3710ProblemCase1

3710ProblemCase1 - n x z-= Step 3 Since the H A here is of...

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MSCI 3710 LECTURE NOTES Dr. Nick Evangelopoulos Hypothesis Testing Problem Case 1 Situation: There is an unknown population mean μ . We know the sample mean x and the sample size n . The population standard deviation σ is given. Then, we do a z test for one mean. (Alternatively, sigma is not given, only the sample st. deviation s is given, but we have a large sample n > 30; then we can do an approximate z test .) Example: Refer to the Everglo example, 2005 text, p.311. We suspect might not be 400 as claimed. A sample of 100 had x = 411. The population st. deviation is known to be σ = 42.5. Test at a = 0.10. Solution: Step1 H 0 : = 400 H A : ≠ 400 Step 2 Given that H 0 is true (which is our assumption anyway), the following quantity is expected to have a standard normal distribution:
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Unformatted text preview: n x z-= Step 3 Since the H A here is of the not-equal-to type, the test is two-tailed. Each tail has a probability equal to a /2. Using a z table we find the critical value to be 1.645. The Decision Rule is Reject H if the observed z value is more extreme than the critical z value. Step 4 The calculated value is: 100 5 . 42 400 411 *-= z = (411 400) / (42.5 / SQRT(100)) = 2.59. Step 5 Conclusion: Reject the null hypothesis H since 2.59 > 1.645, i.e., the calculated value is more extreme than the critical value(s). Therefore, the average lifetime of a light bulb is not equal to 400. a /2 z cr a /2 z cr 0.05 1.645 0.05 1.645 z cr z cr z * = 2.59...
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This note was uploaded on 03/07/2011 for the course DSCI 3710 taught by Professor Dr.jay during the Spring '08 term at North Texas.

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