RI S5 Hypothesis Testing Add Prac Solns

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 ______________________________________________________ Additional Practice Questions for Chapter S5: Hypothesis Testing Page 1 of 16 Additional Practice Questions for Chapter S5: Hypothesis Testing (Solutions) 1 9758/2017/02/Q7 The production manager of a food manufacturing company wishes to take a random sample of a certain type of biscuit bar from the thousands produced one day at his factory, for quality control purposes. He wishes to check that the mean mass of the bars is 32 grams, as stated on the packets. (i) State what it means for a sample to be random in this context. [1] The masses, x grams of a random sample of 40 biscuit bars are summarised as follows.     240 32 7.7 32 11.05n x x      (ii) Calculate unbiased estimates of the population mean and variance of the mass of biscuit bars. [2] (iii) Test at 1% level of significance, the claim that the mean mass of biscuit bars is 32 grams. You should state your hypotheses and define any symbols you use. [5] (iv) Explain why there is no need for the production manager to know anything about the population distribution of the masses of the biscuit bars. [2] Solution (i) For a sample to be random, every biscuit bar has equal chance of being selected into the sample, and the selection of one biscuit bar is independent of the selection of another biscuit bar. (ii) Let 32y x  Then 7.7 0.192540y    and 32 31.8075x y   (note that this is an exact answer) Therefore an unbiased estimate of the population mean is 31.8075 grams.    2 2 2 2 2 7.71 1 11.05 0.245331 39 40y x ys y sn n                     Therefore an unbiased estimate of the population variance is 0.245 grams2 (3.sf) (iii) Let  denote the population mean mass (in grams) of a biscuit bar 0H : 32 1H : 32 Perform a 2-tail test at 1% significance level. Under H0, since 40n is large 0.24533~ N 32,40X     approximately by Central Limit Theorem.

Raffles Institution H2 Mathematics 2025 Year 6 ______________________________________________________________________________________________ ______________________________________________________ Additional Practice Questions for Chapter S5: Hypothesis Testing Page 2 of 16 Using a z-test, p-value 2P 31.8075 0.0140 0.01X    We do not reject 0H and conclude that there is insufficient evidence at 1% level of significance, that the mean mass of a biscuit bar is not 32 grams. (iv) No assumption about the population is needed, as sample size of 40 is large and hence X follows a normal distribution approximately by Central Limit Theorem. 2 SRJC/2006/02/Q31OR(a) To investigate the length of time spent by students in the school canteen during lunch break, a random sample of 90 students was taken. The time, x minutes, spent by each student was measured and it was found that ( 40) 270x  and 2 167095x  . A previous survey indicated that students spent an average of 42.5 minutes in the canteen during lunch break. Test, at the 5% level, whether the average time spent has increased. [6] State what you understand by the expression ‘at the 5% significance level’ in the context of this question. [1] Solution From sample, 4090 270_ x = 43,   2 2 90 431 6 7.69668516709589 90 89s        Let  denote the population mean time spent (in min) by a student in the school canteen during lunch break. H0: 42.5 vs H1: 5.42 Perform a 1-tail test at 5% significance level. Under 0H , since n = 90 is large, 7.6966~ N 42.5,90X     approximately by Central Limit Theorem. Using a z-test, p-valueP 43 0.0437 (3sf)X   Since p-value0.0437 0.05  , we reject 0H and conclude that there is sufficient evidence at 5% significance level that the average time spent has increased. 5% significance level means that there is a probability of 0.05 to conclude that the average time spent by the students in the canteen is more than 42.5 mins when it is 42.5 mins.

Raffles Institution H2 Mathematics 2025 Year 6 ______________________________________________________________________________________________ ______________________________________________________ Additional Practice Questions for Chapter S5: Hypothesis Testing Page 3 of 16 3 9758/2019/02/Q9 A company produces resistors rated at 750 ohms for use in electronic circuits. The production manager wishes to test whether the mean resistance of these resistors is in fact 750 ohms. He knows that the resistances are normally distributed with variance 100 ohms2. (i) Explain whether the manager should carry out a 1-tail test or a 2-tail test. State hypotheses for the test, defining any symbols you use. [2] The production manager takes a random sample of 8 of these resistors. He finds that the resistances, in ohms, are as follows. 742 771 768 738 769 752 742 766 (ii) Find the mean of the sample of 8 resistors. Carry out the test, at the 5% level of significance, for the production manager. Give your conclusion in context. [5] The company also produces resistors rated at 1250 ohms. Nothing is known about the distribution of the resistances of these resistors. (iii) Describe how, and why, a test of the mean resistance of the 1250 ohms resistors would need to differ from that for the 750 ohms resistors. [2] Solution: (i) The manager should carry out a 2-tail test because he wishes to know whether the mean resistance of these resistors is in fact 750 ohms or different from 750 ohms. Let X denote the resistance of a resistor rated at 750 ohms and  the population mean resistance of resistors. Null hypothesis, H0: 750 Alternative hypothesis, H1: 750 (ii) Sample mean resistance, 6048 756.8 xx n   Perform a 2-tail test at 5% significance level. Under H0, 100~ N 750,8X     . Using a z-test, p-value 2P 756 0.0897 (3 sf) 0.05X    . We do not reject 0H . There is insufficient evidence, at 5% level of significance, to conclude that the mean resistance of these resistors is not 750 ohms. (iii) As nothing is known about the distribution of the resistances of the resistors rated at 1250 ohms, a random sample of large enough size (say, 30) must be taken so that Central Limit Theorem can be applied to approximate the distribution of the sample mean resistance of resistors rated at 1250 ohms to a normal distribution.

Raffles Institution H2 Mathematics 2025 Year 6 ______________________________________________________________________________________________ ______________________________________________________ Additional Practice Questions for Chapter S5: Hypothesis Testing Page 4 of 16 4 9233/2004/02/Q30 OR The mean of a random variable X is denoted by . A sample of 50 random observations of X is taken and the results are summarized by 527.1x . (i) It is given that the population variance is 15. Carry out a 2-tail test of the null hypothesis 9.5 , at the 5% significance level. State, giving a reason, whether any assumptions about the population are needed in order for the test to be valid. [4] (ii) It is given instead that 2 6172.31x  . In a 1-tail test of the null hypothesis 11.5 , the alternative hypothesis is accepted. State the alternative hypothesis, and find an inequality satisfied by the significance level of the test. [8] Solution (i) 527.1 10.54250 xx n   To test 0H : 9.5 vs 1H : 9.5 Perform a 2-tail test at the 5% significance level. Under 0H , since 50n is large, 15N 9.5,50X     approximately, by Central Limit Theorem p-value 2P 10.542 0.0571 0.05X     We do not reject 0H and conclude t