Unformatted text preview: ISyE 2028A, Summer 2010 Friday July 9 Midterm 2 Solutions 1. If one wants to develop a 90% confidence interval for the mean μ of a normal population, when the standard deviation σ is known, the confidence level is A. .10 B. .45 C. .90 D. 1.645 ANSWER: C [5 points] 2. If the width of a confidence interval for μ is too wide when the population standard deviation σ is known, which one of the following is the best action to reduce the interval width? A. Increase the confidence level B. Reduce the population standard deviation σ C. Increase the population mean μ D. Increase the sample size n ANSWER: D [5 points] 3. Which of the following expressions are true about a large‐sample upper confidence bound for the population mean μ ? A. μ < x − zα / 2 ⋅ s n B. C. μ < x + zα / 2 ⋅ s n μ < x − zα ⋅ s n D. μ < x + zα ⋅ s n ANSWER: D [5 points] 4. Which of the following statement is false about the chi‐squared distribution with ν degrees of freedom? A. It is a discrete probability distribution with a single parameter ν . B. It is positively skewed (long upper tail) C. It becomes more symmetric as ν increases. ANSWER: A [5 points] 5. The upper limit of a 95% confidence interval for the variance σ 2 of a normal population using a sample of size n and variance value s 2 is given by: 2 A. (n − 1) s 2 / χ.05, n −1 2 B. (n − 1) s 2 / χ.025, n −1 ISyE 2028A, Summer 2010 C. 2 (n − 1) s 2 / χ.95, n −1 Friday July 9 2 D. (n − 1) s 2 / χ.975, n −1 ANSWER: D [5 points] 6. Which of the following statements are not true? A. A test statistic is a function of the sample data on which the decision to reject or not reject the null hypothesis is to be based. B. A rejection region consists of the set of all test statistic values for which the null hypothesis will be rejected. C. A rejection region consists of the set of all test statistic values for which the alternative hypothesis will be rejected. D. A good hypothesis‐testing procedure is one for which the probability of making either type I or type II error is small. ANSWER: C [5 points] 7. Which of the following statements are true? A. When the alternative hypothesis is H a : μ < μo , the null hypothesis H o should be rejected if x is too far to the left of μo . B. When the alternative hypothesis is H a : μ > μo , the null hypothesis H o should be rejected if x is too far to the right of μo . C. When the alternative hypothesis is H a : μ ≠ μo , the null hypothesis H o should be rejected if x is too far to either side of μo . D. All of the above statements are true. ANSWER: D [5 points] 8. Suppose that a two‐tailed test procedure about the population mean μ is performed when the population is normal, but the sample size n is small. The null hypothesis will be rejected at significance level α if the value of the standardized test statistic t is such that A. t ≥ tα , n −1 B. t ≤ −tα , n −1 C. either t ≥ tα / 2, n −1 or t ≤ −tα / 2, n −1 D. −tα / 2, n −1 ≤ t ≤ tα / 2, n −1 ANSWER: C [5 points] 9. A random sample of n = 8 E‐glass fiber test specimens of a certain type yielded a sample mean interfacial shear yield stress of 30.5 and a sample standard deviation of 3.0. Assuming that interfacial shear yield stress is normally distributed, compute a 95% CI for true average stress. ANSWER: [15 points] ISyE 2028A, Summer 2010 Friday July 9 d.f. = n – 1 = 7, so the critical value for a 95% C.I. is t.025,7 = 2.365 = 2.365. The interval is then ⎛ 3.0 ⎞ 30.5 ± (2.365) ⎜ ⎟ = 30.5 ± 2.51(27.99,33.01) . ⎝ 8⎠ 10. A study of the ability of individuals to walk in a straight line reported that accompanying data on cadence (strides per seconds) for a sample of n – 20 randomly selected healthy men: .95 .92 .81 .96 .93 .92 .95 1.00 .93 .78 .86 1.06 1.05 1.06 .92 .96 .85 .85 .81 .92 A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from MINITAB follows. Variable Cadence a. b. N 20 Mean Median StDev SEMean 0.9255 0.9300 0.0809 0.0181 Calculate and interpret a 95% confidence interval for a population mean cadence. Calculate and interpret a 95% prediction interval for the cadence of a single individual randomly selected from this population. ANSWER: [15 points] a. A 95% C.I.: .9255 ± 2.093(.0181) = .9255 ± .0379 ⇒ (.8876,.9634) b. A 95% P.I.: .9255 ± 2.093(.0809) 1 +
1 = .9255 ± .1735 ⇒ (.7520,1.0990) 20 11. State DMV records indicate that of all vehicles undergoing emissions testing during the previous year, 70% passed on the first try. A random sample of 200 cars tested in a particular county during the current year yields 160 that passed on the initial test. Does this suggest that the true proportion for this county during the current year differs from the previous statewide proportion? Test the relevant hypotheses using α = .05. ANSWER: [15 points] Parameter of interest: p = true proportion of cars in this particular county passing emissions ˆ testing on the first try. The sample proportion is p = 160 / 200 = 0.80 H 0 : p = .70 , and H a : p ≠ .70 The test statistic value is z = ˆ p − p0 p0 (1 − p0 ) / n = .80 − .70 .70(.30) / 200 = 3.086 The critical region is either z ≥ 1.96 or z ≤ −1.96 ISyE 2028A, Summer 2010 Friday July 9 Reject H 0 . The data indicates that the proportion of cars passing the first time on emission testing or this county differs from the proportion of cars passing statewide. 12. A plan for an executive traveler’s club has been developed by an airline on the premise that 5% of its current customers would qualify for membership. A random sample of 500 customers yielded 40 who would qualify. a. Using this data, test at level .01 the null hypothesis that the company’s premise is correct against the alternative that it is not correct. b. What is the probability that when the test of part (a) is used, the company’s premise will be judged correct when in fact 10% of all current customers qualify? ANSWER: [15 points] a. P = true proportion of current customers who qualify, H 0 : p = .05 vs H a : p ≠ .05 , ˆ The sample proportion is p = 40 / 500 = .08 The test statistics value is z =
ˆ p − p0 p0 (1 − p0 ) / n =
.08 − .05 (.05)(.95) / 500 = 3.078 The critical region is either z ≥ 2.58 or z ≤ −2.58 , so H 0 is rejected. The company’s premise is not correct. b.
β (.10) = Φ ⎢
⎡ .05 − .10 + 2.58 .05(0.95) / 500 ⎤ ⎡ .05 − .10 − 2.58 .05(0.95) / 500 ⎤ ⎥ − Φ⎢ ⎥ = Φ (−1.85) − Φ ( −5.598) = .0332 0.5(0.95) / 500 0.5(0.95) / 500 ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ISyE 2028A, Summer 2010 Normal table: Friday July 9 Remark: The other half is not necessary. ...
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This note was uploaded on 09/02/2010 for the course ISYE 2028 taught by Professor Shim during the Summer '07 term at Georgia Tech.
 Summer '07
 SHIM

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