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UNIT 8: STATISTICAL HYPOTHESIS TESTING 8.1. Introduction This chapter describes the statistical procedure for testing the hypotheses, which is a very standard procedure that is commonly used by professionals in a wide variety of disciplines. The two major activities of inferential statistics are the estimation of population parameters and hypothesis testing. Hypothesis testing is important because it provides an objective framework for making decisions using probabilistic methods rather than relying on subjective impressions. Definition 8.1.1 A statistical hypothesis is a claim or statement about the property of a population. It is an assertion or conjecture concerning one or more populations. A hypothesis test (or test of significance) is a standard procedure for testing a claim about the property of a population. Example 8.1.2 The following statements are typical of the hypotheses (claims) that can be tested by the procedures we will develop later in this chapter: (i) A computer Engineer claims that the life span of computers produced by a certain Zambian company is less 10 years (ii) Medical researchers claim that the mean body temperature of healthy adults is less than 38β (iii)A food company produces peanuts weighing 336g (on average). Periodically, the quality control department takes samples of peanut packets to determine whether the packaging process is under control. (iv) Mendel claims that under certain circumstances, the percentage of off-spring peas with yellow pods exceeds 25%. Before beginning to study this chapter, we should bare the following rule in mind: Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 1 Rare Event Rule for Inferential Statistics If, under a given assumption, the probability of a particular observed event is exceptionally small, we conclude that the assumption is probably not correct. Following this rule, we test the claim by analyzing the sample data in an attempt to distinguish between results that can easily occur by chance and results that are highly unlikely to occur by chance. 8.2. Basics of Hypothesis Testing In this section, we describe the formal components used in hypothesis testing: null hypothesis, alternative hypothesis, test statistic, critical region, significance level, critical value, π β π£πππ’π, type I error, and type II error. In other words, the objectives of this section are as follows: ο· Given a claim, identify the null hypothesis and the alternative hypothesis, and express both in symbolic form. ο· Given a claim and sample data, calculate the value of the test statistic ο· Given a significance level, identify the critical value(s) ο· Given a value of the test statistic, identify the p β value. ο· State the conclusion of a hypothesis test in simple, non-technical terms ο· Identify the type I error and type II error that could be made when testing a given claim. Null and Alternative Hypotheses Usually, a hypothesis takes the form of a claim, a belief or suspicion. The null hypothesis (denoted by π»0 ) is a statement that the value of a population parameter (such as proportion, mean or standard deviation) is equal to some claimed value. The symbolic statement of the null hypothesis usually contains an equal (=) sign. Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 2 In this sense, we can say that the null hypothesis is a statement asserting no change, no effect or no difference. The alternative hypothesis (denoted by π»1 ππ π»π ) is the statement that the parameter has a value that somehow differs from the null hypothesis. For the methods of this chapter the symbolic form of the alternative hypothesis must use one of the symbols < ππ > ππ β . Below are some typical null and alternative hypotheses in relation to Example 8.1.2. (π)π»0 : π = 10 (ππ)π»0 : π = 38β (πππ)π»0 : π = 336π (ππ£)π»0 : π = 0.25 π»1 : π < 10 π»1 : π < 38β π»1 : π β 336π π»1 : π > 0.225 It should be noted that the above are examples of one-tailed (one β sided) tests while except for (iii) which is a two - tailed (two - sided) test. In two tailed tests, the level of significance, ο‘ is divided equally between the two tails that constitute the critical region. Caution! 1. Although some text books use the symbols such as β€ πππ β₯ in the null hypothesis (π»0 ), most professional journals use only the equal sign and that is what is recommended in this text. We conduct the hypothesis test by assuming that the proportion, mean or variance is equal to some specified value so that we can work with a single distribution having a specific value. 2. If you are conducting a study and want to use a hypothesis test to support your claim, the claim must be worded so that it becomes the alternative hypothesis. This requires that your claim must be expressed using the symbols < ππ > ππ β . You cannot use a hypothesis test to support a claim that some parameter is equal to some specified value. Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 3 Test Statistic The test statistic is a value computed from the sample data, and it is used in making the decision about the rejection of the null hypothesis. It is found by converting the sample statistic (such as the sample proportion, πΜ , or the sample mean, π₯Μ or the sample variance, π 2 ) to a score such as π§, π‘ ππ ο£2 with the assumption that the null hypothesis is true. The test statistic can therefore be used for determining whether there is significant evidence against the null hypothesis. Critical region The critical (rejection) region is the set of all values of the test statistic that cause us to reject the null hypothesis. In other words, the critical region contains the critical value at a given level of significance (and with specified degrees of freedom in a case of the t β test, chi square test or F- test). A critical value is any value that separates the critical region (where we reject the null hypothesis) from the values of the test statistic that do not lead to the rejection of the null hypothesis. The critical value, also called the table value depends on the nature of the null hypothesis, the sampling distribution that applies, and the significance level. When a decision is made about the null hypothesis, there are two possible errors that may be committed. (i) Type I error: Rejection of the null hypothesis when it is actually true (ii) Type II error: Acceptance of the null hypothesis when it is false The table below gives a summary of the possible situations: We decide to reject π»0 Decision We fail to reject π»0 True State of Nature (Reality) π»0 ππ ππππ π π»0 ππ π‘ππ’π Correct decision made Type I error committed Type II error Correct decision made committed Level of significance Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 4 The level of significance (denoted by ο‘) is the probability that the test statistic will fall in the critical region when the null hypothesis is actually true. In other words, ο‘ = P(type I error) and the probability of type II error is given by ο’ = P(type II error). 1 - ο’ is the probability of rejecting π»0 when it is false and it is called the power of the test. If the level of significance is not specified, then it is mandatory to use 5%. Probability value (p β value) This is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true. The null hypothesis is rejected if the p β value is very small (i.e. when p β value < ο‘). Decisions and conclusions From example 8.1.2, we have seen that the original claim sometimes becomes the alternative hypothesis. However, the standard procedure of hypothesis testing requires that we always test the null hypothesis and so, the initial conclusion will always be one of the following: (i) Reject the null hypothesis or (ii) Fail to reject the null hypothesis The decision to reject or fail to reject the null hypothesis is usually made using either the traditional (classical) method, p β value method or based on confidence intervals. In recent years however, use of the traditional method has been declining partly because statistical software packages are often designed for the p β value method. (a) Traditional method: Reject π»0 if the absolute value of the test statistic is greater than the critical value. That is, if the test statistic falls within the critical region. (b) P- value method: Reject π»0 if π β π£πππ’π < πΌ (c) Confidence intervals: Because a confidence interval estimate of a population parameter contains the likely values of that parameter, reject a claim that the population parameter has a value that is not included in the confidence interval. Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 5 The conclusion of rejecting or failure to reject π»0 is fine for those who have done a statistic course, but it is important to use simple and non-technical terms in stating what the conclusion really means. The figure below gives a summary of the wording of the final conclusion. Start Wording of final conclusion Does the original claim contain the condition of equality? Yes Do you reject π»0 ? Yes No (fail to reject H0) No (original claim does not contain equality, so it becomes H1) Do you reject π»0 ? Yes No (fail to reject H0) There is sufficient evidence to warrant rejection of the claim thatβ¦β¦β¦β¦ There is no sufficient evidence to warrant rejection of the claim that β¦.. The sample data support the claim thatβ¦β¦β¦ There is not sufficient sample evidence to support the claim thatβ¦β¦. Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 6 8.3. Tests about one population parameter Now that we have understood the meanings of key concepts regarding hypothesis testing, we can discuss the various tests that can be carried out regarding one population mean, one population proportion and one population variance. The table below gives a summary of the test statistics used in each case. Parameter Population mean Hypotheses H0: ο = ο0 against H1: ο οΎ ο0 or ο οΌ ο0 or ο οΉ ο0 Condition(s) Test statistic Population variance known or population variance unknown with n οΎ 30 Unknown population variance with n ο£ 30 π§= π₯Μ β π0 or π/βπ π§= π₯Μ β π0 π/βπ π₯Μ β π0 π‘= π/βπ Decision Reject H0 if |π§| > π§πΌ for a 1 β tailed test and reject H0 if |π§| > π§πΌ/2 for a 2 β tailed test Reject H0 if |π‘| > π‘πΌ, πβ1 for a 1 β tailed test and reject H0 if |π‘| > π§πΌ,πβ1 for a 2 β 2 Population proportion H0: P = p against H1: P οΎ p or P οΌ p or PοΉp Population variance π»0 : π 2 π»0 : π 2 π»0 : π 2 π»0 : π 2 = > < β π0 2 against π0 2 or π0 2 or π0 2 π§= πΜ β π β ο£2 = ππ π (π β 1)π 2 π2 tailed test Reject H0 if |π§| > π§πΌ for a 1 β tailed test and reject H0 if |π§| > π§πΌ/2 for a 2 β tailed test Reject H0 if |ο£2 | > ο£2 πΌ,πβ1 for a 1 β tailed test and reject H0 if |ο£2 | > ο£2 πΌ,πβ1 for a 2 β 2 tailed test If one needs to make a decision using a p β value method, then it should be noted that π β π£πππ’π = π(π > |π§|)πππ π πππ π‘πππππ πππ π β π£πππ’π = 2π(π > |π§|) πππ π 2 π‘πππππ π‘ππ π‘. In a case of a t β test, π β π£πππ’π = π(ππβ1 > |π‘|)πππ π πππ π‘πππππ πππ π β π£πππ’π = 2π(ππβ1 > |π‘|) For chi β square test, π β π£πππ’π = π(ο£2 πβ1 > |ο£2 |) πππ π β π£πππ’π = 2π(ο£2 πβ1 > |ο£2 |) πππ π 2 π‘πππππ π‘ππ π‘. Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 7 Example 8.3.1 1. A random sample of 100 deaths recorded in the US during the past year showed an average life span of 71.8 years. Assuming a population standard deviation of 8.9 years, does this seem to indicate that the average life span today is greater than 70 years? Use 0.05 level of significance. Working Hypotheses; π»0 : π = 70, πππ π»1 : π > 70 Level of significance; πΌ = 0.05 Test statistic; Since the sample size is greater than 30 and population variance is known, we use z as a test statistic. π§= π₯Μ β π0 π/βπ = 71.8 β 70 = 2.022 8.9 β100 Critical value; ππΌ = π0.05 = 1.645 P β Value (optional); π β π£πππ’π = π(π > |π§|) = π(π > |π§|) = 0.0217 Decision; Since, |π§| > ππΌ , we reject π»0 Note that we could have made a decision using p β value method (i.e. since p - value is less than the level of significance, we reject Ho). Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 8 Conclusion; Rejecting π»0 at 5% level of significance indicates that the sample data supports the claim that the average life span today is greater than 70 years. 2. The Zambian Heart Association recommends that an individualβs cholesterol level be under 200mg per 100ml. The following are the cholesterol readings of 16 women selected randomly from the Kitwe Herat Study: 233 197 192 179 174 217 188 209 196 167 186 221 238 179 196 191 At the 10% level of significance, do these readings suggest that women in Kitwe have cholesterol readings below 200 mg on average? What assumptions are required? Working From the given data set, π₯Μ = 197.6875 πππ π = 20.7066 Hypotheses; π»0 : π = 200, πππ π»1 : π < 200 Level of significance; πΌ = 0.1 Since the sample size is less than 30 and population variance is unknown, we use t as a test statistic. π‘= π₯Μ β π0 π/βπ = 197.6875 β 200 = β0.4467 ο |π‘| = 0.4467 20.7066 β16 Critical value; Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 9 π‘πΌ,πβ1 = π‘0.1,15 = 1.341 ο |π‘| < π‘πΌ,πβ1 P β Value (optional); π β π£πππ’π = π(π15 > |π‘|) = π(π15 > 0.4467), giving π β π£πππ’π > 0.25 ο π β π£πππ’π > 0.1 and so, π β π£πππ’π > πΌ. Decision; Based on either the critical value (traditional) method or p β value method, we fail to reject Ho. Conclusion; Failure to nullify Ho at 10% level of significance indicates that there is no sufficient sample evidence to support the claim that women in Kitwe district have cholesterol readings below 200mg on average. 3. A manufacturer of sports equipment has developed a new synthetic fishing line that he claims has a mean breaking strength of 8kg with a standard deviation of 0.5kg. A random sample of 50 lines is tested and found to have a mean breaking strength of 7.8kg. Is the claim valid at 1% level of significant? Working From the given data set, π₯Μ = 7.8, π = 8, π = 0.5 πππ π = 50 Hypotheses; π»0 : π = 8, πππ π»1 : π β 8 Level of significance; πΌ = 0.01 Since the sample size is greater than 30 and population variance is known, we use π§= π₯Μ β π0 π/βπ = 7.8 β 8 = β2.83 ο |π§| = 2.83 0.5 β50 Critical value; Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 10 ππΌ = π0.005 = 2.576 ο |π§| > ππΌ 2 2 P β Value (optional); π β π£πππ’π = 2π(π > |π§|) = 2π(π > 2.83) = 0.0046, ο π β π£πππ’π < 0.01 and so, π β π£πππ’π < πΌ. Decision; Based on either the critical value (traditional) method or p β value method, we reject Ho. Conclusion; Rejecting Ho at 1% level of significance indicates that there is no sufficient evidence to support the claim that the mean breaking strength was 8kg. 4. A distributor of cigarettes claims that 20% of the smokers in Myami prefer Kent cigarettes. To test the claim, 20 smokers are selected at random and asked what brand they prefer. If 6 of the 20 named Kent as their preference, what conclusion do we draw? Working In this case, we need to carry out the test concerning proportion. π = 20% = 0.2 ο π = 1 β 0.2 = 0.8 πππ πΜ = 6 = 0.3 20 Hypotheses; π»0 : π = 0.2, πππ π»1 : π β 0.2 Level of significance; πΌ = 0.05 Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 11 Test statistic; π§= πΜ β π ππ β π = 0.3 β 0.2 β(0.2)(0.8) 20 = 1.12 Critical value; π0.05 = π0.025 = 1.96 ο |π| < ππΌ 2 2 P β Value (optional); π β π£πππ’π = 2π(π > |π§|) = 2π(π > 1.12) = 0.2628, ο π β π£πππ’π > 0.05 and so, πβ π£πππ’π > πΌ. Decision; Based on either the critical value (traditional) method or p β value method, we fail to reject Ho. Conclusion; Failure to reject Ho at 5% level of significance indicates that there is no sufficient evidence to invalidate the claim that 20% of the smokers in Myami prefer Kent cigarettes. 5. A Cafein content of a certain brand of tea is known to be normally distributed with variance of 1.3 mg. Test this claim using a random sample of 8 packets of tea with standard deviation of 1.8mg at 5% level of significance. Working In this case, π 2 = 1.3, π = 1.8 πππ π = 8 Hypotheses; π»0 : π 2 = 1.3, πππ π»1 : π 2 β 1.3 Level of significance; πΌ = 0.05 Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 12 Test statistic; ο£2 = (π β 1)π 2 (7)1.82 = = 17.446 π2 1.3 Critical value; ο£2 πΌ,πβ1 = ο£2 0.025, 2 7 = 16.013 ο |ο£2 | > ο£2 πΌ,πβ1 2 Decision; Since |ο£2 | > ο£2 πΌ,πβ1 , we reject Ho 2 Conclusion; Rejecting Ho at 5% level of significance indicates that the variance is not equal to 1.3. 8.4. Confidence intervals and hypothesis testing The testing of Ho: ο = οo against H1: ο οΉ οo at ο‘% level of significance is equivalent to computing a (100(1 - ο‘) % confidence interval for ο. In this case, Ho is rejected if οo is not inside the confidence interval. If οo is inside the confidence interval then the null hypothesis is not rejected. If we consider question 3 in example 3.3.1 then, π»0 : π = 8, πππ π»1 : π β 8 At 1% level of significance, we can construct a 99% confidence interval as follows: π₯Μ ± ππΌ/2 π βπ = 7.8 ± π0.005 0.5 β50 = (7.62, 7.98) Since 8 ο (7.62, 7.98), we reject Ho at 1% level of significance and conclude that the manufacturerβs claim is not valid. Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 13 8.5. Tests about two population parameters In this section, we shall only concentrate on the test statistics for each of the three parameters (mean, proportion and variance). The procedure for carrying out the test remains the same as we did in the previous section. 8.5.1. Two population means Test statistics and critical regions for two population means can be summarised as follows: To test π»0 : π1 β π2 = π0 against π»1 : π1 β π2 β π0 or π»1 : π1 β π2 > π0 or π»1 : π1 β π2 < π0 , use; (π)π = π₯Μ 1 β π₯Μ 1 β π0 β π1 2 π2 2 π1 + π2 ππ πππ‘β π1 2 πππ π2 2 πππ ππππ€π πππ π = π₯Μ 1 β π₯Μ 1 β π0 π 1 2 π 2 π1 + π2 β ππ π1 2 πππ π2 2 πππ π’πππππ€π πππ πππ‘β π πππππ π ππ§ππ πππ πππππ‘ππ π‘βππ 30. (ππ) π‘ = π₯Μ 1 β π₯Μ 1 β π0 ππ π1 2 πππ π2 2 πππ π’πππππ€π ππ’π‘ ππ π π’πππ π‘π ππ πππ’ππ. 1 1 π1 + π2 ) πΌπ π‘βππ πππ π, π‘βπ πππππππ ππ πππππππ, πΎ = π1 + π2 β 2 (π1 β 1)π1 2 + (π2 β 1)π2 2 2 πππ π‘βπ ππππππ π£πππππππ, ππ = π1 + π2 β 2 βππ 2 ( (πππ) π‘ = π₯Μ 1 β π₯Μ 1 β π0 2 π 2 π 1 π1 + π2 β ππ π1 2 πππ π2 2 πππ π’πππππ€π πππ πππ‘ πππ’ππ. πΌπ π‘βππ πππ π, π‘βπ πππππππ ππ ππππππ ππ πππ£ππ ππ¦; πΎ = 2 π 2 π 2 ( π1 + π2 ) 1 2 2 π ( π1 ) 1 2 2 π 2 ( π2 ) 2 π1 β 1 + π2 β 1 For paired observations, the mean difference is tested using a t- test with the test value given by; π‘= Μ β π0 π· π€ππ‘β π‘βπ πππππππ ππ πππππππ πππ£ππ ππ¦ πΎ = π β 1. ππ βπ Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 14 Example 8.5.1 1. An experiment was performed to compare the abrasive wear of two laminated materials. 12 pieces of material 1 gave an average (coded) wear of 85 units with a standard deviation of 4 while 10 pieces of material 2 gave an average of 81 and a standard deviation of 5. Can we conclude that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units? Assume that the populations are approximately normal with equal variances. 2. Business schools A and B reported the following summary of GMAT (Graduate Management Apptitude Test) verbal scores. School A B Sample size(n) 201 115 Sample mean 34.75 33.74 Sample variance 48.59 30.68 At 5% level of significance, is there sufficient evidence to believe that there is a difference in the GMAT scores of the two schools? Calculate the p β value associated with this test. Working 1. Let π1 πππ π2 ππ ππππ’πππ‘πππ πππππ πππ πππ‘ππππππ 1 πππ 2 πππ ππππ‘ππ£πππ¦. Then π»0 : π1 β π2 = 2 πππ π»1 : π1 β π1 > 2 Since population variances are unknown but assumed to be equal, then π‘= π₯Μ 1 β π₯Μ 1 β π0 βππ 2 1 1 (π + π ) 1 2 π€βπππ ππ οπ‘ = 2 (π1 β 1)π1 2 + (π2 β 1)π2 2 = = 20.05 π1 + π2 β 2 85 β 81 β 2 β20.05 ( 1 + 1 ) 12 10 = 1.043 π‘0.05,20 = 1.725 ο |π‘| < π‘0.05,20 Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 15 In this case, we fail to reject Ho at 5% level of significance and conclude that there is no sufficient evidence to say that the abrasive wear of material 1 exceeds that of material 2 by more than two units. π β π£πππ’π = 2π(π > 1.42) = 2(0.0778) = π. ππππ 2. Let π1 πππ π2 ππ ππππ’πππ‘πππ ππππ π πππππ πππ π πβππππ π΄ πππ π΅ πππ ππππ‘ππ£πππ¦. Then π»0 : π1 β π2 = 0 πππ π»1 : π1 β π1 β 0 In this case, both sample sizes are greater than 30, so we use; π= π₯Μ 1 β π₯Μ 1 β π0 = 1.42 πππ ππΌ/2 = 1.96 π 2 π 2 β 1 + π1 π2 πππππ |π| < ππΌ/2 we fail to reject Ho at 5% level of significance and conclude that there is no sufficient evidence to believe that the two schools performed differently. 8.5.2. Tests about two population proportions Tests about two population proportions are based on the sampling distribution of the of π1 β π2 . 1. For π»0 : π1 β π2 = 0 ππππππ π‘ π»1 : π1 β π2 β 0, or π1 β π2 > 0 or π1 β π2 < 0, the following test statistic is used. π= πΜ1 β πΜ 2 β 0 1 1 π1 + π2 ) π€βπππ π = βππ ( π₯1 + π₯2 π1 + π2 2. For π»0 : π1 β π2 = π ππππππ π‘ π»1 : π1 β π2 β π, or π1 β π2 > π or π1 β π2 < π, the following test statistic is used. π= πΜ1 β πΜ 2 β π β πΜ1 πΜ1 πΜ2 πΜ2 π1 + π2 Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 16 Example 8.5.2 A poll is taken to compare the proportion of town and county voters favoring the proposal of constructing a chemical plant. Is the proportion of town voters favoring the proposal higher than the proportion of county voters if 120 of 200 town voters and 240 of 500 county voters favour the proposal? Use 2.5% level of significance. Working π»0 : π1 β π2 = 0 (π. π. π1 = π2 )πππ π»0 : π1 β π2 > 0(π. π. π1 > π2 ) In this case; πΜ1 β πΜ2 β 0 π= βππ ( 1 1 + π1 π2 ) πππ πΜ 2 = π€βπππ π = π₯1 + π₯2 120 + 240 120 = = 0.51, πΜ1 = = 0.6 π1 + π2 200 + 500 200 240 = 0.48 ο π = 500 0.6 β 0.48 β 0 β(0.51)(0.49) ( 1 + 1 ) 200 500 = 2.87 π0.025 = 1.96 ο |π| > ππΌ and so, we reject Ho. Rejecting Ho at 2.5% level of significance implies that there is sufficient sample evidence to support the claim that the proportion of town voters favoring the proposal higher than the proportion of county voters. 8.5.3. Tests about the difference in two population variances To test π»0 : π1 2 = π2 2 ππππππ π‘ π»1 : π1 2 β π2 2 or π1 2 > π2 2 or π1 2 < π2 2 use πΉ = π1 2 π2 2 and reject Ho if πΉ > ππΌ (π1 β 1, π2 β 1) for one β tailed test and πΉ > ππΌ/2 (π1 β 1, π2 β 1) for a two- tailed test. Example 8.5.3 Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 17 In the abrasive wear example, we assumed that the two unknown population variances were equal. Were we justified in making that assumption? Use 0.10 level of significance. Material 1: π1 = 12, π₯Μ 1 = 85 πππ π 1 = 4 Material 2: π2 = 10, π₯Μ 2 = 81 πππ π 2 = 5 π»0 : π1 2 = π2 2 ππππππ π‘ π»1 : π1 2 β π2 2 In this case; πΉ= π1 2 π2 2 16 = 25 = 0.64 πππ π0.05 (11, 9) = 3.14+3.07 2 = 3.105 Since πΉ < π0.05 (11, 9) we fail to reject Ho and conclude that the assumption of equal unknown population variances is justified. Activity 1. The average life of 6 car batteries is 30 months with standard deviation of 4 months. The manufacturer claims that an average life is 3 years for his batteries and a customer claims that the manufacturer is exaggerating. If you were in a position of a customer, would you believe the manufacturerβs claim? Test the claim at 5% level of significance. 2. A test was given to a large group of boys who scored on average 64.5. The same test was given to a group of 400 boys who scored on average 62.5 with standard deviation of 12.5. Examine if the difference is significant at 5% level of significance. 3. A poultry farmer is investigating ways of improving the profitability of his operation. Using a standard diet turkeys grow to a mean mass of 4.5kg at age 4 months. A sample of 20 turkeys which were given a special enriched diet had an average mass of 4.8kg after 4 months. The sample standard deviation was 0.5kg. Using 5% level of significance, test whether the new diet is effectively increasing the mass of the turkeys. 4. Aircrew escape systems are powered by a solid propellant. The burning rate of this propellant is an important product characteristic. Specifications require that the mean burning rate must be 50 centimeters per second. We know that the standard deviation of burning rate is π = 2 centimeters per second. The experimenter decides to specify Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 18 a type I error probability of 0.05 and selects a random sample of 25 and obtains a sample average burning rate of 51.3 centimeters per second. What conclusions should be drawn? 5. The mean water temperature downstream from a power plant cooling tower discharge pipe should be no more than 100°F. Past experience has indicated that the standard deviation of temperature is 2°F. The water temperature is measured on nine randomly chosen days, and the average temperature is found to be 98°F. (a) Should the water temperature be judged acceptable with 5% level of significance? (b) What is the P-value for this test? 6. The means of two large samples of sizes 2000 and 1000 are 68.0 and 67.5 respectively. Can the two samples be regarded as drawn from the same population of standard deviation of 2.25. 7. To study the effect of a special study programme, 14 students were selected and paired according to IQ and scholastic performance. One student from each pair was randomly selected to participate in the special programme, while the other student participated in the standard programme. Shortly thereafter, the students took the national exam and obtained the following scores: Special programme 66 82 96 72 78 82 67 Standard programme 60 79 92 73 75 80 69 Is there any difference in the mean scores under the two programmes? Why? 8. An administrator at a large university stated that there was a difference in the mean grade point average of graduating males and females. A random sample of 45 graduating males gave a mean grade point average of 2.10 and a variance of 0.64, while a random sample of 50 graduating females gave a mean grade point average of 2.45 and a variance of 0.70. By constructing a 95% confidence interval or testing the hypothesis at 5% level of significance, would you conclude that the data support the administratorβs belief? 9. Two varieties of maize are being tested in a developing country. 12 test plots are given identical treatment. Six plots are sown in variety 1 and the other six plots in variety 2. In an experiment in which the crop scientist hope to determine whether there is a significant difference between the yields using 5% level of significance. The results were: Variety 1: Variety 2: 1.5 1.6 1.9 1.8 1.2 2.0 1.4 1.8 2.3 1.3 2.3 Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 19 One of the plots planted with variety 2 was accidentally given extra dossal of fertilizer so the result was discarded. Would you conclude that there was a significant difference in the yield between the two varieties of maize? Assume equal population variances. 10. A coin is tossed 256 times and 132 heads are observed. Is there sufficient evidence to conclude that the coin is biased? 11. Twenty people were affected by cholera and out of them only 18 survived. Would you reject the hypothesis that the survival rate if affected by cholera is 85% in favour of the hypothesis that it is more at 5% level of significance? 12. A manufacturing company claims that at least 95% of its products supplied conforms to the specification. Out of a sample of 200 members, 18 are found to be defective. Test the claim at 5% level of significance. 13. A certain geneticist is interested in the proportion of males and females in the population that have a certain minor blood disorder. In a random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appeared to have a disorder. Is there sufficient evidence to conclude that the proportion of females with a minor blood disorder was higher than that of males? Test the hypothesis at 1% level of significance 14. In a random sample of 1000 persons from city A, 400 are found to be consumers of wheat. In another sample of 800 persons from city B, 400 are found to be consumers of wheat. Do these data reveal a significant difference in the proportion of consumers of wheat between the two cities? Lecture Notes on Statistical Hypothesis Testing/Compiled by Angel Mukuka (PhD) 20