In a survey, 32% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so he randomly selected 210 pet owners and discovered that 61 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the α = 0.05 level of significance.

So the p value obtained was a very high value and using the significance level given we have so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of  pet owners contacted by telephone is NOT significantly lower than 0.32

Step-by-step explanation:

Data given and notation

n=210 represent the random sample taken

X=61 represent the number of pet owners contacted by telephone

estimated proportion of  pet owners contacted by telephone

is the value that we want to test

represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportions is lower than 0.32.:  

Null hypothesis:  

Alternative hypothesis:  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided . The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

So the p value obtained was a very high value and using the significance level given we have so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of  pet owners contacted by telephone is NOT significantly lower than 0.32

439400

step-by-step explanation:

130^2 = 16900

16900*26 = 439400

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step-by-step explanation: