Question

According to the 2018 General Social Survey (GSS), 60% of all U.S. households have at least one pet. A veterinarian in Ann Arbor decided to conduct a survey to assess if over 60% of all Ann Arbor households have at least one pet. A 10% significance level was selected to be used. The random sample of 280 Ann Arbor households, resulted in 182 stating they currently have at least one pet. These results are turned over to you to perform the appropriate test. 1) State the appropriate hypotheses to be tested with a complete definition of the parameter of interest. (the significance level has been selected to be 10%). 2) Check the necessary assumption(s) (you may assume that the selected sample is a random sample). 3) Perform the appropriate test (include all supporting computations for the test statistic and p-value) 4) Give your decision and provide a conclusion in context. Your answer needs to be organized well and use labels for your Steps 1, 2, 3, and 4.

Answer 1

Answer:

1) We need to conduct a hypothesis in order to test the claim that more than 60 % of the households have at least 1 pet.:  

Null hypothesis:$p \leq 0.6$  

Alternative hypothesis:$p > 0.6$  

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

2) We assume that we have a random sample.

We calculate $n \hat p = 280 *0.65= 182 \geq 10$ , $n(1-p)= 280*(1-0.65)= 98 \geq 10$ so then the normal approximation makes sense

And we assume that the sample is less than 10% of the population size

3) $z=\frac{0.65 -0.6}{\sqrt{\frac{0.6(1-0.6)}{280}}}=1.708$  

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 $\alpha=0.1$. The next step would be calculate the p value for this test.  

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

$p_v =P(z>1.708)=0.0438$  

4) So the p value obtained was a low value and using the significance level given $\alpha=0.1$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of households with at least one pet is significantly higher than 0.6 or 60%.  

Step-by-step explanation:

Data given and notation

n=280 represent the random sample taken

X=182 represent the number of households with at least one pet

$\hat p=\frac{182}{280}=0.65$ estimated proportion of households with at least one pet

$p_o=0.6$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

1) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that more than 60 % of the households have at least 1 pet.:  

Null hypothesis:$p \leq 0.6$  

Alternative hypothesis:$p > 0.6$  

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

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

2) Check conditions

We assume that we have a random sample.

We calculate $n \hat p = 280 *0.65= 182 \geq 10$ , $n(1-p)= 280*(1-0.65)= 98 \geq 10$ so then the normal approximation makes sense

And we assume that the sample is less than 10% of the population size

3) Conduct the test

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

$z=\frac{0.65 -0.6}{\sqrt{\frac{0.6(1-0.6)}{280}}}=1.708$  

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 $\alpha=0.1$. The next step would be calculate the p value for this test.  

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

$p_v =P(z>1.708)=0.0438$  

4) Conclusion

So the p value obtained was a low value and using the significance level given $\alpha=0.1$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the proportion of households with at least one pet is significantly higher than 0.6 or 60%.