Chapter 9 Review 3.3.2017

Step 1: Decide Proportion or Mean Create Null and alternative hypotheses H0: Ha: Find P-hat Number of “successes” divided by sample size Create null and alternative hypotheses Find X-bar Find Sx Find Degrees of freedom Option 1: “By hand” Find t Tcdf to find p-value Reject or fail to reject null Option 2: Calculator T-test Report t and p-value Reject or fail to reject null Option 1: “by hand” Find Z Normalcdf to find p-value Reject or fail to reject null Option 2: Calculator 1-propZtest Report Z and p-value Reject or fail to reject null

Remember: Significance tests are only useful when we don’t know what the true value is in the population If we already know the population mean/proportion, then we can just say “yes, the null is correct” or “no, the null is wrong—here is what the true value is” Significance tests only useful when we have sample data and want to draw conclusions about the entire population

Example 223 237 244 232 242 225 234 273 228 When Mr. Wetherbee plays golf, he claims that he drives the ball 250 yards, on average. You don’t believe him, and think that his average drive is different from 250 yards. A sample of 10 of his drives is shown to the right. Is there enough evidence to conclude that he is wrong?

Example 223 237 244 232 242 225 234 273 228 When Mr. Wetherbee plays golf, he claims that he drives the ball 250 yards, on average. You don’t believe him, and think that his average drive is different from 250 yards. A sample of 10 of his drives is shown to the right. Is there enough evidence to conclude that he is wrong? YES. P-value=.026. Reject Null

Example 1 A company claims that 60% of graduate students have bought school supplies online. A consumer group believes that this estimate is too low. A random sample of 80 graduate students shows that 55 of them have bought supplies online. Is there enough evidence, at a .05 significance level, to conclude that the company’s claim is wrong?

Example 1 A company claims that 60% of graduate students have bought school supplies online. A consumer group believes that this estimate is too low. A random sample of 80 graduate students shows that 55 of them have bought supplies online. Is there enough evidence, at a .05 significance level, to conclude that the company’s claim is wrong? NO. P-value=.055. Fail to reject the null

Example 3 A high school claims that, on average, each class has 3 people missing per day. Based on your experience, you think that more than 3 people are absent each day. You randomly sample 10 classes over the course of a week. In your sample, you find a mean number of absent students to be 4.1 with a standard deviation of 1.1 At which of the following significance levels would we be able to conclude that the school is wrong (choose all that apply): .1 .05 .01 .005 .001