Question

In a survey, 24 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $31 and standard deviation of $6. Construct a confidence interval at a 80% confidence level.

Answer 1

29.4≤μ≤32.6

Explanation:

The datas given from the questions are as shown:

Number of people n = 24

Mean xbar= $31

Standard deviation σ = $6

Confidence Interval formula is expressed as:

CI = xbar ± Z(σ/√n)

Z value for 80% confidence interval is 1.282

Substituting the values into the Confidence Interval formula will give;

CI = 31 ± 1.282{6/√24}

CI = 31 ± 1.282(1.225)

CI = 31 ± 1.57045

CI = 31+1.57045 and 31-1.57045

CI = (29.42955, 32.57045)

CI = (29.4, 32.6) to 1dp

The confidence interval will be within the range 29.4≤μ≤32.6