document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1179741>Question 1179741</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose a real estate broker is interested in comparing the asking prices of flats in the British cities of Exeter and Cardiff. The broker conducts a small telephone survey in the two cities, asking the prices of flats. A random sample of 21 listings in Exeter resulted in a sample average price of £116,900, with a standard deviation of £2,300. A random sample of 26 listings in Cardiff resulted in a sample average price of £114,000, with a standard deviation of £1,750. The broker assumes that the prices of flats are normally distributed and that the variance in prices in the two cities is about the same. What would he obtain for a 90% confidence interval for the difference in mean prices of mid-range homes between Exeter and Cardiff? Test wheather there is a difference in the mean prices of mid-range homes of two cities for α= 0.10 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850177 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850177\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to solve this problem, including calculating the confidence interval and conducting the hypothesis test:\r<BR>\n" );
document.write( "\n" );
document.write( "**1. Given Information:**\r<BR>\n" );
document.write( "\n" );
document.write( "* **Exeter:**<BR>\n" );
document.write( "    * n1 = 21<BR>\n" );
document.write( "    * x̄1 = £116,900<BR>\n" );
document.write( "    * s1 = £2,300<BR>\n" );
document.write( "* **Cardiff:**<BR>\n" );
document.write( "    * n2 = 26<BR>\n" );
document.write( "    * x̄2 = £114,000<BR>\n" );
document.write( "    * s2 = £1,750<BR>\n" );
document.write( "* Confidence level = 90%<BR>\n" );
document.write( "* Significance level (α) = 0.10<BR>\n" );
document.write( "* Assume normal distribution and equal variances.\r<BR>\n" );
document.write( "\n" );
document.write( "**2. Calculate the Pooled Standard Deviation (sp):**\r<BR>\n" );
document.write( "\n" );
document.write( "Since we assume equal variances, we use the pooled standard deviation:\r<BR>\n" );
document.write( "\n" );
document.write( "sp = √[((n1 - 1) * s1²) + ((n2 - 1) * s2²)) / (n1 + n2 - 2)]\r<BR>\n" );
document.write( "\n" );
document.write( "sp = √[((20 * 2300²) + (25 * 1750²)) / (21 + 26 - 2)]<BR>\n" );
document.write( "sp = √[(105800000 + 76562500) / 45]<BR>\n" );
document.write( "sp = √[182362500 / 45]<BR>\n" );
document.write( "sp = √4052500<BR>\n" );
document.write( "sp ≈ £2,013.08\r<BR>\n" );
document.write( "\n" );
document.write( "**3. Calculate the Standard Error of the Difference (SE):**\r<BR>\n" );
document.write( "\n" );
document.write( "SE = sp * √(1/n1 + 1/n2)<BR>\n" );
document.write( "SE = 2013.08 * √(1/21 + 1/26)<BR>\n" );
document.write( "SE = 2013.08 * √(0.0476 + 0.0385)<BR>\n" );
document.write( "SE = 2013.08 * √0.0861<BR>\n" );
document.write( "SE = 2013.08 * 0.2934<BR>\n" );
document.write( "SE ≈ £590.69\r<BR>\n" );
document.write( "\n" );
document.write( "**4. Find the Critical t-Value:**\r<BR>\n" );
document.write( "\n" );
document.write( "* Degrees of freedom (df) = n1 + n2 - 2 = 21 + 26 - 2 = 45<BR>\n" );
document.write( "* For a 90% confidence interval (α = 0.10, two-tailed), the critical t-value (t*) for df = 45 is approximately 1.68.\r<BR>\n" );
document.write( "\n" );
document.write( "**5. Calculate the Margin of Error (E):**\r<BR>\n" );
document.write( "\n" );
document.write( "E = t* * SE<BR>\n" );
document.write( "E = 1.68 * 590.69<BR>\n" );
document.write( "E ≈ £992.36\r<BR>\n" );
document.write( "\n" );
document.write( "**6. Construct the Confidence Interval:**\r<BR>\n" );
document.write( "\n" );
document.write( "* Difference in sample means (x̄1 - x̄2) = 116,900 - 114,000 = £2,900<BR>\n" );
document.write( "* Lower Bound = (x̄1 - x̄2) - E = 2,900 - 992.36 = £1,907.64<BR>\n" );
document.write( "* Upper Bound = (x̄1 - x̄2) + E = 2,900 + 992.36 = £3,892.36\r<BR>\n" );
document.write( "\n" );
document.write( "**90% Confidence Interval: (£1,907.64, £3,892.36)**\r<BR>\n" );
document.write( "\n" );
document.write( "**7. Hypothesis Test:**\r<BR>\n" );
document.write( "\n" );
document.write( "* **Null Hypothesis (H0):** μ1 - μ2 = 0 (There is no difference in mean prices)<BR>\n" );
document.write( "* **Alternative Hypothesis (H1):** μ1 - μ2 ≠ 0 (There is a difference in mean prices)<BR>\n" );
document.write( "* Significance level (α) = 0.10<BR>\n" );
document.write( "* Test statistic (t): t = (x̄1 - x̄2) / SE\r<BR>\n" );
document.write( "\n" );
document.write( "t = 2900 / 590.69<BR>\n" );
document.write( "t ≈ 4.91\r<BR>\n" );
document.write( "\n" );
document.write( "**8. Find the Critical t-Values:**\r<BR>\n" );
document.write( "\n" );
document.write( "* df = 45<BR>\n" );
document.write( "* For α = 0.10 (two-tailed), the critical t-values are approximately ±1.68.\r<BR>\n" );
document.write( "\n" );
document.write( "**9. Make a Decision:**\r<BR>\n" );
document.write( "\n" );
document.write( "* Calculated t-value (4.91) > critical t-value (1.68).<BR>\n" );
document.write( "* Therefore, we reject the null hypothesis.\r<BR>\n" );
document.write( "\n" );
document.write( "**10. Conclusion:**\r<BR>\n" );
document.write( "\n" );
document.write( "There is sufficient evidence at the α = 0.10 level to conclude that there is a difference in the mean prices of mid-range homes between Exeter and Cardiff.\r<BR>\n" );
document.write( "\n" );
document.write( "**Answers:**\r<BR>\n" );
document.write( "\n" );
document.write( "* 90% Confidence Interval: (£1,907.64, £3,892.36)<BR>\n" );
document.write( "* Hypothesis Test: Reject the null hypothesis. There is a statistically significant difference.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );