Question

The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room at a four-star hotel, beverages, breakfast, taxi fares, and incidental costs. Click on the datafile logo to reference the data. City Daily Living Cost ($) City Daily Living Cost ($) Bangkok 242.87 Mexico City 212.00 Bogota 260.93 Milan 284.08 Cairo 194.19 Mumbai 139.16 Dublin 260.76 Paris 436.72 Frankfurt 355.36 Rio de Janeiro 240.87 Hong Kong 346.32 Seoul 310.41 Johannesburg 165.37 Tel Aviv 223.73 Lima 250.08 Toronto 181.25 London 326.76 Warsaw 238.20 Madrid 283.56 Washington, D.C. 250.61 a. Compute the sample mean (to 2 decimals). b. Compute the sample standard deviation (to 2 decimals). c. Compute a confidence interval for the population standard deviation (to 2 decimals).

Answer 1

`\bar x = 260.1615`

`\sigma = 70.69`

The confidence interval of standard deviation is:
`53.76` to
`103.25`

Explanation:

Given

`n =20`

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

`\bar x = (\sum x)/(n)`

So, we have:

`\bar x = (242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61)/(20)`

`\bar x = (5203.23)/(20)`

`\bar x = 260.1615`

`\bar x = 260.16`

Solving (b): The standard deviation

This is calculated as:

`\sigma = \sqrt{(\sum(x - \bar x)^2)/(n-1)}`

`\sigma = \sqrt{((242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2)/(20 - 1)}`
`\sigma = \sqrt{(94938.80)/(19)}`

`\sigma = √(4996.78)`

`\sigma = 70.69` --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

`c =0.95`

So:

`\alpha = 1 -c`

`\alpha = 1 -0.95`

`\alpha = 0.05`

Calculate the degree of freedom (df)

`df = n -1`

`df = 20 -1`

`df = 19`

Determine the critical value at row
`df = 19` and columns
`(\alpha)/(2)` and
`1 -(\alpha)/(2)`

So, we have:

`X^2_(0.025) = 32.852` ---- at
`(\alpha)/(2)`

`X^2_(0.975) = 8.907` --- at
`1 -(\alpha)/(2)`

So, the confidence interval of the standard deviation is:

`\sigma * \sqrt{(n - 1)/(X^2_(\alpha/2) )` to
`\sigma * \sqrt{(n - 1)/(X^2_(1 -\alpha/2) )`

`70.69 * \sqrt{(20 - 1)/(32.852)` to
`70.69 * \sqrt{(20 - 1)/(8.907)`

`70.69 * \sqrt{(19)/(32.852)` to
`70.69 * \sqrt{(19)/(8.907)`

`53.76` to
`103.25`