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.

The confidence interval with 95%  confidence where the mean amount of caviar (in grams) per tin from that day's production is .

Given to us,

size of the sample, n = 20,

mean of the sample, = 99.8 g

the standard deviation of the sample, σ = 0.9g

As, we know, the confidence level value for 95% confidence is 1.960.

Also, the formula for the confidence interval is given as,

where,

CI=  confidence interval

=  sample mean

z=  confidence level value

σ=  sample standard deviation

n      =  sample size

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

substituting the values,

Hence, the confidence interval with 95%  confidence where the mean amount of caviar (in grams) per tin from that day's production is .

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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

Step(ii):-

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

on simplification, we get

(99.8 -0.4212 , 99.8+0.4212)

(99.3788.100.2212)

Final -

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

(99.3788.100.2212)

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step-by-step explanation: