Additional Information and Full Hypothesis Test Examples Quiz

PDF page 501: • If no level of significance is given, a common standard to use is α = 0.05.

PDF page 501: • When you calculate the p-value and draw the picture, the p-value is the area in the left tail, the right tail, or split

PDF page 501: evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.

PDF page 501: • In a hypothesis test problem, you may see words such as "the level of significance is 1 percent". The "1 percent" is

PDF page 501: the preconceived or preset α.

PDF page 501: • The alternative hypothesis,

PDF page 501: , tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.

PDF page 503: 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: = the mean time to swim the 25-yard freestyle. Distribution for the test: is normal (population standard deviation is known: σ = 0.8) with mean and standard error of ;

PDF page 503: μ = 16.43 comes from H0 and not the data. σ = 0.8, and n = 15.

PDF page 504: μ = 16.43 comes from H0. Our assumption is μ = 16.43.

PDF page 503: p-value = P( < 16) = 0.0187 where the sample mean in the problem is given as 16.

PDF page 504: p-value = 0.0187. The p-value is the area to the left of the sample mean given as 16. Graph:

PDF page 504: Interpretation of the p-value: If H0 is true, there is a 0.0187 probability (1.87 percent), that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87 percent chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event. Compare α and the p-value: α = 0.05 p-value = 0.0187 α > p-value Make a decision: Since α > p-value, reject H0. An alternative approach is to find the z-test corresponding to the sample mean This is .

PDF page 504: 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 506: Problem 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 percent level of significance to determine if the bench press mean is more than 275 pounds. Solution Set up the hypothesis test: Since the problem is about a mean weight, this is a test of a single population mean. This is a right-tailed test.

PDF page 506: Ha: μ > 275 Calculating the distribution needed: Random variable: = the mean weight, in pounds, lifted by the football players. Distribution for the test: It is normal because σ is known.

PDF page 507: Make a decision: Since α < p-value, do not reject H0. Conclusion: At the 2.5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds. The p-value can easily be calculated.

PDF page 508: Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given. You are only given n = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a Student's t-distribution. Use t-distribution. Therefore, the distribution for the test is t with nine degrees of freedom. Calculate the p-value using the Student's t-distribution: First, we compute the sample mean as