Question 1181388
Here's the solution:

**(a) Margin of Error:**

Since the sample size is small (n=15) and the population standard deviation is unknown, we use the t-distribution.  The formula for the margin of error is:

Margin of Error = t * (s / √n)

Where:

* t is the t-score for the desired confidence level and degrees of freedom (df = n - 1 = 15 - 1 = 14)
* s is the sample standard deviation
* n is the sample size

For a 95% confidence level and df = 14, the t-score is approximately 2.145 (you'll find this using a t-table or calculator).

Margin of Error = 2.145 * (0.2 / √15)
Margin of Error = 2.145 * (0.2 / 3.873)
Margin of Error ≈ 2.145 * 0.0516
Margin of Error ≈ 0.11

The margin of error is approximately **0.11 lbs**.

**(b) Interpretation:**

95% of all samples of size 15 have sample means within **0.11** lbs of the population mean.

**(c) 95% Confidence Interval:**

The confidence interval is calculated as:

(x̄ - Margin of Error, x̄ + Margin of Error)

Where x̄ is the sample mean.

(7.46 - 0.11, 7.46 + 0.11)
(7.35, 7.57)

The 95% confidence interval is approximately **7.35 lbs < μ < 7.57 lbs**.

**(d) Effect of Increasing Confidence Level:**

If you increase the confidence level (1 - α), the confidence interval estimate will be *wider*.

*Explanation:* A higher confidence level means you want to be more certain that your interval contains the true population mean. To be more certain, you need to make the interval wider.  A wider interval is more likely to capture the true mean, but it also provides less precise information about the mean's specific value.  Think of it like casting a wider net to catch a fish  -  you're more likely to catch it, but you'll have less idea about exactly where it is.