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​

Algebra ->  Probability-and-statistics -> 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​       Log On


   

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) About Me  (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 H%5B0%5D%3A+p+=+0.39

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.

Summarizing things so far, we know

Null Hypothesis:
H%5B0%5D%3A+p+=+0.39
Alternative Hypothesis
H%5B1%5D%3A+p+%3C+0.39
This is a left tailed test

------------------------------------------------------------------

Next we need to find the test statistic. Use the formula below

z+=+%28phat+-+p%29%2F%28SE%29

where,

phat = sample proportion = x/n = 54/150 = 0.36
p = hypothesized population proportion
SE+=+sqrt%28p%2A%281-p%29%2Fn%29

Let's calculate the standard error (SE) first. Plug in p = 0.39 and the sample size n = 150 to get

SE+=+sqrt%28p%2A%281-p%29%2Fn%29

SE+=+sqrt%280.39%2A%281-0.39%29%2F150%29

SE+=+0.03982461550348

Which then tells us that...

z+=+%28phat+-+p%29%2F%28SE%29

z+=+%280.36+-+0.39%29%2F%280.03982461550348%29

z+=+-0.75330294142773

-------------------------------------------------------------------

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.

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.