Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Foundations of statistics wikipedia , lookup
Bootstrapping (statistics) wikipedia , lookup
Psychometrics wikipedia , lookup
Degrees of freedom (statistics) wikipedia , lookup
Taylor's law wikipedia , lookup
Misuse of statistics wikipedia , lookup
Regression toward the mean wikipedia , lookup
Section 7.3 Hypothesis Testing for the Mean (Small Samples) 1 Section 7.3 Objectives Find critical values in a t-distribution Use the t-test to test a mean μ Use technology to find P-values and use them with a t-test to test a mean μ 2 Finding Critical Values in a tDistribution 1. Identify the level of significance . 2. Identify the degrees of freedom d.f. = n – 1. 3. Find the critical value(s) using Table 5 in Appendix B in the row with n – 1 degrees of freedom. If the hypothesis test is a. left-tailed, use “One Tail, ” column with a negative sign, b. right-tailed, use “One Tail, ” column with a positive sign, c. two-tailed, use “Two Tails, ” column with a negative and a positive sign. 3 Example: Finding Critical Values for t Find the critical value t0 for a left-tailed test given = 0.05 and n = 21. Solution: • The degrees of freedom are d.f. = n – 1 = 21 – 1 = 20. • Look at α = 0.05 in the “One Tail, ” column. • Because the test is lefttailed, the critical value is negative. 4 0.05 -1.725 0 t Example: Finding Critical Values for t Find the critical values t0 and -t0 for a two-tailed test given = 0.05 and n = 26. Solution: • The degrees of freedom are d.f. = n – 1 = 26 – 1 = 25. • Look at α = 0.05 in the “Two Tail, ” column. • Because the test is twotailed, one critical value is negative and one is positive. 5 0.025 -2.060 0 0.025 2.060 t t-Test for a Mean μ (n < 30, Unknown) t-Test for a Mean A statistical test for a population mean. The t-test can be used when the population is normal or nearly normal, is unknown, and n < 30. The test statistic is the sample mean x The standardized test statistic is t. x t s n The degrees of freedom are d.f. = n – 1. 6 Using the t-Test for a Mean μ (Small Sample) In Words 1. State the claim mathematically and verbally. Identify the null and alternative hypotheses. In Symbols State H0 and Ha. 2. Specify the level of significance. Identify . 3. Identify the degrees of freedom and sketch the sampling distribution. d.f. = n – 1. 4. Determine any critical value(s). Use Table 5 in Appendix B. 7 Using the t-Test for a Mean μ (Small Sample) In Words 5. Determine any rejection region(s). 6. Find the standardized test statistic. 7. Make a decision to reject or fail to reject the null hypothesis. 8. Interpret the decision in the context of the original claim. 8 In Symbols x t s n If t is in the rejection region, reject H0. Otherwise, fail to reject H0. Example: Testing μ with a Small Sample A used car dealer says that the mean price of a 2005 Honda Pilot LX is at least $23,900.You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $23,000 and a standard deviation of $1113. Is there enough evidence to reject the dealer’s claim at α = 0.05? Assume the population is normally distributed. (Adapted from Kelley Blue Book) 9 Solution: Testing μ with a Small Sample • • • • • H0: μ ≥ $23,900 Ha: μ < $23,900 α = 0.05 df = 14 – 1 = 13 Rejection Region: • Test Statistic: t -3.026 10 s n 23, 000 23,900 1113 14 • Decision: Reject H0 0.05 -1.771 0 x t 3.026 At the 0.05 level of significance, there is enough evidence to reject the claim that the mean price of a 2005 Honda Pilot LX is at least $23,900 Example: Testing μ with a Small Sample An industrial company claims that the mean pH level of the water in a nearby river is 6.8.You randomly select 19 water samples and measure the pH of each. The sample mean and standard deviation are 6.7 and 0.24, respectively. Is there enough evidence to reject the company’s claim at α = 0.05? Assume the population is normally distributed. 11 Solution: Testing μ with a Small Sample • • • • • H0: μ = 6.8 Ha: μ ≠ 6.8 α = 0.05 df = 19 – 1 = 18 Rejection Region: 0.025 -2.101 0 -1.816 12 0.025 2.101 t • Test Statistic: t x s n 6.7 6.8 0.24 19 1.816 • Decision: Fail to reject H0 At the 0.05 level of significance, there is not enough evidence to reject the claim that the mean pH is 6.8. Example: Using P-values with t-Tests The American Automobile Association claims that the mean daily meal cost for a family of four traveling on vacation in Florida is $118. A random sample of 11 such families has a mean daily meal cost of $128 with a standard deviation of $20. Is there enough evidence to reject the claim at α = 0.10? Assume the population is normally distributed. (Adapted from American Automobile Association) 13 Solution: Using P-values with t-Tests • H0: μ = $118 • Ha: μ ≠ $118 TI-83/84set up: Calculate: Draw: • Decision: 0.1664 > 0.10 Fail to reject H0. At the 0.10 level of significance, there is not enough evidence to reject the claim that the mean daily meal cost for a family of four traveling on vacation in Florida is $118. 14 Section 7.3 Summary Found critical values in a t-distribution Used the t-test to test a mean μ Used technology to find P-values and used them with a t-test to test a mean μ 15