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Unformatted text preview: Math 10 MPS ‐ Homework 6 – Answers to additional problems 1. What are the two types of hypotheses used in a hypothesis test? How are they related? Ho: Null Hypotheses – A statement about a population parameter that is assumed to be true for the purposes of testing Ha: Alternative Hypothesis ‐ A statement about a population parameter that is assumed to be true is the Null Hypothesis is rejected during testing. These two Hypotheses are complements of each other. 2. Describe the two types of error possible in a hypothesis test decision. Type I error: Rejecting a true Ho Type II error: Failing to reject a false Ho True or False? In Exercises 3‐8, determine whether the statement is true of false. If it is false, rewrite it as a true statement. 3. In a hypothesis test, you assume the alternative hypothesis is true. False, you assume the Null Hypothesis is true. 4. A statistical hypothesis is a statement about a sample. False, it is a statement about a population parameter. 5. If you decide to reject the null hypothesis, you can support the alternative hypothesis. True 6. The level of significance is the maximum probability you allow for rejecting a null hypothesis when it is actually true. True 7. A large P‐value in a test will favor a rejection of the null hypothesis. False, a small p‐value supports rejecting the null hypothesis. 8. If you want to support a claim, write it as your null hypothesis. False, to support a claim write it as the alternative hypothesis. Stating Hypotheses In Exercises 9‐14, use the given statement to represent a claim. Write its complement and state which is Ho and which is Ha. 9. Ha: p >.65 Ho: p≤65 10. Ho: μ ≤ 128 Ha: μ >128 11. Ha: σ2 ≠ 5 Ho: σ2 = 5 12. Ho: μ =1.2 Ha: μ ≠1.2 13. Ho: p ≥ 0.45 Ha: p < 0.45 14. Ha: σ < 0.21 Ho: σ ≥ 0.21 15. Additional Problems: The geyser Old Faithful in Yellowstone National Park is claimed to erupt for on average for about three minutes. Thirty‐six observations of eruptions of the Old Faithful were recorded (time in minutes) 1.8 1.82 1.88 1.9 1.92 1.93 1. 2. 1.98 2.37 3.78 4.3 4.53 2.03 2.82 3.83 4.3 4.55 3.87 4.43 4.6 2.05 3.13 2.13 3.27 3.88 4.43 4.6 2.3 3.65 4.1 4.47 4.63 2.35 3.7 4.27 4.47 6.13 Sample mean = 3.394 minutes. Sample standard deviation = 1.168 minutes Test the hypothesis that the mean length of time for an eruption is 3 minutes. General Question a. Why do you think this test is being conducted? We are trying to test the claim that Old Faithful erupts for three minutes. Design a. State the null and alternative hypotheses Ho: μ = 3 Ho: μ ≠ 3 b. What is the appropriate test statistic/model? t ‐Test of mean vs. Hypothesized Value, population standard deviation unknown. t= X − μ0 s n df = 35 c. 3. What is significance level of the test? Not given, so I will choose α = 5% d. What is the decision rule? Critical value method: Reject Ho if t<‐1.96 ot t>1.96 Conduct the test a. Are there any unusual observations that question the integrity of the data or the assumptions of the model? (additional problem only) The value 6.13 appears to be an outlier and increases the standard deviation. b. Is the decision to reject or fail to reject Ho? t= 4. 3.394 − 3 = 2.02 Reject Ho (barely) 1.168 36 Conclusions State a one paragraph conclusion that is consistent with the decision using language that is clearly understood in the context of the problem. Address any potential problems with the sampling methods and address any further research you would conduct. The mean length of time for eruptions of Old Faithful is not three minutes, it appears to be longer. This conclusion is based on 36 measurements taken of Old Faithful. One measurement of 6.13 seems highly unusual and does not seem to fit the data. I would review the data to make sure this result was recorded correctly. 16. Define the following terms – All answers are in glossary of my Online text. 17. A study claims more than 60% of students text‐message frequently. In a poll of 1000 students, 660 students said they text message frequently. Can you support the study’s claim? Conduct the test with α = 1% Ho: p ≤ 0.60 Ha: p > 0.60 α = 1% Model: Z = pˆ − p 0 ( p0 )(1 − p0 ) n Reject Ho if p‐value <.01 660 pˆ = = 0.66 1000 0.66 − 0.60 Z= = 3.87 (0.60)(1 − 0.60) 100 P ( Z > 3.87) ≈ .0000 Reject Ho The study is correct. More than 60% of students text‐message frequently. 18. 15 I‐pod users were asked how many songs were on their I‐pod. Here are the summary statistics of that study: X = 650 s = 200 a. Can you support the claim that the number of songs on a user’s I‐pod is different from 500? Conduct the test with α= 5% . Ho: μ = 500 Ha: μ ≠ 500 Test of mean vs. Hypothesized Value, population standard deviation unknown. t= X − μ0 s n df = 14 Due to the small sample size, we must assume the data is approximately normal (or at least not heavily skewed) for the central limit theorem to apply. Reject Ho if t >2.145 or t<‐2.145 (Two tailed test) t = (650‐500)/(200/sqrt(15)) = 2.90 Æ Reject Ho The mean number of songs on a user’s Ipod is not 500. It is more. b. Can you support the claim that the population standard deviation is under 300? Conduct the test with α= 5% . Ho: σ ≥ 300 Ha: σ < 300 Chi square test of standard deviation vs. Hypothesized Value χ2 = (n − 1) s 2 σ2 df = 14 α = .05 Reject Ho if χ2 <6.571 (lower tailed test) 1. χ 2 = (14) 200 = 6.22 Æ Reject Ho 300 2 2 The standard deviation is under 300. 19. Consider the design procedure in the test you conducted in Question 18a. Suppose you wanted to conduct a Power analysis if the population mean under Ha was actually 550. Use the online Power calculator to answer the following questions. a. Determine the Power of the test. power =.1397 b. Determine Beta. β=1‐power = .8603 c. Determine the sample size needed if you wanted to conduct the test in Question 18a with 95% power n=210 20. The drawing shown diagrams a hypothesis test for population mean design under the Null Hypothesis (top drawing) and a specific Alternative Hypothesis (bottom drawing). The sample size for the test is 200. a. State the Null and Alternative Hypotheses Ho: μ ≥ 8 Ha: μ < 8 b. What are the values of μ0 and μa in this problem? μ0 = 8 Ha: μα = 4 c. What is the significance level of the test? α=.10 d. What is the Power of the test when the population mean = 4? Power = .91 e. Determine the probability associated with Type I error. α=.10 f. Determine the probability associated with Type II error. β=.09 g. Under the Null Hypothesis, what is the probability the sample mean will be over 6? 1−α=.90 h. If the significance level was set at 5%, would the power increase, decrease or stay the same? Power would decrease i. If the test was conducted, and the p‐value was .085, would the decision be Reject or Fail to Reject the Null Hypothesis? Reject Ho because .085<.10 j. If the sample size was changed to 100, would the shaded on area on the bottom (Ha) graph increase, decrease or stay the same? Increase: Beta would increase when sample size decreases. ...
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