Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals 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]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
