Question

The results of 56 randomly selected Verizon airport data speeds were as followed: ?̅=17.60,?=16.02.Construct a 95% confidence interval estimated of μ. Find the (A) Type of distribution. (B) Point Estimate (C) Degrees of freedom (D) Critical Value (E) Margin of error (F) Confidence interval

Answer 1

Solution :

Given that,

= 17.60

s =16.02

n =56

Degrees of freedom = df = n - 1 = 56- 1 =55

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,55 = 2.004 ( using student t table)

Margin of error = E = t/2,df * (s /n)

=2.004 * (16.02 / 56)

= 4.2901

The 95% confidence interval estimate of the population mean is,

- E < < + E

17.60- 4.2901< <17.60 + 4.2901

13.3099 < < 21.8901

(13.3099 , 21.8901 )

this is t-distribution