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

Answer:

95% of the confidence interval of the mean amount of caviar per tin from that day's production

(99.3788.100.2212)

Step-by-step explanation:

Step(i):-

Given that the size of the sample 'n' = 20

Given that mean of the sample = 99.8 g

Given the standard deviation of the sample = 0.9g

Level of significance = 0.05

$t_{\frac{0.05}{2} ,19} = t_{0.025,19} = 2.093$

Step(ii):-

95% of the confidence interval of the mean amount of caviar per tin from that day's production

$(x^{-} - t_{0.025,19} \frac{S}{\sqrt{n} } , x^{-} + t_{0.025,19} \frac{S}{\sqrt{n} })$

$(99.8 - 2.093 \frac{0.9}{\sqrt{20} } , 99.8 + 2.093 \frac{0.9}{\sqrt{20} })$

on simplification, we get

(99.8 -0.4212 , 99.8+0.4212)

(99.3788.100.2212)

Final answer:-

95% of the confidence interval of the mean amount of caviar per tin from that day's production

(99.3788.100.2212)