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MATH 2400. Ch. 15 Notes. Two Types of Statistical Inference. Confidence Intervals Used when your goal is to estimate a population parameter Tests of Significance Used to assess the evidence provided by data about some claim concerning a population. Example of a Test of Significance.
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MATH 2400 Ch. 15 Notes
Two Types of Statistical Inference • Confidence Intervals • Used when your goal is to estimate a population parameter • Tests of Significance • Used to assess the evidence provided by data about some claim concerning a population
Example of a Test of Significance Jordan claims that he makes 75% of his free throws. You tell him to “prove it.” He takes 20 shots from the free throw line and makes 8 of the 20. You conclude that he was lying. What is the probability that he was telling the truth and still makes 8 out the 20 free throws?
Test of Significance, the Basics • Based on asking what would happen if we repeated the sample or experiment many times. • Assume perfectly random SRS from an exactly Normal population. • Assume we know σ.
Example 1 Artificial sweeteners lose their sweetness over time. In a study, Trained testers sipped cola along with drinks of standard sweetness and scored the cola on a “sweetness score” of 1 to 10. The cola was stored for a month at high temperature to simulate storing for 4 months at room temperature. Each taster scored the cola again. This is a matched pairs experiment. Our data represents the difference (score before – score after) in the tasters’ scores. The larger the score, the larger the loss of sweetness.
Example 1 continued… 2.0 0.4 0.7 2.0 -0.4 2.2 -1.3 1.2 1.1 2.3 Most are positive, that is, most tasters found a loss of sweetness. But, the losses are small. Two tasters (the negative scores) though the cola gained sweetness. The average sweetness los is given by the sample mean x̄=1.02. Is this data good evidence that the cola lost sweetness in storage?
Example 1 continued… The reasoning is the same as the free throw example. We make a claim and ask if the data gives evidence against it. We seek evidence that there is a sweetness loss, so the claim we test is that there is nota loss. In that case, μ=0. We can calculate the standard deviation…
Example 1 continued… Let’s consider… • Our case where x̄ = 1.02 and • Another case where some cola already on the market was sampled, and had a mean loss of x̄ = 0.3.
Example 1 continued… For the situation, we assume x̄ = 0 and we just calculated the margin of error to be .316, so Our case: x̄ = 1.02 Another case: x̄ = 0.3 x̄ margin of error = 0 .316 Since Our case, falls well outside of this range, we can say that the evidence shows that the cola did lose sweetness. Since the other case falls within this range, we can not say that the cola lost sweetness.
Stating Hypotheses We start with a careful statement of the claims we want to compare. We look for evidence against a claim, so we start with the claim we seek evidence against, such as “no loss of sweetness.”
Null & Alternate Hypotheses H0 will represent the Null Hypothesis Ha will represent the Alternate Hypothesis Hypotheses always refer to a population, not to a particular outcome. Be sure H0 and Ha are in terms of population parameters. For Example 1: H0: μ = 0 Ha: μ > 0
Example 2 Does the job satisfaction of assembly workers differ when their work is machine paced rather than self-paced. Assign workers either to an assembly line moving at a fixed pace, or to a self-paced setting. All subjects work in both settings, in random order. This is a matched pairs design. After two weeks, the workers take a test of job satisfaction. The response variable is the difference in satisfaction scores, self-paced minus machine-paced.
Example 2 continued. So, we are trying to determine if there is a difference. So, our Null Hypothesis will be that there is no difference. H0: μ = 0 Ha: μ≠ 0 The Hypotheses should be expressed before looking at the data. It is tempting to look at the data and frame the hypotheses to fit what the data shows. The data in this example showed that workers were more satisfied with self-paced work, which should not influence Ha.
Example 3 State the Null and Alternate Hypotheses
Example 4 State the Null and Alternate Hypotheses
Explaining the Null Hypothesis Writing the Null Hypothesis in a way in which we want to find evidence against may seem weird at first. Think about it like a criminal trial. The defendant is “innocent until proven guilty.” That is, the Null is innocent and the prosecution must try to provide convincing evidence against this hypothesis. We are trying to prove the Null false.
P-Value and Statistical Significance Small P-values are evidence against H0 because they say that the observed result would be unlikely to occur if H0 was true. Large P-values fail to give evidence against H0.
Example 1 Revisited The study of sweetness lost tests the hypotheses H0: μ = 0 Ha: μ > 1 Because Ha states that μ > 1, values of x̄ greater than 0 favor Ha over H0. To calculate the p-value… p = P(x̄ > 0.3) = .1711 (by looking at z-table) This is not strong evidence against H0.
Example 1 Revisited… However, for our data which gave x̄ = 1.02… p = P(x̄ > 0.3) = .0006 (by looking at z-table) This is strong evidence against H0and in favor of Ha.
Example 2 Revisited… Consider the job satisfaction example… H0: μ = 0 Ha: μ ≠ 0 Suppose we know that the job satisfaction scores follow the Normal Distribution with σ=60. Data from 18 workers give x̄ = 17. Which means they prefer self-paced on the average. Because the alternative is two-sided. The p-value is the probability of getting an x̄ at least as far from μ=0 in either direction as the observed x̄ = 17.
Example 2 Revisited… So, with σ = 60, n = 18, we get Which means, and, by looking at the z-table… p = 2P(-17< x̄ < 17) = 2(.1151) = .2302. This is not strong evidence against H0.
P-Values… We have said that a p-value of… • 0.0006 was strong evidence against H0. • 0.1711 and .2302 were not strong evidence against H0. There is “no rule” for how small a p-value has to be to reject H0, it’s a matter of judgment.
P-Values However, there are some fixed values that are in common use as standards for evidence against H0. p = 0.05 means that the probability of this happening is 5% when repeated many times p = 0.01 means that the probability of this happening is 1% when repeated many times These are called significance levels.
z Test for a Population Mean Draw an SRS of size n from a Normal population that has unknown mean μ and known standard deviation σ. To test the null hypothesis that μhas a specified value, H0: μ = μ0 Calculate the one-sample z test statistic
Example 6 The systolic blood pressure for males 35 to 44 years of age has mean 128 and st. dev. 15. A large company looks at the medical records of 72 randomly chosen workers in this age group and finds that the mean systolic blood pressure is x̄ = 126.07. Is this evidence that the company’s executives have a different mean systolic blood pressure from the general population?
Example 6 continued… The Null Hypothesis will be no difference and our Alternate Hypothesis will be two-sided because x̄ could be above or below μ. H0: μ = 128 Ha: μ≠ 128 To find the p-value, look for z < -1.09. Our table shows P(z < -1.09) = 0.1379. So, p = 2P(z < -1.09) = (2)(0.1379) = 0.2758 This is not strong evidence that these workers’ blood pressures differ from the rest of the population.
Before Doing a Z Test… • Verify SRS • Normal Distribution • Know σ
Example 7 Here are 6 measurements of the electrical conductivity of an iron rod: 10.08 9.89 10.05 10.16 10.21 10.11 The iron rod is supposed to have conductivity 10.1. Do the measurements give good evidence that the true conductivity is not 10.1? The 6 measurements are an SRS from a population with a Normal distribution with σ = 0.1.
Ti-84 Z Tests STAT, TESTS tab, 1: Z-Test… This is what our screen should look like for Example 1. After hitting “Calculate”
Example 1 Information (1 sided) Set 1Set 2 μ0 = 0 μ0 = 0 σ = 1 σ = 1 x̄ = 1.02 x̄ = 0.3 n = 10 n = 10 p = p =
Example 2 Information (2 sided) μ0 = 0 σ = 60 x̄ = 17 n = 18 p =
Example 6 Information (2 sided) μ0 = 128 σ = 15 x̄ = 126.07 n = 72 p =
Example 7 Information (2 sided) μ0 = 10.1 σ = 0.1 x̄ = 10.083 n = 6 p =