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Answer :
We begin by identifying and verifying the conditions required for inference in a one‐proportion [tex]\( z \)[/tex]-test.
1. Random Condition:
The problem states that a random sample of 100 adults was taken. This meets the requirement that the data come from a random sample of the population.
2. 10% Condition:
The sample of 100 adults should be less than 10% of the total adult population. Since the adult population is very large, the sample of 100 is indeed less than 10% of the population.
3. Large Counts (Success-Failure) Condition:
We check by computing:
[tex]$$ n p_0 = 100 \times 0.25 = 25 $$[/tex]
[tex]$$ n(1-p_0) = 100 \times 0.75 = 75 $$[/tex]
Both [tex]\( 25 \)[/tex] and [tex]\( 75 \)[/tex] are greater than 10, which ensures that the sample counts are large enough to justify the normal approximation.
Since all three conditions are met, we can proceed with the inference using the one-proportion [tex]\(z\)[/tex]-test.
For completeness, here are the key numbers from the analysis:
- Sample size: [tex]\( n = 100 \)[/tex]
- Number of successes: [tex]\( x = 42 \)[/tex] (adults who describe themselves as organized)
- Claimed proportion: [tex]\( p_0 = 0.25 \)[/tex]
- Sample proportion: [tex]\( \hat{p} = \frac{42}{100} = 0.42 \)[/tex]
- Standard error:
[tex]$$ \text{SE} = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.25 \times 0.75}{100}} \approx 0.0433 $$[/tex]
- Test statistic:
[tex]$$ z = \frac{\hat{p} - p_0}{\text{SE}} \approx \frac{0.42 - 0.25}{0.0433} \approx 3.93 $$[/tex]
- The corresponding [tex]\( p \)[/tex]-value for a one-tailed test is approximately [tex]\( 4.32 \times 10^{-5} \)[/tex].
Thus, the conditions for inference are:
- Random: We have a random sample of 100 adults.
- 10% Condition: 100 adults is less than 10% of the total adult population.
- Large Counts:
[tex]$$ n p_0 = 25, \quad n(1-p_0) = 75 $$[/tex]
Both are at least 10.
All conditions are satisfied, so the inference using the one-proportion [tex]\( z \)[/tex]-test is justified.
1. Random Condition:
The problem states that a random sample of 100 adults was taken. This meets the requirement that the data come from a random sample of the population.
2. 10% Condition:
The sample of 100 adults should be less than 10% of the total adult population. Since the adult population is very large, the sample of 100 is indeed less than 10% of the population.
3. Large Counts (Success-Failure) Condition:
We check by computing:
[tex]$$ n p_0 = 100 \times 0.25 = 25 $$[/tex]
[tex]$$ n(1-p_0) = 100 \times 0.75 = 75 $$[/tex]
Both [tex]\( 25 \)[/tex] and [tex]\( 75 \)[/tex] are greater than 10, which ensures that the sample counts are large enough to justify the normal approximation.
Since all three conditions are met, we can proceed with the inference using the one-proportion [tex]\(z\)[/tex]-test.
For completeness, here are the key numbers from the analysis:
- Sample size: [tex]\( n = 100 \)[/tex]
- Number of successes: [tex]\( x = 42 \)[/tex] (adults who describe themselves as organized)
- Claimed proportion: [tex]\( p_0 = 0.25 \)[/tex]
- Sample proportion: [tex]\( \hat{p} = \frac{42}{100} = 0.42 \)[/tex]
- Standard error:
[tex]$$ \text{SE} = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.25 \times 0.75}{100}} \approx 0.0433 $$[/tex]
- Test statistic:
[tex]$$ z = \frac{\hat{p} - p_0}{\text{SE}} \approx \frac{0.42 - 0.25}{0.0433} \approx 3.93 $$[/tex]
- The corresponding [tex]\( p \)[/tex]-value for a one-tailed test is approximately [tex]\( 4.32 \times 10^{-5} \)[/tex].
Thus, the conditions for inference are:
- Random: We have a random sample of 100 adults.
- 10% Condition: 100 adults is less than 10% of the total adult population.
- Large Counts:
[tex]$$ n p_0 = 25, \quad n(1-p_0) = 75 $$[/tex]
Both are at least 10.
All conditions are satisfied, so the inference using the one-proportion [tex]\( z \)[/tex]-test is justified.
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