Suppose that a simple random sample is taken from a normal population having a standard deviation of 15 for the purpose of obtaining a 95% confidence interval for the mean of the population. Complete parts (a) through (c) below.

a. If the sample size is 4, obtain the margin of error.
The margin of error for a sample size of 4 is 14.70. (Round to two decimal places as needed.)

b. Repeat part (a) for a sample size of 36.
The margin of error for a sample size of 36 is 4.90. (Round to two decimal places as needed.)

c. Can you guess the margin of error for a sample size of 324? Explain your reasoning.
The margin of error for a sample size of 324 is 1.63. (Round to two decimal places as needed.)

Explain your reasoning.
Because the sample size increased by a factor of 9 from part (b) to part (c), the margin of error decreased by a factor of 3. (Type whole numbers.)



Answer :

To solve the given problem of determining the margin of error for various sample sizes when constructing a 95% confidence interval for the population mean, let's break down the process step by step:

### Step-by-Step Solution:

#### Step 1: Identify Important Information
- Population standard deviation ([tex]\(\sigma\)[/tex]): 15
- Confidence level: 95%
- Z-score for a 95% confidence level (from standard normal distribution): 1.96

#### Step 2: Formula for Margin of Error
The margin of error (E) for a confidence interval is calculated using the formula:
[tex]\[ E = Z \times \frac{\sigma}{\sqrt{n}} \][/tex]
where:
- [tex]\(Z\)[/tex] is the Z-score for the desired confidence level.
- [tex]\(\sigma\)[/tex] is the population standard deviation.
- [tex]\(n\)[/tex] is the sample size.

#### Part (a): Sample Size = 4
To find the margin of error for a sample size of 4:
[tex]\[ E = 1.96 \times \frac{15}{\sqrt{4}} = 1.96 \times \frac{15}{2} = 1.96 \times 7.5 = 14.70 \][/tex]
So, the margin of error for a sample size of 4 is 14.70.

#### Part (b): Sample Size = 36
To find the margin of error for a sample size of 36:
[tex]\[ E = 1.96 \times \frac{15}{\sqrt{36}} = 1.96 \times \frac{15}{6} = 1.96 \times 2.5 = 4.90 \][/tex]
So, the margin of error for a sample size of 36 is 4.90.

#### Part (c): Sample Size = 324
To find the margin of error for a sample size of 324:
[tex]\[ E = 1.96 \times \frac{15}{\sqrt{324}} = 1.96 \times \frac{15}{18} = 1.96 \times 0.8333 \approx 1.63 \][/tex]
So, the margin of error for a sample size of 324 is 1.63.

Now, let's address the reasoning behind these calculations:

#### Explanation
From part (b) to part (c), the sample size increases from 36 to 324. This change in sample size can be analyzed:
- If sample size increases by a factor, say [tex]\(k\)[/tex], the denominator in the margin of error formula [tex]\(\sqrt{n}\)[/tex] increases by [tex]\(\sqrt{k}\)[/tex].

Comparing sample sizes 36 and 324, we see:
[tex]\[ \frac{324}{36} = 9 \][/tex]
So, the sample size increased by a factor of 9, implying the sample size increased by [tex]\(\sqrt{9}\)[/tex] (i.e., 3) in the margin of error formula.
Hence, the margin of error is divided by 3 when the sample size is increased from 36 to 324.

In summary:
- The sample size increased by a factor of 9.
- The margin of error decreased from 4.90 (for [tex]\(n = 36\)[/tex]) to 1.63 (for [tex]\(n = 324\)[/tex]).
- Therefore, this decrease equates to dividing the margin of error from part (b) by the square root of 9 (which is 3).

So, the explanation for the margin of error change is:
Because the sample size increased by a factor of 9, the margin of error was reduced by a factor of 3.

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