Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
