Respuesta :
Answer:
90% confidence interval for the true mean % cacao is [53.87% , 56.13%].
Step-by-step explanation:
We are given that in order to determine the mean % cacao in its dark chocolate products, quality inspectors sample 36 pieces.
They find a sample mean of 55% with a standard deviation of 4%.
Firstly, the pivotal quantity for 90% confidence interval for the true mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean % cacao = 55%
s = sample standard deviation = 4%
n = sample of pieces = 36
[tex]\mu[/tex] = true mean % cacao
Here for constructing 90% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.
So, 90% confidence interval for the true mean, [tex]\mu[/tex] is ;
P(-1.6895 < [tex]t_3_6[/tex] < 1.6895) = 0.90 {As the critical value of t at 35 degree of
freedom are -1.6895 & 1.6895 with P = 5%}
P(-1.6895 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.6895) = 0.90
P( [tex]-1.6895 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.6895 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90
P( [tex]\bar X-1.6895 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.6895 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90
90% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.6895 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+1.6895 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]0.55-1.6895 \times {\frac{0.04}{\sqrt{36} } }[/tex] , [tex]0.55+1.6895 \times {\frac{0.04}{\sqrt{36} } }[/tex] ]
= [0.5387 , 0.5613]
= [53.87% , 56.13%]
Therefore, 90% confidence interval for the true mean % cacao is [53.87% , 56.13%].
