1

Math 140C Final Exam Fall 2017 Name: Solutions

VERSION C Instructor:

Days and Time of Class:

INSTRUCTIONS

• When capitalized words are separated by slashes ( / ), circle the word(s) that

CORRECTLY / WRONGLY completes the sentence.

• SHOW ALL WORK, including needed formulas and calculator commands. • Please write in pencil and erase anything that you don’t want graded. • Graphing calculators are allowed only if nothing has been stored in them. Books and

notes are NOT allowed.

• You may tear off the formula sheet from the back of the exam and use it as scratch paper, but turn it in with your exam. Nothing on it will be graded.

• Please raise your hand when finished and pack up after your exam has been collected. • No cell phones may be turned on or in sight at any time during the exam. If your phone is visible during

the exam, your papers will be collected and you will receive an F on the exam.

Problem Possible Points Points Scored

1 5

2 4

3 9

4 4

5 9

6 14

7 8

8 12

9 15

10 8

11 12

TOTAL 100

2

1. According to legend, if groundhog Punxsutawney Phil sees his shadow on February 2, the weather will stay wintery. If Phil doesn’t see his shadow, the U.S. will have an early spring. This bar graph gives the results for the last 29 years. www.ncdc.noaa.gov/

customersupport/education-resources/groundhog-day

a. Complete the table so that it gives the same information as in the bar graph.

Weather: Saw Shadow Didn’t See Shadow Total

Early Spring 12 6 18 Wintery 8 3 11

Total 20 9 29

Write answers below as unreduced fractions.

b. In what proportion of all 29 years did Phil see his shadow? 20/29

c. In what proportion of the years that stayed wintery did Phil see his shadow? 8/11

d. If Phil saw his shadow, in what proportion of the years did Phil’s observation correctly predict the weather? 8/20

2. In the U.S., the rate of mothers with at least one chronic condition (such as diabetes or

substance abuse) is 91.8 per 1000 childbirths. There are about 4,000,000 childbirths per year in the U.S. How many of these childbirths involve a mother with at least one chronic condition? Show your work and round your answer to the nearest whole number. journals.lww.com/greenjournal/Citation/2017/12000/Disparities_in_Chronic_Conditions_Among_Women.19.aspx

𝟗𝟏.𝟖𝟏𝟎𝟎𝟎

∙ 𝟒, 𝟎𝟎𝟎, 𝟎𝟎𝟎 = . 𝟎𝟗𝟏𝟖 𝟒, 𝟎𝟎𝟎, 𝟎𝟎𝟎 = 𝟑𝟔𝟕, 𝟐𝟎𝟎

OR

proportion = 𝒑𝒂𝒓𝒕𝒘𝒉𝒐𝒍𝒆

so 𝟗𝟏.𝟖𝟏𝟎𝟎𝟎

= 𝒑𝒂𝒓𝒕𝟒,𝟎𝟎𝟎,𝟎𝟎𝟎

and solving, part = 367,200

02468

101214

Saw Shadow Didn’t See Shadow

Freq

uenc

yEarly Spring Wintery

3

3. These boxplots show the number of deaths that resulted from the last 54 hurricanes to hit the U.S. Half of the hurricanes had a male name and half had a female name. www.statisticsteacher.org/2017/09/15 /lesson-plan-are-female-hurricanes-deadlier-than-male-hurricanes/

a. Sandy was the hurricane that caused the most deaths. About how many people were killed? 159

b. Which gender had the largest interquartile range (IQR)? male

Estimate the IQR for that gender, showing your work.

Q3 – Q1 = 15 – 1 = 14

c. Place one of the following numbers in each blank. One of the numbers will be used twice:

5 11.5 17

The median number of deaths for the 27 hurricanes with female names. 5

The median number of deaths for the 26 hurricanes with female names after Sandy is removed. 5

The mean number of deaths for the 27 hurricanes with female names. 17

The mean number of deaths for the 26 hurricanes with female names after Sandy is removed. 11.5

d. People say that hurricanes with female names tend to have more deaths.

These data DO / DO NOT support that statement because (circle one): A. The hurricanes with female names have the higher median.

B. The largest number of deaths was from a hurricane with a female name.

C. The two distributions have about the same shape, including having the same number of outliers.

D. Except for Sandy, the number of deaths generally are no higher for hurricanes with female names than for hurricanes with male names.

4

4. From January 2000 through December 2015, California had 1,163 large wildfires (meaning that they burned over 300 acres). What was the average number of large wildfires per year? Give your answer to four decimal places.www.dailynews.com/2017/12/06/heres-how-rare-it-is-to-have-wildfires-in-december-in-california/

x =x∑

n = 1,163/16 years = 72.6875 large wildfires per year on average

5. In 2017, SAT-Total scores were

normally distributed with a mean of 1060 and a standard deviation of 195. reports.collegeboard.org/pdf/2017-total-group-sat-suite-assessments-annual-report.pdf

Record your calculator commands and round answers to two decimal places.

a. What percentage of SAT-Total scores were 1200 or higher? Shade the appropriate area on the graph and then find the percentage.

normalcdf (lower, upper,µ,σ )=normalcdf (1200, 99999, 1060, 195)

≈ .2364

= 23.64% OR 24%

b. What score is at the 25th percentile?

invNorm (area,µ,σ )= invNorm (.25, 1060, 195) ≈ 928.47

Write a one-sentence interpretation of the 25th percentile. 25% of all scores are [at or] lower than 928.47 and 75% of all scores are higher than 928.47.

5

6. On a common final exam, CSUN students got a mean of 62.3 with standard deviation 18.5. This histogram shows the distribution of the population of all 367 scores.

a. The shape of this distribution is (circle one)

NORMAL SKEWED LEFT SKEWED RIGHT UNIFORM

b. About what percentage of students got a score of at least 90?

4% = 2% = 6% OR 3.9% + 1.9% = 5.8%

c. About how many students got a score of 50 up to, but not including, 55? about 7.6% of 367 or .076(367) ≈ 27.892 or about 28

This histogram shows an approximate

sampling distribution for the sample mean, made from 400 random samples, each of size 10.

d. The largest sample mean is in the interval from 78 up to, but not including, 80

e. Theoretically, the mean of the sampling distribution is equal to 62.3 OR the mean of the population .

f. Use the appropriate formula to compute the standard error of the sampling distribution, to two decimal places.

SE = σn

= 𝟏𝟖.𝟓𝟏𝟎

≈ 5.85

g. Why isn’t the shape of the sampling distribution approximately normal? (Circle one.)

A. The sample size is only 10. B. The sample size is only 400. C. There are only 10 samples. D. There are only 400 samples. E. The population isn’t normal so a sampling distribution can’t be approximately normal.

6

7. A poll of 1,870 Millennials (Americans age 18–34) found that 73% said they are scared or concerned about what Donald Trump is doing as president. genforwardsurvey.com/assets/uploads/2017/11/NBC-GenForward-November-2017-Toplines-Final.pdf

a. Compute a 95% confidence interval for this situation. If you use the calculator, give the command and inputs. If you use the formula, give it, substitute numbers, get the final answer. Round to three decimal places and write the interval on the line in the form (lower bound, upper bound).

Using 1-PropZInt with x = .73(1870) ≈ 1365, n = 1870 and C-level .95, you get (.70892, .75007).

OR

Using 𝒑 ± z* (𝒑)(𝟏;𝒑)𝒏

= .73 ± 1.96 (.𝟕𝟑)(𝟏;.𝟕𝟑)𝟏𝟖𝟕𝟎

≈ .73 ± .020, you get the

interval (.710, .750) 95% confidence interval: ( .710 , .750 )

b. We are 95% confident that some percentage is in this interval. Circle that percentage.

A. 73%

B. 95%

C. the percentage of all Millennials who would say they are scared or concerned D. the percentage of all Millennials in the sample who said they are scared or concerned

E. the percentage of all Millennials in a different random sample who would say they are scared or concerned

c. Compute the margin of error for this 95% confidence interval and write the confidence interval in the form, observed value ± margin of error.

Show the necessary work here:

The margin of error is half the length of the CI: .750 – .710)/2 = .02

OR The margin of error is upper bound – 𝒑 = .750 – .73 = .02

OR Using 𝒑 ± z* (𝒑)(𝟏;𝒑)𝒏

= .73 ± 1.96 (.𝟕𝟑)(𝟏;.𝟕𝟑)𝟏𝟖𝟕𝟎

≈ .73 ± .020

observed value ± margin of error = .73 ± .02 OR 73% ± 2%

d. If a larger number of Millennials had been polled but the percentage who said they are scared or concerned had been about the same, a 95% confidence interval would be WIDER / NARROWER / THE SAME WIDTH as in part a.

The reason for this is that the VALUE OF Z / ESTIMATED SE for the larger sample size would be LARGER THAN / SMALLER THAN / THE SAME AS for the original sample size.

7

8. Fifty-five percent of all children in the U.S. are vaccinated against flu. In a sample of 291 children who died of flu, 75 had been vaccinated. Is this statistically significant evidence that children who die of flu are less likely to have been vaccinated than are all children? www.nbcnews.com/health/health-news/most-kids-who-died-flu-weren-t-vaccinated-study-finds-n742046

a. What is 𝑝? 75/291 ≈ .2577 n ? 291 x ? 75

b. Write the hypotheses.

Ho: p = .55 Ha: p < .55

c. What does the symbol stand for in your hypotheses in part b? (Circle one.)

A. .26

B. .55

C. the proportion of all children who are vaccinated

D. the proportion of all children in the sample who had been vaccinated

E. the proportion of all children who die of the flu who are vaccinated

d. If the null hypothesis is true, this graph shows the sampling distribution of 𝑝 .

Describe or show where 𝑝 is located on this distribution. 𝒑 =. 𝟐𝟓𝟕𝟕 is located on the number line way to the left of 0.46 OR it’s about where this X is. OR it’s almost 10 SE below the mean of .55.

e. Compute the p-value on your calculator, giving the name of the test you used.

test name: 1-PropZTest p-value: 0

f. In each sentence below circle the best word(s) in each capitalized group to complete a conclusion. Use a significance level of α = .05.

• Because the p-value is LESS / GREATER than .05, the null hypothesis should be ACCEPTED / REJECTED / NOT REJECTED.

• There IS /I S NOT statistically significant evidence that children who die of flu are less likely to have been vaccinated than are all children.

• Having 75 children who are vaccinated out of a sample of 291 who died of flu

IS / IS NOT consistent with the claim that children who die of flu are as likely to have been vaccinated as are all children.

0.64

8

9. Samples of social science and health science students at a large university in South Africa were rated as to how often they used social media (like Facebook and Twitter) during lectures. A larger rating represents more use. The ratings of the 223 social science students had a mean of 6.42 with standard deviation 2.39. The ratings of the 263 health science students had a mean of 6.34 with standard deviation 2.21. Is this statistically significant evidence that social students and health science students have different mean ratings? www.sciencedirect.com/science/article/pii/S0747563217304983

The hypotheses are Ho: 𝜇@ = 𝜇A and Ha: 𝜇@ ≠ 𝜇A where subscript 1 represents the mean for all social science students and the subscript 2 represents the mean for all health science students.

a. Assume the null hypothesis is true:

• The mean of the sampling distribution of x1 – x2( ) is 0 . • Complete the scale on this sampling distribution.

The SE is 0.21.

• Compute x1 – x2( ) and mark it with an arrow on the distribution.

x1 – x2( )= 6.42 – 6.34 = .08

• Shade in the area that represents the p-value.

b. Compute the p-value on your calculator, giving the name of the test you used. Round to four

decimal places

test name: 2-SampTTest p-value: .7037

c. Circle the right words and fill in the blanks to complete this interpretation of the p-value.

If the mean rating of all social science students is EQUAL TO / DIFFERENT FROM the mean rating of all health science students, then there is a .7037 chance of getting a difference of .08 or even larger (in absolute value) in the means from random samples of these sizes.

d. Circle the right word(s) in each capitalized group to complete a conclusion. Use 𝛼 = .05.

We ACCEPT / REJECT / DO NOT REJECT the null hypothesis.

There IS / IS NOT statistically significant evidence that the mean rating of all social science students is different from the mean rating of all health science students.

9

e. For the 223 social science students, the researchers also computed the correlation between a student’s rating on use of social media during lectures and the student’s grade point average (GPA). The correlation was –.174. Circle best word(s) in each capitalized group.

Students who use social media more often during lectures tend to have GPAs that are HIGHER THAN / LOWER THAN / EQUAL TO the GPAs of students who use social media less often.

This relationship is STRONG / MODERATE / WEAK. 10. This table gives the results from a study that investigated the relationship between tattoos and

Hepatitis C. A sample of 626 patients undergoing medical evaluation at a Dallas spinal clinic (for reasons unrelated to hepatitis) were tested for hepatitis C, examined for tattoos, and questioned about where they got their tattoos. University Of Texas Southwestern Medical Center At Dallas. "Tattooing A Major Route Of Hepatitis C Infection, UT Southwestern Researcher Finds." ScienceDaily, 5 April 2001. www.sciencedaily.com/releases/2001/04/010405081407.htm. Published in Medicine.

Tattoo from

Tattoo Parlor Tattoo from

Somewhere Else No Tattoo Total

Hepatitis C 17 8 22 47 No Hepatitis C 35 53 491 579

Total 52 61 513 626

a. Circle the correct word in each capitalized pair to give the hypotheses for a 𝜒A test.

Ho : The two variables are ASSOCIATED / INDEPENDENT.

Ha : The two variables are ASSOCIATED / INDEPENDENT.

ASSOCIATED / INDEPENDENT means that in the population each of the three groups (tattoo parlor, from somewhere else, no tattoo) are equally likely to have hepatitis C.

b. Use the formula to compute the expected count for the cell containing patients who have no tattoo and have hepatitis C. Show your work, including the formula. Round to two decimal places.

E =

row total( ) ⋅ column total( )grand total( ) =

(𝟒𝟕)(𝟓𝟏𝟑)𝟔𝟐𝟔

≈ 38.52

c. The chi-square statistic is 57.9122. The p-value is less than 0.0001. Circle the best words

in each capitalized group to complete a conclusion.

Using a significance level of α = .05, we REJECT / DO NOT REJECT / ACCEPT the null hypothesis.

We DO / DON’T have statistically significant evidence that getting a tattoo and hepatitis C are INDEPENDENT / ASSOCIATED.

10

11. The scatterplot shows the number of high school graduates and the number of GEDs (General Equivalency Diploma) awarded in eleven California counties during a recent year. www.cde.ca.gov/ds/sd/

The line plotted is the regression line.

The regression equation is GEDs = 0.00113 number of graduates + 23.771.

a. Circle the best estimate of the correlation. –.99 –.72 –.19 0 .19 .72 .99

b. Use the regression equation to predict the number of GEDs for a county that has 30,000

students. Show the needed work below.

GEDs = 0.00113 number of graduates + 23.771 = 0.00113 (30,000) + 23.771 = 57.671

c. What is the slope of the regression line? 0.00113

For every additional graduate , the regression line predicts the

number of GEDs will INCREASE / DECREASE by about 0.00113

d. Circle the point on the scatterplot that has the smallest residual.

Estimate the number of GEDs for the county represented by this point. 34 GEDs

11

Math 140 Formulas

sample mean: x =x∑

n

sample standard deviation: s =(x − x )2∑n −1

z-score: z =x − xs

interquartile range: IQR = Q3 – Q1

standard error of sampling distribution of p̂ : SE = p 1− p( )n

standard error of sampling distribution of x : SE = σn

general confidence interval: observed value ± margin of error

confidence interval for p: p̂ ± m where m = z*⋅ p̂ 1− p̂( )

n

confidence interval for µ : x ± m where m = t *sn

general test statistic:

observed value− hypothesized valueSE

one-proportion z-test statistic: z = p̂ − p0

SE where SE =

p0 1− p0( )n

two proportion z-test statistic: z =p̂1 − p̂2( )− 0

SE

where SE = p̂ 1− p̂( ) 1n1

+ 1n2

⎛⎝⎜

⎞⎠⎟

and p̂ = number of successes in both samplesn1 + n2

12

t-test statistic for a mean: t = x − µSEEST

where SEEST =sn

two sample t-test statistic: t =x1 – x2( )− 0SEEST

where SEEST =s12

n1+s22

n2

chi-square statistic: χ 2 =O − E( )E

2

∑ where E =

row total( ) ⋅ column total( )grand total( )

correlation: r =1

n −1x − xsx

⎛⎝⎜

⎞⎠⎟

y − ysy

⎛

⎝⎜⎞

⎠⎟∑ = 1n −1 zx ⋅ zy∑

regression equation: predicted = a + bx

regression coefficients: b = r

sysy

a = y − bx

Calculator Commands

normalcdf(lower,upper,µ,σ )

invNorm(area,µ,σ )