Question

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats. Assume the underlying population of unoccupied seats is normally distributed. Using a t- statistic, construct a 95% confidence interval for the population mean number of unoccupied seats per flight.

Answer 1

The given sample's 95% confidence interval is 11.6 ± 0.5357. I.e., from the lower bound 11.06 to the upper bound 12.14. Its margin error is 0.5357.

How to find the confidence interval?

To find the confidence interval C.I,

• Determine the mean(μ) of the sample
• Determine the standard deviation(σ)
• Determine the z-score for the given confidence level using the z-score table
• Using all these, the confidence interval is calculated by the formula,  C. I = μ ± z(σ/√n) where n is the sample size and z(σ/√n) gives the margin error. So, we can also write C. I = mean ± margin error.

Calculation:

It is given that,

Sample size n = 225

Sample mean μ = 11.6

Standard deviation σ = 4.1

Since 95% confidence interval, z-score is 1.96

So, the required confidence interval is calculated by,

C. I = μ ± z(σ/√n) or mean ± margin error

On substituting,

C. I = 11.6 ± 1.96(4.1/√225)

     = 11.6 ± 1.96 × 0.2733

     = 11.6 ± 0.5357

So, the lower bound = 11.6 - 0.5357 = 11.065 and

the upper bound = 11.6 + 0.5357 = 12.135.

Thus, the confidence interval is from 11.06 to 12.14, and its margin error is 0.0537 for the population mean number of unoccupied seats per flight.

Learn more about confidence intervals here:

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