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MIDTERM EXAM II

ISyE 2028: Basic Statistical Methods Instructor: Tonya Woods 9:20-10:30 am, Wednesday 07/03/2013

1. You have 70 minutes to complete the exam. There are six problems, and please use your time wisely. 2. This must be individual work. It is a closed book exam. Calculators are OK (and strongly recommended), but no computers. 3. Two sheets of notes are allowed (8.5” x 11”, both sides ok). 4. To receive full (or partial) credit, you must show ALL work.

Name (PRINT): Student-ID, 9 digits: Honor Code Certification: I certify that I have abided by the rules of the Georgia Tech honor code for student conduct on exams and by the specific rules on reference materials for this exam, and that I have neither given nor received assistance during this examination. I certify that I have read this statement and understand it. Signature: Problem Score

1 /10

2 /20

3 /16

4 /20

5 /10

6 /24

Total /100

1. (10 points, each 2 points) True/False. Each of the following statements are either TRUE or FALSE. Write “T” in the corresponding blank if the statement is true or “F” in the corresponding blank if the statement is false. (There is no need to show the calculation procedure or to provide explanation.) (a) TRUE Suppose we are performing a one sample hypothesis test on the mean where σ 2 is known (H0 : µ = µ0 versus H1 : µ 6= µ0 ). The closer µ0 is to the true mean µ, the higher the probability of type II error will be. (b) FALSE A 99% confidence interval is narrower than a 90% confidence interval (for the same parameter, using the same sample). (c) FALSE The correct interpretation of a 95% confidence interval for the mean is that there is a 95% chance that the true mean µ falls within the interval. (d) FALSE Suppose we are computing the sample size (n) necessary to limit the margin of error for a confidence interval on a population proportion. Finding the necessary sample size regardless of the value of p (and pb) results in a smaller n than finding the necessary sample size using our point estimator pb. (e) TRUE A type I error occurs when you reject the null hypothesis when the null hypothesis is really true.

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2. (20 points) We are interested in the IQs (Intelligence Quotients) of Georgia Tech undergraduate students in industrial engineering. A random sample of 15 students ¯ = 126. Based on past study of IQs, we know σ = 6. results in X (a) (8 points) Is there evidence to support the claim that the average IQ of Georgia Tech undergraduate students in industrial engineering is greater than 120? Perform a hypothesis test using α = 0.01. (b) (6 points) Answer part (a) using the p-value method. (c) (6 points) Answer part (a) using an appropriate confidence interval. Answer: (a) We are testing H0 : µ = 120 versus H1 : µ > 120. Since the population variance is known, we use the test statistic z0 =

¯ − µ0 X 126 − 120 √ = √ = 3.873 σ/ n 6/ 15

Since we have a one-sided alternative hypothesis, we reject H0 if z0 > zα = z0.01 = 2.33. The test statistic is in the rejection region, so we reject H0 . There is evidence to support the claim. (b) p-value=1 − Φ(z0 ) = 1 − Φ(3.873) = 1 − 0.99995 = 0.00005. Since the p-value is less than the significance level (α = 0.01), we reject the null hypothesis. (c) The corresponding interval is ¯ − zα( √σ ) µ>X n

µ > 126 − 2.33( √615 ) µ > 122.4 Since 120 is not in the interval, we reject H0 . Note that all three methods bring us to the same conclusion.

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3. (16 points) In a production system of rods, the rod diameter is known to follow a normal distribution. A random sample of size 10 is taken. The sample standard deviation, s, is 4.5. (a) (10 points) Find a 95% two-sided confidence interval for σ 2 . (b) (6 points) Find a 95% two-sided confidence interval for σ . Answer: (a) The appropriate interval is σ2 ∈ [

(n − 1)s2 (n − 1)s2 , 2 ] 2 χα/ 2,n−1 χ1−α/2,n−1

2 2 2 2 2 and χα/ 2,n−1 = χ0.025,9 = 19.02, χ 1−α/2,n−1 = χ 0.975,9 = 2.7. Therefore, σ ∈ [9.58, 67.5].

(b) σ =

√

√ √ σ 2 ∈ [ 9.58, 67.5] = [3.09, 8.21]

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4. (20 points) Blood samples from 4 subjects were each divided into 2 parts, to be sent to laboratory A and laboratory B. Each laboratory independently measured the cholesterol levels of the 4 subjects. The results were A B

240 190

210 180

200 215

280 265

where each column represents one of the subjects. (a) (10 points) Is there evidence to support the claim that laboratory A produces higher mean cholesterol measurements than laboratory B? State your assumptions and perform an appropriate t-test. Use the significance level α = 5%. (b) (10 points) Suppose that the same data were obtained in a different way — by taking blood samples from 8 subjects and randomly selecting 4 to send to A and 4 to B. State assumptions and perform an appropriate t-test at the significant level α = 5%. Use the same set of hypotheses as in part (a). Answer: (a) We need to use the paired t-test, since each set of observations from laboratories A and B are from one subject. Therefore, we need to assume the differences (di = Ai −Bi ) in laboratory results are independent and identically normally distributed. We now test H0 : µA = µB versus H1 : µA > µB (equivalently, H0 : µd = 0 versus H1 : µd > 0). The differences are 50, 30, −15, 15 and d¯ = 20, sd = 27.386. The test statistic is √ ¯ √ nd 4(20) t0 = = = 1.4606 sd 27.386 Since we have a one-sided alternative, we reject H0 if t0 > tα,n−1 = t0.05,3 = 2.353. Our test statistic is not in the rejection region, so we fail to reject H0 . There is not evidence to support the claim. (b) In this case, the lab A and B measurements were not taken from the same subject so we should use a two sample t-test. Since the two sample variances are quite different (s2A = 1291.667, s2B = 1441.667), it makes sense to assume unequal population variances. We are testing H0 : µA = µB versus H1 : µA 6= µB and the test statistic is ¯ A¯ − B 232.5 − 212.5 = 0.765 t0 = p 2 =p 2 s A /n + sB /m 1291.667/4 + 1441.667/4 Since we have a one-sided alternative, we reject H0 if t0 > tα,ν . Now ν=

(sA2 /n + sB2 /m)2 s4A n2 (n−1)

+

4 sB m2 (m−1)

=

466944.4 = 5.982 ⇒ 6 34758.39 + 43300.06

and t0.05,6 = 1.943. Because the test statistic does not fall in the rejection region, we fail to reject H0 . There is not evidence to support the claim. 4

5. (10 points) The cooling system in a nuclear submarine consists of an assembly pipe through which a coolant is circulated. Specifications require that weld strength must meet or exceed 150 psi. Suppose the designer decides to test the hypothesis ¯ = 153.7 H0 : µ = 150 versus H1 : µ > 150. A random sample of 20 welds result in X psi. From previous study, we know that the population standard deviation is σ = 11. If the true mean is 151 grams, calculate the probability of type II error for this test. Assume α = 0.05. Answer: For this particular one-sided hypothesis, the power function is √ n(µ − µ0 ) π(µ) = 1 − Φ(zα − ) √σ 20(151 − 150) π(151) = 1 − Φ(1.645 − ) 11 = 1 − Φ(1.238) Then the probability of type II error is β(151) = 1 − π(151) = 1 − (1 − Φ(1.238)) = Φ(1.238) = 0.89

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6. (24 points) For the model Yi = β0 + β1 Xi + ǫi , ǫi ∼ N (0, σ 2 ), the following statistics are computed: P9

i=1

P9

i=1

xi yi = 243 P9

x2i = 324 =

P9

i=1

xi = 0 =

i=1

P9

i=1

yi2

yi

(a) (8 points) Compute the least squares estimates, βb0 and βb1 .

(b) (8 points) Compute SSE, SST, and R2 . Compute σ b 2 , a point estimator of σ 2 . (c) (8 points) Compute a two-sided 95% confidence interval for both βb0 and βb1 . Use the appropriate interval to test for significance of the regression. Answer: (a) We have P Pn Pn n n X ( i=1 yi )( i=1 xi ) X 2 ( ni=1 xi )2 b β1 = ( xi yi − xi − )/( ) n n i=1 i=1 243 − 0 324 − 0 = 0.75

=

βb0 = = = (b) SST =

n X

Y¯ − βb1X¯ 0 − (0.75)(0) 0

yi2

i=1

Sxy =

n X i=1

R2 =

xi yi −

(

−

(

Pn

Pn

i=1

i=1

yi ) 2

n

xi )( n

Pn

= 324

i=1

yi )

= 243

SSE = SST − βb1 Sxy = 324 − (0.75)243 = 141.75

SSR SST − SSE SSE 141.75 = 0.5625 = = 1− = 1− 324 SST SST SST SSE 141.75 σ b2 = = 20.25 = 9−2 n−2 6

(c) Sxx =

n X i=1

x2i

−

(

Pn

i=1

xi )2

n

= 324

q q c2 c 2 σ b b β1 − tα/2,n−2 Sxx ≤ β1 ≤ β1 + tα/2,n−2 Sσxx q q .25 .25 0.75 − 2.365 20324 ≤ β1 ≤ 0.75 + 2.365 20324 0.159 ≤ β1 ≤ 1.341

q q ¯2 b0 + tα/2,n−2 σb2 [ 1 + X¯ 2 ] βb0 − tα/2,n−2 σb2 [ n1 + SXxx ] ≤ β0 ≤ β n Sxx p p 0 − 2.365 20.25[1/9 + 0] ≤ β0 ≤ 0 + 2.365 20.25[1/9 + 0] −3.548 ≤ β0 ≤ 3.548 We may use the confidence interval for β1 and note that 0 is not in the interval. Therefore, we reject H0 . There is evidence of a linear relationship between X and Y .

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