Question

A sports reporter conducts a study of the fans' overall satisfaction with the sporting event after the event is completed. He surveys 151 randomly selected fans on their way out of the event and asks them to rate their satisfaction with the event on a scale from 0 to 10. The average satisfaction rating is 6.4. It is known from previous studies of this type that the standard deviation in satisfaction level is 2.4. Calculate the margin of error and construct the 80% confidence interval for the true mean satisfaction level for the sporting event.

Standard Normal Distribution Table

E=E=

Round to three decimal places if necessary

  <  μ  <  <  μ  <  

Round to three decimal places if necessary

Answer 1

Solution :

Given that,

Z/2 = Z_0.10 = 1.282

Margin of error = E = Z/2 * ( /n)

E = 1.282 * ( 2.4 /  151 )

E = 0.250

At 80% confidence interval estimate of the population mean is,

- E < < + E

6.4 - 0.250 <   < 6.4 + 0.250

( 6.150 <   < 6.650 )