<p>&#160;</p> <p>For a recent year the mean fare to fly from Charlotte, North Ca

Question

<p>&#160;</p>
<p>For a recent year the mean fare to fly from Charlotte, North Carolina to Seattle, Washington on a discount ticket was $267.&#160; A random sample of round-trip discount fares on this route last month gives:</p>
<p>&#160;</p>
<table border="1" cellspacing="0" cellpadding="0">
<tbody>
<tr>
<td width="61" valign="top">
<p>$321</p>
</td>
<td width="61" valign="top">
<p>$286</p>
</td>
<td width="61" valign="top">
<p>$290</p>
</td>
<td width="61" valign="top">
<p>$330</p>
</td>
<td width="61" valign="top">
<p>$310</p>
</td>
<td width="61" valign="top">
<p>$250</p>
</td>
<td width="61" valign="top">
<p>$270</p>
</td>
<td width="61" valign="top">
<p>$280</p>
</td>
<td width="61" valign="top">
<p>$299</p>
</td>
<td width="61" valign="top">
<p>$265</p>
</td>
<td width="61" valign="top">
<p>$291</p>
</td>
<td width="61" valign="top">
<p>$275</p>
</td>
<td width="61" valign="top">
<p>$281</p>
</td>
</tr>
</tbody>
</table>
<p>&#160;</p>
<p>At the point .01 significance level can we conclude that the mean fare has increased?&#160; What is the p-value?</p>

Explanation / Answer

The test hypothesis is
H0: = 267
Ha: > 267


Based on the data,
xbar = 288.308

sd = 22.45879418

The critical t value (from a table), using df = 12, alpha = 0.01, one tail is:

2.681

The test statistic:

t = (x-mu)/(sd/sqrt(N))

t = (288.308-267)/(22.45879418/sqrt(13))

t = 3.42

The p-value is P(t>3.42) is 0.0025

Our test statistic is higher than the critical value, so we can reject the null hypothesis and conclude that the mean fare has increased.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.