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1 Statistics 151 Solution Sample Final Exam Total: 100 points Time: 3 Hour 1. [15 pts ] The lengths of the trout fry in a pond at the fish hatchery are approximately normally distributed with mean μ of 3.4 inches and standard deviation σ of 0.8 inches. a. What proportion of the trout will be greater than 4 inches? Answer: X is N(3.4, .8). So, P (X > 4) = P(Z>.75) = 0.2266 (about 23%). b. 95% of the trout will have a length greater than what value? Answer: P( X > k) = .95 or P( X £ k) = 0.05 k = 3.4 - 1.645· .8 = 2.084. c. How large a sample from this population should be taken if one wants to be 99% sure that the margin of error does not exceed 0.25 inches? 68 95 . 67 25 . 8 . 576 . 2 : Answer 2 2 * = = × = × n m z n s d. Suppose in another pond we are not sure whether the average length m of the trout is the same as or different from 3.4 inches. Assume now that X ~ N (μ, 0.8 2 ). The unknown population mean length of the trout in the second pond needs to be estimated. A random sample of 100 trout is drawn from the population, and it yields a sample mean length of 3.2 inches. Do the data provide any evidence to conclude that the average length of the trout in this pond is less than 3.4 inches? Test appropriate hypotheses using α = 0.01. Answer: Hypotheses: H 0 : m = 3.4, H 0 vs m < 3.4; Test statistic z = - 2.5; p-value = 0.0062; Conclusion: The data provide sufficient evidence to conclude that the average length of the trout in the pond is less than 3.4 inches. 2. [5 pts ] In an experiment, cakes baked from a new cake mix are baked at 300, 320, and 340, and for 1 hour, and 1 hour and 15 minutes. Ten cakes are baked at each combination of temperature and time. A panel of tasters scores each for texture and taste. a. What are the explanatory variables and response variables for this experiment? Answer: Explanatory variables: temperature and time

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Page 1: Statistics 151 Solution Sample Final Examacademic.macewan.ca/burok/Stat151/additional/Final_Exam_sol.pdf · 1 Statistics 151 Solution Sample Final Exam Total: 100 points Time: 3 Hour

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Statistics 151 Solution Sample Final Exam Total: 100 points Time: 3 Hour 1. [15 pts] The lengths of the trout fry in a pond at the fish hatchery are approximately normally distributed

with mean µ of 3.4 inches and standard deviation σ of 0.8 inches.

a. What proportion of the trout will be greater than 4 inches?

Answer: X is N(3.4, .8). So, P (X > 4) = P(Z>.75) = 0.2266 (about 23%).

b. 95% of the trout will have a length greater than what value?

Answer: P( X > k) = .95 or P( X ≤ k) = 0.05 ⇒ k = 3.4 − 1.645×.8 = 2.084.

c. How large a sample from this population should be taken if one wants to be 99% sure that the margin of error does not exceed 0.25 inches?

6895.6725.

8.576.2 :Answer

22*

=⇒=

×=

×≥ nm

zn

σ

d. Suppose in another pond we are not sure whether the average length µ of the trout is the same as or different from 3.4 inches. Assume now that X ~ N (µ, 0.8 2). The unknown population mean length of the trout in the second pond needs to be estimated. A random sample of 100 trout is drawn from the population, and it yields a sample mean length of 3.2 inches. Do the data provide any evidence to conclude that the average length of the trout in this pond is less than 3.4 inches? Test appropriate hypotheses using α = 0.01.

Answer: Hypotheses: H0 : µ = 3.4, H0 vs µ < 3.4; Test statistic z = −2.5; p-value = 0.0062; Conclusion: The data provide sufficient evidence to conclude that the average length of the trout in the pond is less than 3.4 inches.

2. [5 pts] In an experiment, cakes baked from a new cake mix are baked at 300, 320, and 340, and for 1 hour, and 1 hour and 15 minutes. Ten cakes are baked at each combination of temperature and time. A panel of tasters scores each for texture and taste.

a. What are the explanatory variables and response variables for this experiment?

Answer: Explanatory variables: temperature and time

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Response variable: Score for texture and taste

b. Make a diagram to describe the treatments. How many treatments are there? How many cakes are needed?

Answer: There are two factors and 6 treatments. The treatments are given in the following table:

Time Temperature 1 hour 1.25 hour 300 Treatment 1 Treatment 2 320 Treatment 3 Treatment 4 340 Treatment 5 Treatment 6

There are 6 treatments, so 60 cakes will be needed. 3. [10 pts] Many homeowners buy detectors to check for the invisible gas radon in their homes. How

accurate are these detectors? To answer this question, university researchers placed 12 radon detectors in a chamber where they were exposed to 105 picocuries per liter (pCi/l) of radon over 3 days. The detector reading were as follows: 91.9, 97.8, 111.4, 122.3, 105.4, 95.0, 103.8, 99.6, 96.6, 119.3, 104.8, 101.7 a. Calculate the mean and sample standard deviation for this data.

Answer: Sample mean = 104.13, sample standard deviation, s = 9.40.

b. Give a 90% CI for the mean reading µ of all detectors of this type.

Answer: Assuming the readings are approximately normally distributed, a 90% confidence interval is, using a t distribution with 11 degrees of freedom,

)00.109,26.99(87.413.10412

40.9)796.1(13.104* =±=±=±

n

stx .

c. What conclusion can you make from this interval?

Answer: The detectors appear to be accurate.

4. [10 marks] Suppose that 20% of the trees in a forest are infested with a certain type of parasite.

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a. Find the probability that at least one tree will contain parasite in a random sample of 10 trees.

Answer: X is B(10, .2). So, P(X ≥ 1 ) = 1 – P(x = 0) = 1 − .810 = 1 – .1074 = 0.8926 b. Find the probability that not more than 50 trees will contain parasites in a random sample of 300

trees.

Answer: X is B(300, .2). Then µ = np = 60, σ =√np(1-p) = 6.93. Using normal approximation,

P (X ≤ 50) ≈ P( Z ≤ (50−60)/6.93) = P( Z ≤−1.44) = 0.0749 ( Without continuity correction)

With continuity correction: P (X ≤50) ≈ P( Z ≤ (50.5−60)/6.93) = P( Z ≤ −1.37) = 0.0853.

c. After sampling 300 trees, suppose that 72 trees are found to have the parasite. Does this provide strong evidence that the population proportion (p) of infested trees is higher than 20%? Test appropriate hypotheses using α = 0.05.

Answer: Hypotheses: H0: p =.20 vs Ha: p >.20; Test statistic: z = (p-hat − .2)/SE(p-hat) =(.24 − .2)/sqrt((.2*.8)/300) = 1.73; p-value = P(Z ≥ 1.73) = 0.0418; Conclusion: At 5% significant level, we conclude that the percentage of trees infested with parasites is more than 20%.

5. [10 pts] A car dealer specializing in Corvettes enlarged his facilities and offered a number of models for sale during the open house. From his data on price (y in $1000) and age (x in years) of Corvettes, the following data were obtained:

x 1 2 4 5 6 6 10 11 11 12 12 12 12 13 15 y 39.9 32.0 25.0 20.0 16.0 20.0 13.0 13.7 11.0 12.0 20.0 9.0 9.0 12.5 7.0

The regression equation is: Price = 33.73 – 1.862 × Age.

19.245

,816.1175)(,4.268)(,78.499))((,34.17,8.8 22

=

=−=−−=−−== ∑ ∑ ∑SSE

yyxxyyxxyx

a. Is there sufficient evidence to indicate a negative linear relationship between selling price and age?

Test appropriate hypotheses using α = 0.05.

Answer: H0 : β =0 vs Ha: β < 0; Test statistic : t = b/SE(b) = -1.862/.2651 =−7.02 [Note: SE(b)=s/sqrt(268.4), s= sqrt(SSE/n-2)= 4.3429, SE(b) =0.2651], df = 13; p-value < .0005; Conclusion: There is a negative linear relationship between selling price and age.

b. Give a 95% confidence interval for the predicted value for a 5-year old Corvette to be listed in the next week’s paper.

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Answer: 95% Prediction interval is:.

$34350. to14490$)35.34,49.14(39.942.24

4.268)8.85(

151

1)3429.4()16.2(42.24)(

)(11ˆ

2

2

2**

==±=

−++××±=

−−

++××±∑ xx

xxn

sty

c. Is it justifiable to predict the selling price of a 20-year old Corvette from the fitted regression line? Give reasons for your answer.

Answer: No, we cannot predict beyond the range of data. The linear relation cannot hold for cars that old because we know the selling price never goes below.

6. [10 pts] Three models of automobiles were tested for fuel efficiency as follows. Exactly three litres of

gasoline were placed in the gasoline tank of a car. The car was then driven until the fuel was used up. The number of kilometers traveled for each model was recorded for several tests each. The following data were collected.

Car Model A B C Sample size 4 6 8 Sample mean 18 16 20.25 Sample Sd. Dev. 1.633 1.414 2.053

Is there any significant difference in average distance traveled on 3 litres of fuel for these three models? Test appropriate hypotheses using α = 0.01.

Answer: H0 : µ1 = µ2 = µ3 , Ha : Not all three means are equal. Test statistic: F = MSG/MSE

17.315

50.47318

053.27414.15633.13)1(

25.31250.62

13)33.1825.20(8)33.1816(6)33.1818(4

1

)(

33.1818330

18

2222

2222

==−

×+×+×=−−

=

==−

−+−+−=−

−=

===

In

snMSE

I

yynMSG

yny

ii

ii

ii

F = MSG/MSE = [31.25/3.17 = 9.86, with degrees of freedom df1 = 2, df2 = 15. P-value = P( F ≥ 9.86) < 0.01. Conclusion: There is significant difference in average distance traveled on 3 litres of fuel for these three models.

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7. [10 pts] A fruit grower wants to test a new spray that is claimed (by a manufacturer) to reduce the loss due to damage by insects. To this end, the grower performs an experiment with 27 trees in her orchard by treating 12 of those trees with the new spray and the other 15 trees with the standard spray. From the data of fruit yield (in pounds) of those trees, the following summary statistics were found.

New spray Standard Spray Mean yield per tree (lbs)

249 233

Standard deviation

19 45

(a) Do these data provide sufficient evidence to conclude that the mean yield per tree treated with the

new spray exceeds that for trees treated with the standard spray? State the assumptions you make and test at α = 0.05.

Answer: Let µ1 and µ2 are the mean yield per tree treated by the new spray and the standard spray, respectively. Assumptions: We assume that (i) the 12 trees and 15 trees are randomly selected from the orchard and (ii) the yield per tree is distributed normally. Hypotheses : H0 : µ1 = µ2 vs Ha : µ1 > µ2

Test Statistic t = ( x 1 − x 2)/√[s1

2 /n1 + s22 /n2]

= (249 − 233) /√[(19)2 /12 + (45)2 /15] = 1.25 df = 11, which is the smaller of the two (n1−1) and (n2 −1).

P-value : 0.10 < P( t ≥ 1.25) < .15. (p-value > α = 0.05)

Conclusion: The sample data do not support the manufacturer’s claim that the mean yield per tree ( treated by the new spray )will increase by the new spray.

(b) Construct a 95% confidence interval for the difference in mean yields between the new spray and

the standard spray.

Answer: The 95% confidence interval estimate for µ1 - µ2 is = ( x 1 − x 2) ± t*√[s1

2 /n1 + s22 /n2] (df = 11)

=249 − 233 ± 2.201*√[(19)2 /12 + (45)2 /15]

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= 16 ± 28.28 = (−12.28, 44.28)

8. [10 pts] In a study of the relationship between temperament and personality, 49 female high school students who had a high level of reactivity (HRL) and 54 students who had a low level of reactivity (LRL) were classified according to their attitude to group pressure with the following results.

Attitude Submissive Resistant HRL 33 16 LRL 12 42

(a) Is resistance to group pressure significantly lower in the HRL group than the LRL group?

Answer: Let p1 and p2 be the probabilities of resistance for the HRL and LRL groups, respectively.

Hypotheses : H0 : p1 - p2 = 0 Vs. Ha : p1 - p2 < 0

Test statistic : )/1/1/()1(/)( 2121 nnppppz +−−= ))))

1p) = X1/n1= 16/49 = 0.3265, 2p) = X2/n2 = 42/54 = 0.7778,

p) = (X1 + X2)/(n1 + n2) = (16 +42/(49 + 54) = 0.5631

∴ )/1/1/()1(/)( 2121 nnppppz +−−= )))) = −4.61

P-value : P-value = P(Z ≤ -4.61) < 0.0002.

Conclusion : The data strongly support the conclusion that the probability of resistance in the HRL group is significantly lower than that of in the LRL group.

(b) Determine a 99% confidence interval for the difference between the proportions of resistant females in the HRL and LRL populations.

Answer: 99% confidence interval estimate for p1 - p2 is

)2255.,6775.(2258.4513.54

)2222(.7778.49

)6735(.3265*576.24513.

)p̂-ˆ(ˆˆ 21*

21

−−=±−=

+±−=

±− pSEzpp

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9. [10 pts] Suppose the following display represents 500 automobile accidents that occurred in a large city:

No fatalities At least one fatality Involved alcohol 68 142 No alcohol 194 96

Determine whether there is a relationship between alcohol and fatality? Use α = 0.01 in your test. Answer: Hypotheses are: H0 : Fatality is not related to use of alcohol vs Ha: Fatality is related to use of alcohol. Under the null hypothesis of independence, the expected frequencies are: E(1,1) = 110.04, E(1,2) = 99.96, E(2,1) = 151.96, E(2,2) = 138.04 and χ2 = 58.175, df = 1, p-value = 0.0. Conclusion: There is strong evidence to conclude that there is an association between alcohol use and fatality.

10. [10 pts] Does alcohol affect the ability to think? A random sample consisting of 11 automobile drivers was selected to study whether or not alcohol has some effect on time to complete a puzzle. Under one scenario, the person would drink a beverage that contained no alcohol; under another scenario, the person would drink a beverage with alcohol. Each person's time to complete the puzzle was recorded. The following data were obtained:

Driver 1 2 3 4 5 6 7 8 9 10 11

No alcohol 7.1 6.3 6.8 8.4 6.9 8.5 7.3 7.7 8.1 7.4 6.6

With Alcohol 7.4 6.2 6.6 9.3 7.2 8.8 7.6 7.9 8.7 7.9 7.0

Do the data provide any evidence to conclude that more time is required to complete the puzzle after consuming alcohol? Test the appropriate hypotheses at α = 0.01.

Answer: Let µ = µ1 − µ2 = µwithout − µwith = the difference between the average completion times of the puzzle. Hypotheses are: H0 :µ = 0 vs Ha: µ < 0.

Using paired t-procedure, t =-3.49 and p-value < 0.005 (less than α = 0.01) Conclusion: The data provide sufficient evidence to indicate that it requires more time, on average, to complete the puzzle after consuming alcohol.