Question

Suppose that a restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average, with a standard deviation of 0.8 ounces. If you took a sample of the 49 bottles of ketchup what would be the approximate 95% confidence interval for a mean number of ounces of ketchup per bottle in the sample?

Answer 1

Answer:

Confidence Interval = (23.776, 24.224)

Step-by-step explanation:

Restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average.

⇒ Mean = 24 ounces.

Standard Deviation = 0.8

Number of bottles used for sample = 49

⇒ n = 49

Confidence level = 95%

Corresponding z value with 95% confidence level = 1.96

Now, confidence interval is given by the following expression :

$Mean \pm z^*\times \frac{\sigma}{\sqrt{n}}\\\\=24\pm1.96\times \frac{0.8}{\sqrt{49}}\\\\=24\pm 0.224$

Hence, Confidence Interval = (23.776, 24.224)