Given:
Tinterval
(13.046, 22.15)
x= 17.598
Sx=16.01712719
n=50
a) Lower limit = 13.046
Upper limit = 22.15
Round to two decimals, we have:
[tex]13.05Mbps<\mu<22.15Mbps[/tex]b) The best point estimate of μ is: 17.60 Mbps
The margin of error is given by:
[tex]E=\frac{Upper\text{ limit}-Lower\text{ limit}}{2}=\frac{22.15-13.05}{2}=\frac{9.1}{2}=4.55[/tex]So, the margin of error is E = 4.55 Mbps
c) Since the sample size of 50 is larger than that of 30, the distribution of the sample means can be treated as a normal distribution. Therefore, the answer is D.
