Question

3) A sports reporter conducts a study of the fans' overall satisfaction with the sporting event after the event is completed. He surveys 159 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 6.3. It is known from previous studies of this type that the standard deviation in satisfaction level is 2. Calculate the margin of error and construct the 98% confidence interval for the true mean satisfaction level for the sporting event.

E=?

Round to three decimal places if necessary

? <  μ  < ?

Round to three decimal places if necessary

Answer 1

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

From given data, we have

Xbar = 6.3

σ = 2

n = 159

Confidence level = 98%

Critical Z value = 2.3263

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Margin of error = Z*σ/sqrt(n)

Confidence interval = 6.3 ± 2.3263*2/sqrt(159)

Confidence interval = 6.3 ± 0.3690

Margin of error = 0.3690

Lower limit = 6.3 - 0.3690 = 5.931

Upper limit = 6.3 + 0.3690 = 6.669

Confidence interval = (5.931, 6.669)

5.931 < µ < 6.669