Answer :
We are comparing two categorical variables recorded on a random sample of 200 callers: whether the customer “got their way” (Yes/No) and their satisfaction level (with four categories). We want to check the conditions required for a chi‐square test and then choose the correct statements regarding the inference.
Below is a step‐by‐step explanation:
1. Because the data involve two variables measured on each caller in a single sample, the test is one of association rather than one of homogeneity. (A chi‐square test for homogeneity is used when separate samples are drawn from different populations.)
• Thus, statement 1 (“This is a chi-square test for association”) is correct, and statement 2 (“This is a chi-square test for homogeneity”) is not correct.
2. Random Condition:
The manager selected a random sample of callers. This meets the random condition required for inference.
• Therefore, statement 3 is correct.
3. 10% Condition:
The sample size is 200, and it is assumed that 200 is less than 10% of the entire population of callers. This meets the 10% condition, which requires that the sample be no more than 10% of the population when sampling without replacement.
• Hence, statement 4 is correct.
4. Large Counts (Expected Counts) Condition:
The expected counts for each cell in the two-way table are computed using the formula:
[tex]$$
E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Overall Total}}.
$$[/tex]
When these calculations are done for all cells, the smallest expected count turns out to be [tex]$6.5$[/tex].
• Since the rule of thumb is that all expected counts should be at least 5, having a smallest expected count of [tex]$6.5$[/tex] satisfies the large counts condition.
• Thus, the statement that “the Large Counts condition is met because the smallest expected count is 5” is not correct, while the statement that “the Large Counts condition is met because the smallest expected count is [tex]$6.5$[/tex]” is correct.
5. All conditions for the chi-square test (random sampling, 10% condition, and large counts) are met.
• Therefore, statement 7 (“All conditions for inference are met”) is also correct.
In summary, the correct statements are:
1. This is a chi-square test for association.
3. The random condition is met because the manager selected a random sample of callers.
4. The 10% condition is met because 200 is less than 10% of all callers.
6. The Large Counts condition is met because the smallest expected count is [tex]$6.5$[/tex].
7. All conditions for inference are met.
Thus, the correct options are: 1, 3, 4, 6, and 7.
Below is a step‐by‐step explanation:
1. Because the data involve two variables measured on each caller in a single sample, the test is one of association rather than one of homogeneity. (A chi‐square test for homogeneity is used when separate samples are drawn from different populations.)
• Thus, statement 1 (“This is a chi-square test for association”) is correct, and statement 2 (“This is a chi-square test for homogeneity”) is not correct.
2. Random Condition:
The manager selected a random sample of callers. This meets the random condition required for inference.
• Therefore, statement 3 is correct.
3. 10% Condition:
The sample size is 200, and it is assumed that 200 is less than 10% of the entire population of callers. This meets the 10% condition, which requires that the sample be no more than 10% of the population when sampling without replacement.
• Hence, statement 4 is correct.
4. Large Counts (Expected Counts) Condition:
The expected counts for each cell in the two-way table are computed using the formula:
[tex]$$
E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Overall Total}}.
$$[/tex]
When these calculations are done for all cells, the smallest expected count turns out to be [tex]$6.5$[/tex].
• Since the rule of thumb is that all expected counts should be at least 5, having a smallest expected count of [tex]$6.5$[/tex] satisfies the large counts condition.
• Thus, the statement that “the Large Counts condition is met because the smallest expected count is 5” is not correct, while the statement that “the Large Counts condition is met because the smallest expected count is [tex]$6.5$[/tex]” is correct.
5. All conditions for the chi-square test (random sampling, 10% condition, and large counts) are met.
• Therefore, statement 7 (“All conditions for inference are met”) is also correct.
In summary, the correct statements are:
1. This is a chi-square test for association.
3. The random condition is met because the manager selected a random sample of callers.
4. The 10% condition is met because 200 is less than 10% of all callers.
6. The Large Counts condition is met because the smallest expected count is [tex]$6.5$[/tex].
7. All conditions for inference are met.
Thus, the correct options are: 1, 3, 4, 6, and 7.
