document.write( "Question 266304: In Los Angeles there are three network television stations, each with it own evening news program from 6:00 to 6:30 PM. According to a report in this morning’s local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on KNBC (channel 2), 64 watched KABC (channel 7), and 33 viewed KCBS (channel 2). At the 0.05 significance level, is there a difference in the proportion of viewers watching the three channels?\r
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\n" ); document.write( "\n" ); document.write( " 2. Identify the level of significance: Given at 0.05\r
\n" ); document.write( "\n" ); document.write( " 3. Calculate the test statistic\r
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\n" ); document.write( " 4. Formulate the decision rule.\r
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\n" ); document.write( "\n" ); document.write( "5. Make a decision.
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Algebra.Com's Answer #248030 by cr7_2010(1)\"\" \"About 
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State the null and alternative hypothesis.\r
\n" ); document.write( "\n" ); document.write( " HO: Viewers watching the three channels are equal.\r
\n" ); document.write( "\n" ); document.write( "H1: Viewers watching the three channels are not equal.\r
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\n" ); document.write( "\n" ); document.write( " 2. Identify the level of significance: Given at 0.05\r
\n" ); document.write( "\n" ); document.write( "Degrees of freedom (k – 1 = 3 – 1 = 2) at .05 level of significance and based on the right-tail area we have = 5.991\r
\n" ); document.write( "\n" ); document.write( " 3. Calculate the test statistic\r
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\n" ); document.write( "\n" ); document.write( "Category fo fe fo-fe (fo-fe)2 (fo-fe)2/fe
\n" ); document.write( "KNBC 53 50 3 9 0.18
\n" ); document.write( "KABC 64 50 14 196 3.92
\n" ); document.write( "KCBS 33 50 -17 289 5.78
\n" ); document.write( " 3 150 0 X2 = 9.88\r
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\n" ); document.write( " 4. Formulate the decision rule.\r
\n" ); document.write( "\n" ); document.write( "Reject Ho if the computed value of chi-square is greater than 5.991 (critical value)\r
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\n" ); document.write( "\n" ); document.write( "5. Make a decision.\r
\n" ); document.write( "\n" ); document.write( "We reject Ho because the computed Chi-square: 9.88 is greater than 5.991.\r
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