Respuesta :
Answer:
90% confidence interval for the true mean weight of orders is between a lower limit of 103.8645 grams and an upper limit of 116.1355 grams.
Step-by-step explanation:
Confidence interval for true mean weight is given as sample mean +/- margin of error (E)
sample mean = 110 g
sample sd = 14 g
n = 16
degree of freedom = n - 1 = 16 - 1 = 15
confidence level = 90% = 0.9
significance level = 1 - C = 1 - 0.9 = 0.1 = 10%
critical value (t) corresponding to 15 degrees of freedom and 10% significance level is 1.753
E = t × sample sd/√n = 1.753×14/√16 = 6.1355 g
Lower limit of sample mean = sample mean - E = 110 - 6.1355 = 103.8645 g
Upper limit of sample mean = sample mean + E = 110 + 6.1355 = 116.1355 g
90% confidence interval is (103.8645, 116.1355)
Answer:
90% confidence interval for the true mean weight of orders is [103.86 , 116.14].
Step-by-step explanation:
We are given that Past records indicate that the weight of orders is normally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams.
Firstly, the pivotal quantity for 90% confidence interval for the true mean weight of orders 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 = 110 grams
s = sample standard deviation = 14 grams
n = sample of orders = 16
[tex]\mu[/tex] = true population mean
Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.
So, 90% confidence interval for the true mean, [tex]\mu[/tex] is ;
P(-1.753 < [tex]t_1_5[/tex] < 1.753) = 0.90 {As the critical value of t at 15 degree of
freedom are -1.753 & 1.753 with P = 5%}
P(-1.753 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.753) = 0.90
P( [tex]-1.753 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]1.753 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90
P( [tex]\bar X -1.753 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X +1.753 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.90
90% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -1.753 \times {\frac{s}{\sqrt{n} }[/tex] , [tex]\bar X +1.753 \times {\frac{s}{\sqrt{n} }[/tex] ]
= [ [tex]110 -1.753 \times {\frac{14}{\sqrt{16} }[/tex] , [tex]110 +1.753 \times {\frac{14}{\sqrt{16} }[/tex] ]
= [103.86 , 116.14]
Therefore, 90% confidence interval for the true mean weight of orders is [103.86 , 116.14].
