document.write( "Question 1077021: In a​ survey,
\n" ); 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
\n" ); document.write( " randomly selected 150 pet owners and discovered that
\n" ); 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
\n" ); document.write( "\n" ); document.write( "I'm having trouble finding the P value
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Algebra.Com's Answer #691585 by jim_thompson5910(35256)\"\" \"About 
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 \"H%5B0%5D%3A+p+=+0.39\"\r
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\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 \"H%5B1%5D%3A+p+%3C+0.39\". This means we have a left tailed test. This fact is important to figure out the p value.\r
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\n" ); document.write( "\n" ); document.write( "Summarizing things so far, we know\r
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\n" ); document.write( "\n" ); document.write( "Null Hypothesis:
\n" ); document.write( "\"H%5B0%5D%3A+p+=+0.39\"
\n" ); document.write( "Alternative Hypothesis
\n" ); document.write( "\"H%5B1%5D%3A+p+%3C+0.39\"
\n" ); document.write( "This is a left tailed test\r
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\n" ); document.write( "\n" ); document.write( "Next we need to find the test statistic. Use the formula below\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%28phat+-+p%29%2F%28SE%29\"\r
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\n" ); document.write( "\n" ); document.write( "where, \r
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\n" ); document.write( "\n" ); document.write( "\"phat\" = sample proportion = x/n = 54/150 = 0.36
\n" ); document.write( "p = hypothesized population proportion
\n" ); document.write( "\"SE+=+sqrt%28p%2A%281-p%29%2Fn%29\"\r
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\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
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\n" ); document.write( "\n" ); document.write( "\"SE+=+sqrt%28p%2A%281-p%29%2Fn%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"SE+=+sqrt%280.39%2A%281-0.39%29%2F150%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"SE+=+0.03982461550348\"\r
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\n" ); document.write( "\n" ); document.write( "Which then tells us that...\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%28phat+-+p%29%2F%28SE%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%280.36+-+0.39%29%2F%280.03982461550348%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+-0.75330294142773\"\r
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\n" ); document.write( "\n" ); document.write( "The test statistic is approximately \"z+=+-0.75330294142773\". The p value is going to be equal to the area under the standard normal curve to the left of this test statistic. \r
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\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
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\n" ); document.write( "\n" ); document.write( "The p value is approximately 0.2256\r
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\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.
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