Question

Caviar is an expensive delicacy, so companies that package it pay very close attention to the amount of product in their tins. Suppose a company that produces over 1000 tins of caviar per day took an SRS of 20 tins from one day's production. The sample showed a mean of 99.8g of caviar per tin with a standard deviation of 0.9g. The data were roughly symmetric with no outliers.

Required:
Based on this sample, estimate with 95, percent confidence the mean amount of caviar (in grams) per tin from that day's production.

Answer 1

The estimated mean amount of caviar per tin with 95% confidence using a sample mean of 99.8g and a standard deviation of 0.9g is between 99.38g and 100.22g.

Step-by-step explanation:

To estimate the mean amount of caviar per tin with 95% confidence, we need to use the sample mean (μ), the sample standard deviation (s), and the sample size (n) to calculate the margin of error and then construct the confidence interval. Since the sample size is less than 30 and we do not know the population standard deviation, we use the Student's t-distribution.

The mean of the sample is 99.8g, the sample standard deviation is 0.9g, and the sample size is 20. The degree of freedom (df) for the sample is 19 (n-1).

First, we need to find the t-score that corresponds to the 95% confidence level for 19 degrees of freedom. Looking into the t-distribution table or using statistical software, we find that the t-score is approximately 2.093.

The margin of error (ME) is calculated using the formula:

ME = t * (s/√n)

Plugging in the values:

ME = 2.093 * (0.9/√20) ≈ 0.42

Now, we can construct the 95% confidence interval by subtracting and adding the margin of error from the sample mean:

Lower limit

: 99.8g - 0.42g = 99.38g
Upper limit

: 99.8g + 0.42g = 100.22g
The estimated mean amount of caviar per tin from that day's production with 95% confidence is between 99.38g and 100.22g.