Question

In their reporting on this era of house-bound isolation, CTV Canada kept noting many people with bad haircuts. Hashtags like #pandemicbangs and #coronacuts are trending on social media. CTV Canada decided to take an informal poll to see how many people are getting their hair cut at home. You have taken their data and divided it based on age groups to see if the likelihood that a person cut's their own hair depends on their age. You summarized the data below. Haircut at home? Ages 35-55 18-34 55+ Total 304 Yes No Total 1,416 1,928 3,344 689 993 856 1238 2,094 2,576 3,855 6,431 Part A: What percentage of young adults (18-34) cut their own hair at home? [2 marks]

Answer 1

Part A. The probability of a randomly selected adult is young adult (18-34) is

$\begin{align*} P(\text{18-34})&=\frac{\text{number of young adults in the sample}}{\text{total number of adults in the sample}}\\ &=\frac{3344}{6431} \end{align*}$

The probability of a randomly selected adult is young adult (18-34) and cut their own hair is

$\begin{align*} P(\text{18-34}\cap \text{Yes})&=\frac{\text{number of young adults who cut their own hair in the sample}}{\text{total number of adults in the sample}}\\ &=\frac{1416}{6431} \end{align*}$

The proportion of young adults (18-34) who cut their own hair is sames as the conditional probability of a randomly selected adult cut their own hair given that the person is young adult (18-34) is

P(Yes 18-34) using the formula for conditional probability P(18-34 n Yes) P(18-34) 1416 6431 3344 6431 1416 3344 = 0.4234

ans: The percentage of young adults (18-34) who cut their own hair is 42.34%

Part B) The probability of a randomly selected adult cut their own hair at home is

P(18-34) number of adults who cut their own hair in the sample total number of adults in the sample 2576 6431

The probability of a randomly selected adult is 35 years or older and cut their own hair is

$\begin{align*} P(\text{35+}\cap \text{Yes})&=\frac{\text{number of 35+ years or older who cut their own hair in the sample}}{\text{total number of adults in the sample}}\\ &=\frac{304+856}{6431}\\ &=\frac{1160}{6431} \end{align*}$

Among those who cut their own hair at home, the proportion of 35 years or older is sames as the conditional probability of a randomly selected adult is 35 years or older given that they cut their own hair at home

$\begin{align*} P(\text{35+}\mid \text{Yes})&=\frac{P(\text{35+}\cap \text{Yes})}{P(\text{Yes})}\quad\text{using the formula for conditional probability}\\ &=\frac{\frac{1160}{6431}}{\frac{2576}{6431}}\\ &=\frac{1160}{2576}\\ &=0.4503 \end{align*}$

ans: Among those who cut their own hair at home, the percentage of 35 years or older is 45.03%

Part C) We want to test if a person's willingness to cut their own hair depends on their age

Part 1) The hypotheses are

ans:

B. Ho: A person's willingness to cut their own hair is independent of their age HA: A person's willingness to cut their own hair depends on their age

part 2) The degrees of freedom are (number of rows-1)*(number of columns-1) = (2-1)*(3-1)=2

ans: The degrees of freedom are 2

part 3) The chi-square test for homogeneity/independence is a right tailed test

The right critical value for significance level $\begin{align*} \alpha=0.05 \end{align*}$ is

$\begin{align*} P(\chi^2>\chi^2_c)=\alpha=0.05 \end{align*}$

Using the chi-square tables for df=2 and the area under the right tail=0.05, we get the critical value=5.991

ans: The critical value is 5.991

Part 4) We will reject the null hypothesis, if the chi-square test statistic is greater than the critical value.

Here, the test statistic is 44.75 and it is greater than 5.991. Hence we reject the null hypothesis.

ans: Reject the null hypothesis. The chi-square test statistic is greater than the critical value.

Part 5)

ans: A

A. Yes, there is enough evidence to conclude that a person's decision for a haircut at home depends on his/her age