Question

If n=30, ¯ x (x-bar)=43, and s=15, construct a confidence interval at a 95% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.

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Answer 1

Solution :

Given that,

= 43

s =15

n =30

Degrees of freedom = df = n - 1 = 30- 1 =29

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,29 =2.045 ( using student t table)

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

= 2.045* (15 / 30)

= 5.60

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

- E < < + E

43 - 5.60 < <43 + 5.60

37.40 < < 48.60

( 37.40 , 48.60 )