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Neko [114]
3 years ago
13

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

cupied 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. Descriptive statistics for the data are as follows: "x-bar" = 11.596 and "s" = 4.103.
How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?
Mathematics
1 answer:
Bingel [31]3 years ago
6 0

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when M = 0.3, \sigma = 4.103. So

M = z*\frac{\sigma}{\sqrt{n}}

0.3 = 1.96*\frac{4.103}{\sqrt{n}}

0.3\sqrt{n} = 1.96*4.103

\sqrt{n} = \frac{1.96*4.103}{0.3}

(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}

n = 718.57

Rouding up

We need a sample size of at least 719

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