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Chapter 10 – Hypothesis Testing. What is a hypothesis? A statement about a population that may or may not be true. What is hypothesis testing? A statistical test to prove or disprove a hypothesis. At the end of the test, either the hypothesis is rejected or not rejected.

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## Chapter 10 – Hypothesis Testing

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**Chapter 10 – Hypothesis Testing**• What is a hypothesis? A statement about a population that may or may not be true. • What is hypothesis testing?A statistical test to prove or disprove a hypothesis. At the end of the test, either the hypothesis is rejected or not rejected.**Steps in hypothesis testing**• There are 5 steps in hypothesis testing: • Step 1 – pattern of population distribution (normal) • Step 2 – Formulation of hypothesis Null Hypothesis (H0)Alternate Hypothesis (H1)**Steps in hypothesis testing**• Step 3 - Level of significance (α ) Alpha is the probability of rejecting a null hypothesis when it is true. Alpha is also known as the level of risk. It is described in terms of percent or decimals (5% or 0.05, 1% or 0.01, and so on.) • Step 4 - Test statisticIt is a quantity that is used to compare with critical Z from appendix D to determine if a null hypothesis is to be rejected or not.**Steps in hypothesis testing**• This quantity is the z-statistic _ Z = (x – m) / (a / Ön) . Step 5 - Decision Rule: It is a rule by which a null hypothesis is rejected or not rejected. It always takes the form: Reject H0 if Z-statistic is > Critical Z or < - Critical Z Critical Z-value is obtained from appendix D.**Steps in hypothesis testing**Decision: Reject H0 or Do not reject H0 Hypothesis tests can be one-tailed or two-tailed depending on how the alternate hypothesis is written.**Let’s take an example**• Suppose: H : μ = 100 0 H : μ > 100 1 This is a one tailed test. • Another example: H : μ = 100 0 H : μ < 100 1**An example of a two tailed test**• Suppose: H : μ = 100 0 H : μ ≠ 100 1 • This is a two tailed test. • Exercises from book: Problem 7(P age 239), Problem 8 (Page 240), Problem 9 (Page 240)**Exercises from Book**• Problem 7 (pg 283): • a. ONE-TAILED b. ONE-TAILED • c. ONE-TAILED d. ONE-TAILED • e. TWO-TAILED • Problem 8 (pg 283): • a. Ho : μ = 500, H1 : μ ≠ 500 • b. Ho : μ = 500, H1 : μ > 500 • c. Ho : μ = 500, H1 : μ < 500**Exercises from book**• Problem 9 (pg 283) • Ho : μ = 60; H1 : μ > 60 • Problem 10 (pg 283) • Ho : μ = 1000; H1 : μ ≠ 1000 • Ho : μ ≤ 1000; H1 : μ > 1000**Testing for population mean**• Such testing can be made under 2 situations: (1) When population standard deviation is known and (2) when population standard deviation is unknown. This chapter discusses the first.Refer to example problem 10-2 (page 241) • Two hypotheses are: H : μ = 10 0 H : μ < 10 1**Testing for population mean**• Alpha ( α ) is 5% (or 0.05). Critical Z from appendix D is –1.645 • Reject H0 if Z-statistic < - 1.645 Test statistic (Z-statistic) _ X – μ 8.8 - 10 Z = -------- = ---------- = - 2.0 σ / √ n 2.4 / 4**Testing for population mean**Decision: Reject H0 • What does it mean? The new manufacturing process reduces tar content of cigarettes. • Example problem 10-4.**Testing for population mean**H : μ = 1000 0 H : μ > 1000 1 α = 0.025 Critical Z = 1.96 (From Appendix D) _ X – μ 1050 - 1000 Test Statistic Z = --------- = -------------- = 2.5 σ / n 100 / 5 Decision: Reject H 0 What does it mean? The new manufacturing process increases life of fuses.**This is a two tailed test**Example Problem 10-7 (page 248) H : μ = 100 0 H : μ ≠ 100 1 Critical Z’s are – 2.575 and + 2.575 Decision Rule: Reject H0 if Z-statistic > + 2.575 or < - 2.575 _ X - 100 Z-statistic = -------------- = 1.667 Decision: Do not reject H √ (225 / 25) 0 IQs of the school district students are no different from the national average**Example problem 13**• Problem 13 (page 251)H : μ ≥ 30 0 H : μ < 30 1Critical Z = - 2.326Decision Rule: Reject H0 if Z-statistic ≤ - 2.326 27 - 30Z-Statistic = ------------- = - 5.0 6 / √ 100 Decision: Reject H0 • What does it mean? Vendor’s claim that mean weight of his chickens is at least 30 ounces is rejected.**Testing for difference between 2 means**• These tests are performed when one wants to compare 2 groups of populations. • For example, is these a difference between average GPAs of students of two universities? Or, are average monthly incomes the same between two groups of people? • The testing procedure is similar to what has been presented in the previous section.**Testing for difference between 2 means**• The null hypothesis is always of the form:H : μ = μ • 0 1 2 • The alternative hypothesis can be either of the 3 forms: • μ > μ or μ < μ or μ ≠ μ 1 2 1 2 1 2**Testing for difference between 2 means**Formula 10-1, p. 252 The decision rule is the same as before Example problem 10-8 (page 252-253) H : μ = μ H : μ ≠ μ 0 1 2 1 1 2**Testing for difference between 2 means**0.8 – 1.0 • Z- statistic = --------------------------- = - 1.0 √ (0.36/25) + (0.64/25) • At α = 0.05, critical Z = -1.96 and +1.96 Decision: Do not reject H 0 • Interpretation: The avg. nicotine contents of two brands of cigarettes are equal.**Problem 4, pg 255**• Is average hourly output of male workers less than that of female workers? • This is a test involving difference of means between two population groups. • Data are given in the table Ho: μ1 = μ2; H1: μ1 < μ2 (μ1 indicates output of males)**Z-statistic =**(150-153)/√[(70/36) + (74/36)] = -1.5 • At α = 0.05, critical Z is -1.645 • Decision: Do not reject Ho • Interpretation: Average outputs of male and female workers are the same.**Testing for population proportion**• In some cases, we are interested in population proportions rather than population means. For example, are more than 50% of CSULB students females? Or, is a new medicine more than 60% effective? • Two hypothesis are: H : p = a given value 0 H : p > a given value OR 1 p < a given value OR p ≠ a given value Example 10-11 (Page 257-258)**The test statistic**• Z-statistic = (X – n.p)/√[(n.p(1-p)] • The decision rule is the same as before.**Example problem 10-12 (pg 259)**• A TV program attracts 50% audience. A new anchorperson has been hired. Does the new hire increase audience level? Use 5% level of significance. • Ho: p = 0.5; H1: p > 0.5 • Given n = 100; critical Z = 1.645 • P = 55/100**Problem 10-12, cont.**• Z-statistic = (55-50)/√[(100)(0.5)(0.5)] = 1 • Decision: Do not reject Ho • Interpretation: • The audience level has remained the same (at 50%) after hiring the new anchorperson.**Problem 6, pg 261**• An auto manufacturer claims that 20% of customers prefer his products. In a sample of 100 customers, 15 indicate their preferences for his products. At 5% level of significance, can we support his claim? • Ho: p = 0.2; H1: p ≠ 0.2**Problem 6, cont.**• Z-statistic = (15-20)/√[(20)(0.08)] = -1.25 • Decision: Do not reject Ho. • Interpretation • The auto manufacturer’s claim cannot be disputed.**Hypothesis Testing**• Strength of rejection (p-value) • When we reject a null hypothesis, there is a quantity that describes the strength of rejection. This quantity is called the p-value. • The lower the p-value, the greater the strength of refection. In other words, the lower the p-value, the greater the strength in the rejection.**What is p-value?**• It is the area to right right of the calculated value of Z-statistic when the area of rejection is on the right side of the normal curve. When the area of rejection is on the left side of the normal curve, the p-value is the area to the left of the calculated value of the Z-statistic. If there are two area of rejection, add the area to the right of Z-statistic and the area to the left of Z-statistic.**How do we find the P-value?**• Let’s take an example problem: • Test if the mean waiting time is less than 3 minutes. • α = 0.05 • μ = 3, σ= 1, n=50, X=2.75 • Ho: μ = 3; H1: μ < 3 • Decision rule: Reject Ho if z-statistic < critical Z • Z-statistic = (2.75 - 3)/(1/√50) = -1.768 • Critical Z - -1.645 so we reject Ho. • Interpretation: Mean waiting time is < 3 minutes.**What is the p-value?**• Find the area to the left of -1.768. • This area is the p-value. • P-value = 1.0 – 0.9616 = 0.0384

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