document.write( "Question 1077021: In a survey,
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document.write( "39% of the respondents stated that they talk to their pets on the answering machine or telephone. A veterinarian believed this result to be too high, so
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document.write( " randomly selected 150 pet owners and discovered that
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document.write( "54 of them spoke to their pet on the answering machine or telephone. Does the veterinarian have a right to be skeptical? Use the α=0.1 level of significance.\r
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document.write( "I'm having trouble finding the P value \n" );
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Algebra.Com's Answer #691585 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! Let p = population proportion of people who talk to their pets on the answering machine or telephone. The initial survey claims that p = 0.39, so the null hypothesis is \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "In contrast, the alternative hypothesis is where the veterinarian is making the claim that the value of p is too high. Therefore, the veterinarian believes p should be lower than 0.39, so the alternative hypothesis is \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Summarizing things so far, we know\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Null Hypothesis: \n" ); document.write( " \n" ); document.write( "Alternative Hypothesis \n" ); document.write( " \n" ); document.write( "This is a left tailed test\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "------------------------------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Next we need to find the test statistic. Use the formula below\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "where, \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "p = hypothesized population proportion \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let's calculate the standard error (SE) first. Plug in p = 0.39 and the sample size n = 150 to get\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Which then tells us that...\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "-------------------------------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The test statistic is approximately \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We can use a table or a calculator to find this approximate area. I recommend using calculator which reports approximately 0.2256\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The p value is approximately 0.2256\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since the p value is larger than alpha = 0.1, this means that we fail to reject the null. We must accept that p = 0.39. We don't have enough evidence to overturn it. \n" ); document.write( " |