To determine the 95% confidence interval for the mean number of ounces of ketchup per bottle, recall that the confidence interval can be calculated using the following formula:
[tex]\[ \text{CI} = \overline{x} \pm (z \times \text{SE}) \][/tex]
where:
- [tex]\(\overline{x}\)[/tex] is the sample mean,
- [tex]\(z\)[/tex] is the z-score corresponding to the desired confidence level (for a 95% confidence level, [tex]\(z \approx 1.96\)[/tex]),
- [tex]\(\text{SE}\)[/tex] is the standard error of the mean, which is calculated as [tex]\(\frac{\sigma}{\sqrt{n}}\)[/tex].
Given the data:
- Mean ([tex]\(\overline{x}\)[/tex]) = 24 ounces
- Standard deviation ([tex]\(\sigma\)[/tex]) = 0.8 ounces
- Sample size ([tex]\(n\)[/tex]) = 49 bottles
First, calculate the standard error (SE) of the mean:
[tex]\[ \text{SE} = \frac{0.8}{\sqrt{49}} = \frac{0.8}{7} \approx 0.114 \][/tex]
Next, use the z-score for a 95% confidence level, which is [tex]\(z = 1.96\)[/tex], to calculate the margin of error (ME):
[tex]\[ \text{ME} = z \times \text{SE} = 1.96 \times 0.114 \approx 0.224 \][/tex]
Now, calculate the confidence interval:
[tex]\[ \text{CI} = 24 \pm 0.224 \][/tex]
So, the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle is:
[tex]\[ (23.776, 24.224) \][/tex]
This gives us:
[tex]\[ 24 \pm 0.224 \][/tex]
Comparing this result with the options provided, the most appropriate choice is:
D. [tex]\(24 \pm 0.229\)[/tex]
