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To answer these questions about standard error and normality, we need to follow a specific procedure for each part. The standard error for a proportion is calculated using the formula:
[tex]\[ \text{Standard Error} = \sqrt{\frac{\pi (1 - \pi)}{n}} \][/tex]
We also need to verify the normality condition, which states that the sample size and the proportion should satisfy:
[tex]\[ n \pi \geq 5 \quad \text{and} \quad n (1 - \pi) \geq 5 \][/tex]
Let's discuss each part step-by-step.
### (a) Given: [tex]\( n = 27 \)[/tex], [tex]\( \pi = 0.32 \)[/tex]
1. Standard Error Calculation:
[tex]\[ \text{Standard Error} = \sqrt{\frac{0.32 \times (1 - 0.32)}{27}} = 0.0898 \][/tex]
2. Normality Condition:
[tex]\[ n \pi = 27 \times 0.32 = 8.64 \quad (\text{>= 5}) \][/tex]
[tex]\[ n (1 - \pi) = 27 \times 0.68 = 18.36 \quad (\text{>= 5}) \][/tex]
Both conditions are satisfied, so normality can be assumed.
### Result:
- Standard Error: 0.0898
- Normality: True
### (b) Given: [tex]\( n = 58 \)[/tex], [tex]\( \pi = 0.45 \)[/tex]
1. Standard Error Calculation:
[tex]\[ \text{Standard Error} = \sqrt{\frac{0.45 \times (1 - 0.45)}{58}} = 0.0653 \][/tex]
2. Normality Condition:
[tex]\[ n \pi = 58 \times 0.45 = 26.1 \quad (\text{>= 5}) \][/tex]
[tex]\[ n (1 - \pi) = 58 \times 0.55 = 31.9 \quad (\text{>= 5}) \][/tex]
Both conditions are satisfied, so normality can be assumed.
### Result:
- Standard Error: 0.0653
- Normality: True
### (c) Given: [tex]\( n = 119 \)[/tex], [tex]\( \pi = 0.39 \)[/tex]
1. Standard Error Calculation:
[tex]\[ \text{Standard Error} = \sqrt{\frac{0.39 \times (1 - 0.39)}{119}} = 0.0447 \][/tex]
2. Normality Condition:
[tex]\[ n \pi = 119 \times 0.39 = 46.41 \quad (\text{>= 5}) \][/tex]
[tex]\[ n (1 - \pi) = 119 \times 0.61 = 72.59 \quad (\text{>= 5}) \][/tex]
Both conditions are satisfied, so normality can be assumed.
### Result:
- Standard Error: 0.0447
- Normality: True
### (d) Given: [tex]\( n = 359 \)[/tex], [tex]\( \pi = 0.004 \)[/tex]
1. Standard Error Calculation:
[tex]\[ \text{Standard Error} = \sqrt{\frac{0.004 \times (1 - 0.004)}{359}} = 0.0033 \][/tex]
2. Normality Condition:
[tex]\[ n \pi = 359 \times 0.004 = 1.436 \quad (\text{< 5}) \][/tex]
[tex]\[ n (1 - \pi) = 359 \times 0.996 = 357.564 \quad (\text{>= 5}) \][/tex]
The first condition is not satisfied, so normality cannot be assumed.
### Result:
- Standard Error: 0.0033
- Normality: False
In summary:
[tex]\[ \begin{array}{|c|c|c|c|} \hline & n & \pi & \text{Standard Error} & \text{Normality} \\ \hline (a) & 27 & 0.32 & 0.0898 & \text{True} \\ \hline (b) & 58 & 0.45 & 0.0653 & \text{True} \\ \hline (c) & 119 & 0.39 & 0.0447 & \text{True} \\ \hline (d) & 359 & 0.004 & 0.0033 & \text{False} \\ \hline \end{array} \][/tex]
[tex]\[ \text{Standard Error} = \sqrt{\frac{\pi (1 - \pi)}{n}} \][/tex]
We also need to verify the normality condition, which states that the sample size and the proportion should satisfy:
[tex]\[ n \pi \geq 5 \quad \text{and} \quad n (1 - \pi) \geq 5 \][/tex]
Let's discuss each part step-by-step.
### (a) Given: [tex]\( n = 27 \)[/tex], [tex]\( \pi = 0.32 \)[/tex]
1. Standard Error Calculation:
[tex]\[ \text{Standard Error} = \sqrt{\frac{0.32 \times (1 - 0.32)}{27}} = 0.0898 \][/tex]
2. Normality Condition:
[tex]\[ n \pi = 27 \times 0.32 = 8.64 \quad (\text{>= 5}) \][/tex]
[tex]\[ n (1 - \pi) = 27 \times 0.68 = 18.36 \quad (\text{>= 5}) \][/tex]
Both conditions are satisfied, so normality can be assumed.
### Result:
- Standard Error: 0.0898
- Normality: True
### (b) Given: [tex]\( n = 58 \)[/tex], [tex]\( \pi = 0.45 \)[/tex]
1. Standard Error Calculation:
[tex]\[ \text{Standard Error} = \sqrt{\frac{0.45 \times (1 - 0.45)}{58}} = 0.0653 \][/tex]
2. Normality Condition:
[tex]\[ n \pi = 58 \times 0.45 = 26.1 \quad (\text{>= 5}) \][/tex]
[tex]\[ n (1 - \pi) = 58 \times 0.55 = 31.9 \quad (\text{>= 5}) \][/tex]
Both conditions are satisfied, so normality can be assumed.
### Result:
- Standard Error: 0.0653
- Normality: True
### (c) Given: [tex]\( n = 119 \)[/tex], [tex]\( \pi = 0.39 \)[/tex]
1. Standard Error Calculation:
[tex]\[ \text{Standard Error} = \sqrt{\frac{0.39 \times (1 - 0.39)}{119}} = 0.0447 \][/tex]
2. Normality Condition:
[tex]\[ n \pi = 119 \times 0.39 = 46.41 \quad (\text{>= 5}) \][/tex]
[tex]\[ n (1 - \pi) = 119 \times 0.61 = 72.59 \quad (\text{>= 5}) \][/tex]
Both conditions are satisfied, so normality can be assumed.
### Result:
- Standard Error: 0.0447
- Normality: True
### (d) Given: [tex]\( n = 359 \)[/tex], [tex]\( \pi = 0.004 \)[/tex]
1. Standard Error Calculation:
[tex]\[ \text{Standard Error} = \sqrt{\frac{0.004 \times (1 - 0.004)}{359}} = 0.0033 \][/tex]
2. Normality Condition:
[tex]\[ n \pi = 359 \times 0.004 = 1.436 \quad (\text{< 5}) \][/tex]
[tex]\[ n (1 - \pi) = 359 \times 0.996 = 357.564 \quad (\text{>= 5}) \][/tex]
The first condition is not satisfied, so normality cannot be assumed.
### Result:
- Standard Error: 0.0033
- Normality: False
In summary:
[tex]\[ \begin{array}{|c|c|c|c|} \hline & n & \pi & \text{Standard Error} & \text{Normality} \\ \hline (a) & 27 & 0.32 & 0.0898 & \text{True} \\ \hline (b) & 58 & 0.45 & 0.0653 & \text{True} \\ \hline (c) & 119 & 0.39 & 0.0447 & \text{True} \\ \hline (d) & 359 & 0.004 & 0.0033 & \text{False} \\ \hline \end{array} \][/tex]
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