Question 1170235: In a random sample of 20 days, a newspaper dealer sold on the average
345 newspapers per day, with a standard deviation of 20 newspapers.
Construct a 95% confidence interval for the true number of newspapers
she sells per day.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's construct the 95% confidence interval for the true number of newspapers sold per day.
**Understanding the Problem**
* We have a sample of 20 days (n = 20).
* The sample mean (x̄) is 345 newspapers.
* The sample standard deviation (s) is 20 newspapers.
* We want to construct a 95% confidence interval.
**Key Concepts**
* **t-distribution:** Since the population standard deviation is unknown and the sample size is small (n < 30), we will use the t-distribution.
* **Degrees of freedom (df):** df = n - 1 = 20 - 1 = 19.
* **Confidence level:** 95%, which means α = 1 - 0.95 = 0.05.
* **t-value:** We need to find the t-value for a 95% confidence interval with 19 degrees of freedom.
**Calculations**
1. **Find the t-value:**
* We need the t-value for α/2 = 0.025 with 19 degrees of freedom.
* Using a t-table or a calculator, we find that t(0.025, 19) ≈ 2.093.
2. **Calculate the standard error (SE):**
* SE = s / √n = 20 / √20 ≈ 20 / 4.472 ≈ 4.472
3. **Calculate the margin of error (ME):**
* ME = t * SE = 2.093 * 4.472 ≈ 9.369.
4. **Construct the confidence interval:**
* Confidence interval = x̄ ± ME
* Confidence interval = 345 ± 9.369
5. **Determine the interval's endpoints:**
* Lower bound: 345 - 9.369 ≈ 335.631
* Upper bound: 345 + 9.369 ≈ 354.369
**Answer**
The 95% confidence interval for the true number of newspapers she sells per day is approximately (335.631, 354.369). We can say with 95% confidence that the true number of newspapers sold per day lies within this range.
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