IDNLearn.com: Where your questions are met with thoughtful and precise answers. Discover in-depth and trustworthy answers to all your questions from our experienced community members.
Sagot :
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.
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.
