Answer :
Certainly! Let's go through each part of the question step-by-step to provide a detailed solution.
### (a) Finding a 90% Confidence Interval for the Population Mean Price (per 100 pounds)
Given:
- Sample mean price, [tex]\(\bar{x} = \$6.88\)[/tex]
- Sample size, [tex]\(n = 43\)[/tex]
- Population standard deviation, [tex]\(\sigma = \$1.98\)[/tex]
- Confidence level, [tex]\(90\%\)[/tex]
First, we need to find the z-score corresponding to a 90% confidence level. For a two-tailed test with [tex]\(\alpha = 0.10\)[/tex] (because [tex]\(100\% - 90\% = 10\%\)[/tex]), the z-score is approximately [tex]\(1.645\)[/tex].
Now, the standard error (SE) of the mean is calculated as:
[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{1.98}{\sqrt{43}} \approx 0.3011 \][/tex]
The margin of error (ME) can be found by multiplying the z-score by the standard error:
[tex]\[ \text{ME} = z_{\alpha/2} \times \text{SE} = 1.645 \times 0.3011 \approx 0.4967 \][/tex]
The confidence interval is then:
[tex]\[ \text{Lower limit} = \bar{x} - \text{ME} = 6.88 - 0.4967 \approx 6.38 \][/tex]
[tex]\[ \text{Upper limit} = \bar{x} + \text{ME} = 6.88 + 0.4967 \approx 7.38 \][/tex]
So, the 90% confidence interval for the population mean price per 100 pounds is:
- Lower limit: \[tex]$6.38 - Upper limit: \$[/tex]7.38
- Margin of error: \[tex]$0.50 (rounded to two decimal places) ### (b) Finding the Sample Size for a 90% Confidence Level with a Maximal Error of Estimate \(E = 0.27\) Given: - Desired maximal error of estimate, \(E = 0.27\) - Population standard deviation, \(\sigma = 1.98\) - Confidence level, \(90\%\) To find the required sample size (\(n\)), we use the formula: \[ n = \left( \frac{z_{\alpha/2} \cdot \sigma}{E} \right)^2 \] Using the z-score \(1.645\): \[ n = \left( \frac{1.645 \cdot 1.98}{0.27} \right)^2 \approx 145.70 \] Since the sample size must be a whole number, we round up to the nearest whole number: - Required sample size: 146 farming regions ### (c) Finding a 90% Confidence Interval for the Population Mean Cash Value of 15 Tons of Watermelon Given: - Sample mean price, \(\bar{x} = \$[/tex]6.88\) per 100 pounds
- Sample size, [tex]\(n = 43\)[/tex]
- Population standard deviation, [tex]\(\sigma = \$1.98\)[/tex]
- Confidence level, [tex]\(90\%\)[/tex]
- Total weight, [tex]\(15\)[/tex] tons = [tex]\(15 \times 2000 = 30000\)[/tex] pounds
Convert the sample mean price to the total price for 30000 pounds:
[tex]\[ \text{Sample mean total cash value} = \bar{x} \times \frac{30000}{100} = 6.88 \times 300 = \$2064 \][/tex]
Using the margin of error calculated in part (a) for 100 pounds:
[tex]\[ \text{ME for 30000 pounds} = \text{ME for 100 pounds} \times \frac{30000}{100} = 0.4967 \times 300 \approx 149.00 \][/tex]
The confidence interval for the total cash value is:
[tex]\[ \text{Lower limit} = 2064 - 149.00 \approx 1915.00 \][/tex]
[tex]\[ \text{Upper limit} = 2064 + 149.00 \approx 2213.00 \][/tex]
So, the 90% confidence interval for the total cash value of 15 tons of watermelon is:
- Lower limit: \[tex]$1915.00 - Upper limit: \$[/tex]2213.00
- Margin of error: \$149.00 (rounded to two decimal places)
I hope this detailed solution helps you understand the process! If you have any further questions, feel free to ask.
### (a) Finding a 90% Confidence Interval for the Population Mean Price (per 100 pounds)
Given:
- Sample mean price, [tex]\(\bar{x} = \$6.88\)[/tex]
- Sample size, [tex]\(n = 43\)[/tex]
- Population standard deviation, [tex]\(\sigma = \$1.98\)[/tex]
- Confidence level, [tex]\(90\%\)[/tex]
First, we need to find the z-score corresponding to a 90% confidence level. For a two-tailed test with [tex]\(\alpha = 0.10\)[/tex] (because [tex]\(100\% - 90\% = 10\%\)[/tex]), the z-score is approximately [tex]\(1.645\)[/tex].
Now, the standard error (SE) of the mean is calculated as:
[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{1.98}{\sqrt{43}} \approx 0.3011 \][/tex]
The margin of error (ME) can be found by multiplying the z-score by the standard error:
[tex]\[ \text{ME} = z_{\alpha/2} \times \text{SE} = 1.645 \times 0.3011 \approx 0.4967 \][/tex]
The confidence interval is then:
[tex]\[ \text{Lower limit} = \bar{x} - \text{ME} = 6.88 - 0.4967 \approx 6.38 \][/tex]
[tex]\[ \text{Upper limit} = \bar{x} + \text{ME} = 6.88 + 0.4967 \approx 7.38 \][/tex]
So, the 90% confidence interval for the population mean price per 100 pounds is:
- Lower limit: \[tex]$6.38 - Upper limit: \$[/tex]7.38
- Margin of error: \[tex]$0.50 (rounded to two decimal places) ### (b) Finding the Sample Size for a 90% Confidence Level with a Maximal Error of Estimate \(E = 0.27\) Given: - Desired maximal error of estimate, \(E = 0.27\) - Population standard deviation, \(\sigma = 1.98\) - Confidence level, \(90\%\) To find the required sample size (\(n\)), we use the formula: \[ n = \left( \frac{z_{\alpha/2} \cdot \sigma}{E} \right)^2 \] Using the z-score \(1.645\): \[ n = \left( \frac{1.645 \cdot 1.98}{0.27} \right)^2 \approx 145.70 \] Since the sample size must be a whole number, we round up to the nearest whole number: - Required sample size: 146 farming regions ### (c) Finding a 90% Confidence Interval for the Population Mean Cash Value of 15 Tons of Watermelon Given: - Sample mean price, \(\bar{x} = \$[/tex]6.88\) per 100 pounds
- Sample size, [tex]\(n = 43\)[/tex]
- Population standard deviation, [tex]\(\sigma = \$1.98\)[/tex]
- Confidence level, [tex]\(90\%\)[/tex]
- Total weight, [tex]\(15\)[/tex] tons = [tex]\(15 \times 2000 = 30000\)[/tex] pounds
Convert the sample mean price to the total price for 30000 pounds:
[tex]\[ \text{Sample mean total cash value} = \bar{x} \times \frac{30000}{100} = 6.88 \times 300 = \$2064 \][/tex]
Using the margin of error calculated in part (a) for 100 pounds:
[tex]\[ \text{ME for 30000 pounds} = \text{ME for 100 pounds} \times \frac{30000}{100} = 0.4967 \times 300 \approx 149.00 \][/tex]
The confidence interval for the total cash value is:
[tex]\[ \text{Lower limit} = 2064 - 149.00 \approx 1915.00 \][/tex]
[tex]\[ \text{Upper limit} = 2064 + 149.00 \approx 2213.00 \][/tex]
So, the 90% confidence interval for the total cash value of 15 tons of watermelon is:
- Lower limit: \[tex]$1915.00 - Upper limit: \$[/tex]2213.00
- Margin of error: \$149.00 (rounded to two decimal places)
I hope this detailed solution helps you understand the process! If you have any further questions, feel free to ask.
