Homework

A recent survey of 273 randomly selected teens living in Massachusetts asked about social media. Sixty-three said that they routinely use a certain app to share pictures. The researchers want to determine if there is good evidence that more than 30 percent of teens use this app. The alternative hypothesis is as follows:

1. p < 0.30
2. p ≤ 0.30
3. p ≥ 0.30
4. p > 0.30

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are as follows:

1. H_o: $\overline{x}$ = 4.5, H_a : $\overline{x}$ > 4.5
2. H_o: μ ≥ 4.5, H_a: μ < 4.5
3. H_o: μ = 4.75, H_a: μ > 4.75
4. H_o: μ = 4.5, H_a: μ > 4.5

For Statements A–J in Exercise 9.66, answer the following in complete sentences.

1. State a consequence of committing a Type I error.
2. State a consequence of committing a Type II error.

A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.

1. at least 20 percent, when, in fact, it is less than 20 percent.
2. 20 percent, when, in fact, it is 20 percent.
3. less than 20 percent, when, in fact, it is at least 20 percent.
4. less than 20 percent, when, in fact, it is less than 20 percent.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The Type I error is

1. to conclude that the current mean hours per week is higher than 4.5, when, in fact, it is higher.
2. to conclude that the current mean hours per week is higher than 4.5, when, in fact, it is the same.
3. to conclude that the mean hours per week currently is 4.5, when, in fact, it is higher.
4. to conclude that the mean hours per week currently is no higher than 4.5, when, in fact, it is not higher.

The National Institute of Mental Health published an article stating that in any one-year period approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

1. Is this a test of one mean or proportion?
2. State the null and alternative hypotheses.
   H₀: ____________________ H_a: ____________________
3. Is this a right-tailed, left-tailed, or two-tailed test?
4. What symbol represents the random variable for this test?
5. In words, define the random variable for this test.
6. Calculate the following:
   1. x = ________________
   2. n = ________________
   3. $p^{\prime }$ = _____________
7. Calculate σ_x = __________. Show the formula setup.
8. State the distribution to use for the hypothesis test.
9. Find the p-value.
10. At a pre-conceived α = 0.05, give your answer for each of the following:
    1. Decision:
    2. Reason for the decision:
    3. Conclusion (write out in a complete sentence):

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1 percent level?

In 1955, Life Magazine reported that the 25-year-old mother of three worked, on average, an 80-hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. Eighty-one women were surveyed with the following results. The sample mean was 83; the sample standard deviation was 10. Does it appear that the mean work week has increased for women at the 5 percent level?

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5, 4, 7, 3, 6, 4, 5, 3, 6, 3, 8, 5. Conduct a hypothesis test of your belief.

According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7 percent girls). Suppose you don’t believe the reported figures of the percentage of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percentage of girls born in China is 46.7?

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks 10 engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70, 45, 55, 60, 65, 55, 55, 60, 50, 55.

Use the Initial Public Offering data (see Appendix C: Data Sets) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample.”

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle, and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety-six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi

Scene: The great library of the castle, in which Hamlet does his lessons

Act I

The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.

POLONIUS: If thou dost insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What a piece of work is man, how noble in reason, how infinite in faculties”? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.

Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.

Act II

The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.

POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)

(Curtain falls)

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation. “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters that have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names that end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, coworkers’, and acquaintances’ names that I could remember. Following are the names. Some are repeats. Test to see if the proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68 percent also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68 percent represents California. Note: For more accurate results, use more California community colleges and this past year's data.

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the United States is less than four years old.

A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.
At a 1 percent level of significance, what is an appropriate conclusion?

1. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20 percent.
2. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20 percent.
3. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20 percent.
4. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20 percent.

A research group is studying a particular infectious disease. In 2011 at least 18 percent of nursing home residents had the disease. An Introduction to Statistics class in Daviess County, KY, conducted a hypothesis test at the nursing home (approximately 1,200 residents) to determine if the local nursing home's incidence was lower. One hundred fifty residents were chosen at random and surveyed. Of the 150 residents surveyed, 82 have the disease. Use a significance level of 0.05 and, using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.

Driver error can be listed as the cause of approximately 54 percent of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using α = 0.05, is the AAA proportion accurate?

For Americans using library services, the American Library Association claims that at most 67 percent of patrons borrow books. The library director in Owensboro, KY, feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use α = 0.01 level of significance. What is the possible proportion of patrons who do borrow books from the Owensboro Library?

A survey in the New York Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX, chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the α = 0.10 level, is the Austin, TX, commute significantly less than the mean commute time for the 15 largest U.S. cities?

According to the New York Times Almanac the mean family size in the United States is 3.18. A sample of a college math class resulted in the following family sizes:
5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2
At α = 0.05, is the class’s mean family size greater than the national average? Does the Almanac result remain valid? Why?