Chapter 12.1, Problem 10E

The paper “Sociochemosensory and Emotional Functions” (Psychological Science [2009]: 1118–1124) describes an experiment to determine if college students can identify their roommates by smell. Forty-four female college students participated as subjects in the experiment. Each subject was presented with a set of three t-shirts that were identical in appearance. Each of the three t-shirts had been slept in for at least 7 hours by a person who had not used any scented products (like scented deodorant, soap, or shampoo) for at least 48 hours prior to sleeping in the shirt.

One of the three shirts had been worn by the subject’s roommate. The subject was asked to identify the shirt worn by her roommate. This process was then repeated with another three shirts, and the number of times out of the two trials that the subject correctly identified the shirt worn by her roommate was recorded. The resulting data are summarized in the accompanying table.

1. a. Can a person identify her roommate by smell? If not, the data from the experiment should be consistent with what we would have expected to see if subjects were just guessing on each trial. That is, we would expect that the probability of selecting the correct shirt would be 1/3 on each of the two trials.

   Calculate the proportions of the time we would expect to see 0, 1, and 2 correct identifications if subjects are just guessing. (Hint: 0 correct identifications occurs if the first trial is incorrect and the second trial is incorrect.)

2. b. Use the three proportions calculated in Part (a) to carry out a test to determine if the numbers of correct identifications by the students in this study are significantly different from what would have been expected by guessing. Use α = 0.05. (Note: One of the expected counts is just a bit less than 5. For purposes of this exercise, assume that it is OK to proceed with a goodness-of-fit test.)

To determine

Calculate the proportion of time expected to see 0, 1, and 2 correct identifications if subjects are just guessing.

Answer to Problem 10E

When the subjects are just guessing, the proportion of time expected to see 0, 1, and 2 correct identifications are $p(0)=\frac{4}{9}$, $p(1)=\frac{4}{9}$, and $p(2)=\frac{1}{9}$ respectively.

Explanation of Solution

Calculation:

It is given that if subjects were just guessing on each trial, the probability of selecting the correct shirt would be $\frac{1}{3}$.

Thus, the probability of selecting the wrong shirt is $1-\frac{1}{3}=\frac{2}{3}$.

The proportion of 0 correct identifications expected in two trials is shown below.

$\begin{array}{c}p(0)=P(\text{wrong shirt})\times P(\text{wrong shirt)}\\ ={\left(\frac{2}{3}\right)}^2\\ =\frac{4}{9}\end{array}$

The proportion of 1 correct identifications expected in two trials is shown below.

$\begin{array}{c}p(1)=\frac{2!}{1!\left(2-1\right)!}\times \left(\frac{1}{3}\right)\times \left(\frac{2}{3}\right)\\ =\frac{2}{1!1!}\times \left(\frac{2}{9}\right)\\ =\frac{4}{9}\end{array}$

The proportion of 2 correct identifications expected in two trials is shown below

$\begin{array}{c}p(2)=P(\text{correct identification)}\times P(\text{correct identification)}\\ ={\left(\frac{1}{3}\right)}^2\\ =\frac{1}{9}\end{array}$

To determine

Test whether the numbers of correct identifications by the students in this study are significantly different from what would have been expected by guessing at 0.05 level of significance.

Answer to Problem 10E

The numbers of correct identifications by the students in this study are significantly different from those that would have been expected by guessing.

Explanation of Solution

The given data represents the number of correct identifications of shirt worn by her roommate in two trials.

The expected counts can be calculated using the formula, $\text{Expected count}=n\times \text{hypothesized proportion}$, where n is the total observed frequency and the hypothesized proportions observed in Part (a). The following table shows the calculation of expected counts.

Number of correct identification	Observed Frequency	Expected counts
0	21	$44\times \frac{4}{9}=19.5556$
1	10	$44\times \frac{4}{9}=19.5556$
2	13	$44\times \frac{1}{9}=4.8889$
Total	44	44

The nine step hypotheses testing procedure to test goodness-of-fit is given below.

1. The proportion of correct identifications are $p_1$, $p_2$, and $p_3$.

2. Null hypothesis:

$H_0:$$p_1=\frac{4}{9}$, $p_2=\frac{4}{9}$, and $p_3=\frac{1}{9}$.

3. Alternative hypothesis:

$H_a:$ At least one of the population proportions is not equal to its corresponding hypothesized proportion.

4. Significance level:

$\alpha =0.05$

5. Test statistic:

${\chi }^2={\displaystyle \sum \frac{{(\text{observed count}-\text{expected count})}^2}{\text{expected count}}}$

6. Assumptions:

• Randomness assumption is not necessary, as the question is only to test whether the observed counts differ from expected by guessing.
• From the table above, it is observed that one of the expected counts is a little less than 5. However, as per the instruction, the goodness-of-fit test can be done.

7. Calculation:

Software procedure:

Step-by-step procedure to obtain the test statistics and P-value using the MINITAB software:

• Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
• In Observed counts, enter the column of Observed count.
• In Category names, enter the column of Number of correct identification.
• Under Test, select the column of Proportion in Proportions specified by historical counts.
• Click OK.

Output using the MINITAB software is given below:

From the output, ${\chi }^2=18.233$.

8. P-value:

From the MINITAB output, $df=2$ and P-value is 0.000.

9. Conclusion:

Decision rule:

• If P-value is less than or equal to the level of significance, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

Conclusion:

Here the level of significance is 0.05.

Here, P-value is less than the level of significance.

That is, $0.000<0.05$.

Therefore, reject the null hypothesis. Hence, the numbers of correct identifications by the students in this study are significantly different from those that would have been expected by guessing.