Respuesta :
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].
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