2019 H2 Prelim (Hypothesis Testing) Qns with Sol

2019 H2 MA Prelim Compilation - Hypothesis Testing (17 Questions with Solutions) ACJC JC2 Prelim 9758/2019/02/Q7 A concert promoter claims that the mean price of a ticket to a pop concert is $200. A media company collected information on ticket price, $ x , for 50 randomly chosen people who bought pop concert tickets. The results are summarised as follows. ( 200) 450 x    2 ( 200) 55 000 x    (i) Test, at the 4% level of significance, the concert promoter’s claim that the mean price of a ticket to a pop concert is $200. You should state your hypotheses and define any symbols you use. [5] (ii) The media company took another random sample of n tickets, and found that the average ticket price for this sample is $206. If the standard deviation of ticket price is now known to be $32.25, find the maximum value of n such that there is insufficient evidence at the 4% level of significance to reject t he concert promoter’s claim. [2] ACJC JC2 Prelim 9758/2019/02/Q7 (Solutions) 7(i) ( 200) 200 209 50 x x        2 2 450 1 50950 55000 49 50 49 s            or 1039.8 (5 s.f.) Let μ be the mean price of all concert tickets. To test 0 H : μ 200  against 1 H : μ 200  at the 4% level of significance. Under 0 H , Z =   200 ~N 0,1 50 X s  or 50950 ~N 200, 49(50) X       approximately by central limit theorem since sample size, 50, is large. Value of test statistic 1.97 z  p -value = 0.0484 > 0.04 (do not reject 0 H ) There is insufficient evidence at the 4% level of significance that the mean concert ticket price is not $200. (ii) Now given the standard deviation is 32.25 Under 0 H , 2 32.25 ~N 200, X n       by CLT since n is large

For H 0 to not be rejected, the test statistic, z , must not be in critical region. Hence 2.0537 2.0537 z    i.e. 206 200 2.0537 2.0537 32.25 n     i.e. 6 2.0537 2.0537 32.25 n    i.e. 11.039 11.039 n    Hence largest value of n is 121. ASRJC JC2 Prelim 9758/2019/02/Q10 Along a 3km stretch of a road, the speed in km/h of a vehicle is a normally distributed random variable T . Over a long period of time, it is known that the mean speed of vehicles traveling along that stretch of the road is 90.0 km/h. To deter speeding, the traffic governing body introduced a speed monitoring camera. Subsequently, the speeds of a random sample of 60 vehicles are recorded. The results are summarised as follows.   2 5325, 2000. t t t      (i) Find unbiased estimates of the population mean and variance, giving your answers to 2 decimal places. [2] (ii) Test, at the 5% significance level, whether the speed-monitoring camera is effective in deterring the speeding of vehicles on the stretch of road. [4] In another sample of size n ( n > 30) that was collected independently, it is given that t = 89.0. The result of the subsequent test using this information and the unbiased estimate of the population variance in part (i) is that the null hypothesis is not rejected. Obtain an inequality involving n , and hence find the largest possible value n can take. [4] ASRJC JC2 Prelim 9758/2019/02/Q10 (Solutions) 10 (iii) Unbiased estimate of the population mean = 5325 88.75 60 t   2% -2.0537 2.0537 2%

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.

CJC JC2 Prelim 9758/2019/02/Q9 The security guard of a particular school claims that the average speed of the cars in the school compound is greater than the speed limit 25 km/h. To investigate the security guard’s claim, the traffic police randomly selected 50 cars and the speed was recorded. The total speed and the standard deviation of the 50 cars are found to be 1325 km/h and 7.75 km/h respectively. (i) Find the unbiased estimates of the population mean and variance. [2] (ii) Test at 5% level of significance whether there is sufficient evidence to support the security guard’s claim. [4] It is now known that the speed of the cars is normally distributed with mean 25 km/h and standard deviation of 6 km/h. (iii) A new sample of n cars is obtained and the sample mean is found to be unchanged. Using this sample, the traffic police conducts another test at 10% level of significance and concludes that the security guard’s claim is valid. Find the set of values that n can take. [3] (iv) What is the speed exceeded by 75% of the cars? [1] CJC JC2 Prelim 9758/2019/02/Q9 (Solutions) 9 (i) Let X be the random variable denoting the speed of a car in the school compound. Unbiased estimate of population mean 1325 26.5 50   Unbiased estimate of population variance 2 50 7.75 61.288 61.3(3sf) 49        (ii) 0 H : 25   1 H : 25   , where  is the population mean of X. Under 0 H , Since n is large, by CLT, 61.28 0 ~ 2 N 8 5, 5 X       approx. 26.5 25 61.288 50 test z   Since -value 0.0877345 0.05 p   , we do not reject 0 H and conclude that we have insufficient evidence at 5% level of significance that the mean speed of the cars is more than 25. (iii)   6 ~ N 25,3 X 6 ~ N 3 25, X n       26.5 25 0.25 36 test z n n   

Since 0 H is rejected at 10 % level of significance, 1.28155, test z  0.25 1.28155 26.278 { : , 27} n n n n n     (iv)   25,36 P( ) 0.75 21.0 (3 s.f.) ~ N X X t t    DHS JC2 Prelim 9758/2019/02/Q9 The time taken, T (in minutes), for a 17-year-old student to complete a 5-km run is a random variable with mean 30. After a new training programme is introduced for these students, a random sample of n students is taken. The mean time and standard deviation for the sample are found to be 28.9 minutes and 4.0 minutes respectively. (a) Find the unbiased estimate of the population variance in terms of n . [1] (b) Using 40, n  (i) carry out a test at the 10% significance level to determine if the mean time taken has changed. State appropriate hypotheses for the test and define any symbols you use. [4] (ii) State what it means by the p -value in this context. [1] (iii) Give a reason why no assumptions about the population are needed in order for the test to be valid. [1] (c) The trainer claims instead that the new training programme is able to improve the mean of T , 30 minutes, by at least 5%. The school wants to test his claim. (i) Write down the null and alternative hypothesis. [1] 1.28155 0.10