Question

roblem #1 ravelocity would like to estimate the current average domestic airfare. The data contained in tomesticairfare.xlsx show the airfares for 24 randomly selected tickets for travel within the United tates. Constructa 99% confidence interval with these data to estimate the average domestic airfare. a What is the margin of error for this sample (when using Excel, please provide the formula used (2pts) b)

Answer 1

Ans. We have given the airfares of randomly selected sample of n=24 tickets for travel within the United States.

(a) We need to construct a 99% confidence interval for the true average domestic airfare.

The formula to compute the confidence interval is : $\bar{x}\pm t_{n-1}^{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}$

where,

$\bar{x}:$ sample mean, sample standard deviation, sample size

Sample Mean $(\bar{x})$ :

$\bar{x}=\frac{\Sigma{x}}{n}$

356 275 371 384 457 326 4 14 367 362 286 104 136 320 244 370 2 15 322 409 303 489 251 361 337 265 24

21.833333 ae 321 321.833333 321.8333

Sample standard deviation :

$s=\sqrt{\frac{\Sigma(x_{i}-\bar{x})^{2}}{n-1}}$

$\Sigma(x_{i}-\bar{x})^{2}=189051.3$

Σ(zi-i)2 /189051·3 24 1 90.662129575 90.6621 rl

s 90.6621

Now, since we have to calculate the 99% confidence interval, which means

α 0.01

$t_{n-1}^{\frac{\alpha}{2}}=t_{23}^{\frac{0.01}{2}}=2.807336\approx 2.8073$

99% Confidence interval:

Lower limit: $\bar{x}- t_{n-1}^{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}=321.8333-2.8073\left ( \frac{90.6621}{\sqrt{21}} \right )=269.8804976\approx 269.8805$

Upper limit: $\bar{x}+t_{n-1}^{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}=321.8333+2.8073\left ( \frac{90.6621}{\sqrt{21}} \right )=373.7861024\approx 373.7861$

So the 99% confidence interval for true population average domestic airfare is (269.8805, 373.7861)

Or we can say that if the true population average domestic airfare is $\mu$, then we say that we are 99% confident that the confidence interval will capture the true population domestic airfare $(\mu)$, i.e., $269.8805<\mu< 373.7861$ .

(b). The formula to calculate the Margin of Error for the sample is: $ME=t_{n-1}^{\frac{\alpha}{2}}\left ( \frac{s}{\sqrt{n}} \right )$

$t_{n-1}^{\frac{\alpha}{2}}=t_{23}^{\frac{0.01}{2}}=2.8073$

s= 90.6621 and n=24

$ME=t_{n-1}^{\frac{\alpha}{2}}\left ( \frac{s}{\sqrt{n}} \right )=2.8073\left ( \frac{90.6621}{\sqrt{24}} \right )$

$ME=51.95280243\approx 51.9528$

The margin of error for this sample is ME=51.9528