Question 1186384: A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 20 greeting cards indicates a mean value of $3.55 and a standard deviation of $0.54.
Construct a 95% confidence interval estimate for the mean value of all greeting cards in the store’s inventory.
If the manager wants to estimate, with 95% confidence, the mean value of all greeting cards to within ± $0.03 sample error, what sample size is needed(assuming normal)?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**1. Constructing the 95% Confidence Interval:**
* **Sample mean (x̄):** $3.55
* **Sample standard deviation (s):** $0.54
* **Sample size (n):** 20
* **Confidence level:** 95%
Since the sample size is small (n < 30) and the population standard deviation is unknown, we use the t-distribution.
* **Degrees of freedom (df):** n - 1 = 20 - 1 = 19
* **Critical t-value:** For a 95% confidence level and 19 degrees of freedom, the critical t-value is approximately 2.093 (you can find this using a t-table or calculator).
* **Standard error:** SE = s / √n = 0.54 / √20 ≈ 0.1208
* **Margin of error:** ME = t * SE = 2.093 * 0.1208 ≈ 0.2529
* **Confidence interval:** x̄ ± ME = 3.55 ± 0.2529
Therefore, the 95% confidence interval is approximately **($3.2971, $3.8029)**.
**2. Determining the required sample size:**
* **Desired margin of error (E):** $0.03
* **Confidence level:** 95% (z-score = 1.96, since we are assuming normality for sample size calculation)
* **Estimated standard deviation (s):** $0.54 (we use the sample standard deviation as an estimate)
The formula for sample size is:
n = (z * s / E)²
n = (1.96 * 0.54 / 0.03)²
n = (1.96 * 18)²
n = (35.28)²
n ≈ 1244
Therefore, a sample size of approximately **1244** greeting cards is needed to estimate the mean value to within ±$0.03 with 95% confidence.
|
|
|
