Question

According to the 2017 Corporate Travel Index compiled by Business Travel News, the average daily cost for business travel in the United States rose to $321 per day (Executive Travel website). The file Travel contains sample data for an analogous study on the estimated daily living costs for an executive traveling to various international cities. The estimates include a single room at a four-star hotel, beverages, breakfast, taxi fares, and incidental costs.
City | Daily Living Cost ($) City Daily Living Cost ($)
City 1 242.87 City 6 211.00
City 2 260.93 City 7 284.08
City 3 194.19 City 8 139.16
City 4 260.86 City 9 436.72
City 5 355.36 City 10 240.87
(a) Compute the sample mean (in dollars). (Round your answer to the nearest cent.)
(b) Compute the sample standard deviation (in dollars). (Round your answer to the nearest cent.)
(c) Compute a 95% confidence interval (in dollars) for the population standard deviation. (Round your answers to the nearest cent.)

Answer 1

Answer:

1) X[bar]= $262.60

2) S= $83.55

3) [$57.47; $152.55]

Step-by-step explanation:

Hello!

The objective is to study the average daily cost for business travel in the United States. To test this a sample of the estimated living expenses to 10 international cities was taken. The study variable is X: estimated living costs (including a single room at 4 stars hotel, beverages, breakfast, taxi fares, and incidental costs) on an international city.

Sample data:

n=10

242.87, 260.93, 194.19, 260.86, 355.36, 211.00, 284.08, 139.16, 436.72, 240.87

∑Xi= 2626.04

∑Xi²= 752439.39

1)

The sample mean is calculated by adding all the observed costs and divide it by the sample size:

X[bar]= ∑Xi/n = 2626.04/10= $262.60

2)

To calculate the standard deviation of the sample you have to calculate the sample variance first using the formula:

S²= $\frac{1}{n-1}[∑Xi^2-\frac{(∑Xi)^2}{n} ]$

S²= $\frac{1}{9}[752439.39-\frac{(2626.04)^2}{10} ]$

S²= 6981.1979

S= √S² = √6981.1979 = $83.55

3)

To calculate a Confidence Interval for the population standard deviation, you have to first calculate the Confidence Interval for the population variance and then calculate it's square root to obtain the interval for the standard deviation.

This happens because to study a population parameter you need a statistic that follows the following conditions: 1) have known distribution, 2) include the population parameter and it's point estimator, 3) the only incognito of the statistic equation must be the parameter under study. There is no statistic that meets these conditions, but the chi-square statistic meets these conditions and allows you to study the population variance. And this is only possible if the study variable has a normal distribution.

Assuming that the daily living costs have a normal distribution, the formula of the Confidence Interval for the population variance is:

$[\frac{(n-1)S^2}{X^2_{(n-1); 1- \alpha/2} } ; \frac{(n-1)S^2}{X^2_{(n-1); \alpha/2} } ]$

$[\frac{9*6981.1979}{19.023} } ; \frac{9*6981.1979}{2.700} } ]$

[$²3302.88; $²23270.66]

So with a confidence level of 95% you'd expect that the confidence interval

[$²3302.88; $²23270.66] will contain the population variance of the estimated living costs (including single room at 4 stars hotel, beverages, breakfast, taxi fares and incidental costs) on an international city.

Now you calculate the square root of the interval:

[√$²3302.88; √$²23270.66]

[$57.47; $152.55]

And at the same level, you'd expect that the interval [$57.47; $152.55] contains the population standard deviation of the estimated living costs (including a single room at 4 stars hotel, beverages, breakfast, taxi fares, and incidental costs) on an international city.

I hope it helps!