Question 1181114: (a) An advertising company wants to estimate with 98% confidence interval, the number of times a website is hit during an hour. It is determined that = 26. How large a sample should the company take, if it wishes that the margin of error should not exceed 10? [04]
(b) If we take a sample from an infinite population, what will happen to the standard error of the mean when the sample size is increased from 60 to 240?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **(a) Sample size calculation:**
1. **Identify the given information:**
* Confidence level = 98%
* Margin of error (E) = 10
* Population standard deviation (σ) = 26
2. **Find the critical z-value:**
For a 98% confidence level, α = 1 - 0.98 = 0.02. α/2 = 0.01. The critical z-value (z*) corresponding to 0.01 in the tail of the standard normal distribution is approximately 2.33.
3. **Use the sample size formula:**
n = (z* * σ / E)²
4. **Plug in the values:**
n = (2.33 * 26 / 10)²
n = (60.58 / 10)²
n = 6.058²
n ≈ 36.7
5. **Round up:** Since the sample size must be a whole number, always round up to the nearest integer. Therefore, the company should take a sample of at least 37.
**(b) Effect of sample size on standard error:**
The standard error of the mean (SEM) is calculated as:
SEM = σ / √n
Where:
* σ = population standard deviation
* n = sample size
If the sample size is increased from n₁ = 60 to n₂ = 240, let's see what happens to the SEM.
* Initial SEM (SEM₁) = σ / √60
* New SEM (SEM₂) = σ / √240
We can rewrite √240 as √(4 * 60) = 2√60
So, SEM₂ = σ / (2√60) = (1/2) * (σ / √60) = (1/2) * SEM₁
Therefore, when the sample size is increased from 60 to 240 (a four-fold increase), the standard error of the mean is *halved*.
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