Defining the null and alternative hypotheses For the purposes of this class, the null hypothesis represents the status quo and will always be of the form Ho: μ = μo The choice of the alternative hypothesis depends on the purpose of the hypothesis test. 11/29/2018

Defining the null and alternative hypotheses If the primary concern is deciding whether the population mean is different from some specified value, then the alternative takes the form: Ha : μ ≠ μo If the primary concern is deciding whether the population mean is less than some specified value, then the alternative takes the form: Ha : μ < μo 11/29/2018

Defining the null and alternative hypotheses If the primary concern is deciding whether the population mean is greater than some specified value, then the alternative takes the form: Ha : μ > μo A hypothesis test is called two-tailed if the alternative hypothesis takes the form “different from” and is called one-tailed if the alternative hypothesis is of the form “greater than” or “less than”. 11/29/2018

Making a decision How do we decide whether or not to reject the null hypothesis in favor of the alternative hypothesis? Take a random sample from the population If the sample data are consistent with the null hypothesis, then do not reject the null hypothesis If the sample data are inconsistent with the null hypothesis, then reject the null hypothesis and conclude that the alternative hypothesis is true. 11/29/2018

Hypothesis Testing Procedure Set up the null and alternative hypotheses. Define the test statistic Define the rejection region based on the significance level Calculate the value of the test statistic and determine whether or not to reject the null hypothesis State your conclusion in terms of the original problem 11/29/2018

Define the test statistic For tests regarding the mean of a population when the population standard deviation, σ, is known, the test statistic is given by 𝑥 − 𝜇 𝑜 𝜎/ 𝑛 where 𝜇 𝑜 is the value specified by the null hypothesis. Note that the test statistic will have a standard normal distribution if the null hypothesis is true. 11/29/2018

Define the rejection region The rejection region is the set of values for the test statistic that leads to rejection of the null hypothesis. The rejection region will depend on (1) the requested significance level and (2) the form of the alternative hypothesis (one-tailed or two-tailed). 11/29/2018

Type I and Type II Errors Null is true Null is false Decision: Do not reject the null Correct decision Incorrect decision Type II Error Reject the null Type I Error We want to guard against making a type I error. That is, we want to insure that the probability of making a type I error is small. The significance level (α) of a hypothesis test is the probability of making a type I error. 11/29/2018

Defining the Critical Values Suppose that a hypothesis test is to be performed at a specified significance level. Then, the critical value(s) must be chosen so that if the null hypothesis is true, the probability that the test statistic will fall into the rejection region is equal to α. (See illustrations) P 513 Weiss 𝑧 0.10 𝑧 0.05 𝑧 0.025 𝑧 0.01 𝑧 0.005 1.28 1.645 1.96 2.33 2.575 11/29/2018

Drawing conclusions Suppose that a hypothesis test is conducted using a small significance level. Then we can draw one of the two following conclusions: If the null hypothesis is rejected, we conclude that the alternative hypothesis is true. We can also say that the test results are “statistically significant” at the α level. If the null hypothesis is not rejected, we conclude that the data do not provide sufficient evidence to support the alternative hypothesis. We can also say that the test results are “not statistically significant” at the α level. 11/29/2018

Example A company that produces snack foods uses a machine to package 454-gram bags of pretzels. We will assume that the net weights are normally distributed and that the population standard deviation of all such weights is 7.8 grams. A random sample of 25 bags of pretzels has an average net weight of 450. At the 5% significance level, do the data provide sufficient evidence to conclude that the packaging machine is not working properly? 11/29/2018

Example The RR Bowker Company of New york collects information on the retail prices of books. In 1993, the mean retail price of all history books was $40.69. Based on a sample of 40 history books taken this year, the mean price for the sample is $44.12. The population standard deviation is known to be $7.61. At the 1% significance level, do the data provide sufficient evidence to conclude that this year’s mean retail price of all history books has increased over the 1993 mean of $40.69? p. 515 11/29/2018

Example According to the Food and Nutrition Board of the National Academy of Sciences, the recommended daily allowance of calcium for adults is 800mg. A random sample of 18 people with incomes below the poverty level yields an average daily calcium intake of 747.4mg. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean calcium intake of people with incomes below the poverty level is less than the recommended daily allowance of 800mg? Assume σ=188mg. Also do this example in Excel and discuss p-values 11/29/2018

P-values A p-value is the probability of obtaining a value more extreme as defined by the alternative hypothesis) than the value of the test statistic. The p-value can be used as follow to arrive at a conclusion for a hypothesis test: Reject the null hypothesis if p-value < α Fail to reject the null hypothesis if p-value ≥ α Show illustration 11/29/2018

Hypothesis test for the mean of a normal population (σ unknown) For tests regarding the mean of a normal population when the population standard deviation, σ, is NOT known, the test statistic is given by 𝑡= 𝑥 − 𝜇 𝑜 𝑠/ 𝑛 For a specified significance level, α, the critical value for the test is given by 𝑡 𝛼/2 with n-1 degrees of freedom for 2-tailed tests and 𝑡 𝛼 with n-1 degrees of freedom for 1-tailed tests Note that this test is referred to as the one-sample t-test. Show critical regions (p.550) 11/29/2018

Example According to the US Energy Information Administration, the mean energy expenditure of all US households in 1993 was $1282. That same year, 36 randomly selected households living in single family detached homes reported the energy expenditures shown in the attached Excel table. At the 5% significance level, do the data provide sufficient evidence to conclude that in 1993, households living in single family detached homes spent more for energy, on average, than the national average of $1282? Also do this example in Excel and discuss p-values 11/29/2018

Independent Samples (σ’s unknown) Suppose that X is a normally distributed random variable that is measured on each of two populations where the mean for the first population is 𝜇 1 and 𝜎 1 is not known and the mean of the second population is 𝜇 2 and and 𝜎 2 is unknown. Then, for independent samples of sizes 𝑛 1 and 𝑛 2 from the two populations: 11/29/2018

Test Statistic The test statistic is given by 𝑡= 𝑥 1 − 𝑥 2 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 t follows has approximately a t-distribution with degrees of freedom as defined by equation 9.7 in the book. 11/29/2018

Example Example 9.5 p 313 11/29/2018

Independent Samples (σ’s unknown, equal) Suppose that X is a random variable that is measured on each of two populations where the mean for the first population is 𝜇 1 and the mean of the second population is 𝜇 2 . Assume that 𝜎 1 = 𝜎 2 = σ, which is unknown. Then, for independent samples of sizes 𝑛 1 and 𝑛 2 from the two populations, the pooled standard deviation is given by 𝑠 𝑝 2 = 𝑛 1 −1 𝑠 1 2 +( 𝑛 2 −1) 𝑠 2 2 𝑛 1 + 𝑛 2 −2 11/29/2018

Test Statistic The test statistic is given by 𝑡= 𝑥 1 − 𝑥 2 𝑠 𝑝 2 𝑛 1 + 𝑠 𝑝 2 𝑛 2 t follows a t-distribution with ( 𝑛 1 + 𝑛 2 −2) degrees of freedom. 11/29/2018

When can we assume equal variances? Assume equal variances based on past experience Formal test to show equal variances If you can assume equal variances, you now have a test statistic with an exact distribution and you have a more powerful test (e.g., it is more likely that we will be able to reject the null hypothesis when in fact it is false) 11/29/2018

Example Suppose that you’re a financial analyst for Charles Schwab. You want to see if there is a difference in dividend yield between stocks listed on the NYSE and NASDAQ. Assuming that the variances are equal, conduct a hypothesis test at the 5% significance level. NYSE NASDAQ n 21 25 Sample mean 3.27 2.53 Sample std dev 1.30 1.16 11/29/2018

Example Test Statistic: Critical Value: 2.015 (interpolate) Decision: Reject the null hypothesis Conclusion: Conclude that the data provide significant evidence that there is a difference between the dividend yield from the NYSE and the NASDAQ 11/29/2018

Comparing Two or More Population Means Two different options: Independent samples – the sample selected from one population has no effect or bearing on the sample selected from the other population Dependent samples – there is some sort of pairing or matching between the observations from the two populations (e.g., before and after measurements) 11/29/2018

Paired Samples Suppose that we want to decide whether a newly developed gasoline additive increases gas mileage. Paired samples remove a source of extraneous variation which results in smaller sampling error and a more powerful test Pair consists of a car driven with the additive and then without the additive Analyze the difference between gas mileages for each car with and without the additive 11/29/2018

Mileage without additive Paired Samples Car Mileage with additive Mileage without additive Paired difference 1 25.7 24.9 0.8 2 20.0 18.8 1.2 3 28.4 27.7 0.7 4 13.7 13.0 5 17.8 1.0 6 12.5 11.3 7 27.8 0.6 8 8.1 8.2 -0.1 9 23.1 0.0 10 10.4 9.9 0.5 11/29/2018

Test Statistic The test statistic is the same as the test statistic for the 1-sample t-test, except that it is based on the sample differences 𝑡= 𝑑 − 𝐷 0 𝑠 𝑑 / 𝑛 t follows has a t-distribution with n-1 degrees of freedom 11/29/2018