Answer :
Let's work through each part of the question step by step. The problem involves understanding the statistical concepts of sampling distribution, calculating the standard error, and finding probabilities using the normal distribution.
### Part (a)
Sampling distribution for a sample of 540 St. Paulites:
1. Expected value of [tex]\(\bar{p}\)[/tex]: The expected value [tex]\(E(\bar{p})\)[/tex] is the same as the population proportion [tex]\(p\)[/tex]. Given that [tex]\(p = 0.33\)[/tex]:
[tex]\[ E(\bar{p}) = 0.33 \][/tex]
2. Standard error of [tex]\(\bar{p}\)[/tex]: The standard error [tex]\(\sigma_{\bar{p}}\)[/tex] for the sample proportion is calculated using the formula:
[tex]\[ \sigma_{\bar{p}} = \sqrt{\frac{p(1 - p)}{n}} \][/tex]
For a sample size [tex]\(n = 540\)[/tex]:
[tex]\[ \sigma_{\bar{p}} = \sqrt{\frac{0.33 \times (1 - 0.33)}{540}} = \sqrt{\frac{0.33 \times 0.67}{540}} = 0.0202 \][/tex]
So, the sampling distribution of [tex]\(\bar{p}\)[/tex] for a sample size of 540 is:
[tex]\[ E(\bar{p}) = 0.33, \quad \sigma_{\bar{p}} = 0.0202 \][/tex]
### Part (b)
Probability that the sample proportion will be within 0.02 of the population proportion for a sample of 540:
1. Margin of error: We are asked to find the probability that the sample proportion is within 0.02 of the population proportion [tex]\(0.33\)[/tex].
2. Z-score calculation: The Z-score for the margin can be calculated as follows:
[tex]\[ Z = \frac{margin}{\sigma_{\bar{p}}} = \frac{0.02}{0.0202} = 0.990 \][/tex]
3. Using the standard normal distribution table: We need to find the probability that Z is between [tex]\(-0.990\)[/tex] and [tex]\(0.990\)[/tex]:
[tex]\[ P(-0.990 \leq Z \leq 0.990) \][/tex]
From the standard normal distribution table, we find that the cumulative probability for Z = 0.990 is approximately 0.677:
[tex]\[ P(-0.990 \leq Z \leq 0.990) = 0.677 \][/tex]
### Part (c)
Sampling distribution for a sample of 240 St. Paulites:
1. Expected value of [tex]\(\bar{p}\)[/tex]: The expected value [tex]\(E(\bar{p})\)[/tex] is still the population proportion [tex]\(p\)[/tex]. Given that [tex]\(p = 0.33\)[/tex]:
[tex]\[ E(\bar{p}) = 0.33 \][/tex]
2. Standard error of [tex]\(\bar{p}\)[/tex]: The standard error [tex]\(\sigma_{\bar{p}}\)[/tex] for the sample proportion is calculated using the same formula, but with [tex]\(n = 240\)[/tex]:
[tex]\[ \sigma_{\bar{p}} = \sqrt{\frac{0.33 \times (1 - 0.33)}{240}} = \sqrt{\frac{0.33 \times 0.67}{240}} = 0.0304 \][/tex]
So, the sampling distribution of [tex]\(\bar{p}\)[/tex] for a sample size of 240 is:
[tex]\[ E(\bar{p}) = 0.33, \quad \sigma_{\bar{p}} = 0.0304 \][/tex]
### Part (d)
Probability that the sample proportion will be within 0.02 of the population proportion for a sample of 240:
1. Margin of error: We are asked to find the probability that the sample proportion is within 0.02 of the population proportion [tex]\(0.33\)[/tex].
2. Z-score calculation: The Z-score for the margin can be calculated as follows:
[tex]\[ Z = \frac{margin}{\sigma_{\bar{p}}} = \frac{0.02}{0.0304} = 0.658 \][/tex]
3. Using the standard normal distribution table: We need to find the probability that Z is between [tex]\(-0.658\)[/tex] and [tex]\(0.658\)[/tex]:
[tex]\[ P(-0.658 \leq Z \leq 0.658) \][/tex]
From the standard normal distribution table, we find that the cumulative probability for Z = 0.658 is approximately 0.4901:
[tex]\[ P(-0.658 \leq Z \leq 0.658) = 0.4901 \][/tex]
In summary, for part (d), the probability that the sample proportion will be within 0.02 of the population proportion is 0.4901.
To recap:
a. For [tex]\(n = 540\)[/tex]:
- [tex]\(E(\bar{p}) = 0.33\)[/tex]
- [tex]\(\sigma_{\bar{p}} = 0.0202\)[/tex]
b. The probability within 0.02 for [tex]\(n = 540\)[/tex] is 0.677
c. For [tex]\(n = 240\)[/tex]:
- [tex]\(E(\bar{p}) = 0.33\)[/tex]
- [tex]\(\sigma_{\bar{p}} = 0.0304\)[/tex]
d. The probability within 0.02 for [tex]\(n = 240\)[/tex] is 0.4901
### Part (a)
Sampling distribution for a sample of 540 St. Paulites:
1. Expected value of [tex]\(\bar{p}\)[/tex]: The expected value [tex]\(E(\bar{p})\)[/tex] is the same as the population proportion [tex]\(p\)[/tex]. Given that [tex]\(p = 0.33\)[/tex]:
[tex]\[ E(\bar{p}) = 0.33 \][/tex]
2. Standard error of [tex]\(\bar{p}\)[/tex]: The standard error [tex]\(\sigma_{\bar{p}}\)[/tex] for the sample proportion is calculated using the formula:
[tex]\[ \sigma_{\bar{p}} = \sqrt{\frac{p(1 - p)}{n}} \][/tex]
For a sample size [tex]\(n = 540\)[/tex]:
[tex]\[ \sigma_{\bar{p}} = \sqrt{\frac{0.33 \times (1 - 0.33)}{540}} = \sqrt{\frac{0.33 \times 0.67}{540}} = 0.0202 \][/tex]
So, the sampling distribution of [tex]\(\bar{p}\)[/tex] for a sample size of 540 is:
[tex]\[ E(\bar{p}) = 0.33, \quad \sigma_{\bar{p}} = 0.0202 \][/tex]
### Part (b)
Probability that the sample proportion will be within 0.02 of the population proportion for a sample of 540:
1. Margin of error: We are asked to find the probability that the sample proportion is within 0.02 of the population proportion [tex]\(0.33\)[/tex].
2. Z-score calculation: The Z-score for the margin can be calculated as follows:
[tex]\[ Z = \frac{margin}{\sigma_{\bar{p}}} = \frac{0.02}{0.0202} = 0.990 \][/tex]
3. Using the standard normal distribution table: We need to find the probability that Z is between [tex]\(-0.990\)[/tex] and [tex]\(0.990\)[/tex]:
[tex]\[ P(-0.990 \leq Z \leq 0.990) \][/tex]
From the standard normal distribution table, we find that the cumulative probability for Z = 0.990 is approximately 0.677:
[tex]\[ P(-0.990 \leq Z \leq 0.990) = 0.677 \][/tex]
### Part (c)
Sampling distribution for a sample of 240 St. Paulites:
1. Expected value of [tex]\(\bar{p}\)[/tex]: The expected value [tex]\(E(\bar{p})\)[/tex] is still the population proportion [tex]\(p\)[/tex]. Given that [tex]\(p = 0.33\)[/tex]:
[tex]\[ E(\bar{p}) = 0.33 \][/tex]
2. Standard error of [tex]\(\bar{p}\)[/tex]: The standard error [tex]\(\sigma_{\bar{p}}\)[/tex] for the sample proportion is calculated using the same formula, but with [tex]\(n = 240\)[/tex]:
[tex]\[ \sigma_{\bar{p}} = \sqrt{\frac{0.33 \times (1 - 0.33)}{240}} = \sqrt{\frac{0.33 \times 0.67}{240}} = 0.0304 \][/tex]
So, the sampling distribution of [tex]\(\bar{p}\)[/tex] for a sample size of 240 is:
[tex]\[ E(\bar{p}) = 0.33, \quad \sigma_{\bar{p}} = 0.0304 \][/tex]
### Part (d)
Probability that the sample proportion will be within 0.02 of the population proportion for a sample of 240:
1. Margin of error: We are asked to find the probability that the sample proportion is within 0.02 of the population proportion [tex]\(0.33\)[/tex].
2. Z-score calculation: The Z-score for the margin can be calculated as follows:
[tex]\[ Z = \frac{margin}{\sigma_{\bar{p}}} = \frac{0.02}{0.0304} = 0.658 \][/tex]
3. Using the standard normal distribution table: We need to find the probability that Z is between [tex]\(-0.658\)[/tex] and [tex]\(0.658\)[/tex]:
[tex]\[ P(-0.658 \leq Z \leq 0.658) \][/tex]
From the standard normal distribution table, we find that the cumulative probability for Z = 0.658 is approximately 0.4901:
[tex]\[ P(-0.658 \leq Z \leq 0.658) = 0.4901 \][/tex]
In summary, for part (d), the probability that the sample proportion will be within 0.02 of the population proportion is 0.4901.
To recap:
a. For [tex]\(n = 540\)[/tex]:
- [tex]\(E(\bar{p}) = 0.33\)[/tex]
- [tex]\(\sigma_{\bar{p}} = 0.0202\)[/tex]
b. The probability within 0.02 for [tex]\(n = 540\)[/tex] is 0.677
c. For [tex]\(n = 240\)[/tex]:
- [tex]\(E(\bar{p}) = 0.33\)[/tex]
- [tex]\(\sigma_{\bar{p}} = 0.0304\)[/tex]
d. The probability within 0.02 for [tex]\(n = 240\)[/tex] is 0.4901
