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INDE 3364 Instructor: Yaping Wang 1 HW1 (For Review Statistics I) 1. Multiple Choices (1) A two-sided confidence interval on the mean of a normal distribution with known variance 𝜎 2 is found to be 𝑥̅ − 1.18 𝜎 √𝑛 ≤ 𝜇 ≤ 𝑥̅ + 1.18𝜎/√? . What is the confidence level? A. 88.1% B. 76.2% C. 23.8% D. None of the above. (2) The percentage of titanium in an alloy used in aerospace castings is measured for 41 randomly selected parts. The sample standard deviation is ? = 0.3 . Assume the percentage of titanium in an alloy is normally distributed. Test the hypothesis 𝐻 0 : 𝜎 2 = 0.4 vs 𝐻 1 : 𝜎 2 ≠ 0.4 using ? = 0.05 . Which of the following is INCORRECT? A. The test statistic equals to 9. B. The fail-to-reject region of this test is 24.43 ≤ 𝜒 0 2 ≤ 59.34 C. We should reject the null hypothesis at the significance level of 0.05. D. None of the above. (3) Suppose you are doing an upper-tail t test on mean of a normal distribution with sample size ? = 12 . The test statistic ? 0 = 2.85 . Find the P-value range for this test. A. 0 .005 ≤ P-value ≤ 0 .01 B. 0.01 ≤ P-value ≤ 0.02 C. 0.99 ≤ P-value ≤ 0.995 D. None of the above. (4) In an upper-tail t-test on the mean of a normal distribution with unknown variance, the t test statistics is found to be t subscript 0 equals 2.49 based on a random sample of size 25. Which of the following statements is CORRECT regarding the P-value range and conclusion of this test? A. 0.01 < P-value < 0.025, so we fail to reject the null hypothesis at the significance level of 0.05. B. 0.01 < P-value < 0.025, so we reject the null hypothesis at the significance level of 0.05. C. 0.005 < P-value < 0.01, so we fail to reject the null hypothesis at the significance level of 0.05. D. 0.005 < P-value < 0.01, so we reject the null hypothesis at the significance level of 0.05.

INDE 3364 Instructor: Yaping Wang 2 (5) An Izod impact strength test was performed on 16 specimens of PVC pipe. The sample mean is 1.25 psi and the sample standard deviation is ? = 0.25 psi. Suppose that you as an engineer want to use the data from this experiment to support a claim that the mean impact strength of PVC pipe exceeds the ASTM standard of 1 psi. Assume the impact strength follows a normal distribution. Which of the following is INCORRECT? A. This problem tries to test on the mean of a normal distribution with variance unknown. B. For this problem, you should use an upper-tail t-test: 𝐻 0 : 𝜇 = 1 ?? 𝐻 1 𝜇 > 1 . C. The test statistic equals to 4 and we fail to reject the null hypothesis at the significance level of 0.05. D. None of the above. (6) A hypothesis is to test that a normal population mean equals 5 versus the alternative that the normal population mean is not equal to 5 with known variance 𝜎 2 = 4. Suppose the sample mean of a random sample of size ? = 25 turns out to be 5.8 (i.e., 𝑥̅ = 5.8). Which of the following statements if CORRECT regarding this hypothesis test? A. The test statistic 𝑧 0 = 1.2 B. One fails to reject the null hypothesis because the test statistics falls in the fail-to-rejection region. C. This is a CASE 2 hypothesis test we discussed in class. D. None of the above. (7) Which of the following statements is FALSE regarding hypothesis test? A. One can answer a hypothesis testing question by constructing an appropriate confidence interval. B. The CASE 4 hypothesis test we studied in class is a large sample hypothesis test on population proportion 𝑝 . C. For t-test in CASE 2, one can choose an appropriate OC curve chart to determine the required sample size to obtained a desired level of power of test. D. None of the above. Free Response Questions. Show ALL your work (formulas, calculations, final solution or other forms of answer as required) to get credit. 2. The waiting time 𝑋 follows a normal distribution with mean of 8.2 minutes and variance of 1.5 2 ?𝑖????? 2 . By the sampling distribution of sample mean ( 𝑋 ̅ ) of normal population, the average waiting time of 49 customers, 𝑋 ̅ = 1 49 ∑ 𝑋 𝑖 49 𝑖=1 , will follow a normal distribution with mean of 𝜇 𝑋 ̅ = 𝜇 = 8.2 minutes and standard deviation 𝜎 𝑋 ̅ = 𝜎 √𝑛 = 1.5 √49 = 0.2143 minutes. In other words, 𝑋 ̅ ~𝑁(8.2, (0.2143) 2 ) .

INDE 3364 Instructor: Yaping Wang 3 By normal standardization of 𝑋 ̅ , we get (a) 𝑃(𝑋 ̅ < 10) = 𝑃(? < 10−8.2 0.2143 ) = Φ (8.4) ≈ 1 (b) 𝑃(5 ≤ 𝑋 ̅ ≤ 10) = 𝑃 ( 5−8.2 0.2143 ≤ ? ≤ 10−8.2 0.2143 ) = Φ (8.4) − Φ (−14.932) ≈ 1 − 0 = 1 (c) 𝑃(𝑋 ̅ < 6) = 𝑃(? < 6−8.2 0.2143 ) = Φ (−10.27) ≈ 0 3. Let 𝑋 𝑖 denote the yearly claim of 𝑖 𝑡ℎ policy holder and each 𝑋 𝑖 follows the same unknown distribution with mean of 320 and standard deviation of 540, 𝑖 = 1, 2, … , 25,000 . With ? = 25,000 ≫ 30 , by Central Limit Theorem we know that 𝑋 ̅ = 1 25000 ∑ 𝑋 𝑖 25000 𝑖=1 approximately follows a normal distribution with mean 320 and variance of 540 2 25000 (i.e., standard deviation 540 √ 25,000 = 54 √250 ). The requested probability that the total yearly claim exceeds 8.3 million is 𝑃 (∑ 𝑋 𝑖 25000 𝑖=1 > 8.3 × 10 6 ) = 𝑃 (𝑋 ̅ > 8.3 × 10 6 25000 ) = 𝑃(𝑋 ̅ > 332) = 𝑃 (? > 332 − 320 54 √250 ) ≈ 1 − Φ(3.51) ≈ 0.0023 Therefore, there are only 2.3 chances out of 10,000 that the total yearly claim will exceed 8.3 million. Alternatively , by Central Limit Theorem we know that ∑ 𝑋 𝑖 25000 𝑖=1 approximately follows a normal distribution with mean 320×25,000 = 8× 10 6 and standard deviation 540 ∗ √25,000 = 8.5381 × 10 4 . Therefore, 𝑃 (∑ 𝑋 𝑖 25000 𝑖=1 > 8.3 × 10 6 ) 4. A standard procedure is expected to produce washers with very small deviation in their thickness. Suppose that 10 such washers were randomly chosen and measured, and the sample variance ? 2 is 1.366 × 10 −5 𝑖??ℎ?? 2 . Construct a 90% lower confidence bound for the standard deviation of the thickness of a washer produced by this procedure. (? − 1)? 2 𝜒 0.1,9 2 ≤ 𝜎 2 From upper percentage points table of chi-square distribution, we have 𝜒 0.1,9 2 = 14.68 .

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INDE 3364 Instructor: Yaping Wang 4 9 × 1.366 × 10 −5 14.68 ≤ 𝜎 2 9 × 1.366 × 10 −5 14.68 ≤ 𝜎 2 8.375 × 10 −6 ≤ 𝜎 2 Therefore, 2.894 × 10 −6 ≤ 𝜎 5. News agencies are taking polls to try to figure out the general population’s opinions on an election poll. Suppose a female candidate wants to see how much percentage of people will vote for her. Suppose a survey of 500 randomly selected respondents shows that 125 of them will vote for her in the next election. (1) Obtain a 95% two-sided CI for the proportion of the population that will vote for this candidate. A point estimate of the population proportion 𝑝 that will vote for this candidate is 𝑝̂ = 125 500 = 0.25 . A 95% two-sided CI for p is computed as 𝑝̂ − 𝑧 0.025 √ 𝑝̂(1 − 𝑝̂) ? ≤ 𝑝 ≤ 𝑝̂ + 𝑧 0.025 √ 𝑝̂(1 − 𝑝̂) ? 0.25 − 1.96 √ 0.25(0.75) 500 ≤ 𝑝 ≤ 0.25 + 1.96 √ 0.25(0.75) 500 0.212 ≤ 𝑝 ≤ 0.288 (2) If you are asked to test the hypothesis: 𝐻 0 : 𝑝 = 0.2 vs 𝐻 1 : 𝑝 ≠ 0.2. Based on the CI obtained in part (1), will you reject 𝐻 0 at the significance level of 0.05? Why? Since the hypothesized value 𝑝 0 = 0.2 is not contained in the 95% two-sided CI obtained in part (1), we should reject 𝐻 0 at 0.05 significance level. 6. A cereal company claims that the population mean sodium content in a box of cereal is 130 mg . A random sample of 25 boxes are taken and the sample mean 𝑥̅ = 129.8 mg . It is known from past history that 𝜎 = 2 ?? . Assume that the sodium content in a box of cereal is normally distributed. We are interested in testing whether there is sufficient evidence against the advertising on the box, i.e., testing the following hypotheses: 𝐻 0 : 𝜇 = 130 ?? vs 𝐻 1 : 𝜇 ≠ 130 ??

INDE 3364 Instructor: Yaping Wang 5 (1) Use fixed significance level approach ( ? = 0.01 ) to test whether we should reject the null hypothesis. 𝑧 0 = 𝑥̅ − 𝜇 0 𝜎/√? = 129.8 − 130 2/√25 = −0.5 Critical values: ±𝑧 𝛼/2 = ±𝑧 0.005 = ±2.58 Because 𝑧 0 = −0.5 falls into the fail to reject region of [-2.58, 2.58], we fail to reject 𝐻 0 at significance level of 0.01. (2) Test the same hypotheses at the significance level of ? = 0.01 using the P-value approach. Do you get the same conclusion as part (1)? Because this is a two-sided z-test, so 𝑃 -value = 2(1 − Φ(|−0.5|) = 2(1 − 0.691462) = 0.617076 Since P-value is greater than the significant level of 0.01, we fail to reject 𝐻 0 as in part (1). (3) What is the required sample size to reject the null hypothesis with probability of 0.9 if the true mean sodium content is 130.5 mg ? Power of test = 0.9 ➔ ? = 0.1 . Mean shift 𝛿 = 𝜇 − 𝜇 0 = 130.5 − 130 = 0.5 Because this is a two-sided z-test on the mean of Normal distribution with variance known, The required sample size is ? = (𝑧 ? 2 +𝑧 ? ) 2 𝜎 2 𝛿 2 = (𝑧 0.05 +𝑧 0.1 ) 2 2 2 0.25 2 =238.79 ≈ 238 where 𝑧 0.05 = 2.58, ??? 𝑧 0.1 = 1.28 . 7. A study on the reaction time of children with cerebral palsy reports a mean of 1.6 seconds on a particular task. A researcher believes that the reaction time can be reduced by using a motivating set of directions. A set of ten equivalent children is randomly selected and they complete the same task with the motivating set of directions. Assume the reaction time is normally distributed. The reaction times for the ten children are recorded as follows: 1.4, 1.8, 1.1, 1.4, 1.3, 1.6, 0.9, 1.5, 1.9, and 1.2 seconds. The sample mean and sample standard deviation are 1.41 seconds and 0.307 seconds, respectively. Test the claim of the researcher using the fixed significance level approach with ? = 0.05 . • State the null and alternative hypothesis. 𝐻 0 : 𝜇 = 1.6 sec

INDE 3364 Instructor: Yaping Wang 6 𝐻 1 : 𝜇 < 1.6 sec • Is this a z-test on the mean of t-test on the mean? Why? t-test because this is a test on the mean of normal distribution • Compute test statistics for the test. ? 0 = 𝑥̅ − 𝜇 0 ?/√? = 1.41 − 1.6 0.307/√25 = −1.957 • Will you reject the null hypothesis? Why? Yes, we should reject the null hypothesis because ? 0 is less than the critical value of −? 𝛼,𝑛−1 = −? 0.05,9 = −1.833. 8. A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally distributed with standard deviation of 0.25 volt , and the manufacturer wishes to test 𝐻 0 : 𝜇 = 5 volts versus 𝐻 1 : 𝜇 ≠ 5 volts , using ? = 9 units. Find the Type I Error probability ? if the fail-to-reject region is defined as 4.85 ≤ 𝑥̅ ≤ 5.15.  = P(Reject 𝐻 0 |𝐻 0 is true) = 𝑃(𝑋 ̅ < 4.85|𝜇 = 5) + 𝑃(𝑋 ̅ > 5.15|𝜇 = 5) = 𝑃 (? < 4.85 − 5 0.25 √9 ) + 𝑃(? > 5.15 − 5 0.25 √9 ) = P(Z  − 1.8) + P(Z > 1.8) = P(Z  − 1.8) + (1 − P(Z  1.8) = 2(1 −Φ(1.8)) = 2(1- − 0.9641) = 0.0718 9. Consider the probability density function (PDF), ?(𝑥) = 1 𝜃 2 𝑥? − 𝑥 𝜃 , 0 ≤ 𝑥 ≤ ∞, 0 < 𝜃 < ∞ Find the maximum likelihood estimator for 𝜃 . First, let’s write out the likelihood function of 𝜃,

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INDE 3364 Instructor: Yaping Wang 7 𝐿(𝜃) = ∏ 𝑥 𝑖 ? − 𝑥 𝑖 𝜃 𝜃 2 𝑛 𝑖=1 = 𝜃 −2𝑛 (∏ 𝑥 𝑖 ) 𝑛 𝑖=1 ? − ∑ 𝑥 𝑖 𝑛 𝑖=1 𝜃 Then, take a natural logarithm of the above likelihood function and we get ln 𝐿(𝜃) = ∑ ln (𝑥 𝑖 ) 𝑛 𝑖=1 − ∑ 𝑥 𝑖 𝑛 𝑖=1 𝜃 − 2? ∗ ln 𝜃 Take derivative w. r. t. 𝜃, then we get ∑ 𝑥 𝑖 𝑛 𝑖=1 𝜃 2 − 2𝑛 𝜃 = 0 ➔ 𝜃 ̂ = ∑ 𝑥 𝑖 𝑛 𝑖=1 2𝑛

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