statistics chapter 2 1 homework answers

Statistics Chapter 2 Homework

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Chapter 2: Descriptive Statistics

Chapter 2 Practice

Bringing it together from 2.4.

Santa Clara County, CA, has approximately 27,873 Japanese Americans. Their ages are as follows:

• Construct a histogram of the Japanese American community in Santa Clara County, CA. The bars will not be the same width for this example. Why not? What impact does this have on the reliability of the graph?
• What percentage of the community is under age 35?
• Which box plot most resembles the information above?

• For graph, check student’s solution.
• 49.7% of the community is under the age of 35.
• Based on the information in the table, graph (a) most closely represents the data.

Bringing It Together from 2.5

Javier and Ercilia are supervisors at a shopping mall. Each was given the task of estimating the mean distance that shoppers live from the mall. They each randomly surveyed 100 shoppers. The samples yielded the following information.

• How can you determine which survey was correct?
• Explain what the difference in the results of the surveys implies about the data.

Use the following information to answer the next three exercises : We are interested in the number of years students in a particular elementary statistics class have lived in California. The information in the following table is from the entire section.

What is the IQR ?

What is the mode?

Is this a sample or the entire population?

• entire population

Bringing It Together from 2.7

Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows:

• Find the sample mean [latex]\overline{x}[/latex].
• Find the approximate sample standard deviation, s .

Forty randomly selected students were asked the number of pairs of sneakers they owned. Let X = the number of pairs of sneakers owned. The results are as follows:

• Find the sample mean [latex]\overline{x}[/latex]
• Find the sample standard deviation, s
• Construct a histogram of the data.
• Complete the columns of the chart.
• Find the first quartile.
• Find the median.
• Find the third quartile.
• Construct a box plot of the data.
• What percent of the students owned at least five pairs?
• Find the 40 th percentile.
• Find the 90 th percentile.
• Construct a line graph of the data
• Construct a stemplot of the data

Following are the published weights (in pounds) of all of the team members of the San Francisco 49ers from a previous year.

177; 205; 210; 210; 232; 205; 185; 185; 178; 210; 206; 212; 184; 174; 185; 242; 188; 212; 215; 247; 241; 223; 220; 260; 245; 259; 278; 270; 280; 295; 275; 285; 290; 272; 273; 280; 285; 286; 200; 215; 185; 230; 250; 241; 190; 260; 250; 302; 265; 290; 276; 228; 265

• Organize the data from smallest to largest value.
• The middle 50% of the weights are from _______ to _______.
• If our population were all professional football players, would the above data be a sample of weights or the population of weights? Why?
• If our population included every team member who ever played for the San Francisco 49ers, would the above data be a sample of weights or the population of weights? Why?
• the population mean, μ .
• the population standard deviation, σ .
• the weight that is two standard deviations below the mean.
• When Steve Young, quarterback, played football, he weighed 205 pounds. How many standard deviations above or below the mean was he?
• That same year, the mean weight for the Dallas Cowboys was 240.08 pounds with a standard deviation of 44.38 pounds. Emmitt Smith weighed in at 209 pounds. With respect to his team, who was lighter, Smith or Young? How did you determine your answer?
• 174; 177; 178; 184; 185; 185; 185; 185; 188; 190; 200; 205; 205; 206; 210; 210; 210; 212; 212; 215; 215; 220; 223; 228; 230; 232; 241; 241; 242; 245; 247; 250; 250; 259; 260; 260; 265; 265; 270; 272; 273; 275; 276; 278; 280; 280; 285; 285; 286; 290; 290; 295; 302

• 205.5, 272.5
• 0.84 std. dev. below the mean

One hundred teachers attended a seminar on mathematical problem solving. The attitudes of a representative sample of 12 of the teachers were measured before and after the seminar. A positive number for change in attitude indicates that a teacher’s attitude toward math became more positive. The 12 change scores are as follows:

3 8 –1 2 0 5 –3 1 –1 6 5 –2

• What is the mean change score?
• What is the standard deviation for this population?
• What is the median change score?
• Find the change score that is 2.2 standard deviations below the mean.

Refer to [link] determine which of the following are true and which are false. Explain your solution to each part in complete sentences.

• The medians for all three graphs are the same.
• We cannot determine if any of the means for the three graphs is different.
• The standard deviation for graph b is larger than the standard deviation for graph a.
• We cannot determine if any of the third quartiles for the three graphs is different.

In a recent issue of the IEEE Spectrum , 84 engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days. Let X = the length (in days) of an engineering conference.

• Organize the data in a chart.
• Find the median, the first quartile, and the third quartile.
• Find the 65 th percentile.
• Find the 10 th percentile.
• The middle 50% of the conferences last from _______ days to _______ days.
• Calculate the sample mean of days of engineering conferences.
• Calculate the sample standard deviation of days of engineering conferences.
• Find the mode.
• If you were planning an engineering conference, which would you choose as the length of the conference: mean; median; or mode? Explain why you made that choice.
• Give two reasons why you think that three to five days seem to be popular lengths of engineering conferences.

A survey of enrollment at 35 community colleges across the United States yielded the following figures:

6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825; 2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681; 3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771; 2861; 1263; 7285; 28165; 5080; 11622

• Organize the data into a chart with five intervals of equal width. Label the two columns “Enrollment” and “Frequency.”
• If you were to build a new community college, which piece of information would be more valuable: the mode or the mean?
• Calculate the sample mean.
• Calculate the sample standard deviation.
• A school with an enrollment of 8,000 would be how many standard deviations away from the mean?

• Check student’s solution.

Use the following information to answer the next two exercises. X = the number of days per week that 100 clients use a particular exercise facility.

The 80 th percentile is _____

The number that is 1.5 standard deviations BELOW the mean is approximately _____

• Cannot be determined

Suppose that a publisher conducted a survey asking adult consumers the number of fiction paperback books they had purchased in the previous month. The results are summarized in the [link] .

• Are there any outliers in the data? Use an appropriate numerical test involving the IQR to identify outliers, if any, and clearly state your conclusion.
• If a data value is identified as an outlier, what should be done about it?
• Are any data values further than two standard deviations away from the mean? In some situations, statisticians may use this criteria to identify data values that are unusual, compared to the other data values. (Note that this criteria is most appropriate to use for data that is mound-shaped and symmetric, rather than for skewed data.)
• Do parts a and c of this problem give the same answer?
• Examine the shape of the data. Which part, a or c, of this question gives a more appropriate result for this data?
• Based on the shape of the data which is the most appropriate measure of center for this data: mean, median, or mode?

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N ( 244 , 15 50 ) N ( 244 , 15 50 )

As the sample size increases, there will be less variability in the mean, so the interval size decreases.

X is the time in minutes it takes to complete the U.S. Census short form. X ¯ X ¯ is the mean time it took a sample of 200 people to complete the U.S. Census short form.

CI: (7.9441, 8.4559)

The level of confidence would decrease, because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases.

• x ¯ x ¯ = 2.2

X ¯ X ¯ is the mean weight of a sample of 20 heads of lettuce.

EBM = 0.07 CI: (2.1264, 2.2736)

The interval is greater, because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals.

The confidence level would increase.

(24.52,36.28)

We are 95 percent confident that the true mean age for winter Foothill College students is between 24.52 and 36.28.

The error bound for the mean would decrease, because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean.

X is the number of hours a patient waits in the emergency room before being called back to be examined. X ¯ X ¯ is the mean wait time of 70 patients in the emergency room.

CI: (1.3808, 1.6192)

• x ¯ x ¯ = 151
• s x s x = 32
• n – 1 = 107

X ¯ X ¯ is the mean number of hours spent watching television per month from a sample of 108 Americans.

CI: (142.92, 159.08)

(2.93, 3.59)

We are 95 percent confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.

The error bound would become EBM = 0.245. This error bound decreases, because as sample sizes increase, variability decreases, and we need less interval length to capture the true mean.

It would decrease, because the z -score would decrease, which would reduce the numerator and lower the number.

X is the number of successes where the woman makes the majority of the purchasing decisions for the household. P ′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household.

CI: (0.5321, 0.6679)

EBM : 0.0679

X is the number of successes where an executive prefers a truck. P ′ is the percentage of executives sampled who prefer a truck.

CI: (0.19432, 0.33068)

EBM : 0.0707

The sampling error means that the true mean can be 2 percent above or below the sample mean.

P ′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election.

CI: (0.62735, 0.67265);

EBM: 0.02265

the number of girls, ages 8 to 12, in the 5 p.m. Monday night beginning ice-skating class

P ′ ~ N ( 0.8 , ( 0.8 ) ( 0.2 ) 80 ) P ′ ~ N ( 0.8 , ( 0.8 ) ( 0.2 ) 80 )

CI = (0.72171, 0.87829).

(0.72; 0.88)

With 92 percent confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72 percent and 88 percent.

The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error.

• X is the height of a Swedish male, and is the mean height from a sample of 48 Swedish males.
• Normal. We know the standard deviation for the population, and the sample size is greater than 30.
• CI: (70.151, 71.49)
• EBM = 0.849
• The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean.
• x ¯ x ¯ = 23.6
• X is the time needed to complete an individual tax form. X ¯ X ¯ is the mean time to complete tax forms from a sample of 100 customers.
• N ( 23.6 , 7 100 ) N ( 23.6 , 7 100 ) because we know sigma.
• (22.228, 24.972)
• EBM = 1.372
• It will need to change the sample size. The firm needs to determine what the confidence level should be and then apply the error bound formula to determine the necessary sample size.
• The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval.
• According to the error bound formula, the firm needs to survey 206 people. Because we increase the confidence level, we need to increase either our error bound or the sample size.
• X is the number of letters a single camper will send home. X ¯ X ¯ is the mean number of letters sent home from a sample of 20 campers.

N 7.9 ( 2.5 20 ) 7.9 ( 2.5 20 )

• CI: (6.98, 8.82)
• The error bound and confidence interval will decrease.
• x ¯ x ¯ = $568,873
• CL = 0.95, α = 1 – 0.95 = 0.05, z α 2 z α 2 = 1.96 EBM = z 0.025 σ n z 0.025 σ n = 1.96 909200 40 909200 40 = $281,764

Alternate solution:

Using the TI-83, 83+, 84, 84+ Calculator

• Press STAT and arrow over to TESTS .
• Arrow down to 7:ZInterval .
• Press ENTER .
• Arrow to Stats and press ENTER .
• σ : 909,200
• x ¯ x ¯ : 568,873
• Arrow down to Calculate and press ENTER .
• The confidence interval is ($287,114, $850,632).
• Notice the small difference between the two solutions—these differences are simply due to rounding error in the hand calculations.
• We estimate with 95 percent confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637.

Use the formula for EBM , solved for n : n =   z 2 σ 2 E B M 2 n =   z 2 σ 2 E B M 2

From the statement of the problem, you know that σ = 2.5, and you need EBM = 1.

z = z 0.035 = 1.812.

(This is the value of z for which the area under the density curve to the right of z is 0.035.)

n =   z 2 σ 2 E B M 2 = 1.812 2 2.5 2 1 2   ≈   20.52 . n =   z 2 σ 2 E B M 2 = 1.812 2 2.5 2 1 2   ≈   20.52 .

You need to measure at least 21 male students to achieve your goal.

• CI: (6244, 11,014)
• It will become smaller.
• x ¯ x ¯ = 2.51
• s x s x = 0.318
• The effective length of time for a tranquilizer
• The mean effective length of time of tranquilizers from a sample of nine patients
• We need to use a Student’s t -distribution, because we do not know the population standard deviation.
• CI: (2.27, 2.76)
• Check student's solution.
• If we were to sample many groups of nine patients, 95 percent of the samples would contain the true population mean length of time.

x ¯ = $ 251 , 854.23 ; x ¯ = $ 251 , 854.23 ;

s =   $ 521 , 130.41 . s =   $ 521 , 130.41 .

Note that we are not given the population standard deviation, only the standard deviation of the sample.

There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29.

CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04.

α 2 = 0.02 t α 2 = t 0.02 α 2 = 0.02 t α 2 = t 0.02 = 2.150.

E B M = t α 2 ( s n ) = 2.150 ( 521 , 130.41 30 )   ~   $ 204 , 561.66 . E B M = t α 2 ( s n ) = 2.150 ( 521 , 130.41 30 )   ~   $ 204 , 561.66 .

x ¯ x ¯ - EBM = $251,854.23 - $204,561.66 = $47,292.57.

x ¯ x ¯ + EBM = $251,854.23 + $204,561.66 = $456,415.89.

We estimate with 96 percent confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89.

Alternate Solution

The difference between solutions arises from rounding differences.

• X is the number of unoccupied seats on a single flight. X ¯ X ¯ is the mean number of unoccupied seats from a sample of 225 flights.
• We will use a Student’s t-distribution, because we do not know the population standard deviation.
• CI: (11.12 , 12.08)
• CI: (7.64, 9.36)
• The sample should have been increased.
• Answers will vary.
• The sample size would need to be increased, because the critical value increases as the confidence level increases.

X = the number of people who believe that the president is doing an acceptable job;

P ′ = the proportion of people in a sample who believe that the president is doing an acceptable job.

• N ( 0.61 , ( 0.61 ) ( 0.39 ) 1200 ) N ( 0.61 , ( 0.61 ) ( 0.39 ) 1200 )
• CI: (0.59, 0.63)
• Check student’s solution.
• (0.72, 0.82)
• (0.65, 0.76)
• (0.60, 0.72)
• Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.
• We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families.
• We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families.
• X = the number of adult Americans who believe that crime is the main problem; P′ = the proportion of adult Americans who believe that crime is the main problem.
• Because we are estimating a proportion, that P′ = 0.2 and n = 1,000, the distribution we should use is N ( 0.2 , ( 0.2 ) ( 0.8 ) 1000 ) N ( 0.2 , ( 0.2 ) ( 0.8 ) 1000 ) .
• CI: (0.18, 0.22)
• One way to lower the sampling error is to increase the sample size.
• The stated ± 3 percent represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3 percent. Thus, they estimate the percentage of adult Americans who the percentage of adult Americans who that crime is the main problem to be between 18 percent and 22 percent.
• p′ = (0 .55 + 0 .49) 2 (0 .55 + 0 .49) 2 = 0.52; EBP = 0.55 – 0.52 = 0.03
• No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.

STAT TESTS A: 1-PropZinterval with x = (0.52)(1,000), n = 1,000, CL = 0.75.

Answer is (0.502, 0.538).

• Yes, this interval does not fall below 0.50, so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75 percent confidence.

CL = 0.95; α = 1 – 0.95 = 0.05; α 2 α 2 = 0.025; z α 2 z α 2 = 1.96. Use p ′ = q ′ = 0.5.

n =   z α 2 2 p ′ q ′ E B P 2 =   1.96 2 ( 0.5 ) ( 0.5 ) 0.05 2 = 384.16 . n =   z α 2 2 p ′ q ′ E B P 2 =   1.96 2 ( 0.5 ) ( 0.5 ) 0.05 2 = 384.16 .

You need to interview at least 385 students to estimate the proportion to within 5 percent at 95 percent confidence.

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• Authors: Barbara Illowsky, Susan Dean
• Publisher/website: OpenStax
• Book title: Statistics
• Publication date: Mar 27, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/statistics/pages/1-introduction
• Section URL: https://openstax.org/books/statistics/pages/8-solutions

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