2. Section 8-3 Testing the Difference Between Means (Dependent Samples) We can conduct the hypothesis test on two dependent samples if ALL of the following conditions are met: 1)The samples must be randomly selected. 2)The samples must be dependent (paired). 3)Both populations must be normally distributed. If all of these conditions are met, we will use a t-distribution with n – 1 degrees of freedom (n is the number of data pairs).

3. Steps to Using the t-Test for the Difference Between Means (Dependent Samples) State H 0 and H a. Identify α Identify the degrees of freedom (d.f. = n-1) Determine the critical valueUse the t-distribution chart OR InvT function on the calculator Determine the rejection regionTo the left, right, or both sides of the critical value. Find the differences between data pairs STAT Edit, L1 and L2, highlight L3; type in L1-L2

5. Example 1 on page 463— A golf club manufacturer claims that golfers can lower their scores by using the manufacturer’s newly designed golf clubs. Eight golfers are randomly selected, and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are again asked to give their most recent score. The scores for each golfer are shown in a table below. Assuming the golf scores are normally distributed, is there enough evidence to support the manufacturer’s claim at α = 0.10? Golfer12345678 Old Clubs8984968274928591 New Clubs83 928476918091

7. Example 1 on page 463— 2 nd VARS 4 Enter 1 - α for right tail test and 7 for df. Calculate 1.415 (same as the chart!!) Step 5:Determine the rejection region: The rejection region is t > 1.415.

8. Example 1 on page 463— Step 6:Determine the standardized test statistic and/or the p-value. STAT Edit Enter Sample 1 data values into L1 Enter Sample 2 data values into L2 Highlight L3 and type in L1-L2. This will automatically put the differences into L3 Golfer12345678 Old Clubs8984968274928591 New Clubs83 928476918091

10. Example 2 (Page 465) A state legislator wants to determine whether her performance rating (0-100) has changed from last year to this year. The following table shows the legislator’s performance rating from the same 16 randomly selected voters for last year and this year. At α = 0.01, is there enough evidence to conclude that the legislator’s performance rating has changed? Assume the performance ratings are normally distributed. Voter12345678910111213141516 Last Year60547884912550656881754562795863 This Year564870608540 5580757850 855360

12. Example 2 (Page 465) 2 nd VARS 4 Enter α/2 for two tail test and 15 for df. Calculate -2.947 (same as the chart!!) Again, remember to use both the positive and the negative of the value given. Step 5:Determine the rejection region: The rejection region is t > 2.947 and t < -2.947.

13. Example 2 (Page 465) Step 6:Determine the standardized test statistic and/or the p-value. STAT Edit Enter Sample 1 data values into L1 Enter Sample 2 data values into L2 Highlight L3 and type in L1-L2. This will automatically put the differences into L3. Voter12345678910111213141516 Last Year60547884912550656881754562795863 This Year564870608540 5580757850 855360

15. Assignments: Classwork:Page 466 #1-8 All Homework:Pages 466-469 #10-20 Evens