Notes on Hypothesis Testing Problems with Solutions | DSCI 3710, Study notes of Humanities

Download Notes on Hypothesis Testing Problems with Solutions | DSCI 3710 and more Study notes Humanities in PDF only on Docsity! DSCI 3710 LECTURE NOTES Dr. Nick Evangelopoulos Hypothesis Testing Problem Case 7 (z test for one-sample proportions) Situation: There is an unknown population proportion p. We have information on a sample coming from this population, and we know the sample size n, and how many sampled items share a quality of interest. Then, we do a z test for one proportion. Example: Refer to the Payroll Checks example, text, pp.413, 416, & 422. Out of 250 audited checks, 14 were found with errors. Does this evidence support the claim that the proportion is greater than 0.03? Test at a = 0.10. Solution: Step1 H0: p <= 0.03 HA: p > 0.03 Step 2 Given that H0 is true, the following quantity is expected to have a z distribution: n pp pp z )1( ˆ    , where p̂ is the sample proportion. Step 3 Since the HA here is of the “greater-than” type, the test is right-tailed. The tail probability is a = 0.10. Using a z table we find the critical value to be 1.28. The Decision Rule is “Reject H0 if the observed t value is more extreme than the critical z value”. Here, this means: reject H0 if z-calculated > 1.28. Step 4 The calculated value is: 250 )97.0)(03.0( 03.0056.0  z = (0.056 – 0.03) / SQRT((0.03)*(0.97)/250) = 2.41. Note that, in the above formula, 250 14 ˆ  n x p = 0.056. Step 5 Conclusion: Reject the null hypothesis H0 since the calculated value (2.41) is more extreme than the critical (1.28). Therefore, we have significant evidence that the proportion of checks with errors exceeds 3%. a z cr 0.10 1.28 zcr z* = 2.41