Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
1 Formula for one-sample z-test To test the null hypothesis of H0 : µ = µ0 against one of the alternatives from below: HA : µ > µ0 (Upper-tailed alternative) HA : µ < µ0 (Lower-tailed alternative) HA : µ 6= µ0 (Two-tailed alternative) We use the test statistic of z = X − µ0 √ , σ/ n and with significance level α, HA rejection region p-value P ower upper-tailed z > zα 1 − Φ(z) 1 − Φ(zα − µA ) lower-tailed z < −zα Φ(z) Φ(−zα − µA ) z < −zα/2 or z > zα/2 2 1 − Φ(|z|) 1 − Φ(z α2 − µA ) + Φ(−z α2 − µA ) two-tailed µA = µ−µ √0 σ/ n 2 HW 7 on One-sample Inference Name: Questions 1. A random sample of 100 lightning flashes in a certain region resulted in a sample average radar echo duration of .91 sec. Assume that the observations are random sample from the normal distribution with (true) standard deviation of .34 sec. (a) with α = 5%, test the null hypothesis of µ = 1 sec against lower-tail alternative. Picture below represents the behavior of test statistic z under H0 and HA . i. mean of curve under H0 is 0 ii. mean of curve under HA is (formula) iii. SD of curve under H0 is 1 iv. SD of curve under HA is 1 µ−µ √0 σ/ n −zα = −1.65 v. critical value is vi. We reject H0 if test statistic is greater/less than the CV. Less vii. Label α, β, and Power on the picture below: (b) Calculate the test statistic and report your conclusion with signifiance level of 5%. Test statistic z is z= x̄ − µ0 .91 − 1 √ = √ = −2.64 σ/ n .34/ 100 Since the test statistic is less than CV, we reject H0 . There is an evidence that true mean duration of the lightning µ is less than 1 second. 3 (c) Calculate P-value from above test statistic. Since this is lower-tailed test, P-value is Φ(−2.64) = .004. (d) Calculate probability of false negative of above test, when µ = .9. P( false negative) = β. Using the formula from the first page, µ − µ0 .9 − 1 √ √ 1 − Φ − zα − = 1 − Φ − 2.33 − σ/ n .34/ 100 = 1 − Φ(.61) = .73 So if true µ is .9, then this test will reject H0 with probability .27. Since α=5%, we have zα = 1.65, and above solution is wrong. It should be .9 − 1 √ β = 1 − Φ − 1.65 − = 1 − Φ(1.29) = 0.0985. .34/ 100 I gave everybody +1 point on this. 2. Suppose we are interested in modeling the distribution of tread life for a premium brand of radial tire. Assuming that the life of the tires are normaly distributed with unkown mean µ and standard deviation σ = 5120, the company has decided to test randomly choose n tires and test their lifetime. (a) With α = 5%, we wish to test µ = 40000 miles against upper-tail alternative. Picture below represents the behavior of test statistic z under H0 and HA . i. mean of curve under H0 is 0 ii. mean of curve under HA is (formula) iii. SD of curve under H0 is 1 iv. SD of curve under HA is 1 v. critical value is µ−µ √0 σ/ n zα = 1.65 vi. We reject H0 if test statistic is greater/less than the CV. Greater vii. Label α, β, and Power on the picture below: (b) Suppose in the actual experiment, random sample of size n = 100 were tested, and resulted in sample mean of 43563 miles. Perform the test, and report P-value and conclusion. 4 Test statistic z is z= x̄ − µ0 43564 − 40000 √ = √ = 6.96 σ/ n 5120/ 100 Since test stat is greater than the critical value, we can reject H0 , and say there is strong evidence that true mean lifetime of the tire µ is greater than 40000 miles. (c) What is the p-value of the above test? Since this is upper-tailed test, P-value is 1 − Φ(6.96) ≈ 0. In the z−table, 6.96 is out of chart. So that means Φ(6.96) is very close to 1. So the p-value is very close to 0. (d) Calculate the sample size n we need in order to have probability of false negative of above test equal to .1 when µ = 41000. This is upper-tailed test, so our β is µ − µ0 41000 − 4000 √ √ β = Φ zα − = Φ 1.65 − = .1 σ/ n 5120/ n Since from the z-table we know that Φ(−1.28) = .1, we have equation 1.65 − 41000 − 40000 √ = −1.28 5120/ n Solving the equation for n, we get n = 225.05. 5 3. An article in the Journal of Heat Transfer (Trans. ASME, Sec. C, 96. 1974. p. 59) described a new method of measuring the thermal conductivity of Armco iron. Using a temperature of 100F and a power input of 550 watts, the 10 measurements of thermal conductivity (in Btu/hr-ft-F) resulted in sample mean of 41.92. Assuming the measurements were random sample from Normal distribution with mean µ and SD of .3. (a) Test the null hypothesis that µ = 42 against two-sided alternative with significance level of 5%. i. Critical Values are and ii. Rejection regions are and iii. Test Statistic is iv. P-value is . ±1.96 . z < −1.96 and z > 1.96 -.8432 . .4 v. State your conclusion: Test statistic z is z= 41.92 − 42 x̄ − µ0 √ = √ = −.8432 σ/ n .3/ 10 Since this is two-tailed test, P-value is 2(1 − Φ(| − .8432|)) = 2(.20) = .4. This is very high p-value. If you use α = 5%, (or any other reasonable α) you wouldn’t be able to reject H0 . There’s no evidence that µ is different from 42. The test is inconclusive.