RI 2025 Y6T2W9Mathfocus Summary+on+Hypothesis+Testing

Page 1 of 4 RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 Term 2 Revision (Summary) Topic: Hypothesis Testing Summary for Hypothesis Testing Performing a Hypothesis Test Step 1 Understand the given question and write down the null hypothesis 0H and the alternative hypothesis 1H Step 2 Write down the level of significance  (usually given in the question) Step 3 Decide on the test statistic to be used and determine its distribution Step 4 Use the GC to calculate the p-value Step 5 Reject 0H if p-value , OR Do not reject 0H if p-value Write down the conclusion in the context of the question Example: From records, the IQ of university students is normally distributed with a mean of 118. Students from ABC University claim that the mean IQ of the students in their university is greater than 118. A sample of 50 students was tested and the mean IQ was found to be 121. Test, at 5% significance level, whether the sample provides significant evidence to support the students’ claim, assuming that (a) the IQ of university students is normally distributed with standard deviation, , where 12 , (b) the IQ of university students is normally distributed and an unbiased estimate of the population variance, calculated based on the sample of 50 students, is 97.454, (c) an unbiased estimate of the population variance for the IQ of university students, calculated based on the sample of 50 students, is 97.454. Solution Let X be the IQ and be the mean IQ of a student in ABC University. Step 1: To test 0H : 118 vs 1H : 118 Remember to always define X and any notations used if they are not defined in the question. Read question carefully to decide >, < or ≠.

Page 2 of 4 Step 2: Perform a 1-tail test at 5% level of significance. For part a: Sample from a Normal population of known variance Step 3: Under 0H , 2 0~ N ,X n     , where 0 118  and 12 . From the sample, 121x , 50n . Step 4: Using a z-test, -valuep  P 121X  0.0385 (3 s.f.) Step 5: Since -valuep 0.0385 0.05, we reject 0H and conclude that there is sufficient evidence, at 5% level of significance, to support the claim that the mean IQ of students in ABC University is greater than 118. If test is performed at 1% level of significance, Step 5: Since -valuep 0.0385 0.01, we do not reject 0H and conclude that there is insufficient evidence, at 1% level of significance, to support the claim that the mean IQ of students in ABC University is greater than 118. [Notice the only difference are the phrases in bold, the end of the sentence is to describe the alternative hypothesis H1.] For part b: Large sample from a Normal population of unknown variance Step 3: Under 0H , 2 0~ N , sX n    approximately, with 0 118  . From the sample, 121x , 97.454s Step 4: Using a z-test, -valuep P 121X  0.0158 (3 s.f.) If H1 involves > or <, it is a 1-tail test If H1 involves ≠, it is a 2-tail test.

Page 3 of 4 Step 5: Since -valuep 0.0158 0.05, we reject 0H and conclude that there is sufficient evidence, at 5% level of significance, to support the claim that the mean IQ of students in ABC University is greater than 118. For part c: Large sample from a non-Normal population of unknown variance Step 3: Under 0H , since 50n is large, 2 0~ N , sX n    approximately, by Central Limit Theorem, with 0 118  . From the sample, 121x , 97.454s Step 4 and Step 5 are identical to those in part b. General Tips - Take note of the nature of the “claim” in the question and the respective conclusion e.g. if the claim is “the mean IQ is equal to 118 or at least 118 or at most 118”, then when 0H : 118 is rejected, there is “sufficient evidence at … to conclude that the claim is invalid”; whereas if the claim is “the mean IQ is higher or lower than 118”, then when 0H : 118 is rejected, there is “sufficient evidence at … to conclude that the claim is valid”. - If 0H is rejected at 5% level of significance for a certain p-value, 0H will also be rejected at 10% level of significance or any level higher than 5%. On the contrary, 0H may or may not be rejected at 1% or any level lower than 5%. Definitions 1. The level of significance (or significance level) of a hypothesis test, denoted by , is defined as the probability of rejecting 0H when 0H is true e.g. “5% level of significance” means “there is a 0.05 probability of wrongly concluding that the mean IQ of the students is more than 118 when in fact it is 118”. 2. The -valuep P( 121)X  in this context refers to the probability of getting a sample with average IQ at least 121. This is also the value we will put down on our script if question asked for the smallest level of significance at which 0H can be rejected in favour of 1H .

Page 4 of 4 Important Cases where conclusion is given and we are asked to find: Case 1: Level of significance % Carry on as if we are performing a test. After finding the p-value: If given 0H is rejected, p-value 100  ; if given 0H is not rejected, p-value 100  . Case 2: Sample mean x(No standardization required) Example: For a test at 5% level of significance and under 0H , 2 0~ N ,X n     , where 0 118  and 12 . From the sample, 50n . If given 0H is rejected, use: [GC invNorm: key in 12standard deviation = 50, not 12] 1-tail 1e.g. H : 118 2-tail 1e.g. H : 118 p-value 0.05 P 0.05X x  Using GC, P 120.791 0.05X  121x p-value 0.05 where  2P , if 118value =2P , if 118 X x xp X x x       0H is rejected    P 0.025, if 118 P 0.025, if 118 X x x X x x         Case 3: 2 2 0 or or or n s  All steps are similar to that of Case 2 above except we have to standardize in order to solve for the unknown parameter. Example: (to find 0 ) 1-tail 1 0e.g. H : ; from sample, 121x   -value=P 121 0.05p X   0 2 121P 0.0512 50 Z           Using GC, P 1.6449 0.05Z  0 2 121 1.6449 12 50   