Respuesta :
Answer:
•Lower confidence limit = $155.2109
•Upper confidence limit ,= $174.5891
Step-by-step explanation:
We are given:
U = 17
x' = $165.40
S.d = $21.70
a = 10℅ (90℅ confidence interval)
For s.d(x'), we have:
S/√u = [tex] \frac{21}{\sqrt{7}}[/tex]
= 5.2630
Therefore, S.E (x') = $5.2630
At a= 0.10 (90℅ confidence level)
and degree of freedom = 16 (17-1)
[tex]Thus, t_1_6_, _0_._1_0 = 1.746[/tex]
90℅ confidence interval all around sample mean, will be:
(x' ± [tex]t_(_n_-_1)[/tex]; a S.E (x')
= (165.4±(174.6)(5.2630)
(165.5±9.189)
($156.2109, $175.5891)
Answer:
The 90% confidence interval around this sample mean is between $127.51 and $203.29.
Step-by-step explanation:
We are in posession of the sample's standard deviation, so we use the students t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 17 - 1 = 16
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 16 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95([tex]t_{95}[/tex]). So we have T = 1.7459
The margin of error is:
M = T*s = 1.7459*21.70 = 37.89
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 165.40 - 37.89 = $127.51
The upper end of the interval is the sample mean added to M. So it is 165.40 + 37.89 = $203.29
The 90% confidence interval around this sample mean is between $127.51 and $203.29.
