The construction of a hypothesis test for a population parameter can be stated as follows. There is a variable X associated with a given population, and there is a hypothesis about a certain parameter θ of that population. For example, we state that the true value of θ is 20. A random sample of elements from this population is collected, and with it we wish to verify whether or not this hypothesis holds.As we have seen previously, we begin our analysis by clearly stating the hypothesis we are putting to the test, and we call it the null hypothesis, writing:\[ H_0 : θ = 20 \]Next, it is also useful to state the hypothesis that will be considered acceptable if H0 is rejected. This hypothesis is called the alternative hypothesis, and its statistical characterization will depend on the degree of knowledge we have about the problem being studied. The most general alternative would be:\[ H_1 : θ ≠ 20 \]We could also have alternatives of the form:\[ H_1: θ < 20 \text{or} H_1: θ > 20 \]depending on the information provided by the problem.Whatever decision is made, we have seen that we are subject to making errors. To simplify the terminology, we introduce the definitions:Type I error: rejecting the null hypothesis when it is true.\[ P(\text{type I error})=P(\text{reject H_0|H_0 is true}) \]Type II error: not rejecting H0 when H0 is false.\[ P(\text{type II error})=P(\text{not reject H_0|H_0 is false}) \]The objective of the hypothesis test is to say whether hypothesis H0 is acceptable or not. Operationally, this decision is made by considering a critical region. If the observed value of the statistic belongs to this region, we reject H0; otherwise, we do not reject H0. This region is constructed so that the probability of rejecting H0 when it is true (type I error) is equal to a probability defined a priori. CR is called the critical region or rejection region of the test. An important fact to emphasize is that the critical region is always constructed under the assumption that H0 is true. Determining type II error is more difficult, since we usually do not specify fixed values for the parameter under the alternative hypothesis.The probability of committing a type I error (or error of the first kind) is an arbitrary value and is called the significance level of the test. The sample result is more significant for rejecting H0 the smaller this level is. That is, the smaller the significance level, the lower the probability of obtaining a sample with a statistic belonging to the critical region, making it unlikely to obtain a sample from the population for which H0 is true and have the test reject it. Usually, the value of α is set at 5%, 1%, or 0.1%.Steps for Constructing a Hypothesis TestBelow we provide a sequence that can be used systematically for any hypothesis test.Step 1. Set which hypothesis H0 is to be tested and which is the alternative hypothesis H1.Step 2. Use statistical theory and the available information to decide which statistic (estimator) will be used to test hypothesis H0. Obtain the properties of this statistic (distribution, mean, standard deviation).Step 3. Set the probability α of committing a type I error and use this value to construct the critical region (decision rule). Remember that this region is constructed for the statistic defined in step 2, using the parameter values hypothesized by H0.Step 4. Use the sample observations to calculate the value of the test statistic.Step 5. If the value of the statistic calculated from the sample data does not belong to the critical region, do not reject H0; otherwise, reject H0.Whenever we carry out hypothesis tests, we will try to clearly distinguish these five steps. Finally, a comment about H0 and type I error. We should take as H0 the hypothesis which, if rejected, would lead to a type I error that is more important to avoid.Let us look at an example due to Neyman (1978). Suppose an experiment is conducted to determine whether or not a product A is carcinogenic. After the test is performed, we may conclude: (i) A is carcinogenic or (ii) A is not carcinogenic. Each of these conclusions may be wrong, and we have the two types of error already mentioned, depending on which hypothesis is H0. From the product user’s point of view, the hypothesis to be tested should be:\[ H_0: \text{A is carcinogenic} \]because the probability of error in rejecting this hypothesis, if it is true, must be very small.Number of TailsThere are 3 types of tests:One-tailed (left-tailed)\[ H_0:μ=μ0 \]\[ H_1:μ<μ0\]Reject \( Z{calc} < -Z\alpha \)One-tailed (right-tailed)\[ H_0:μ=μ0\]\[ H_1:μ>μ0\]Reject \(Z{calc} > Z\alpha \)Two-tailed\[ H_0:μ=μ0\]\[ H_1:μ≠μ0\]Reject \(Z{calc}>Z{\frac {\alpha }{2}} \)Tests on the Mean of a Population with Known VarianceLet us now look at an application of the five steps defined in the previous section to test the hypothesis that the mean of a population μ is equal to a fixed number μ0, assuming the variance σ of this population is known.ExamplesAn automatic machine for filling coffee packages fills them according to a normal distribution, with mean μ and variance always equal to 500 grams and 400 grams, respectively. The machine was set to μ = 500 g. We periodically wish to collect a sample of 16 packages and check whether production is under control, that is, whether μ = 500 g or not. If one of these samples had a mean of 492 g, would you stop production to adjust the machine at a 1% significance level?Step 1. The hypotheses of interest to us are:\[ H0:μ=500g\]\[H1:μ≠500g \]because the machine may become misadjusted upward or downward.Step 2. The population mean is 500 g and its standard deviation is equal to:\[\sqrt{\frac{400}{16}}=\sqrt{25}=5 \]Therefore, the standard deviation is equal to 5.Step 3. based on the alternative hypothesis, we see that H0 must be rejected when x is very small or very large (we say that we have a two-tailed test). We will thus have a 0.5% probability on each side:The first point of the CR will be:\[ z1=−2.58 => x1 = 500−5 * z1=>x1=487.1 \]The second:\[ z2=2.58=>x2=500 + 5 * z2=>x2=512.9 \]It is therefore known that the critical region will be: {x belongs to CR | x < 487.1 or x > 512.9 }Step 4. The relevant information from the sample is its mean, which in this particular case is x = 492.Step 5. Since x does not belong to the critical region, our conclusion will be not to reject H0. That is, the deviation of the sample mean from the mean proposed by H0 can be considered as due only to the random selection of the packages.In the ceramics industry, the strength of samples of ceramic bodies is systematically evaluated after the firing process. From these evaluations, it is known that a certain type of body has approximately normal mechanical strength, with a mean of 53 MPa and variance of 16 MPa². After changing some raw material suppliers, the goal is to verify whether there was a change in quality. A sample of 15 ceramic body specimens showed a mean equal to 50 MPa. What is the conclusion at the 5% significance level?Step 1. We will perform a two-tailed test, because before the sample was observed, we did not know whether the strength should be greater or less than 53. Thus:\[H_0:μ=53\]\[H_1:μ≠53 \]Step 2. The population mean is 53 and its standard deviation will be 4 divided by the square root of 15.Step 3. based on the alternative hypothesis, we see that H0 must be rejected when x is very small or very large (we say that we have a two-tailed test). We will thus have a 2.5% probability on each side:What is the probability that the sample mean is 50?\[z=\frac{50−53}{\frac{4}{\sqrt{15}}}=−2.9 \]With z of -2.9, its p-value is 0.79%, which is a lower probability of occurring than 2.5%.Therefore, because the probability of a sample with this mean occurring is very small, we can reject H0 that the strength of the sample set is similar to that of the population.Would you have surgery with this doctor?Exercises1) A cigarette company announces that the average nicotine level of the cigarettes it manufactures is at the recommended level of 23 mg per cigarette. A laboratory performs six analyses of this level, obtaining: 27, 24, 21, 25, 26, 22. It is known that the nicotine level is normally distributed, with variance equal to 4 mg2. Can the manufacturer’s claim be accepted at the 10% level?2) The consumer of a certain product accused the manufacturer, saying that more than 20% of the units manufactured are defective. To confirm his accusation, he used a sample of size 50, where 27% of the parts were defective. Show how the manufacturer could refute the accusation. Use a significance level of 10%. (standard deviation = square root of (p * (1 - p))3) A manufacturer guarantees that 90% of the equipment it supplies to a factory complies with the required specifications. Examination of a sample of 200 pieces of this equipment revealed 25 defective ones. Test the manufacturer’s claim at the 5% and 1% levels. (standard deviation = square root of (p * (1 - p))4) The tensile strength of stainless steel produced at a certain plant remained stable, with an average strength of 72 kg/mm2 and a standard deviation of 2.0 kg/mm2 with normal distribution. Recently the machine was adjusted. In order to determine the effect of the adjustment, 10 samples were tested. The tests showed an average strength of 75 kg/mm2. Consider that the standard deviation did not change. Based on these data, answer: with a significance level of 5%, is it possible to state that the mean value has not changed?5) The DeBug Company sells an insect repellent that claims to be effective for a minimum of 400 hours. An analysis of nine randomly chosen items showed an average effectiveness of 380 hours. Test the company’s claim against the alternative that the duration is less than 400 hours, at the 1% significance level, if the population standard deviation is 60 hours (consider normal distribution).6) A buyer, upon receiving a large batch of parts from a supplier, decided to inspect 200 of them. He also decided that the batch will be rejected if he is convinced, at the 5% significance level, that the proportion of defective parts in the batch is greater than 4%. What will his decision be (accept or reject the batch) if eleven defective parts were found in the sample? (standard deviation = square root of (p * (1 - p))7) A buyer, upon receiving a large batch of parts from a supplier, decided to inspect 200 of them. He also decided that the batch will be rejected if he is convinced, at the 5% significance level, that the proportion of defective parts in the batch is greater than 4%. What will his decision be (accept or reject the batch) if 14 defective parts were found in the sample? (standard deviation = square root of (p * (1 - p))Answer Key1) We can reject the null hypothesis.2) In order not to reject the client’s hypothesis, he should have observed a quantity of at least 27.24% defective items.3) In no situation can the null hypothesis be rejected.4) Ho is rejected, that is, it is not possible to state so.5) It is not possible to reject the null hypothesis.6) It is not possible to reject the null hypothesis.7) The null hypothesis can be rejected.
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