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Answer :
We begin with Keith’s contingency table of observed counts:
[tex]$$
\begin{array}{|l|c|c|}
\hline
& \text{Major} & \text{Minor} \\
\hline
\text{Happy} & 49 & 24 \\
\hline
\text{Sad} & 27 & 25 \\
\hline
\end{array}
$$[/tex]
There are a total of 125 songs in the sample.
Below is the step‐by‐step process of checking the conditions for inference:
1. Type of Test
Since Keith selected one random sample and is classifying songs based on two categorical variables (Emotion and Key), the appropriate test is a chi-square test to determine if there is an association between the variables.
- Statement: “This is a chi-square test for association.” is true.
- Note: Although some might mention a chi-square test for homogeneity, that test is used when comparing separate groups. Here, a single random sample is used to test independence/association.
2. Random Condition
Keith selected a random sample of songs. Therefore, the sample is representative of the population, and this condition is met.
- Statement: “The random condition is met because Keith selected a random sample of songs.” is true.
3. 10% Condition
The sampling condition requires that the sample size is less than 10% of the population. With 125 songs, it is assumed that [tex]$125$[/tex] is less than 10% of all songs in the universe.
- Statement: “The 10% condition is met because [tex]$125 < 10\%$[/tex] of all songs.” is true.
4. Large Counts (Expected Counts) Condition
The expected counts for each cell of a [tex]$2 \times 2$[/tex] table are calculated using the formula:
[tex]$$
\text{Expected count} = \frac{(\text{Row total})(\text{Column total})}{\text{Grand total}}
$$[/tex]
The totals are:
- Row totals:
- Happy: [tex]$49 + 24 = 73$[/tex]
- Sad: [tex]$27 + 25 = 52$[/tex]
- Column totals:
- Major: [tex]$49 + 27 = 76$[/tex]
- Minor: [tex]$24 + 25 = 49$[/tex]
We calculate the expected counts:
[tex]$$
\begin{aligned}
E[\text{Happy, Major}] &= \frac{73 \times 76}{125} \approx 44.384,\\[1mm]
E[\text{Happy, Minor}] &= \frac{73 \times 49}{125} \approx 28.616,\\[1mm]
E[\text{Sad, Major}] &= \frac{52 \times 76}{125} \approx 31.616,\\[1mm]
E[\text{Sad, Minor}] &= \frac{52 \times 49}{125} \approx 20.384.
\end{aligned}
$$[/tex]
The smallest expected count is approximately [tex]$20.384$[/tex], which is well above the minimum requirement of [tex]$5$[/tex].
- Statement: “The Large Counts condition is met because the smallest expected count is 5.” is false since 5 is not the actual smallest count.
- Statement: “The Large Counts condition is met because the smallest expected count is [tex]$20.384$[/tex].” is true.
5. Conclusion on All Conditions
Every required condition (random sampling, 10% condition, and all expected counts greater than 5) has been satisfied.
- Statement: “All conditions for inference are met.” is true.
---
Final Answer:
The correct statements are 1, 3, 4, 6, and 7.
[tex]$$
\begin{array}{|l|c|c|}
\hline
& \text{Major} & \text{Minor} \\
\hline
\text{Happy} & 49 & 24 \\
\hline
\text{Sad} & 27 & 25 \\
\hline
\end{array}
$$[/tex]
There are a total of 125 songs in the sample.
Below is the step‐by‐step process of checking the conditions for inference:
1. Type of Test
Since Keith selected one random sample and is classifying songs based on two categorical variables (Emotion and Key), the appropriate test is a chi-square test to determine if there is an association between the variables.
- Statement: “This is a chi-square test for association.” is true.
- Note: Although some might mention a chi-square test for homogeneity, that test is used when comparing separate groups. Here, a single random sample is used to test independence/association.
2. Random Condition
Keith selected a random sample of songs. Therefore, the sample is representative of the population, and this condition is met.
- Statement: “The random condition is met because Keith selected a random sample of songs.” is true.
3. 10% Condition
The sampling condition requires that the sample size is less than 10% of the population. With 125 songs, it is assumed that [tex]$125$[/tex] is less than 10% of all songs in the universe.
- Statement: “The 10% condition is met because [tex]$125 < 10\%$[/tex] of all songs.” is true.
4. Large Counts (Expected Counts) Condition
The expected counts for each cell of a [tex]$2 \times 2$[/tex] table are calculated using the formula:
[tex]$$
\text{Expected count} = \frac{(\text{Row total})(\text{Column total})}{\text{Grand total}}
$$[/tex]
The totals are:
- Row totals:
- Happy: [tex]$49 + 24 = 73$[/tex]
- Sad: [tex]$27 + 25 = 52$[/tex]
- Column totals:
- Major: [tex]$49 + 27 = 76$[/tex]
- Minor: [tex]$24 + 25 = 49$[/tex]
We calculate the expected counts:
[tex]$$
\begin{aligned}
E[\text{Happy, Major}] &= \frac{73 \times 76}{125} \approx 44.384,\\[1mm]
E[\text{Happy, Minor}] &= \frac{73 \times 49}{125} \approx 28.616,\\[1mm]
E[\text{Sad, Major}] &= \frac{52 \times 76}{125} \approx 31.616,\\[1mm]
E[\text{Sad, Minor}] &= \frac{52 \times 49}{125} \approx 20.384.
\end{aligned}
$$[/tex]
The smallest expected count is approximately [tex]$20.384$[/tex], which is well above the minimum requirement of [tex]$5$[/tex].
- Statement: “The Large Counts condition is met because the smallest expected count is 5.” is false since 5 is not the actual smallest count.
- Statement: “The Large Counts condition is met because the smallest expected count is [tex]$20.384$[/tex].” is true.
5. Conclusion on All Conditions
Every required condition (random sampling, 10% condition, and all expected counts greater than 5) has been satisfied.
- Statement: “All conditions for inference are met.” is true.
---
Final Answer:
The correct statements are 1, 3, 4, 6, and 7.
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