Question

A researcher wants to find out if educaiton level and frequency of exercise for adults are independent of each other. His data is shown in the table below: Timos exercised HS diploma AA degree Bachelors degree Masters Degree weekly 8 10 11 12 1-2 11 22 31 12 3-4 21 16 10 6 5+ 9 10 7 4 0 Use the information to complete the following: A) What type of test should be used? B) State the null and alternative hypothesis. C) Compute the test statistic D) Compute the p-value E) Using a significance level of 0.05, state the conclusion of the hypothesis test.

Answer 1

HYPOTHESIS TEST-

We have to perform Chi-square test for independence between two random variables education level and frequency of exercise for adults.

We have to test for null hypothesis

$\tiny H_0:\text{Education level and frequency of exercise for adults are independent.}$

against the alternative hypothesis

$\tiny H_1:\text{Education level and frequency of exercise for adults are not independent.}$

Observed frequencies $\tiny \left(f_{ij}\right)$ are as follows.

Observed frequencies HS diploma 8 0 Times 1-2 exercised 3-4 weekly 5+ Column total 11 21 9 49 Education level Row total AA degree Bachelor degree Masters Degree 10 11 12 41 22 31 12 76 16 10 6 53 10 7 4 30 58 59 34 200

Under null hypothesis, we obtain expected frequencies $\tiny \left(e_{ij}\right)$ by multiplying corresponding row total and column total and dividing it by grand total as follows.

Expected frequencies 0 Times exercised weekly 1-2 3-4 HS diploma 10.0450 18.6200 12.9850 7.3500 49.0000 Education level Row total AA degree Bachelor degree Masters Degree 11.8900 12.0950 6.9700 41.0000 22.0400 22.4200 12.9200 76.0000 15.3700 15.6350 9.0100 53.0000 8.7000 8.8500 5.1000 30.0000 58.0000 59.0000 34.0000 200.0000 5+ Column total

Our Chi-square test statistic is given by

$\tiny \chi^2=\sum_{i=1}^{m}\sum_{j=1}^{n}\frac{\left(f_{ij}-e_{ij}\right)^2}{e_{ij}}$

Here,

Number of rows $\tiny m=4$

Number of columns $\tiny n=4$

Corresponding calculations (chi square component for each cell, which are to be added later) are as follows.

Chi_square Row total HS diploma 0.4163 3.1184 Education level AA degree Bachelor degree Masters Degree 0.3004 0.0991 3.6300 0.0001 3.2835 0.0655 0.0258 2.0309 1.0056 0.1943 0.3867 0.2373 0.5206 5.8003 4.9383 0 Times 1-2 exercised 3-4 weekly 5+ Column total 4.4459 6.4675 4.9473 0.3704 8.8524 8.0096 1.1886 20.1115

$\tiny \therefore \chi^2_{calculated}=\sum_{i=1}^{3}\sum_{j=1}^{3}\frac{\left(f_{ij}-e_{ij}\right)^2}{e_{ij}}=20.1115$

Degrees of freedom $\tiny v=(m-1)(n-1)=9$

$\tiny \therefore \text{p-value}=P\left(\chi^2_{9}>20.1115\right)=0.01723663$ [Using R-code '1-pchisq(20.1115,9)']

Level of significance $\tiny \alpha=0.05$

We reject our null hypothesis if $\tiny \text{p-value}<\alpha$

Here, we observe that $\tiny \text{p-value}=0.01723663\nless0.05=\alpha$

So, we reject our null hypothesis.

ANSWERS-

A.

We have to perform Chi-square test for independence.

B.

Null hypothesis is $\tiny H_0:\text{Education level and frequency of exercise for adults are independent.}$

Alternative hypothesis is $\tiny H_1:\text{Education level and frequency of exercise for adults are not independent.}$

C.

Test statistic is $\tiny \chi^2=20.1115$

D.

$\tiny \text{p-value}=0.01723663$

E.

Based on the given data we can conclude that there is significant evidence that education level and frequency of exercise for adults are not independent.