The assumptions of a t-test and the power of a t-test Section 11.1.4 Lesson 11.1.4

Starter 11.1.4 Seven students took an SAT prep course after doing poorly on the test. Here are their before (row 1) and after (row 2) scores: Use a matched-pairs t-test on your calculator to determine if they improved their scores in a statistically significant way. 1020 1100 1110 1000 990 1050 1060 1030 1120 1040 1160

Today’s Objectives Students will state the assumptions needed to do a t-test or t-interval and verify whether they are met. Students will calculate the power of a t test. California Standard 20.0 Students are familiar with the t distributions and t test and understand their uses.

Assumptions of Procedures Each of the procedures in this unit requires that certain conditions be true. The AP readers will expect you to name the procedure you will use, state the conditions (or assumptions) needed, and verify that they are met. Start a page in your notes labeled “Assumptions” Create two columns: “Procedure” and “Assumptions” As we go through each procedure, add to this list. The assumptions for a t-test or t-interval are: The sample is an SRS of the population. The population distribution is approximately normal.

Priority of assumptions Valid sampling method Assumption of SRS more important than assumption of normal population. Approximately normal population Sample size less than 15 t procedures OK if data close to normal Verify by graphing (boxplot or normal probability plot) Sample size at least 15 t procedures OK if no outliers or strong skew Large sample size (30 or more) t procedures OK even if skewed Central Limit Theorem justifies

The power of a t test Like the z test, the power of a t test is the probability that we properly reject Ho under a given alternative value for the mean. Step1: Write the rule for rejecting Ho in terms of Ë. Or: What set of Ë values would cause you to reject the null? Step 2: Find P(rejecting Ho) assuming some given mean that falls in Ha Note: We assume σ = s for this part

Example 11.5 20 teachers are enrolled in a summer course. The course director wants to detect an improvement of 2 points on pre- and post-test scores. State the hypotheses: Ho: μ=0 Ha: μ>0 Let α = .05 Assuming s = 3, find the power of the t test at the 5% significance level against the alternative μ = 2.

Example continued Reject Ho if t > upper 5% of t19 From the table, t* = 1.729 Note: You can also use the T-Interval screen to find the border.

Example concluded Power is P(reject Ho when Ha true). So find P(Ë>1.160) if μ = 2 and σ = 3. normalcdf(1.160, 999, 2, 3 / √20) = .89 Notice that we use normalcdf instead of tcdf because we are assuming σ = s and if we know σ we use the z distribution instead of t. So a true difference of 2 points will produce results significant at the 5% level in 89% of all samples.

Example Extension Given all the same facts, what is the power of the test to detect an improvement of 3 points? Still reject Ho if Ë ≥ 1.160 P(Ë ≥ 1.160 | μ = 3) = normalcdf(1.160, 999, 3, 3/√20) = .997 So the probability of detecting improvement is 99.7% if the improvement is 3 points or more.

Today’s Objectives Students will state the assumptions needed to do a t-test or t-interval and verify whether they are met. Students will calculate the power of a t test. California Standard 20.0 Students are familiar with the t distributions and t test and understand their uses.

Homework Read pages 605 – 611 Do problems 15 – 19