AMS310 Fall 2018 HW5 - Homework PDF

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AMS 310

Prof. Rispoli

Homework #5 Due Date: 11:59 pm, 11/18/2018

Reminder: Please show all steps and do not just give the final answers. Material is based on Chapters 7 and 8 from Ahn. Please upload your homework to Blackboard and you have two attempts to upload correctly for each homework. 1. If the variance of a normal population is 3, what is the 95th percentile of the variance of a random sample of size 15? a) Find the percentile using the distribution table. Hint, use the χ2 distribution. 23.685

b) Find the percentile using R. > qchisq(.95, df=14) [1] 23.68479 2. In a scientific study, a statistical test yielded a p-value of 0.067. Which of the following is the correct decision? Explain your answer. a) Reject H0 at  = 0.05 and reject it for  = 0.01. b) Reject H0 at  = 0.05 and but not for  = 0.01. c) Reject H0 at  = 0.01 and but not for  = 0.05. d) Do not reject H0 at  = 0.05 and do not reject it for  = 0.01. D is the correct answer. If 𝑃 < 𝛼( 0.05 𝑜𝑟 0.01), then Reject 𝐻0 3. A printer company claims that the mean warm-up time of a certain brand of printer is 15 seconds. An engineer of another company is conducting a statistical test to show this is an underestimate. a) State the testing hypothesis. 𝐻0 ≤ 15, 𝐻1 > 15

b) The test yielded a p-value of 0.035. What would be the decision of the test if  = 0.05? The null hypothesis is rejected, and the company is correct.

AMS 310

Prof. Rispoli

c) Suppose a further study establishes that the true mean warm-up time is 14 seconds. Did the engineer make the correct decision? If not, what type of error did he/she make? No, type I error.

4. The incubation period of Middle East respiratory syndrome is known to have a normal distribution with a mean of 8 days and a standard deviation of 3 days. Suppose a group of researchers claimed that the true mean is shorter than 8 days. A test is conducted using a random sample of 20 patients. It is known that  = 3. a) Formulate hypotheses for this test. 𝐻0 ≥ 8, 𝐻1 < 8

 ≤ 6}. Suppose the test failed to reject H0. If the true mean b) Consider a rejection region {X incubation period is 5 days, what is the Type II error probability? P(𝑇𝑦𝑝𝑒 𝐼𝐼 𝑒𝑟𝑟𝑜𝑟)

= 𝑃(𝑥 > 6 | 𝜇 = 5, 𝜎 = 3)

= 𝑃(𝑥 > 6 | 𝜇 = 5, 𝜎 = 3) = 𝑃(

𝑥 −5 3 √30

>

= 1 − Φ(

3

6−5 3

√30

√30

)

)

=0.068 5. The shelf life of a certain ointment is known to have a normal distribution. A sample of size  = 36.1 and s = 3.7. 120 tubes of ointment gives X

a) Construct a 95% confidence interval of the population mean shelf life. 𝑍𝛼 > 1.96 = 36.1 ± 1.96 2

3.7

√120

= (35.4379, 36.762)

b) Suppose a researcher believed that before the experiment  = 4. What would be the required sample size to estimate the population mean to be within 0.5 months with 99% confidence? n=(

𝑍𝑎/2 𝜎 2 ) 𝐸

=(

𝑍0.005 𝜎 2 ) 𝐸

=(

2.85∗4 2 ) 0.5

= 426.1

AMS 310

Prof. Rispoli

6. The foot size of each of 16 men was measured, resulting in the sample mean of 27.32 cm. Assume that the distribution of foot sizes is normal with  = 1.2 cm. a) Test if the population mean of men’s foot sizes is 28.0 cm using  = 0.01. 𝐻0 = 28, 𝐻1 ≠ 28

𝑧=

27.32−28 1.2 √16

= −2.27

𝑃 = 2𝑃 (𝑧 < 𝑧𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 ) = 2𝑃(𝑍 < −2.27) = 2(0.0116) = 0.023

0.023 > 0.01, so, cannot reject 𝐻0 . Conclude that the population mean is not equal to 28. b) If  = 0.01 is used, what is the probability of a type II error when the population mean is 27.0 cm? 𝛽(27) = Φ (2.576 +

28−27 1.2 4

) − Φ (−2.576 +

28−27 1.2 4

) = Φ (5.91) − Φ(0.76) = 0.2266

c) Find the sample size required to ensure that the type II error probability (27) = 0.1 when  = 0.01. 𝑛=(

𝜎(𝑍𝛼 +𝑍𝛽 ) 2

𝜇0

−𝜇′

)2 = (

1.2(𝑍0.005 +𝑍0.1 ) 2 ) 28−27

=(

1.2(2.576+1.28) 2 ) 1

= 4.652 ≈ 22

7. The mean weight loss of n = 16 grinding discs after a certain length of time in a mill is 3.42 grams with a standard deviation of 0.68 grams. Construct a 99% confidence interval for the true mean weight loss. Assume a normal distribution. Z = 2.75(for 99%) CI = μ ±

𝜎

√𝑛

= 3.42 ± 2.575 (

0.68 4

) = 3.42 ± 0.4378 = (2.982, 3.858)

8. A trucking firm is suspicious of the claim that the average lifetime of certain tires is at least 28,000 miles. To check the claim, the firm puts 40 of theses on its trucks, and gets a mean lifetime of 27,463 miles with a standard deviation of 1,348 miles. Test the claim using 0.01? Use the 5 step method and label all steps. a. 𝐻0 ≤ 2800, 𝐻1 > 2800

b. Z =

27463−28000 1348 √10

= −2.519, |−𝑍| = 2.5195

AMS 310

Prof. Rispoli c. 𝑍𝛼 = 2.33

d. Not reject to 𝐻0

e. There is sufficient evidence to support the claim

9. A computer company claims that the batteries in its laptops last 4 hours on average. A consumer report firm gathered a sample of 16 batteries and conducted tests on this claim. The sample mean was 3 hours 50 minutes, and the sample standard deviation was 20 minutes. Assume that the battery time distribution as normal. a) Test if the average battery time is shorter than 4 hours at  = 0.05. Use the 5-step method. 𝐻0 ≥ 240𝑚, 𝐻1 < 240𝑚 σ=

20 4

=5

degree of freedom = 15 𝑃 (𝑍 <

230−240 5

) = 𝑃 (𝑍 < −2) = 0.032

0.032 < 0.05, so, reject to 𝐻0

b) Construct a 95% confidence interval of the mean battery time. 𝐶𝐼 = 230 ± 𝑍𝛼 2

𝜎

√𝑛

= 230 ± 2.131 × 5 = (219 .34, 240.66)

c) If you were to test H0: µ =240 minutes vs. H1: µ ≠ 240 minutes, what would you conclude from your result in part (b)? Conclude that the battery life is 240. d) Suppose that a further study establishes that, in fact, e population mean is 4 hours. Did the test in part (c) make a correct decision? If not, what type of error did it make? Part(c) makes a right decision

10. Suppose that we want to estimate the true proportion of defectives in a very large shipment of adobe bricks, and that we want to be at least 95% confident that the error is at most 0.04. How large a sample will we need if: a) We have no idea what the true proportion might be.

AMS 310

Prof. Rispoli

1−0.95 2

= 0.025

𝑍0.025 = 1.96

𝑛=(

𝑍𝛼 ×𝜎 2

𝐸

1.96 2

)2 × 𝑝 × (1 − 𝑝) = ( 0.04) × 0.5 × 0.5 = 601

b) We know that the true proportion is approximately 0.12? 𝑛=(

𝑍𝛼 ×𝜎 2

𝐸

1.96 2

)2 × 𝑝 × (1 − 𝑝) = ( 0.04) × 0.12 × 0.88 = 254

11. In a random sample of 600 cars making a right turn at a certain intersection, 157 pulled into the wrong lane. Test the hypothesis that actually 30% of all drivers make this mistake at the given intersection. Use the level of significance. a)  = 0.05 𝐻0 : 𝑝 = 0.3, 𝐻1 : 𝑝 ≠ 0.3

𝑧=

0.262−0.3 √0.3×

0.7 600

= −2.049

𝑃 = 2𝑃 (𝑍 < −2.05) = 2 × 0.0202 = 0.0404

0.05 > 0.0404, So, reject to 𝐻0 b)  = 0.01 0.01 < 0.0404, Not reject.

12. The process used to grind certain silicon wafers to the proper thickness is acceptable only if , the population standard deviation of the thickness of dice cut from the wafers, is at most 0.50 mil. Use  = 0.05 to test the null hypothesis  = 0.50 against the alternative  > 0.50, if the thickness of 15 dice cut from such wafers have a standard deviation of 0.64 mil. Use the 5 step method. 𝐻0 : 𝜎 ≤ 0.5, 𝐻1 : 𝜎 > 0.5 𝜒2 =

(𝑛−1)𝑠2 𝜎0

2

=

14×0.642 0.25

= 22.94

𝑃 = 𝑃 (𝜒14 2 > 22.94) = 0.0613

0.0613 > 0.05, so, the test is insignificant, and we cannot reject 𝐻0 13. Suppose the average lifetime of a certain type of car battery is known to be 60 months. Consider conducting a two-sided test on it based on a sample of size 25 from a normal distribution with a population standard deviation of 4 months.

AMS 310

Prof. Rispoli

a) If the true average lifetime is 62 months and  =0.01, what is the probability of a type II error? β(62) =Ф(𝑍𝑎/2 +

μ0 −𝜇 ′ 𝜎/√𝑛

μ0 −𝜇 ′

) -Ф(−𝑍𝑎/2 +

𝜎/√𝑛

)= Ф(0.076)- Ф(-5.08)=0.528-0=0.528

b) What is the required sample size to satisfy and the type II error probability of (62) = 0.1? n=(

𝜎(𝑍𝑎 +𝑍𝛽 ) 2

μ0 −𝜇 ′

4(𝑍0.005 +𝑍0.1 ) 2 4(2.58+1.28) 2 ) =( ) =59.598 μ0 −𝜇 ′ 60−62

)2=(

The required sample size is 60 14. For the data below construct a 95% confidence interval for the population mean. 53.4 51.6 48.0 49.8 52.8 51.8 48.8 43.4 48.2 51.8 54.6 53.8 54.6 49.6 47.2 𝑥 = 50.63

s= 3.157636 𝑛 = 15

CI = μ ±

𝜎

√𝑛

= 50.6 ± 2.145

3.158 √15

= 50.6 ± 1.749 = (48.851, 52.349)

15. A phone manufacturer wishes to estimate the proportion of people who want to purchase a cell phone which costs more than $800. Find the required sample size to yield a 90% confidence interval whose length is below 0.02. 1 𝑧𝑎/2 2 ) 𝐸

n=4(

1 1.645 2 ) =1691.27 0.02

= 4(

The required sample size is 1692...