Chapter 15, Problem 15P

a.

To determine

To Find: If the proportion of participants who claim to remember broken glass differ significantly from group to group for the given question.

Answer to Problem 15P

Reject null hypothesis and conclude that proportion of participants who claim to remember broken glass differ significantly from group to group.

Explanation of Solution

Given info:

A sample of 150 students were involved in a study based on “the response they gave regarding the broken glass or verb used for speed”. The distribution is given in the question. Use $\alpha =0.05$ to test the claim.

	Yes	No	Total
Smashed into	16	34	50
Hit	7	43	50
Control (not asked)	6	44	50
Total	29	121	150

Calculation:

Step 1: Null Hypothesis and Alternate Hypothesis are:

$H_0:$ Proportion of participants who claim to remember broken glass does not differ significantly from group to group.

$H_1:$ Proportion of participants who claim to remember broken glass differ significantly from group to group

Step 2: For the given sample, degrees of freedom equals:

$\begin{array}{c}df=(R-1)(C-1) \text{where }R\text{ equals number of rows and }C\text{ equals columns}\\ =(3-1)(2-1)\\ =2\end{array}$

With $\alpha =0.05$ and $df=2$, the critical value (CV)  is obtained from the ${\chi }^2-table$ as

${\chi }^2=5.991$

Step 3: ${\chi }^2-\text{statistic}$ is calculated as:

${\chi }^2={\displaystyle \sum \frac{{(f_o-f_e)}^2}{f_e}}$

The formula to calculate expected frequency is:

$f_e=\frac{f_c{f}_r}{n}\mathrm{...}\text{where }{f}_r\text{ is row frequency and }{f}_c\text{ is column frequency}$

Substitute $n=600$ in the above formula and compute respective values of expected frequencies:

For the category “smashed into”, the expected frequencies are:

$\begin{array}{c}{f}_{e,yes}=\frac{29\times 50}{150}\\ f_{e,no}=\frac{121\times 50}{150}\end{array}$

For the category “Hit”, the expected frequencies are:

$\begin{array}{c}{f}_{e,yes}=\frac{29\times 50}{150}\\ f_{e,no}=\frac{121\times 50}{150}\end{array}$

Similarly, for the category “control”, the expected frequencies are:

$\begin{array}{c}{f}_{e,yes}=\frac{29\times 50}{150}\\ f_{e,no}=\frac{121\times 50}{150}\end{array}$

The contingency table is :

	Yes	No	Total
Smashed into	16 (9.66)	34 (40.33)	50
Hit	7 (9.66)	43 (40.33)	50
Control (not asked)	6 (9.66)	44 (40.33)	50
Total	29	121	150

Here the values within the braces are the expected frequencies.

Finally substitute the values in the ${\chi }^2$-statistics formula as:

$\begin{array}{c}{\chi }^2=\frac{{(16-9.66)}^2}{9.66}+\frac{{(7-9.66)}^2}{9.66}+\frac{{(6-9.66)}^2}{9.66}+\frac{{(34-40.33)}^2}{40.33}+\frac{{(43-40.33)}^2}{40.33}+\frac{{(44-40.33)}^2}{40.33}\\ =\frac{40.19}{9.66}+\frac{7.07}{9.66}+\frac{13.39}{9.66}+\frac{40.06}{40.33}+\frac{7.12}{40.33}+\frac{13.46}{40.33}\\ =6.27+1.50\\ =7.77\end{array}$

Step 4: Rejection rule: Reject when ${\chi }^2-\text{statistic}>CV$. Since ${\chi }^2-\text{statistic}(=7.77)>\text{critical value}(=5.991)$, reject the null hypothesis.

Step 5: Based on the results of hypothesis test, there is sufficient evidence to reject the null hypothesis at $\alpha =0.05$.

Hence, reject null hypothesis and conclude that proportion of participants who claim to remember broken glass differ significantly from group to group.

b.

To determine

To Find: The value of Cramer’s V for the given question.

Answer to Problem 15P

The value of Cramer’s V is 0.227.

Explanation of Solution

Calculations:

The formula for Cramer’s V is:

$Cramer’sV=\sqrt{\frac{{\chi }^2}{(d{f}^{\ast })n}}$

Here,

$\begin{array}{l}d{f}^{\ast }=\min (R-1,C-1)i.e.\text{ minimum of either rows or columns}\\ \because \text{Number of columns }\left(=2\right)<\text{Number of rows }\left(=3\right)\\ \therefore d{f}^{\ast }=1\end{array}$

Substitute 7.77 for ${\chi }^2$, 150 for $n$ and 1 for $d{f}^{\ast }$,

$\begin{array}{c}Cramer’sV=\sqrt{\frac{7.77}{1\times 150}}\\ =0.227\end{array}$

Hence, the value of Cramer’s V is 0.227.

c.

To determine

To Describe: How does the phrasing of the question influenced the participants memories.

Answer to Problem 15P

The phrasing of question influenced the participants memories little bit.

Explanation of Solution

Cramer’s V is used as post-test to determine strengths of association once the chi-square has determined significance.

A value of 0.227 indicates a small effect. That is, a little association between the groups.

d.

To determine

How would the outcome of hypothesis and Cramer’s V will be written in the report.

Answer to Problem 15P

Report will be ${\chi }^2(2,n=150)=7.77,p<0.05,V=0.227$ .

Explanation of Solution

The result showed that the proportion of participants who claim to remember broken glass differ significantly from group to group.

${\chi }^2(2,n=150)=7.77,p<0.05,V=0.227$