To determine the [tex]$95\%$[/tex] confidence interval for the mean number of ounces of ketchup per bottle in the sample, we follow these steps:
1. Identify the given data:
- Population mean ([tex]\(\mu\)[/tex]) = 24 ounces
- Population standard deviation ([tex]\(\sigma\)[/tex]) = 0.2 ounces
- Sample size ([tex]\(n\)[/tex]) = 49 bottles
- Confidence level = 95%
2. Determine the [tex]\(z\)[/tex]-score for a 95% confidence level:
The [tex]\(z\)[/tex]-score for a 95% confidence interval (which corresponds to [tex]\( \frac{1 + 0.95}{2} \)[/tex]) is approximately 1.96.
3. Calculate the standard error of the mean (SEM):
The standard error of the mean is calculated using the formula:
[tex]\[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
\][/tex]
Substituting the given values:
[tex]\[
\text{SEM} = \frac{0.2}{\sqrt{49}} = \frac{0.2}{7} \approx 0.0286
\][/tex]
4. Calculate the margin of error (ME):
The margin of error is given by:
[tex]\[
\text{ME} = z \times \text{SEM}
\][/tex]
Using the [tex]\(z\)[/tex]-score of 1.96:
[tex]\[
\text{ME} = 1.96 \times 0.0286 \approx 0.057
\][/tex]
5. Determine the confidence interval:
The confidence interval is centered around the sample mean, which in this case, is assumed to be the population mean (24 ounces). Thus, the confidence interval is:
[tex]\[
\mu \pm \text{ME} = 24 \pm 0.057
\][/tex]
So, the correct answer is:
C. [tex]\(24 \pm 0.057\)[/tex]
