SOLUTION: In a​ survey,
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​
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Question 1077021: In a survey,
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
randomly selected 150 pet owners and discovered that
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.
I'm having trouble finding the P value
Answer by jim_thompson5910(35256) (Show Source): 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
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 . This means we have a left tailed test. This fact is important to figure out the p value.
Summarizing things so far, we know
Null Hypothesis:
Alternative Hypothesis
This is a left tailed test
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Next we need to find the test statistic. Use the formula below
where,
= sample proportion = x/n = 54/150 = 0.36
p = hypothesized population proportion
Let's calculate the standard error (SE) first. Plug in p = 0.39 and the sample size n = 150 to get
Which then tells us that...
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The test statistic is approximately . The p value is going to be equal to the area under the standard normal curve to the left of this test statistic.
We can use a table or a calculator to find this approximate area. I recommend using calculator which reports approximately 0.2256
The p value is approximately 0.2256
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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