To determine the [tex]$95\%$[/tex] confidence interval for the mean number of ounces of ketchup per bottle in a sample of 49 bottles, follow these steps:
### Step 1: Identify the given data
- Population mean ([tex]\(\mu\)[/tex]): 24 ounces
- Population standard deviation ([tex]\(\sigma\)[/tex]): 0.4 ounces
- Sample size ([tex]\(n\)[/tex]): 49 bottles
- Confidence level: 95%
### Step 2: Find the z-score for the 95% confidence level
Since we are dealing with a 95% confidence interval, we need to find the z-score that corresponds to the upper 2.5% of the standard normal distribution (since it's a two-tailed test, we use 0.975). The z-score for 95% confidence is approximately 1.96.
### Step 3: Calculate the standard error of the mean
The standard error (SE) of the sample mean is given by:
[tex]\[
SE = \frac{\sigma}{\sqrt{n}}
\][/tex]
Plugging in the values:
[tex]\[
SE = \frac{0.4}{\sqrt{49}} = \frac{0.4}{7} = 0.0571
\][/tex]
### Step 4: Calculate the margin of error
The margin of error (ME) is calculated as:
[tex]\[
ME = z \times SE
\][/tex]
Using the z-score of 1.96 we obtained:
[tex]\[
ME = 1.96 \times 0.0571 \approx 0.112
\][/tex]
### Step 5: Construct the confidence interval
The confidence interval for the population mean is computed as:
[tex]\[
\text{Confidence Interval} = \left( \mu - ME , \mu + ME \right)
\][/tex]
For our data:
[tex]\[
\text{Lower bound} = 24 - 0.112 = 23.888
\][/tex]
[tex]\[
\text{Upper bound} = 24 + 0.112 = 24.112
\][/tex]
### Result
Therefore, the 95% confidence interval for the mean number of ounces of ketchup per bottle is approximately [tex]\( [23.888, 24.112] \)[/tex].
Expressing the result in the form [tex]\( 24 \pm ME \)[/tex]:
[tex]\[
\text{Confidence Interval} \approx 24 \pm 0.114
\][/tex]
The correct answer to the given question is:
D. [tex]\(24 \pm 0.114\)[/tex]
