Question

a researcher wanted to study the tendency of peoples choices to be influenced by their environment, so she conducted an experiment. the researcher recruited 90 volunteers. she randomly assigned 30 to sit in a room with red walls, 30 to sit in a room with yellow walls, and 30 to sit in a room with orange walls. after sitting in the room for a while doing mundane activities, she offered them all a snack of strawberries, bananas, or oranges. for each individual, she recorded which room the participant was in and which snack they chose. the data are displayed in the table.
the researcher would like to know if these data provide convincing evidence that the distribution of snack choice differs for the various room colors in the population of all volunteers like these. are the conditions for inference met?
no, the random condition is not met.
no, the 10% condition is not met.

Explanation:

Step1: Check Random Condition

The researcher randomly assigned volunteers to rooms. So, the random condition is met.

Step2: Check 10% Condition

We assume the population of all volunteers like these is much larger than 90 (since \(90\times10 = 900\), and if the population \(N\geq900\), the 10% condition \(n\leq0.1N\) is met. Here \(n = 90\)). So, the 10% condition is likely met.

Step3: Check Large Counts Condition

For a chi - square test of homogeneity (which is appropriate here to test if the distribution of snack choice differs by room color), the expected count formula is \(E_{ij}=\frac{\text{Row Total}\times\text{Column Total}}{\text{Grand Total}}\).

• Row totals: Strawberries: \(16 + 7+9=32\), Bananas: \(10 + 16+9 = 35\), Oranges: \(4+7 + 12=23\)
• Column totals: Red: \(16+10 + 4=30\), Yellow: \(7+16+7 = 30\), Orange: \(9+9+12=30\)
• Grand total \(n=90\)
• For example, \(E_{11}=\frac{32\times30}{90}=\frac{960}{90}\approx10.67\), \(E_{12}=\frac{32\times30}{90}\approx10.67\), \(E_{13}=\frac{32\times30}{90}\approx10.67\), \(E_{21}=\frac{35\times30}{90}\approx11.67\), \(E_{22}=\frac{35\times30}{90}\approx11.67\), \(E_{23}=\frac{35\times30}{90}\approx11.67\), \(E_{31}=\frac{23\times30}{90}\approx7.67\), \(E_{32}=\frac{23\times30}{90}\approx7.67\), \(E_{33}=\frac{23\times30}{90}\approx7.67\)
• All expected counts (\(E_{ij}\geq5\))

Answer:

No, the problem statement is incomplete. But if we assume the options are about the three common conditions (random, 10% and large counts) for a chi - square test of homogeneity:

• Random condition: Met (random assignment)
• 10% condition: Likely met (assuming population \(N\geq900\))
• Large counts: Met (all expected counts \(\geq5\))

If the original options were:
A. No, the random condition is not met.
B. No, the 10% condition is not met.
C. No, the large counts condition is not met.
D. Yes, all conditions are met.

Based on our analysis (random - met, 10% - likely met, large counts - met), if we assume the options in the user's problem (even though the third option in the user's text is cut off) and assuming the correct logic of checking conditions for a chi - square test of homogeneity, if the third option was “No, the large counts condition is not met” (but our calculation shows large counts are met), and if the user's options have an error, but if we go by standard condition checking:
The random condition (random assignment) is met. The 10% condition (since \(n = 90\), and if we assume the population of all such volunteers is more than \(900\)) is met. For the large counts, since all expected counts are \(\geq5\) (calculated above), the answer is that all conditions are met. But since the user's options are cut off, if we assume the intended correct answer based on standard test (chi - square test of homogeneity conditions) and the given data (random assignment, expected counts calculation) the answer is that all conditions are met. But if we follow the user's provided (cut - off) options:
If the third option was “No, the large counts condition is not met” (incorrect as per calculation), but if we assume the user made a typo and the correct answer is that all conditions are met (if that was an option). But as per the given (cut - off) options in the user's problem (assuming the third option is wrong as per our calculation), there is an issue. But if we follow the strict condition checking:
Random: yes (random assignment), 10%: yes (assuming population is large enough, which is a usual assumption in experiments), large counts: yes (expected counts \(\geq5\)). So the answer is that all conditions are met. But since the user's options are not fully visible, if we assume the intended answer (based on common test question structure for such problems) and condition checking, if the options were:
A. No, the random condition is not met.
B. No, the 10% condition is not met.
C. No, the large counts condition is not met.
D. Yes, all conditions are met.
The answer is D. Yes, all conditions are met.