Additional Information and Full Hypothesis Test Examples Quiz

PDF page 522: • If no level of significance is given, a common standard to use is α = 0.05.

PDF page 522: • The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.

PDF page 522: • The alternative hypothesis, Ha , tells you if the test is left, right, or two-tailed. It is the key to conducting the

PDF page 522: • Ha never has a symbol that contains an equal sign.

PDF page 522: Ho: μ = 5, Ha: μ < 5

PDF page 522: Test of a single population mean. Ha tells you the test is left-tailed. The picture of the p-value is as follows:

PDF page 524: Ha: μ < 16.43 For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed. Determine the distribution needed: ̄ Random variable: X = the mean time to swim the 25-yard freestyle. ̄ Distribution for the test: X is normal (population standard deviation is known: σ = 0.8) ̄ ⎛

PDF page 524: Calculate the p-value using the normal distribution for a mean: ̄ p-value = P( x < 16) = 0.0187 where the sample mean in the problem is given as 16.

PDF page 525: p-value = 0.0187 (This is called the actual level of significance.) The p-value is the area to the left of the sample mean is given as 16. Graph:

PDF page 525: Make a decision: Since α > p-value, reject H0. This means that you reject μ = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles. Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. The p-value can easily be calculated.

PDF page 524: μ = 16.43 comes from H0 and not the data. σ = 0.8, and n = 15.

PDF page 525: μ = 16.43 comes from H0. Our assumption is μ = 16.43.

PDF page 525: Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 16.43 for μ0 (null hypothesis), .8 for σ, 16 for the sample mean, and 15 for n. Arrow down to μ : (alternate hypothesis) and arrow over to < μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0187) but it also calculates the test statistic (z-score) for the sample mean. μ < 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to Draw(instead of Calculate). Press ENTER. A shaded graph appears with z = -2.08 (test statistic) and p = 0.0187 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.

PDF page 526: HISTORICAL NOTE (EXAMPLE 9.11) The traditional way to compare the two probabilities, α and the p-value, is to compare the critical value (z-score from α) to the test statistic (z-score from data). The calculated test statistic for the p-value is –2.08. (From the Central Limit ̄ x − μ ̄

PDF page 527: A college football coach records the mean weight that his players can bench press as 275 pounds, with a standard deviation of 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1). Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds.

PDF page 527: p-value = P( x > 286.2) = 0.1323 .