Question
The head librarian at the Bloemfontein Library has asked her assistant for an interval estimate of the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day. She asked her assistant for a 95% confidence interval, approximately how large a sample did her assistant use to determine the interval estimate?
125
11
13
4
125
11
13
4
Answers
GPT-4o mini
To determine the sample size that the assistant used to construct the 95% confidence interval, we can use the formula for the confidence interval of the mean when the population standard deviation is known:
\[
\text{Confidence Interval} = \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right)
\]
Where:
- \(\bar{x}\) is the sample mean,
- \(Z\) is the Z-score corresponding to the confidence level (for 95%, Z is approximately 1.96),
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.
The interval estimate given is from 740 to 920, so we can find the sample mean \(\bar{x}\) and the margin of error (ME).
1. **Calculate \(\bar{x}\) (the midpoint of the interval):**
\[
\bar{x} = \frac{740 + 920}{2} = \frac{1660}{2} = 830
\]
2. **Calculate the margin of error (ME):**
\[
\text{ME} = \frac{920 - 740}{2} = \frac{180}{2} = 90
\]
Now we know:
- \(\text{ME} = 90\)
- \(\sigma = 150\)
- \(Z \approx 1.96\) for a 95% confidence level.
3. **Set up the equation for the margin of error:**
\[
\text{ME} = Z \left( \frac{\sigma}{\sqrt{n}} \right)
\]
Substituting the known values:
\[
90 = 1.96 \left( \frac{150}{\sqrt{n}} \right)
\]
4. **Solve for \(n\):**
Rearranging the equation to find \(n\):
\[
\sqrt{n} = \frac{1.96 \cdot 150}{90}
\]
First calculate \(\frac{1.96 \cdot 150}{90}\):
\[
\sqrt{n} = \frac{294}{90} \approx 3.267
\]
Now square both sides:
\[
n = (3.267)^2 \approx 10.67
\]
Since \(n\) must be a whole number, we round up to the nearest whole number:
\[
n \approx 11
\]
Thus, the assistant used a sample size of approximately **11**.
The answer is **11**.
\[
\text{Confidence Interval} = \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right)
\]
Where:
- \(\bar{x}\) is the sample mean,
- \(Z\) is the Z-score corresponding to the confidence level (for 95%, Z is approximately 1.96),
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.
The interval estimate given is from 740 to 920, so we can find the sample mean \(\bar{x}\) and the margin of error (ME).
1. **Calculate \(\bar{x}\) (the midpoint of the interval):**
\[
\bar{x} = \frac{740 + 920}{2} = \frac{1660}{2} = 830
\]
2. **Calculate the margin of error (ME):**
\[
\text{ME} = \frac{920 - 740}{2} = \frac{180}{2} = 90
\]
Now we know:
- \(\text{ME} = 90\)
- \(\sigma = 150\)
- \(Z \approx 1.96\) for a 95% confidence level.
3. **Set up the equation for the margin of error:**
\[
\text{ME} = Z \left( \frac{\sigma}{\sqrt{n}} \right)
\]
Substituting the known values:
\[
90 = 1.96 \left( \frac{150}{\sqrt{n}} \right)
\]
4. **Solve for \(n\):**
Rearranging the equation to find \(n\):
\[
\sqrt{n} = \frac{1.96 \cdot 150}{90}
\]
First calculate \(\frac{1.96 \cdot 150}{90}\):
\[
\sqrt{n} = \frac{294}{90} \approx 3.267
\]
Now square both sides:
\[
n = (3.267)^2 \approx 10.67
\]
Since \(n\) must be a whole number, we round up to the nearest whole number:
\[
n \approx 11
\]
Thus, the assistant used a sample size of approximately **11**.
The answer is **11**.