RI T2W9 Hypo+Testing %28Soln%29

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 _______________________________ Y6 H2 Math Term 2 Revision W9 Hypothesis Testing Page 1 of 4 Term 2 W9 Revision`: Hypothesis Testing Questions 1 DHS JC2 Mid-Year CT 9758/2018/02/Q1 The government of a particular country reported that the mean monthly water consumption of households is 317.9 m . This is followed by a nationwide water conservation campaign to encourage households to reduce their water consumption. Subsequently after the campaign, the water consumption 3 mx of a random sample of 50 households is recorded and their data are summarised as follows. ( 15) 103,x  2( 15) 599.x  (i) Find unbiased estimates of the population mean and variance. [2] (ii) Test at the 5% significance level whether the campaign is effective. State, giving a reason, whether any assumption about the distribution of the monthly water consumption of households is needed in order for the test to be valid. [4] (iii) Explain the meaning of ‘5% significance level’ in the context of the question. [1] Solution : (i) An unbiased estimate for the population mean is ( 15) 10315 15 17.0650 50 xx      An unbiased estimate for the population variance is   2 2 2 2( 15)1 ( 15)50 1 50 1 10359949 50 7.8943 (to 5 s.f.) = 7.89 (to 3 s.f.) xs x                (ii) Let X represent the water consumption (in cubic metres) of a household with population mean . To test H0:  = 17.9 vs H1:  <17.9 Perform 1-tail test at 5% significance level

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________ _______________________________ Y6 H2 Math Term 2 Revision W9 Hypothesis Testing Page 2 of 4 Under H0, 7.8943~ N , 51 07.9X     approximately, by Central Limit Theorem, since n = 50 is large. From the sample, 6.17.0x Using z-test, p-value = P 17.06X = 0.0173 (3 s.f.) Since p-value <0.05, we reject H0 and conclude that there is sufficient evidence at 5% significance level that the campaign was effective. No assumptions on the distribution of X are required as the sample size is large, by Central Limit Theorem, the distribution of the sample mean (X) is approximately normal. (iii) The 5% significance level means that there is a probability of 0.05 that we conclude that the mean water consumption of households has decreased, when it is actually unchanged at 317.9 m . 2 RI Prelim 9758/2020/02/Q7 In a factory, machines pack sugar into bags of 1 kg each on average, with variance 2 kg2. The manufacturer is concerned that the machines are putting too much sugar into the bags and decides to carry out a hypothesis test. A random sample of 8 bags are selected and their total mass is 8.4 kg. (i) Stating a necessary assumption, carry out a test of the manufacturer’s concern at the 5% significance level if 0.08 . [5] (ii) Use an algebraic method to calculate the range of values of 2 for which the null hypothesis would not be rejected at the 5% significance level. [3] Solution : (i) Let Xbe the mass of a bag of sugar in kg, and be the population mean mass of sugar in a 1 kg bag. The necessary assumption is X follows a normal distribution. 0 1 H : 1 H : 1     Perform a 1-tail test at 5% significance level. Under H0, 20.08~ N 1,8X     8.4 1.058x  Using a z-test, p-value P 1.05 0.0385 0.05X    Since p-value< 0.05, we reject 0H and conclude that there is sufficient evidence, at the 5% significance level, to support the manufacturer’s concern.

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________ _______________________________ Y6 H2 Math Term 2 Revision W9 Hypothesis Testing Page 3 of 4 (ii) Under H0, 2 ~ N 1,8X     Since H0 is not rejected at 5% significance level,    2 2 2 2 2 2 -value 0.05 P 1.05 0.05 1.05 1P 0.05 8 Since P 1.6449 0.05, 80.05 1.6449 0.058 1.64485 0.0073918 0.00739 (3sf) p X Z Z                           

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________ _______________________________ Y6 H2 Math Term 2 Revision W9 Hypothesis Testing Page 4 of 4 3 ACJC Prelim 9758/2019/02/Q7 modified A concert promoter claims that the mean price of a ticket to a pop concert is $200. A media company took a large random sample of n tickets, and found that the average ticket price for the sample is $206. If the standard deviation of ticket price is known to be $32.25, find the minimum value of n such that there is sufficient evidence at the 4% level of significance to reject the concert promoter’s claim. [4] Solution : Let $X be the price of a pop concert ticket and µ be the population mean of X. 0H : 200μ vs 1H : 200μ Perform a 2-tail test at 4% significance level Under 0H , 232.25~N 200, X n     approximately by Central Limit Theorem since n is large From the sample, 206x To reject H0, p-value = 2P 206X  0.04  P 206 0.02X  206 200P 0.0232.25Z n         Since  P 2.0537 0.02Z  , 6 2.053732.25 n  121.85n Minimum value of n is 122. 0.02 2.0537 0 Alternatively,  P 206 0.02X  Using GC table of values, n P 206X 121 0.0204 > 0.02 122 0.0199 < 0.02 Minimum value of n is 122.