Question

A random group of apartments was selected from a city to analyze the number of bedrooms they have, with the results as shown:No. of bedrooms: 1 bedroom 2 bedrooms 3 bedroomsNo. of apartments: 22 28 10Based on this sample, is there evidence to reject the hypothesis that the apartments are equally distributed between 1-bedroom, 2-bedroom, and 3-bedroom apartments? Use a significance level of α = .05?

Answer 1

Answer:

Step-by-step explanation:

Hello!

The study variable of this exercise is X: the number of bedrooms per apartment in the city, categorized: 1 bedroom, 2 bedrooms, 3 bedrooms.

The claim is that the apartments are equally distributed in 1-bedroom, 2-bedroom and 3-bedroom, if they equally distributed, then the probability of happening is equal, symbolically P(1b)=P(2b)=P(3b)= 1/3

To test if the sample follows this distribution, you have to apply a Chi-Square Goodness to Fit-test, where the hypotheses are:

H₀: P(1b)=P(2b)=P(3b)= 1/3

H₁: The data is not consistent with the distribution.

α: 0.05

The statistic is:

$X^2= sum(\frac{(O_i-E_i)^2}{E_1} )~X^2_{k-1}$

Where

Oi: Observed frequency of the i-category.

Ei: Expected frequency of the i-category

k: number of categories of the variable

First step is to calculate the expected frequencies for each category:

$Ei= n * Pi$

Where Pi is the theoretical proportion for the i-category.

$E_{P1b}= 60*1/3= 20$

$E_{P2b}= 60*1/3= 20$

$E_{P3b}= 60*1/3= 20$

Sample:

1 bedroom 22

2 bedrooms  28

3 bedrooms 10

n= 22+28+10= 60

Note if the summatory of the expected frequencies is equal to the sample size ∑Ei=n, if it is not so (by a big difference) then you should check your calculations.

Second step, calculate the statistic:

$X^2= (\frac{(22-20)^2}{20} )+(\frac{(28-20)^2}{20} )+(\frac{(10-20)^2}{20} )= 8.4$

I'll decide using the critical value approach. This test is always one-tailed to the right, this means that you will reject the null hypothesis to big values of the statistic. The critical value is:

$X^2_{k-1;1-\alpha }= X^2_{2;0.95}= 5.991$

If X²≥5.991, then you decide to reject the null hypothesis.

If X²<5.991, then you do not reject the null hypothesis.

The value of the statistic is greater than the critical value so the decision is to reject the null hypothesis. Using a significance level of 5% you have enough evidence to say that the number of bedrooms per apartment in the city are not equally distributed.

I hope it helps!