Today’s Agenda

  • Using Chi-Square table
  1. Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of \(60\) managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as:
Monday Tuesday Wednesday Thursday
number of absences 15 12 9 9

Suppose there are \(60\) absences in an average week. Test the goodness of fit of this data to a uniform distribution with a significance level of \(.05\).

  1. Suppose it is known that the distribution of males age \(18\) to \(24\) in the U.S. population is as follows:
Marital Status Percent
never married \(31\)
married \(56.5\)
widowed \(2.5\)
divorced/separated \(10\)

From a random sample of \(400\) mean ages \(18\) to \(24\), the following data is collected:

Marital Status Count
never married \(140\)
married \(238\)
widowed \(2\)
divorced/separated \(20\)

Perform a goodness of fit test with significance level of \(.05\).

  1. One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population):
Number of Televisions Percent
\(0\) \(10\)
\(1\) \(16\)
\(2\) \(55\)
\(3\) \(11\)
\(4+\) \(8\)

A random sample of 600 families in North Carolina gave the following results:

Number of Televisions Count
\(0\) \(66\)
\(1\) \(119\)
\(2\) \(340\)
\(3\) \(60\)
\(4+\) \(15\)

At the \(1 \%\) significance level, does it appear that the distribution of number of televisions in North Carolina is different from the distribution for the American population as a whole?

  1. Suppose that \(600\) thirty-year-olds were surveyed to determine whether or not there is a relationship between the highest education completed and salary. Conduct a test of independence.
Salary No HS diploma HS College Masters
\(< \$30,000\) \(15\) \(25\) \(10\) \(5\)
\(30\)k\(-40\)k \(20\) \(40\) \(70\) \(30\)
\(40\)k\(-50\)k \(10\) \(20\) \(40\) \(55\)
\(50\)k\(-60\)k \(5\) \(10\) \(20\) \(60\)
\(> \$60,000\) \(0\) \(5\) \(10\) \(150\)

\(H_0:\)

\(H_A:\)

Correlation

For the following section we will use data from: “Learning Statistics with R” by Danielle Navarro:

load("./parenthood.Rdata")
load("./pearson_correlations.RData")
load("./effort.Rdata")
  • This data captures how grumpy Danielle is, how much she slept in a day, and how much her baby slept in a day.

Consider the following plots:

ggplot(parenthood, aes(x=dan.sleep, y=dan.grump)) + geom_point()

ggplot(parenthood, aes(x=baby.sleep, y=dan.grump)) + geom_point()

Correlation coefficient

The Pearson correlation coefficient \(r_{XY}\) is a standardized covariance measure:

\[r_{XY} = \frac{1}{N-1} \sum_{i=1}^{N} \frac{X_i-\bar{X}}{s_X}\frac{Y_i-\bar{Y}}{s_Y} \]

where \(s_x\) and \(s_y\) are the sample standard deviations.

The R code for the Pearson correlation coefficient is:

Properties of \(r_{XY}\)

Interpreting the Pearson correlation coefficient

Below is data with various Pearson correlation coefficients:

ggplot(outcomes, aes(x=V1,y=V2))+geom_point() + facet_wrap(~pearson)

  • Exactly what constitutes as a strong correlation depends on the context.
  • You can, however, use these general guidlines:
Correlation Strength Direction
\(-1\) to \(-0.9\) Very Strong Negative
\(-0.9\) to \(-0.7\) Strong Negative
\(-0.7\) to \(-0.4\) Moderate Negative
\(-0.4\) to \(-0.2\) Weak Negative
\(-0.2\) to \(0\) Negligible Negative
\(0\) to \(0.2\) Negligible Positive
\(0.2\) to \(0.4\) Weak Positive
\(0.4\) to \(0.7\) Moderate Positive
\(0.7\) to \(0.9\) Strong Positive
\(0.9\) to \(1\) Very Strong Positive