document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197932>Question 1197932</A>: <I><SPAN STYLE=\"word-break: break-all;\"> . The National Cable and Telecommunications Association recently reported that <BR>\n" );
document.write( "the mean number of HDTVs per household in the United States is 2.30 with a standard <BR>\n" );
document.write( "deviation of 1.474 sets. A sample of 100 homes in Boise, Idaho, revealed the following <BR>\n" );
document.write( "sample information.<BR>\n" );
document.write( "Tutorial #21 <BR>\n" );
document.write( "in Connect<BR>\n" );
document.write( "Final PDF to printer<BR>\n" );
document.write( "NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 577<BR>\n" );
document.write( "lin39470_ch15_542-578.indd 577 10/09/19 07:51 AM<BR>\n" );
document.write( "Number of HDTVs Number of Households<BR>\n" );
document.write( "0 7<BR>\n" );
document.write( "1 27<BR>\n" );
document.write( "2 28<BR>\n" );
document.write( "3 18<BR>\n" );
document.write( "4 10<BR>\n" );
document.write( "5 or more  10<BR>\n" );
document.write( " Total 100<BR>\n" );
document.write( " At the .05 significance level, is it reasonable to conclude that the number of HDTVs per <BR>\n" );
document.write( "household follows a normal distribution? (Hint: Use limits such as 0.5 up to 1.5, 1.5 up to <BR>\n" );
document.write( "2.5, and so on.) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #848342 by </B><A HREF=/tutors/aboutme.mpl?userid=onyulee><B>onyulee(41)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=onyulee><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=848342\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> **1. Define Hypotheses**\r<BR>\n" );
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document.write( "* **Null Hypothesis (H0):** The number of HDTVs per household in Boise follows a normal distribution with a mean of 2.30 and a standard deviation of 1.474.<BR>\n" );
document.write( "* **Alternative Hypothesis (H1):** The number of HDTVs per household in Boise does not follow a normal distribution.\r<BR>\n" );
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document.write( "**2. Determine Expected Frequencies**\r<BR>\n" );
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document.write( "* **Divide the data into intervals:** <BR>\n" );
document.write( "    * 0.5 - 1.5<BR>\n" );
document.write( "    * 1.5 - 2.5<BR>\n" );
document.write( "    * 2.5 - 3.5<BR>\n" );
document.write( "    * 3.5 - 4.5<BR>\n" );
document.write( "    * 4.5 - 5.5<BR>\n" );
document.write( "    * 5.5 and above\r<BR>\n" );
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document.write( "* **Calculate the z-scores for the interval boundaries:**<BR>\n" );
document.write( "    * For example, for the first interval (0.5 - 1.5):<BR>\n" );
document.write( "        * z1 = (0.5 - 2.30) / 1.474 = -1.22<BR>\n" );
document.write( "        * z2 = (1.5 - 2.30) / 1.474 = -0.54\r<BR>\n" );
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document.write( "* **Use the standard normal distribution table (z-table) to find the area under the curve for each interval.**<BR>\n" );
document.write( "* **Multiply the area under the curve for each interval by the sample size (100) to get the expected frequency for that interval.**\r<BR>\n" );
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document.write( "**3. Calculate the Chi-Square Test Statistic**\r<BR>\n" );
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document.write( "* **For each interval:**<BR>\n" );
document.write( "    * Calculate the difference between the observed frequency and the expected frequency.<BR>\n" );
document.write( "    * Square the difference.<BR>\n" );
document.write( "    * Divide the squared difference by the expected frequency.<BR>\n" );
document.write( "* **Sum the values calculated for each interval.** This sum is the chi-square test statistic.\r<BR>\n" );
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document.write( "**4. Determine the Degrees of Freedom**\r<BR>\n" );
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document.write( "* Degrees of freedom (df) = k - p - 1 <BR>\n" );
document.write( "    * where k is the number of intervals (6 in this case)<BR>\n" );
document.write( "    * and p is the number of parameters estimated from the sample (0 in this case, as we are using the population mean and standard deviation)<BR>\n" );
document.write( "    * df = 6 - 0 - 1 = 5\r<BR>\n" );
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document.write( "**5. Find the Critical Value**\r<BR>\n" );
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document.write( "* Use a chi-square distribution table to find the critical value at the 0.05 significance level with 5 degrees of freedom.\r<BR>\n" );
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document.write( "**6. Compare the Test Statistic to the Critical Value**\r<BR>\n" );
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document.write( "* If the calculated chi-square test statistic is greater than the critical value, reject the null hypothesis. <BR>\n" );
document.write( "* If the calculated chi-square test statistic is less than or equal to the critical value, fail to reject the null hypothesis.\r<BR>\n" );
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document.write( "**7. Conclusion**\r<BR>\n" );
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document.write( "* If you reject the null hypothesis, you can conclude that the number of HDTVs per household in Boise does not follow a normal distribution.<BR>\n" );
document.write( "* If you fail to reject the null hypothesis, you cannot conclude that the number of HDTVs per household in Boise does not follow a normal distribution.\r<BR>\n" );
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document.write( "**Note:**\r<BR>\n" );
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document.write( "* This is a general outline of the process. You would need to use statistical software or a calculator to perform the calculations and find the critical value.<BR>\n" );
document.write( "* This analysis assumes that the sample is representative of the population of households in Boise.\r<BR>\n" );
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document.write( "**Disclaimer:** This explanation provides a general framework. The specific calculations and interpretations may vary depending on the software used and the exact values obtained.<BR>\n" );
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