Question

Use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places.

A recent debate about where in the United States skiers believe the skiing is best prompted the following survey. Test to see if the best ski area is independent of the level of the skier.

U.S. Ski Area	Beginner	Intermediate	Advanced
Tahoe	$20$	$30$	$40$
Utah	$10$	$30$	$60$
Colorado	$10$	$40$	$50$

Table$11.43$

Short answer

From the given information in the question, it can be concluded that the Ski area and level of skier are not independent. so we reject the null hypothesis.

Step-by-step solution

Given Information

A recent debate about where in the United States skiers believe the skiing is best prompted the following survey. Test to see if the best ski area is independent of the level of the skier.

Step 2: The table with data 

The Table with data is given by

Ski Area	Beginner	Intermediate	Advanced	Total
Tahoe	$20$	$30$	$40$	$90$
Utah	$10$	$30$	$60$	$100$
Colorado	$10$	$40$	$50$	$100$
Total	$40$	$100$	$150$	$290$

The table with expected values

The table with expected values:

Ski Area	Beginner	Intermediate	Advanced	Total
Tahoe	$12.41379$	$31.03448$	$46.55172$	$90$
Utah	$13.79310$	$34.48276$	$51.72414$	$100$
Colorado	$13.79310$	$34.48276$	$51.72414$	$100$
Total	$40$	$100$	$150$	$290$

Step 4: Hypotheses Test

We want to test these hypotheses:

H_$0$ : Ski area is independent of the level of the skier

H_$1$: Ski area depends on the level of the skier

Since there are $3$ rows and $3$columns, the number of degrees of freedom is $(3-1)(3-1)=4$

Test statistics is given by

${\chi }^2=\frac{(20-12.4{)}^2}{12.4}+\frac{(10-13.8{)}^2}{13.8}+\frac{(10-13.8{)}^2}{13.8}+$

$+\frac{(30-31{)}^2}{31}+\frac{(30-34.5{)}^2}{34.5}+\frac{(40-34.5{)}^2}{34.5}+$

$+\frac{(40-46.6{)}^2}{46.6}+\frac{(60-51.7{)}^2}{51.7}+\frac{(50-51.7{)}^2}{51.7}$

$=4.7+0.03+0.93+1.05+0.59+$

$+1.332+1.046+0.877+0.056$

$=10.57$

Step 5: Using the applet

$\text{Using the applet, we get that}p\text{-value is}0.03185\text{:}$

Step 6: Result

$\text{Taking}\alpha =0.05\text{, we can see that}$

$p=0.03185<0.05$

This means that we reject the null hypothesis. We conclude that the ski area and level of the skier are not independent.