M117 Sample Final - for Betsy Review - with answers-revised-final PDF

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MAT 117 CRN SAMPLE FINAL

DR. MIRANDA 20 DEC 2018

1 (4 pts). The following histogram represents the number of books Lisa Simpson reads per month. What is the frequency of books Lisa read between MAR 1 and MAY 31?

A. B. C. D.

3 24 33 13

Answer: C Explanation: Add the frequency (number) of books for each month: MAR = 11, APR = 13, and MAY = 9. So, the total is 11+13+9 or 33. 2 (4 pts). In a landmark study, the “hero return,” a new measure of performance being studied by Marvel University Professor Anna Marconi, was determined for several superhero and villains. The covariance of the hero return between Professor Xavier and Magneto is –0.108. The standard deviation of the rates of hero return is 0.96 for Professor Xavier and 0.11 for Magneto. The correlation of the hero return between Professor Xavier and Magneto is closest to _____. A. –0.12 B. 0.12 C. –1.02 D. 1.02 Answer: C Explanation: The sample correlation coefficient is defined as

r xy=

S xy −0.108 = =−1.02 (0.96)×(0.11) Sx S y

3 (4 pts). The following data represents the actual daily playing time, in hours, for 11 X-Box One gamers playing Fortnite: Battle Royale. 5.1 19.1 21.6 9.5 20.3 18.0 4.6 5.2 11.9 19.7 18.5 What percentile is the daily playtime of 15.6 hours? A. 38th

Pace University B. C. D.

MAT 117 CRN SAMPLE FINAL

DR. MIRANDA 20 DEC 2018

50th 54th 64th

Answer: B Explanation: Arrange data in ascending order. 4.6

5.1

5.2 9.5

11.9 15.6

18.5 19.1

19.7 20.3

21.6

L p ≈ 6 ; n=11 To find

p , we rearrange

L p= (n+1 )

p 100

to

p=L p

100 600 100 = =6 × =50 11+1 12 n+1

4 (8 pts). Restaurants in London, Paris, and New York want diners to experience eating in pitch darkness to heighten their senses of taste and smell (Vanity Fair, December 2011). Suppose 400 people were asked, “If given the opportunity, would you eat at one of these restaurants?” The accompanying contingency table, cross-classified by age, would be produced.

Would you eat in a pitch dark restaurant?

AGE GROUP 18–29

30–44

45–64

65+

TOTALS

Yes

50

39

76

65

230

No

50

61

24

35

170

100

100

100

100

400

TOTALS

a. Fill in the table. Use this information to create a joint probability table.

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Yes No Total

MAT 117 CRN SAMPLE FINAL 18− 29 0.125 0.125 0.250

30− 44 0.0975 0.1525 0.250

45− 64 0.19 0.06 0.250

DR. MIRANDA 20 DEC 2018 65+ 0.1625 0.0875 0.250

Total 0.575 0.425 1.000

b. What is the probability that a respondent would eat at one of these restaurants? We find the relative frequency of people who answered yes to the question, “Would you eat in a pitch-dark restaurant?” P(Yes) = (50 + 39 + 76 + 65)/400 = 0.575 c. What is the probability that a respondent would eat at one of these restaurants or is in the 30–44 age bracket? P(Yes OR 30 – 44) = (50 + 39 + 76 + 65 + 61)/400 = 0.7275 d. Given that the respondent would eat at one of these restaurants, what is the probability that he or she is in the 30–44 age bracket? P(30 – 44|Yes) = 39/(50 + 39 + 76 + 65) = 0.1696 e. Determine if patron age and restaurant attendance are independent? Explain using probabilities. Not independent because P(30 – 44) = 0.25 and so P(30 – 44) ≠ P(30 – 44│Yes) 5 (4 pts). According to ScreenRant Analytics, 6.28% of web traffic for screenrant.com comes from social media. A random sample of eight site-visits were selected. What is the probability that fewer than three of the site-visits came from social media? A. 0.718 B. 0.783 C. 0.989 D. 0.825 Answer: C Explanation: For a binomial random variable X, the probability of x successes in n Bernoulli trials is calculated as

P ( X=x )=

n! n! p x (1−p)n −x p x q n−x = x ! ( n− x ) ! x ! ( n−x ) !

Therefore, in this case, we need to find:

P ( X z )=1−P ¿

P ( X >0.10)= 1− P ( X ≤0.10 )=1−.6879 =0 .0764.

7 (4 pts). Selection bias occurs when _______. A. the population has been divided into strata B. portions of the population are excluded from the consideration for the sample C. cluster sampling is used instead of stratified random sampling D. those responding to a survey or poll differ systematically from the non-respondents Answer: B Explanation: Selection bias refers to a systematic underrepresentation of certain groups from consideration for a sample. 8 (4 pts). The average ticket price for a movie ticket was $8.97 in 2017, with a standard deviation of $1.89. If a random sample of 50 movie tickets from various areas is taken, what's the probability that the mean cost of a movie ticket is between $9.00 and $9.50? A. B. C. D.

0.5517 0.4168 0.4323 0.1515

Answer: C Explanation: As a general guideline for the central limit theorem, the normal distribution approximation ´ is normal, we can transform it into a standard normal random is justified when n ≥30 . If X variable as

z=

´x −μ σ √n

, and any value of

´x on

´x −μ σ . Compute P($ 9.00 ≤ ´x ≤ $ 9.50) . n √ x´2−μ 9.50−8.97 0.53 = =1.983 z 2= = 0.2673 σ 1.89 √n √ 50 z=

´ X

has a corresponding value z on Z given by

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z 1=

MAT 117 CRN SAMPLE FINAL

DR. MIRANDA 20 DEC 2018

x´1−μ 9.00−8.97 0.03 = =0.1122 = 1.89 0.2673 σ √n √ 50

Use z table. Note that

P ( $ 9.00 ≤ Z ≤ $ 9.50 )=P ( Z ≤ $ 9.50 )−P ( Z ≤ $ 9.00 )=0.9761−0.5438 =0.4323 .

9 (4 pts). Suppose that, on average, mathematicians earn approximately µ = $84,000 per year in the United States. Assume that the distribution for mathematicians’ yearly earnings is normally distributed and that the standard deviation is σ = $18,000. What is the probability that the average salary of four randomly selected mathematicians is less than $80,000? A. 0.2200 B. 0.3707 C. 0.6293 D. 0.3300 Answer: D Explanation: If

´ X−μ Z= σ √n

X´

and any value of ´x

´x −μ Z= σ √n

is normal, we can transform it into a standard normal random variable as

on

´ X

has a corresponding value

z

on

Z

given by

P( X´ p0. When testing the population proportion, the value of the test statistic is computed as

z=

´p− p0 . For a left-tailed test, the p-value is computed as P ( Z ≤ z) . √ p0 (1− p0 )/ n

The decision rule is to reject the null hypothesis if the p-value < α and not reject the null hypothesis if the p-value ≥ α.

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z-score

¿

MAT 117 CRN SAMPLE FINAL

0.225−0.14 √0.14 (1−0.14 )/ 40

DR. MIRANDA 20 DEC 2018

= 1.5493

In this case, you would check the z-table of 1.55 = 0.0607 p-value = 1- = 0.0607 13 (4 pts). A concert analyst is interested in optimizing a band’s vendor list by showing that the mean attendance of the band’s concerts exceeds 350. (HINT: this is the analyst’s hypothesis). A sample of 36 distinct concert venue managers is surveyed and the analysist found that the mean attendance for the band was 360. The population standard deviation is assumed to equal 40. If the analyst wants to prove her hypothesis with a 95% confidence level (i.e., at a 5% significance level), what kind of decision should she make. A. She rejects H 0 , she can conclude that the mean attendance of the band’s concerts exceeds 350 B. She rejects H 0 , she cannot conclude that the mean attendance of the band’s concerts exceeds 350 C. She does not reject H 0 , she can conclude that the mean attendance of the band’s concerts exceeds 350 D. She does not reject H 0 , she cannot conclude that the mean attendance of the band’s concerts exceeds 350 Answer: D Explanation: When testing if the population mean and standard deviation is known, the value of the test statistic is computed as z=

x´ −μ 0 . σ /√ n

For a right-tailed test, the p-value is computed as P(Z ≥ z). The decision rule is to reject the null hypothesis if the p-value < α and not reject the null hypothesis if the p-value ≥ α z-score =

360 −350 =15.000 . 40/ √ 36

14 (4 pts). A NYC museum has two membership levels for the summer season. The elite level allows daily visits for entire 90-day summer season, and the express level, which allows 24 weekend visits for the entire 90-day summer season. The cost of the elite level is $180, while the cost of the express level is $96. What type of hypothesis test should we use to test if whether the elite level is a better bargain than the express level if the cost of the weekday visit is $10? A. A hypothesis test for p1− p2 . B. A hypothesis test for μ1 − μ2 . C. A matched-pairs hypothesis test. D. We are unable to conduct a hypothesis test because independent random samples of each group could not be collected. Answer: A Explanation: A “bargain” is a qualitative response, so a test about the difference of two proportions should be done. 15 (6 pts). The following table shows the annual returns (in percent) for the Fidelity Utilities fund from 2006 through 2010.

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MAT 117 CRN SAMPLE FINAL

Year Utilities 2006 30.52 2007 10.83 2008 −34.55 2009 11.08 2010 17.33 Mean 7.042 a. Calculate the point estimate of σ 2 . b. Construct a 95% confidence interval of

DR. MIRANDA 20 DEC 2018

σ2 .

Answer: a.

2 ∑ ( x i−´x ) s= 2

n−1

=604.40 b. [216.96, 4990.74]

2 2 2 2 2 ( 30.52 −7.042 ) + (10.83 −7.042 ) +( −34.55−7.042 ) +( 11.08 −7.042 ) + ( 17.33−7.042) = 5−1

Explanation: A point estimate of the population variance is computed as

2

s=

∑ ( x i−´x )2 n−1

.A

100 (1−α ) % confidence interval of the population variance σ 2 is computed as 2 2 ( n−1) s ( n−1 ) s ( 5−1 ) (604.40) ( 5−1) (604.40) , 2 = =[216.96, 4990.74] , 2 11.143 . 484417 χ 1−α /2 ,df χ α / 2 ,df

[

] [

]

16 (4 pts). A card-dealing machine deals spades (1), hearts (2), clubs (3), and diamonds (4) at random as if from an infinite deck. In a randomness check, 1,600 cards were dealt and counted. The results are shown below. Observe Suit d Spades

410

Hearts

405

Clubs

370

Diamond 415 s To test if the poker-dealing machine deals cards at random, the null/alternative hypotheses are: A. B. C. D.

H 0 : p 1= p 2= p3= p 4=0 ; H A : Not all population proportions are equal to (0,25) H 0 : p 1= p 2= p3= p 4=¿ 0.25; H A : Not all population proportions are equal to (0,25) H 0 : p 1= p 2= p3= p 4=¿ 1; H A : Not all population proportions are equal to (0,25) H 0 : p 1= p 2= p3= p 4=0.2 ; H A : Not all population proportions are equal to (0,25)

Answer: B

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MAT 117 CRN SAMPLE FINAL

DR. MIRANDA 20 DEC 2018

Explanation: When setting up the competing hypotheses for a multinomial experiment, we have essentially two choices. We can set all population proportions equal to some specific value, or equal to each-other. The sum of the category probabilities for a multinomial experiment is

p1+ p 2+…+ pk =1

Therefore, for four suits, we have

p1+ p 2+ p3 + p 4=1 , so

p1= p2= p3 =p 4=0.25

17 (9 pts). The following table shows the cross-classification of accounting practices (either straight line, declining balance, or both) and country (either France, Germany, or United Kingdom). Observed Franc German e y

United Kingdom

TOTAL 61

Straight Line

20

11

30

Declining Balance

14

16

15

Both

15

23

13

51

TOTAL

49

50

58

157

45

Expected TOTAL

Straight Line 19.03821656

19.42675159

United Kingdom 19.03821656

Declining Balance

14.04458599

14.33121019

16.62420382

45

Both

15.91719745

16.24203822

18.84076433

51

France

TOTAL

49

Germany

50

58

61

157

a. Set up the competing hypotheses to determine if accounting practice and country are dependent. b. Calculate the value of the test statistic and determine the degrees of freedom. c. Compute the p-value. Does the evidence suggest market capitalization and objective are dependent at the 1% significance level? Answer: a. H0: Accounting practice and country are independent; HA: Accounting practice and country are dependent b.

( o ij−eij )

2

(20−19.04 )2 ( 11−19.43) 2 ( 30−19.04 ) 2 ( 14−14.04 )2 ( 16 −14.33 ) 2 ( 15−16.62 + + + + + 19.04 19.43 19.04 14.04 14.33 16.62 e ij The degrees of freedom are df = (3−1 ) (3−1 ) =4 χ =∑ i ∑ j 2 df

=

c. The p-value is between 0.01 and 0.05. To find the p-value, you go to Chi-Square Table and find where

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MAT 117 CRN SAMPLE FINAL

DR. MIRANDA 20 DEC 2018

11.205 would lie between given df =4. Because the p-value is less than the significance level, we reject the null hypothesis and we conclude that the accounting practice and country are dependent. Explanation: The null hypothesis for the chi-square test for independence implies that two variables are independent and the alternative hypothesis implies that two variables are dependent. For a contingency table with r rows and c columns, the test statistic for a test of independence is computed as

( o ij −eij )

2

χ =∑ i ∑ j 2 df

e ij

with df = (r−1 ) (c−1 ) .

The critical value for the right-tailed test

can be found using χ df2 distribution table. The p-value can be found using the χ df2 distribution table, Excel or R. Having the p-value defined, the decision rule is to reject the null hypothesis if the p-value is less than the significance level α; do not reject the null hypothesis if the p-value is greater than the significance level α. The chi-square test of independence is valid only when the expected frequencies in each cell of the contingency table are five or more.

χ α2 ,df

18 (4 pts). Over the past 30 years, the sample standard deviations of the rates of return for stock X and Stock Y were 0.20 and 0.12, respectively. The sample covariance between the returns of X and Y is 0.0096. To determine whether the correlation coefficient is significantly different from zero, the appropriate hypotheses are A) H0: μ = 0, HA: μ ≠ 0 B) H0: ρxy = 0, HA: ρxy ≠ 0 C) H0: μ = 1, HA: μ ≠ 1 D) H0: ρxy = 1, HA: ρxy ≠ 1 Answer: B Explanation: The competing hypotheses are H0: ρxy = 0, HA: ρxy ≠ 0. 19 (4 pts). A statistics student is asked to estimate y = β0 + β1x + ε. She calculates the following values: = 440, = −568, Which of the following is the sample regression equation?

,

= 1,120, n = 11.

A) = 9.2 - 1.29x B) = 9.2 + 1.29x C) = 60.80 - 1.29x D) = 60.80 + 1.29x Answer: C Explanation: The slope b1 and the intercept b0 of the simple regression equation are computed as b1 =

and b0 = − b1 . The simple linear regression equation is = b0 + b1x.

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MAT 117 CRN SAMPLE FINAL

DR. MIRANDA 20 DEC 2018

20 (4 pts). Consider the following sample regression equation = 150 – 20x, where y is the demand for Product A (in 1,000s) and x is the price of the product (in $). The slope coefficient indicates that if ________. A) the price of Product A increases by $1, then we predict the demand to decrease 20 B) the price of Product A increases by $1, then we predict the demand to increase by 20 C) the price of Product A increases by $1, then we predict the demand to decrease by 20,000 D) the price of Product A increases by $1, then we predict the demand to increase by 20,000 Answer: C Explanation: The slope parameter b1 indicates that if x increases by 1 unit then y is predicted to change by b1....