SAJC 2021 H1 Chapter 6 Math Hypothesis Testing Check your understanding Teacher v1

2021 SAJC JC1 H2 Maths Chapter 6 : Hypothesis Testing 1 Check Your Understanding (Hypothesis Testing) Formulating Null and Alternative Hypothesis ( No need to perform the hypothesis testing) 1. For each of the following parts, justify why a z -test (z-test means a test where the test statistic is normally distributed) can be used by writing down the following (i) The null and alternative hypotheses; (ii) If the sample size is considered large or small; (iii) The population variance , or if unavailable, the unbiased estimate of the population variance ; (iv) If the population is normally distributed or not (if not, also indicate whether an assumption for the population to be normally distribu ted is needed to complete the test); (v) If the Central Limit Theorem is to be used when performing the hypothesis test. You do NOT need to actually perform the hypothesis test (a) The heights of boys in SAJC are normally distributed with a mean 𝜇 cm and standard deviation 5 cm. It is claimed that 𝜇 = 171. However, a random sample of 50 SAJC boys has mean height 173 cm. Test, at the 2% significance level, whether the mean height of SAJC boys in general is above 171 cm. (b) The heights of boys in SAJC are distributed with a mean μ cm and standard deviation 5 cm. It is claimed that μ = 171. However, a random sample of 50 SAJC boys has mean height 169 cm. Test, at the 2% significance level, whether the mean height of SAJC boys in general is below 171 cm. (c) The heights of boys in SAJC are normally distributed with a mean μ cm. It is claimed that μ = 171. However, a random sample of 50 SAJC boys has mean height 173 cm with standard deviation 5 cm. Test, at the 2% significance level, whether the mean height of SAJC boys in general is not equal to 171 cm. (d) The heights of boys in SAJC are normally distributed with a mean μ cm and standard deviation 5 cm. It is claimed that μ = 171. However, a random sample of 5 SAJC boys has mean height 173 cm. Test, at t he 2% significance level, whether the mean height of SAJC boys in general is above 171 cm. (e) The heights of boys in SAJC are distributed with a mean μ cm. It is claimed that μ = 171. However, a random sample of 50 SAJC boys has mean height 173 cm with st andard deviation 5 cm. Test, at the 2% significance level, whether the mean height of SAJC boys in general is not equal to 171 cm. (f) The heights of boys in SAJC are distributed with a mean μ cm and standard deviation 5 cm. It is claimed that μ = 171. However, a random sample of 5 SAJC boys has mean height 169 cm. Test, at the 2% significance level, whether the mean height of SAJC boys in general is not equal to 171 cm. 2 2s

2021 SAJC JC1 H2 Maths Chapter 6 : Hypothesis Testing 2 Solution: (a) , Large sample size, Population is normally distributed, use a z-test, do not need to use CLT (b) , Large sample size, Population is not stated to be normally distributed (do not need to assume normality of population), use a z-test, use CLT (c) , Large sample size, Population is normally distributed, use a z-test, do not need to use CLT (d) , Small sample size, Population is normally distributed, use a z-test, do not need to use CLT (e) , Large sample size, Population is not stated to be normally d istributed (do not need to assume normality of population), use a z-test, use CLT (f) , Small sample size, Population is not stated to be normally distributed (need to assume normality of population), use a z-test, do not need to use CLT Sample size large/small, Population normal, Population variance known 2. [N2011/2/Q10 Modified] In a factory, the time in minutes for an employee to install an electronic component is a normally-distributed continuous random variable T. The standard deviation of T is 5.0 and under ordinary conditions the expected value of T is 38.0. After background music is introduced into the factory, a sample of n components is taken and the mean time taken for randomly chosen employees to install them is found to be t minutes. A test is carried out, at the 5% significance level, to determine whether the mean time taken to install a component has been reduced. (i) State appropriate hypotheses for the test, defining any symbols you use. (ii) Given that n=50, state the set of values of for which the result of the test would be to reject the null hypothesis. (iii) It is given instead that =37.1, and the result of the test is that the null hypothesis is not rejected. Obtain an inequality involving n, and hence find the set of values that n can take. Solution: 01 171,:: 171HH  22 5  01 171,:: 171HH  22 5  01 171,:: 171HH  22 50 549s  01 171,:: 171HH  22 5  01 171,:: 171HH  22 50 549s  01 171,:: 171HH  22 5  t t

2021 SAJC JC1 H2 Maths Chapter 6 : Hypothesis Testing 3 (i) Let refers to the population mean time taken for the population for employees to install one component. Test 0 : 38.0H   against 1 : 38.0H  at 5% level of significance Under H0, since n is large, 25~ (38, )TN n .  2 ~ 0,1TZN n    (ii) Critical region  : 1.64485zz  Given that 50n , 2 38 5 50 cal tz  , Since H0 is rejected, 2 1.64485 38 1.64485 5 50 36.8 calz t t     The set of values for (iii) Critical region  : 1.64485zz  Given that 2 37.1 38 5 calz n  .   | 0 36.8t t t    37.1t  Examiners’ Report: (i) Most candidates correctly define and in terms of . Although many candidates went on to explain what and were (which were not required), very few defined and a very small minority defined it correctly as the population mean time taken for the population for employees to install one component. (ii) A few candidates used 0.05 rather than a z-value. Others used +1.645 rather than . Still others used the wrong inequality or multiplied by a factor . (iii) Candidates found this part rather challenging. Candidates made sign errors manipulating an inequality involving negatives and square root. For full credit, candidates were expected to realise that n was an integer. 0H 1H  0H 1H  1.645 1 n n

2021 SAJC JC1 H2 Maths Chapter 6 : Hypothesis Testing 4 Since H0 is not rejected, 2 1.64485 37.1 38 1.64485 5 0.9 1.644855 9.1377 83.5 calz n n n n        The set of values for 3. [SAJC 2013 Prelim P1/10 (Modified)] A manufacturer claims that his new dieting pill helps people lose weight. A random sample of 20 people took the pill for a month and the loss in weight (initial weight – final weight) after a month, x kg, was summarized as follow Suppose that the population follows a normal distribution and the population standard deviation is known to be 1.6 kg. The manufacturer claims that the average weight loss for people taking the pill is at least . A test at the 10% level of significance indicates that the manufacturer’s claim is valid. Find the largest value of . Solution: The population mean is unknown, hence we need to standardise the normal distribution. Solution: 20.49 1.024520x  Let X be the r.v. “loss in weight” for a randomly chosen person who took the pill for a month and be mean loss weight loss. Test 00:H  against 10:H  at 10% level of significance Under H ,  2 ~ 0,1XZN n    (If the manufacturer’s claim is valid, it means that we do not reject H ) Critical region  : 1.2816zz   | 0 83n n n     20.49x 0 0  0 2 0 1.6~, 20XN   0

2021 SAJC JC1 H2 Maths Chapter 6 : Hypothesis Testing 5 2 1.0245 1.6 20 o calz  . Since H0 is not rejected, 2 1.2816 1.0245 1.64485 1.6 20 1.0245 0.45850 1.483 cal o o o z          Largest value of is 1.48. Sample size large, Population non-normal 4. A shopkeeper complains that the average weight of chocolate bars of a