1. Chapter 10 One-Sample Test of Hypothesis

2. Example The Jamestown steel company manufactures and assembles desks and other office equipment at several plants in western New York State. The weekly production of the Model A325 desk at the Fredonia Plant is normally distributed, with a mean of 200 and a SD of 16. Recently, because of market expansion, new production methods have been introduced and new employees hired. The vice president of manufacturing would like to investigate whether there has been a change in the weekly production of the Model A325 desk. To put another way, is the mean number of desks produced at Fredonia Plant different from 200 at the.01 significance level.

4. Example (cont’d) Step 1: H 0 : µ = 200; H 1 : µ ≠ 200 (2-tailed test) Step 2: α =.01 Step 3: z statistic Step 4: this is a 2-tailed test, so divide.01 by 2 (.005 for each tail). Area where H 0 is not rejected is.99. Get critical z value from Appendix D (for.5000 -.005 =.4950). It is 2.58. Decision rule: reject H 0 and accept H 1 if the computed z value is not between -2.58 and +2.58. Do not reject H 0 is z value is between -2.58 and +2.58.

5. Example (cont’d) Step 5: take a sample, compute z, apply decision rule, and make a decision. The mean number of desks produced last year was 203.5, in 50 weeks: 1.55 does not fall in the rejection area, so we do not reject H0. We conclude that the mean in not different from 200.

7. Do We Prove H 0 ? Not really. We fail to disprove the null hypothesis. It is not the same thing as proving it true. The significance level should be set before gathering sample evidence and not changed based on sample evidence.

8. Example The Jamestown steel company manufactures and assembles desks and other office equipment at several plants in western New York State. The weekly production of the Model A325 desk at the Fredonia Plant is normally distributed, with a mean of 200 and a SD of 16. Recently, because of market expansion, new production methods have been introduced and new employees hired. The vice president of manufacturing would like to investigate whether there has been an increase in the weekly production of the Model A325 desk. To put another way, can we conclude that the mean number of desks produced at Fredonia Plant in the last 50 weeks more than 200 at the.01 significance level.

9. Example (cont’d) This is a one tailed test. H 0 : µ≤200 H 1 : µ>200 Critical z value is 2.33, found by subtracting.01 from.5000 (.4900) and locating critical z in Appendix D.

11. p-value in Hypothesis Testing How confident are we in rejecting the null hypothesis? A process that compares the probability of getting a value of the test statistic as extreme as the value actually obtained, called p-value, with the significance level. If the p-value is smaller than the significance level, H 0 is rejected. If it is larger than the significance level, H 0 is not rejected.

12. Interpreting the Weight of Evidence Against H 0 p-value is a way to express the likelihood that H 0 is false. If the p-value is less than:.10, we have some evidence that H 0 is not true..05, we have strong evidence that H 0 is not true..01, we have very strong evidence that H 0 is not true..001, we have extremely strong evidence that H 0 is not true.

13. p-value Computation Referring to the Jamestown manufacturing company example, we did not reject the null hypothesis. The computed z value of 1.55 fell in the do not reject region of -2.58 and +2.58. The probability of finding a z value of 1.55 or more is.0606 (.5000 -.4394). p-value is 2(.0606) or.1212. It is greater than the significance level of.01 decided upon initially. So H 0 is not rejected.

15. Testing for µ: Large Sample, Unknown Population SD In most cases in business the population SD (σ) is unknown. σ has to be based on prior studies or estimated by the sample SD (s). As long as sample size, n, is at least 30, we can use the formula (ch.8):

16. Example The Thompson’s Store issues its own credit card. The credit manager wants to find whether the mean monthly unpaid balance is more than $400. The level of significance is set at.05. A random check of 172 unpaid balances revealed the sample mean is $407 and the SD of the sample is $38. Should the credit manager conclude that the population mean is greater than $400, or is it reasonable that the difference of $7 is due to chance.

17. Example (cont’d) The null and alternate hypotheses are: H 0 : µ ≤ $400 H 1 : µ > $400 (one-tailed test) The critical z value is 1.65 (for sig. level of.05). The computed z value is:

19. Example (cont’d) Computed z value (2.42) > critical z value (1.65); therefore, we reject H 0 and accept H 1. The credit manager can conclude that the mean unpaid balance is greater than $400. The p-value provides additional insight into the decision. We find the probability of a z value greater than 2.42. It is (.5000 -.4922) =.0078. We conclude that the likelihood of finding a z value of 2.42 or larger when H 0 is true is.78%. Therefore, it is unlikely that H 0 is true.

20. Tests Concerning Proportions Follow the same procedure for hypothesis testing. The only difference is in computing z. The formula is: π is the population proportion; p is the sample proportion; n is the sample size; σ p is the standard error of the proportion.

21. Testing for µ: Small Sample, Unknown Population SD Follow same procedures as before. Conduct a test of hypothesis using the t distribution. with n-1 degrees of freedom.

22. Homework 33, 37, 39 (pg. 349), 43, 45 (pg. 350), 53, 55 (pg. 351).