2023 S7 Hypo Test BMQ APQ (RVHS)

2023 JC2 H2 Mathematics (9758) River Valley High School Mathematics Department Page 1 of 39 STATISTICS 7 TUTORIAL (SOLUTIONS) Hypothesis Testing H2 Maths Topic: Hypothesis testing Purpose of LE: LE:  Relate the notion of null and alternative hypothesis to the statement “innocent until proven guilty” To enhance understanding of the concept of null and alternative hypothesis and the asymmetry of these hypotheses by relating it to a concrete analogy which is easier to appreciate. Remarks:  Teachers can pose the question during lecture or tutorial to bring out the above idea, by asking students to compare the null and alternative hypothesis against the Singapore legal system “innocent until proven guilty”.  Amount of time required: 5 – 10 minutes Details of Activity Present the reading to the students. In pairs, discuss the concept of null and alternative hypothesis that you have just learned with the idea of “innocent until proven guilty” used in the Singapore legal system. The Singapore Legal System In Singapore, “The presumption of innocence is widely recognised as the cornerstone of the criminal justice system. Under the presumption, the Prosecution has a “duty” to prove the accused’s guilt, “subject to” the insanity defence and statutory exceptions. Hence, if there remains a “reasonable doubt … on the whole of the case”, the prosecution has not established its case and the accused is acquitted.” (http://www.eastasiareview.org/issues/2013/articles/2013_Low.pdf) (Available as of 11 Jul 2016) Notes for teachers The hypothesis test works like a criminal trial: defendants are presumed innocent until proven guilty. Imagine the following scenario: A suspect was put on trial for a criminal offence. The Prosecution thought that he was guilty. H0: Suspect was innocent H1: Suspect was guilty The Prosecution would try to show the judge, beyond reasonable doubt, that the suspect was guilty (in other words, there was enough evidence to “reject the null hypothesis”). If the Prosecution did not do his job well, then he would not have enough evidence to convict the suspect. Although the judge might still think that the suspect was guilty, but the judge could not legally prove it. In this case, with the given evidence, the jury would “fail to reject the null hypothesis”.

2023 JC2 H2 Mathematics (9758) River Valley High School Mathematics Department Page 2 of 39 Basic Mastery Question 1. The solution to the following problem is provided below the question. Highlight the errors in the solution and amend these errors. A manufacturer claims that the life span of an electrical component it produces is normally distributed with mean 22 hours and standard deviation 6 hours. A random sample of 50 such electrical components has a mean life span of 20.2 hours. Test, at the 1% significance level, whether the mean life span of the electrical components has changed. Solution: To test H 0 : 0 22  against H1 : 0 22  Under H 0, 22~ (0,1)6 / 50 xZ N  2-tail test at 1% level of significance. Using GC, p-value 0.0339 , we do not reject H0. There is insufficient evidence that 0 22  . Ans: Test 0 1H : 22 against H : 22   0 22Under H , ~ (0,1)6 / 50 XZ N  2-tail test at 1% level of significance From GC, p-value = 0.0339 > 0.01, so we do not reject 0H There is insufficient evidence, at the 1 % significance level, that the mean life span has changed.

2023 JC2 H2 Mathematics (9758) River Valley High School Mathematics Department Page 3 of 39 2. A supermarket manager investigated the lengths of time that customers spent shopping in the store. The time, x minutes, spent by each of a random sample of 150 customers was measured, and it is found that 2871x and   029602x . Test, at the 5% level of significance, the hypothesis that the mean time spent shopping by customers is 20 minutes, against the alternative that it is less than this. What is the purpose of the question? Can you express the problem in mathematical terms? Let  minutes be the mean length of time that customers spent shopping in the store. Test 0 1H : 20 against H : 20   One-tail test at 5 % level of significance. What are the information given? What additional information do you need? From the sample data, n 150 22871; 60029x x   2871 19.14150 xx n   2 2 2 2 ( )1 1 (2871)[ ] [60029 ] 34.0811 149 150 xs x n n      What are the mathematical concepts used? Test statistic: Under 0H ,  20 N 0,1 approximately since is large 150 XZ ns   What is the interpretations of the calculated test statistic and p- value? Method 1: p – value = 0.0355 < 0.05 We reject0H . Method 2: Critical region to reject 0H : 1.645z calculated 19.14 20 1.80434.081 150 z   Since calculatedz lies within critical region, we reject0H . There is sufficient evidence at 5 % level of significance that the mean time spent shopping by customers is less than 20 minutes. What are the assumptions? Are the assumptions valid? There is no need to assume that X is normally distributed as n = 150 is large and henceX follows a normal distribution approximately.

2023 JC2 H2 Mathematics (9758) River Valley High School Mathematics Department Page 4 of 39 3. The principal of a private college claims that graduates from his school have an average starting salary of $2050. A private body checks the claim by interviewing a random sample of 60 graduates from the school. The data obtained is summarized below, where x denotes the monthly salary per person. ( 2000) 2740x  and 2( 2000) 162001x  Carry out an appropriate test at the 3% significance level to determine whether the principal is overestimating his claim. Do we need to assume that X follows a normal distribution? Solution: ( 2000) 27402000 2000 2045.6666760 60 xx       2 2 2 2 ( 2000)1 ( 2000)1 60 1 2740162001 624.9887059 60 xs x n                  Let µ be mean starting salary of graduates from the college. o 1Test H : 2050 against H : 2050    at 3% significance level Test statistic: Under H0, since n = 60 is large, 262498870X N 2050, 60     approximately 2050~ N(0,1)/ 60 XZ s   approximately Using GC, p-value=0.089693 > 0.03, we do not reject oH . There is insufficient evidence at 3% significance level that the principal is overestimating his claim. There is no need to assume that X follows a normal distribution as n = 60 is large, therefore X follows a normal distribution approximately.

2023 JC2 H2 Mathematics (9758) River Valley High School Mathematics Department Page 23 of 39 Or 2 10911 6320s  9(vi) Since 2s from Germsfree is smaller, hence the test statistic value from the test for Germsfree will be more negative. Therefore 1 2p p . Additional Practice Questions 1(i) x = 90 90 + 30 = 31 s2 =   2901 203789 90       = 1947 89 or 21.9 (to 3 sig figs) Unbiased estimate means the mean of the estimate is expected to equal to the parameter when collected infinitely(many times). E() = p where  is estimate and p is parameter as well as having the minimum variance as compared to the other estimators. (ii) Test0 1: 30 vs : 30 H H   and perform a upper-tailed z-test at 5% significance level. Under 0H , X ~ (30, 2 90  ). As 2 is unknown, estimate with s2 = 1947 89 The test Statistic, Z = s n X  ~ N(0,1) approx.. Using GC, p-v