1. 1 Section 8.5 Testing a claim about a mean (σ unknown) Objective For a population with mean µ (with σ unknown), use a sample to test a claim about the mean. Testing a mean (when σ known) uses the t-distribution

2. 2 Notation

3. 3 (1) The population standard deviation σ is unknown (2) One or both of the following: Requirements The population is normally distributed or The sample size n > 30

4. 4 Test Statistic Denoted t (as in t-score) since the test uses the t-distribution.

5. 5 People have died in boat accidents because an obsolete estimate of the mean weight (of 166.3 lb.) was used. A random sample of n = 40 men yielded the mean x = 172.55 lb. and standard deviation s = 26.33 lb. Do not assume the population standard deviation  is known. Test the claim that men have a mean weight greater than 166.3 lb. using 90% confidence. What we know: µ 0 = 166.3 n = 40 x = 172.55 s = 26.33 Claim: µ > 166.3 using α = 0.1 Note: Conditions for performing test are satisfied since n >30 Example 1

6. 6 What we know: µ 0 = 166.3 n = 40 x = 172.55 s = 26.33 Claim: µ > 166.3 using α = 0.1 H 0 : µ = 166.3 H 1 : µ > 166.3 right-tailed test Initial Conclusion: Since t in critical region, Reject H 0 Final Conclusion: Accept the claim that the mean weight is greater than 166.3 lb. t in critical region (df = 39) Using Critical Regions Example 1 t α = 1.304 t = 1.501 Test statistic: Critical value:

7. 7 Stat → T statistics → One sample → with summary Calculating P-value for a Mean (σ unknown)

8. 8 Then hit Next Enter the Sample mean (x) Sample std. dev. (s) Sample size (n) Calculating P-value for a Mean (σ unknown)

9. 9 Then hit Calculate Select Hypothesis Test Enter the Null:mean (µ 0 ) Select Alternative (“ ”, or “≠”) Calculating P-value for a Mean (σ unknown)

10. 10 Test statistic (t) P-value Calculating P-value for a Mean (σ unknown) The resulting table shows both the test statistic (t) and the P-value Initial Conclusion Since P-value < α (α = 0.1), reject H 0 Final Conclusion Accept the claim the mean weight greater than 166.3 Ib

11. 11 Using StatCrunch Using the P-value Example 1 Stat → T statistics→ One sample → With summary Null: proportion= Alternative Sample mean: Sample std. dev.: Sample size: ● Hypothesis Test 172.55 37.8 40 166.3 > P-value = 0.0707 What we know: µ 0 = 166.3 n = 40 x = 172.55 s = 26.33 Claim: µ > 166.3 using α = 0.1 Initial Conclusion: Since P-value < α, Reject H 0 Final Conclusion: Accept the claim that the mean weight is greater than 166.3 lb. H 0 : µ = 166.3 H 1 : µ > 166.3

12. 12 P-Values A useful interpretation of the P-value: it is observed level of significance Thus, the value 1 – P-value is interpreted as observed level of confidence Recall: “Confidence Level” = 1 – “Significance Level” Note: Only useful if we reject H 0 If H 0 accepted, the observed significance and confidence are not useful.

13. 13 P-Values From Example 1: P-value = 0.0707 1 – P-value = 0.9293 Thus, we can say conclude the following: The claim holds under 0.0707 significance. or equivalently… We are 92.93% confident the claim holds

14. 14 Loaded Die When a fair die (with equally likely outcomes 1-6) is rolled many times, the mean valued rolled should be 3.5 Your suspicious a die being used at a casino is loaded (that is, it’s mean is a value other than 3.5) You record the values for 100 rolls and end up with a mean of 3.87 and standard deviation 1.31 Using a confidence level of 99%, does the claim that the dice are loaded? What we know: µ 0 = 3.5 n = 100 x = 3.87 s = 1.31 Claim: µ ≠ 3.5 using α = 0.01 Note: Conditions for performing test are satisfied since n >30 Example 2

15. 15 H 0 : µ = 3.5 H 1 : µ ≠ 3.5 What we know: µ 0 = 3.5 n = 100 x = 3.87 s = 1.31 Claim: µ ≠ 3.5 using α = 0.01 two-tailed test Example 2 t in critical region (df = 99) Test statistic: Critical value: z = 3.058 z α = -2.626 z α = 2.626 Using Critical Regions Initial Conclusion: Since P-value < α, Reject H 0 Final Conclusion: Accept the claim the die is loaded.

16. 16 Using StatCrunch Using the P-value Example 2 Null: proportion= Alternative Sample mean: Sample std. dev.: Sample size: ● Hypothesis Test 3.87 1.31 100 3.5 ≠ P-value = 0.0057 Initial Conclusion: Since P-value < α, Reject H 0 Final Conclusion: Accept the claim the die is loaded. H 0 : µ = 3.5 H 1 : µ ≠ 3.5 What we know: µ 0 = 3.5 n = 100 x = 3.87 s = 1.31 Claim: µ ≠ 3.5 using α = 0.01 We are 99.43% confidence the die are loaded Stat → T statistics→ One sample → With summary