that there is sufficient evidence at the 5% level of significance that the engine is efficient. [3] (iii) The engineers collected five more data points: 56, 34, 63, 50, 54. The engineers found that when all 75 data points are considered, they can no longer claim that the engine is efficient at the  % level of significance. Show that the unbiased estimate for the population variance, for the set of 75 data points is 426.378, correct to 3 decimal places. Find the greatest integer value of  . [4] RVHS JC2 Prelim 9758/2019/02/Q10 (Solutions) Question 10 [9] Hypo Test (i) 27 46 45.61428 45.6 (to 3s.f.) 70 x        2 2 27 1 30939 448.2403727 448 (to 3 s.f.) 69 70 s              (ii) Let 0  be the mean number of hours the engineer claimed. Test H 0 :  = 0  against H 1 :  > 0  at 5% significance level Test statistic: Under H 0 ,   0 ~ N 0,1 / 70 X Z s    To reject H 0 ,

0 0 0 0 1.644853632 / 70 1.644853632 70 41.45197671 41.5 (to 3 s.f.) x s s x           Thus, the greatest mean number of hours the engineers should claim is 41.5. (iii) Based on n = 70, 27 46 70 3193 70 x             Based on n = 75,       2 2 2 2 2 2 2 2 2 3193 56 34 63 50 54 46 75 46 30939 (56 46) (34 46) (63 46) (50 46) (54 46) 31552 426.3783783784 426.378 (3.s.f) 74 x x x x x x s                               Test H 0 :  = 41.5 against H 1 :  > 41.5 at  % significance level Test statistic: Under H 0 ,   41.5 ~ N 0,1 / 75 X Z s   By GC, p -value = 0.029559 Do not reject H 0 , 100 0.029559 100 2.9559 p value          The greatest integer value of  is 2. [ If 0 41.4   , p -value = 0.026649, then the greatest integer value of  is 2.

If 0 41.45197671   , p -value = 0.028230374, then the greatest integer value of  is 2. ] SAJC JC2 Prelim 9758/2019/02/Q11 The time T seconds required for a computer to boot up, from the moment it is switched on, is a normally distributed random variable. The specifications for the computer state that the population mean time should not be more than 30 seconds . A Quality Control inspector checks the boot up time using a sample of 25 randomly chosen computers. A particular sample yielded t   802.5 and 2 26360.25 t   . (i) Calculate the unbiased estimates of the population mean and variance. [2] (ii) What do you understand by the term “unbiased estimate”? [1] (iii) Test, at the 5% level of significance level, whether the specification is being met. Explain in the context of the question, the meaning of “5% level of significance”. [5] (iv) Find the range of values of t such that the specification will be met in the test carried out in part (iii) . [1] (v) A new Quality Control policy is that when the specification is not met, all the computers will be sent back to the manufacturer for upgrading. The inspector tested a second random sample of 25 computers, and the boot up time, y seconds, of each computer is measured, with 32.4 y  . Using a hypothesis test at the 5% level of significance, find the range of values of the population standard deviation such that the computers will not be sent back for upgrading. [3] SAJC JC2 Prelim 9758/2019/02/Q11 (Solutions) 11(i) Let T be the random variable “ time taken in seconds for a computer to boot up”, with population mean  . Unbiased estimate of the population mean, 802.5 32.1 25 t   Unbiased estimate of the population variance, 2 2 1 802.5 26360.25 25 24 25 s          (ii) A statistic is said to be an unbiased estimate of a given parameter when the mean of the sampling distribution of the statistic can be shown to be equal to the parameter being estimated. For example, E( ) X   . (iii) Test H 0 : 30   against 1 : 30 H   at the 5% level of significance. Under H 0 , 25 ~ (30, ) 25 T N . Using GC,

t = 32.1 gives rise to z calc = 2.1 and p -value = 0.0179 Since p -value = 0.0179  0.05, we reject H 0 and conclude that there is sufficient evidence at the 5% significance level that the specification is not being met (or the computer requires more than 30 seconds to boot up). “5% significance level” is the probability of wrongly concluding that the mean boot up time for the computer is more than 30 seconds when in fact it is not more than 30 seconds. (iv) The critical value for the test is 31.645. For the specification to be met, H 0 is not rejected. 31.6 (3 s.f.) t  Since t > 0, Answer is 0 31.6. t   (v) Under H 0, 2 ~ 30, 25 Y N        (30) ~ N(0,1) 25 Y Z    Using a 1 – tailed z test, calc 5 32.4 30 12 z      , crit 1.64485 z  In order not to reject H 0, 12 1.64485   7.2955 7.30     TJC JC2 Prelim 9758/2019/02/Q10 (a) Two random samples of different sample sizes of households in the town of Aimek were taken to find out the mean number of computers per household there. The first sample of 50 households gave the following results. Number of computers 0 1 2 3 4 Number of households 5 12 18 10 5 The results of the second sample of 60 households were summarised as follows. 118 y   2 314 y   , where y is the number of computers in a household.

(i) By combining the two samples, find unbiased estimates of the population mean and variance of the number of computers per household in the town. [4] (ii) Describe what you understand by ‘population’ in the context of this question. [1] (b) Past data has shown that the working hours of teachers in a city are normally distributed with mean 48 hours per week. In a recent study, a large random sample of n teachers in the city was surveyed and the number of working hours per week was recorded. The sample mean was 46 hours and the sample variance was 131.1 hours 2 . A hypothesis test is carried out to determine whether the mean working hours per week of teachers has been reduced. (i) State appropriate hypotheses for the test. [1] The calculated value of the test statistic is z = −1.78133 correct to 5 decimal places. (ii) Deduce the conclusion of the test at the 2.5 % level of significance. [2] (iii) Find the value of n . [3] (iv) In another test, using the same sample, there is significant evidence at the %  level that there is a change in the mean working hours per week of teachers in that city. Find the smallest possible integral value of  . [2] TJC JC2 Prelim 9758/2019/02/Q10 (Solutions) 10(a)(i) Let X be the number of computers in each household in the first sample. Using GC, 98 fx   , 2 254 fx   Unbiased estimate of the population mean 98 118 108 ( 1.96) 50 60 55      Unbiased estimate of the population variance     2 98 118 1 7912 254 314 ( 1.32) 110 1 110 5995                (a)(ii) The population in this question refers to all the households in the town. (b)(i) 0 H : 48   1 H : 48   (ii) Critical region =   : 1.960 z z   Since z cal = 1.781  > −1.960, we do not reject H 0 OR p -value =   P 1.781 0.0375 Z    > 0.025, we do not reject H 0 There is insufficient evidence at 2.5% level of significance that the mean working hours per week of teachers in the city has been reduced. (iii)   2 131.1 1 n s n   Given z = 1.781  46 48 1.781 131.1 1 n            [Turn over

131.1 2 1.781 1 n     1.781 131.1 1 2 n     105 n  (iv) p -value for the two tailed test =   2 P 1.781 0.0749 Z     (accept 0.075) For H 0 to be rejected, p - value ≤ 100  7.49   Smallest value of  = 8 TMJC JC2 Prelim 9758/2019/02/Q7 MHL bakery sells mini breads that weighs an average of 45 grams each. A customer claims that the bakery is overstating the average weight of mini breads. To test this claim, a random sample of 80 mini breads are selected from MHL bakery and the weight, x grams, of each mini bread is measured. The results are summarised as follows. 80 n  3571 x   2 159701 x   Calculate unbiased estimates of the population mean and variance of the weight of mini breads. [2] Test, at the 4% level of significance, whether there is sufficient evidence to support the customer’s claim. [4] From past records, it is known that the weights of the mini breads from MHL bakery are normally distributed with standard deviation 1.5 grams. To further investigate the customer’s claim, the bakery records the weights of another 20 randomly selected mini breads and the average weight for the second sample is k grams. Based on the combined sample of 100 mini breads, find the range of values of k such that the customer’s claim is valid at the 4% level of significance. [4] TMJC JC2 Prelim 9758/2019/02/Q7 (Solutions) Qn Solution Mark Scheme 7 Hypothesis Testing [10]

Let X be the weight of a randomly chosen mini bread (in grams). Let  denote the population mean weight of mini breads (in grams) 3571 Unbiased estimate of population mean, 80 44.6375 x     2 2 3571 1 Unbiased estimate of population variance, 159701 79 80 3.8036 3.80 (3 s.f.) s             0 1 H : 45 H : 45     Under 0 H , since 80 n  is large, by Central Limit Theorem, 3.8036 ~ N 45, 80 X       approximately Test Statistic: Z = 45 3.8036 80 X  Level of significance : 4 % Reject H 0 if p - value < 0.04 Using GC, p -value = 0.0482 Since p -value = 0.0482 0.04  , we do not reject 0 H and conclude that there is insufficient evidence, at the 4% level of significance, that the population mean weight is less than 45 grams. Thus, the customer’s claim is not supported at the 4% significance level. Sample mean based on combined sample = 20 3571 20 80 20 100 x k k      0 1 H : 45 H : 45     Under H 0 ,   2 2 1.5 ~ N 45,1.5 ~ N 45, 100 X X        Test statistic: 2 45 1.5 100 X Z   Level of significance : 4 % Reject H 0 if value 1.7507 z   

2 45 value 1.5 100 x z    Since there is sufficient evidence that the customer’s claim is valid at 4% level of significance, H 0 is rejected 2 45 1.7507 1.5 100 44.737395 3571 20 44.737395 100 45.137 45.1 (3 s.f) x x k k k         VJC JC2 Prelim 9758/2019/02/Q9 Exposure to Volatile Organic Compounds (VOCs), which have been identified in indoor air, is suspected as a cause for headaches and respiratory symptoms. Indoor plants have not only a positive psychological effect on humans, but may also improve the air quality. Certain species of indoor plants were found to be effective removers of VOCs. A commonly known VOC is Benzene . The following data gives the benzene levels, x (in ppm) in 40 test chambers containing the indoor plant Epipremnum aureum . , , . The initial mean Benzene level (in ppm) without Epipremnum aureum was found to be 26.0. (i) Test, at the 5% level of significance, the claim that the mean Benzene level, (in ppm), has decreased as a result of the indoor plant Epipremnum aureum . You should state your hypotheses clearly. [5] (ii) State, giving a reason, whether there is a need to make any assumptions about the population distribution of the Benzene level in order for the test to be valid. [2] The Benzene levels of another 50 test chambers containing the indoor plant Epipremnum aureum were recorded, The sample mean is ppm and the sample variance is 8.33 ppm 2 . (iii) The acceptance region of a test of the null hypothesis is . State the alternative hypothesis and find the level of significance of the test. [4] 40 n    26.0 30.1 x       2 26.0 214.61 x     x 26.0   25.1 x 

(iv) If the null hypothesis is , where , would the significance level of a test with the same acceptance region in part (iii) be larger or smaller than that found in part (iii) ? Give a reason for your answer. [2] VJC JC2 Prelim 9758/2019/02/Q9 (Solutions) 9i 0 1 H : 26.0 H : 26.0     Level of significance: 5% Test Statistic: Since n is sufficiently large, by CLT, X is approximately normal.   0 26.0 When H is true N 0,1 X Z S n   approx Computation:   2 2 30.1 40, 26 25.2475 40 30.1 1 214.61 4.9220 39 40 n x s                 value 0.015969 0.0160 p    Conclusion: Since value 0.0160 p   < 0.05, 0 H is rejected at the 5% level of significance. Therefore, there is sufficient evidence to conclude that the mean benzene level has decreased as a result of the indoor plant Epipremnum aureum at the 5% level of significance. 9ii There is no need to assume the population distribution of the benzene level because n = 40 is sufficiently large, so by Central Limit Theorem, the sample mean benzene level, X , follows a normal distribution approximately . 9iii 0 1 H : 26.0 H : 26.0     Test Statistic: Since n is sufficiently large, by Central Limit Theorem, X is approximately normal.   0 26.0 When H is true N 0,1 X Z S n   approx. Computation:   2 50 8.33 8.5 49 s   0    0 26.0  

Rejection region is 25.1 x  Level of significance     0 0 P 25.1 when H is true OR P 2.1828 when H is true 0.014524 X Z      Level of significance of this test is 1.45%. 9iv 0 0 0 1 0 H : , where 26.0 H :         Level of significance   0 P 25.1 when X        P 25.1 when 26.0 X     (refer to diagram above) Level of significance is smaller than that in (iii) . YIJC JC2 Prelim 9758/2019/02/Q10 A previous study revealed that the average time taken to assemble a certain type of electrical component is at least 15 minutes. The manager wants to investigate if the results of the study is valid. A random sample of 40 components is taken and the times taken to assemble the components are summarised in the following table: Time to assemble a component (min) 9 10 12 13 15 16 17 18 Number of components 1 6 3 8 5 7 8 2 (i) Find unbiased estimates of the population mean and variance. [2] (ii) Test at the 5% level of significance whether the results of the study is valid. You should state your hypotheses and define any symbols you use. [5] (iii) Explain why the manager is able to conduct the test without knowing anything about the distribution of the times taken to assemble the electrical components. [1] (iv) Explain what is meant by the phrase “5% level of significance” in this context. [1] 0.05 25.1 26.0

The manager claims that the average time taken to assemble another type of electrical component is 30 minutes. A random sample of 50 components of this type is chosen and the time taken to assemble each component is recorded. The mean and standard deviation of the sample are 29.7 minutes and k minutes respectively. Find the range of possible values of k if a test at the 8% significance level shows that there is sufficient evidence that the manager’s claim is valid. [4] YIJC JC2 Prelim 9758/2019/02/Q10 (Solutions) 10(i) An unbiased estimate of  is x  14.2 2 2 An unbiased estimate of is 7.0871 7.09 s    10(ii)  is the population mean time taken to assemble an electrical component. H 0 : 15   (results of study) H 1 : 15   Under H 0 , the test statistic is ~ N(0,1) X Z S n    approximately (by CLT), where 15   , 2 7.0871 s  , n = 40, 14.2 x  . From GC, p -value = 0.0287 Since the p- value = 0.0287  0.05, we reject H 0 and conclude that at the 5% level, there is sufficient evidence that the results of the study is not valid. 10(iii) Since sample size is large, by Central Limit Theorem, the sample mean time taken to assemble an electrical component will be approximately normal. 10(iv) “5% level of significance” means that there is a probability of 0.05 of concluding that the results of the study is not valid when the mean time taken to assemble an electrical component is indeed 15 minutes. H 0 : 30   (claim) H 1 : 30   Under H 0 , the test statistic is ~ N(0,1) X Z S n    approximately (by CLT), where 30   , 2 2 50 49 s k  , n = 50, 29.7 x  . Critical Regions: z  1.750686 or z  1.750686  Given that H 0 is not rejected

1.750686 1.750686 0.250098 0.3 and 0.3 0.250098 1.19953 1.19 2 95 . 7 3 9 7 30 k k k k k             1.20 k   (3 s.f)