Answer:
The marginal error is 2.355
Confidence interval is [tex](34.645,\ 39.355)[/tex]
The LB and UB are 34.645 and 39.3555 respectively.
Step-by-step explanation:
Consider the provided information.
n = 49, s = 8.2 c = 95%and [tex]\bar x=37[/tex]
Degree of freedom is: n-1 = 49-1 = 48.
[tex]\alpha =\frac{1-c}{2}= \frac{1-0.95}{2}\\\alpha = \frac{0.05}{2}= 0.025[/tex]
From the table [tex]t_{\alpha /2}=2.010635[/tex]
Marginal error is:
[tex]E=t_{\alpha /2}\times \frac{s}{\sqrt{n}}[/tex]
[tex]E=2.0106\times \frac{8.2}{\sqrt{49}}[/tex]
[tex]E=2.0106\times \frac{8.2}{7}[/tex]
[tex]E=2.0106\times 1.171[/tex]
[tex]E=2.355[/tex]
Hence, the marginal error is 2.355.
Part (B)
95% confidence interval estimate
[tex](\bar x-E,\ \bar x+E)[/tex]
[tex](37-2.355,\ 37+2.355)[/tex]
[tex](34.645,\ 39.355)[/tex]
Thus, the LB and UB are 34.645 and 39.3555 respectively.