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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.

Sagot :

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

[tex]t_{\frac{0.05}{2} ,19} = t_{0.025,19} = 2.093[/tex]

Step(ii):-

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

[tex](x^{-} - t_{0.025,19} \frac{S}{\sqrt{n} } , x^{-} + t_{0.025,19} \frac{S}{\sqrt{n} })[/tex]

[tex](99.8 - 2.093 \frac{0.9}{\sqrt{20} } , 99.8 + 2.093 \frac{0.9}{\sqrt{20} })[/tex]

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)

The confidence interval with 95%  confidence where the mean amount of caviar (in grams) per tin from that day's production is [tex]\bold{99.8 \pm 0.9344}[/tex].

Given to us,

size of the sample, n = 20,

mean of the sample, [tex]\bar{x}[/tex] = 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.

[tex]z = 1.960[/tex]

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

[tex]CI = \bar{x} \pm z \dfrac{s}{\sqrt{n}}[/tex]

where,

CI =  confidence interval

[tex]\bar{x}[/tex] =  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,

[tex]CI = \bar{x} \pm z \dfrac{\sigma}{\sqrt{n}}[/tex]

substituting the values,

[tex]\begin{aligned}CI &= 99.8 \pm (1.960) \dfrac{0.9}{\sqrt{20}}\\&=99.8 \pm 0.9344\\\end{aligned}[/tex]

Hence, the confidence interval with 95%  confidence where the mean amount of caviar (in grams) per tin from that day's production is [tex]\bold{99.8 \pm 0.9344}[/tex].

To know more visit:

https://brainly.com/question/2396419

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