Question
A random sampling of 50 bags of oranges is weighed. The mean weight is 3.02 pounds with a standard deviation of 0.39. Find the 95 percent confidence interval.(1 point)
Responses
0.28, 0.5
0.28, 0.5
49.89, 50.11
49.89, 50.11
2.91, 3.13
2.91, 3.13
2.63, 3.41
Responses
0.28, 0.5
0.28, 0.5
49.89, 50.11
49.89, 50.11
2.91, 3.13
2.91, 3.13
2.63, 3.41
Answers
GPT-4o mini
To calculate the 95% confidence interval for the mean weight of the bags of oranges, we can use the formula for the confidence interval (CI) for the mean when the population standard deviation is unknown:
\[
CI = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right)
\]
Where:
- \(\bar{x}\) = sample mean
- \(t^*\) = t-value for the desired confidence level (from the t-distribution table)
- \(s\) = sample standard deviation
- \(n\) = sample size
Given:
- \(\bar{x} = 3.02\) pounds
- \(s = 0.39\) pounds
- \(n = 50\)
### Step 1: Find the t-value
For a 95% confidence level and degrees of freedom \(df = n - 1 = 50 - 1 = 49\), we can look up the t-value. For a two-tailed test, \(t^* \approx 2.009\) (you can confirm this with a t-table).
### Step 2: Calculate the standard error (SE)
\[
SE = \frac{s}{\sqrt{n}} = \frac{0.39}{\sqrt{50}} \approx \frac{0.39}{7.071} \approx 0.055
\]
### Step 3: Calculate the margin of error (ME)
\[
ME = t^* \cdot SE \approx 2.009 \cdot 0.055 \approx 0.1105
\]
### Step 4: Calculate the confidence interval
\[
CI = \bar{x} \pm ME = 3.02 \pm 0.1105
\]
Lower limit:
\[
3.02 - 0.1105 \approx 2.9095 \quad \text{(approximately 2.91)}
\]
Upper limit:
\[
3.02 + 0.1105 \approx 3.1305 \quad \text{(approximately 3.13)}
\]
### Conclusion
The 95% confidence interval for the mean weight of the bags of oranges is approximately:
\[
(2.91, 3.13)
\]
So, the correct response is:
**2.91, 3.13**
\[
CI = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right)
\]
Where:
- \(\bar{x}\) = sample mean
- \(t^*\) = t-value for the desired confidence level (from the t-distribution table)
- \(s\) = sample standard deviation
- \(n\) = sample size
Given:
- \(\bar{x} = 3.02\) pounds
- \(s = 0.39\) pounds
- \(n = 50\)
### Step 1: Find the t-value
For a 95% confidence level and degrees of freedom \(df = n - 1 = 50 - 1 = 49\), we can look up the t-value. For a two-tailed test, \(t^* \approx 2.009\) (you can confirm this with a t-table).
### Step 2: Calculate the standard error (SE)
\[
SE = \frac{s}{\sqrt{n}} = \frac{0.39}{\sqrt{50}} \approx \frac{0.39}{7.071} \approx 0.055
\]
### Step 3: Calculate the margin of error (ME)
\[
ME = t^* \cdot SE \approx 2.009 \cdot 0.055 \approx 0.1105
\]
### Step 4: Calculate the confidence interval
\[
CI = \bar{x} \pm ME = 3.02 \pm 0.1105
\]
Lower limit:
\[
3.02 - 0.1105 \approx 2.9095 \quad \text{(approximately 2.91)}
\]
Upper limit:
\[
3.02 + 0.1105 \approx 3.1305 \quad \text{(approximately 3.13)}
\]
### Conclusion
The 95% confidence interval for the mean weight of the bags of oranges is approximately:
\[
(2.91, 3.13)
\]
So, the correct response is:
**2.91, 3.13**