Slidio

• Home
• Profile

The same survey that asked 1,200 U.S. college students about their body perception also asked the following question:

"With whom do you find it easiest to make friends?" (opposite sex, same sex or no difference).

In this activity we will use the collected data to:

• learn how to tally our data into a table of counts and percents.
• learn how to produce a pie chart.

Let us start by creating a frequency table of the data. We will use the tally function in the mosaic package to achieve this.

x       variable name
data    dataset name
format  count, proportion or percent
margins should marginal distributions be displayed?

The number of facebook friends a person has is normally distributed with mean 190 and standard deviation 36. Answer the next three questions based on this.

Sandra Guerra Thompson, a Professor of Law and Criminal Justice and Institute Director at the University of Houston Law Center, wrote Beyond a Reasonable Doubt? Reconsidering Uncorroborated Eyewitness Identification Testimony, in 2008 in the U. C. Davis Law Review.

This article reviews the overwhelming scientific evidence that establishes that eyewitnesses are notoriously inaccurate in identifying strangers, especially under the conditions that exist in many serious offenses such as robbery. Many of the factors that tend to decrease the accuracy of an identification are intrinsic to a witness’ abilities, and not the product of inappropriate suggestion by the police.

We know, for example, that eyewitnesses identify a known wrong person (a “filler” or “foil”) in approximately 20% of all real criminal lineups.

Using the data that 20% of eyewitnesses identify a known wrong person, we wish to create a probability distribution and answer some probability questions.

# Let's look at the shape of the full model surface. 
# expand.grid makes it easy to generate an evenly spaced 
# grid of predictors over the entire design space.
grid <- expand.grid(
 class = unique(mpg$class),
 displ = seq(min(mpg$displ), max(mpg$displ), 
 length = 20)
)
grid$consump <- predict(mod1, newdata = grid)
qplot(displ, consump, data = mpg) + 
 geom_line(data = grid, colour = "grey50") + 
 facet_wrap(~ class)
qplot(displ, resid, data = mpg) + facet_wrap(~ class)

Did I Get?

According to pollingreport.com, a poll conducted by FOX News estimated that on the issue of abortion, roughly 44% of registered voters in the United States are pro-choice, and roughly 47% are pro-life (the remaining 9% are undecided). What is the probability that two simultaneously and randomly chosen registered voters are pro-life?

What is the sample space for this experiment?

A researcher wished to compare the average daily hotel room rates between San Francisco and Los Angeles. The researcher obtained an SRS of 27 hotels in downtown San Francisco and found the sample mean to be \$156, and a standard deviation of \$18. The researcher also obtained an independent SRS of 24 hotels in downtown Los Angeles and found the sample mean to be \$143, with a standard deviation of \$10.

You can use the factorial function in R to compute this coefficient using the formula. An easier way to achieve the same is to directly use the choose function.

choose(5, 3)
## [1] 10
factorial(5)/(factorial(3)*factorial(2))
## [1] 10

Suppose \(X \sim Bin(n, p)\) follows a binomial distribution with n trials and success probability p. For large values of n, we can approximate the distribution of \(X\) using a normal distribution.

\[X \sim N(\mu = np, \sigma = \sqrt{np(1-p)})\]

You ask an SRS of 200 adults if they are concerned about nutrition when eating away from home. Let \(X\) represent the total count of people responding "Yes". If the national proportion \(p = 0.4\) holds in your area, what is the probability that \(X\) lies between 75 and 85?

We can approximate the distribution of \(X\) using a normal distribution with mean \(\mu = np\) and standard deviation \(\sigma = np(1-p)\).

We can do all the computations using R, as shown below.

n = 200; p = 0.4
mu = n * p
sigma = sqrt(n * p * (1 - p))
pnorm(85, mu, sigma) - pnorm(75, mu, sigma)
## [1] 0.5295

You can visualize the probabilities here

To identify, if a setting fits the binomial distribution, you need to verify four things.

1. Does every trial have two outcomes?
2. Do we have a fixed number of trials?
3. Are the trials independent?
4. Is the probability of success constant across all trials?

Let \(\mu_{LA}\) and \(\mu_{SF}\) represent the average daily hotel room rates for the two cities. The null hypothesis is that there is no difference between the average rates for these two cities, while the alternate hypothesis is that there is a difference. Accordingly,

\[ \begin{aligned} H_0: \mu_{LA} - \mu_{SF} & = 0 \\ H_a: \mu_{LA} - \mu_{SF} & \neq 0 \\ \end{aligned} \]

The NCAA considers a student a "partial qualifier" eligible to practice and receive an athletic scholarship, but not compete, if the combined SAT score is at least 720. What proportion of all students who take the SAT would be partial qualifiers?

A Pennsylvania research rm conducted a study in which 30 drivers (of ages 18 to 82 years old) were sampled, and for each one, the maximum distance (in feet) at which he/she could read a newly designed sign was determined. The goal of this study was to explore the relationship between a driver's age and the maximum distance at which signs were legible, and then use the study's findings to improve safety for older drivers.

The form of the relationship is its general shape. When identifying the form, we try to find the simplest way to describe the shape of the scatterplot. There are many possible forms. Here are a couple that are quite common:

The medical director of a large company is concerned about the effects of stress on the company’s younger executives. According to the National Center for Health Statistics, the mean systolic blood pressure for males 35 to 44 years of age is 128 and the standard deviation in this population is 15. The medical director examines the records of 72 executives in this age group and finds that their mean systolic blood pressure is 129.93. Is this evidence that the mean blood pressure for all the company’s young male executives is higher than the national average? Use \(\alpha = 0.05\).

Given

• \(\bar{x} = 129.93\)
• \(\sigma = 15\)
• \(n = 72\)
• \(\mu_0 = 128\)

The medical director is interested in evidence that supports his hypothesis that the mean blood pressure for the company's male executives is higher than the national average (which is 128). Hence, this would be the alternate hypothesis.

\[\begin{aligned} H_0: \mu & = 128 \\ H_a: \mu & > 128 \\ \end{aligned}\]

Step 2. Identify Test Statistic

We are testing a hypothesis about a single mean. Moreover, we DONT know the standard deviation of the population. Hence, we use a 1-sample t-test with \(T \sim t(df = n - 1)\).

Below is a boxplot of yearly income by marital status for individuals in the United States.

Click p to see the R code used to create the plot.

1. Slides
2. Screencasts
3. Slidecasts
4. Integrated

x <- rnorm(100)
stem(x)
  The decimal point is at the |

  -2 | 3
  -1 | 9665421110000
  -0 | 9988877776666665544444433332221111111
   0 | 011123344555555555777777888999
   1 | 0000012334566789
   2 | 269

The nth percentile of an observation variable is the value that cuts off the first n percent of the data values when it is sorted in ascending order.

Example

Find the 32nd, 57th and 98th percentiles of the eruption durations in the data set faithful.

Solution

We use the quantile function to compute the percentiles of eruptions with the desired percentage ratios.

> quantile(faithful$eruptions, c(0.32, 0.57, 0.98))
##   32%   57%   98% 
## 2.395 4.133 4.933

The interquartile range of an observation variable is the difference of its upper and lower quartiles. It is a measure of how far apart the middle portion of data spreads in value.

We can use the IQR function to compute the interquartile range of a variable x.

> x <- c(62, 63, 64, 70, 72, 76, 77, 81, 81)
> IQR(x)
## [1] 13

It is easy to draw a boxplot in R using the bwplot function in the lattice package. Let us draw a boxplot of the variable mpg in the data set mtcars.

bwplot(~ mpg, data = mtcars)

General Multiplication Rule