0.11 Ch 11: the chi-square distribution  (Page 2/2)

The following table shows a random sample of 100 hikers and the area of hiking preferred.

Hiking preference area

Gender	The Coastline	Near Lakes and Streams	On Mountain Peaks
Female	18	16	11
Male	16	25	14

• $H_o$ : Gender and preferred hiking area are independent.
• $H_a$ : Gender and preferred hiking area are not independent

The distribution for the hypothesis test is $x_2^2$ .

The df's are equal to: $(\mathrm{rows}-1)(\mathrm{columns}-1)=(2-1)(3-1)=2$

The chi-square statistic is calculated using ${\sum }_{(2-3)}\frac{(0-E{)}^2}{E}$

Each expected (E) value is calculated using $\frac{(\text{rowtotal})(\text{columntotal})}{\text{totalsurveyed}}$

The first expected value (female, the coastline) is $\frac{\text{45}\cdot \text{34}}{\text{100}}=\text{15}\text{.}3$

The expected values are: 15.3, 18.45, 11.25, 18.7, 22.55, 13.75

The chi-square statistic is:

${\sum }_{(2-3)}\frac{(0-E{)}^2}{E}=$

$\frac{(\text{18}-\text{15}\text{.}3{)}^2}{\text{15}\text{.}3}+\frac{(\text{16}-\text{18}\text{.}\text{45}{)}^2}{\text{18}\text{.}\text{45}}+\frac{(\text{11}-\text{11}\text{.}\text{15}{)}^2}{\text{11}\text{.}\text{25}}+\frac{(\text{16}-\text{18}\text{.}7{)}^2}{\text{18}\text{.}7}+\frac{(\text{25}-\text{22}\text{.}\text{55}{)}^2}{\text{22}\text{.}\text{55}}+\frac{(\text{14}-\text{13}\text{.}\text{75}{)}^2}{\text{13}\text{.}\text{75}}$

$=1\text{.}\text{47}$

Calculator instructions

The TI-83/84 series have the function $x^2$ -Test in STAT TESTS to preform this test. First, you have to enter the observed values in the table into a matrix by using 2nd MATRIX and EDIT [A]. Enter the values and go to $x^2$ -Test. Matrix [B] is calculated automatically when you run the test.

Probability Statement: $p$ -value $=0\text{.}\text{4800}$ (A right-tailed test)

Since $\alpha$ is less than 0.05, we do not reject the null.

There is not sufficient evidence to conclude that gender and hiking preference are not independent.