Question

A Fiji Travel Data Center survey reported that Fijians stayed an average of 7.5 nights when they went on vacation around the country. For the purpose of the survey, the sample size was 1500. Assume the population standard deviation was 0.8; find a point estimate of the population mean. In addition, find the 95% confidence interval of the true mean.

Answer 1

Given:

A Fiji Travel Data Center survey reported that Fijians stayed an average of 7.5 nights.

For the purpose of the survey, the sample size was 1500.

Assume the population standard deviation was 0.8

To find:

Find a point estimate of the population mean. In addition, find the 95% confidence interval of the true mean.

Solution:

From given, we have,

Mean  = μ = 7.5

The sample size = N = 1500

Standard deviation = σ = 0.8

Sample variance = σ² = $\dfrac{\sum (x-\bar{x})^2}{N-1}$

From given, we have,

σ = 0.8

∴ Sample variance = σ² = 0.8² = 0.64

The 95% confidence interval of the true mean is given by a formula,

μ = M ± Z(sM)

where:   M = sample mean

Z = Z statistic determined by confidence level

sM = standard error = √(s2/n)

M = 7.5

Z = 1.96

sM = √(0.82/1500) = 0.02

μ = M ± Z(sM)

μ = 7.5 ± 1.96*0.02

μ = 7.5 ± 0.04

∴ 95% confidence interval = 7.5 ± 0.04