According to a study by the American Pet Food Dealers Association, 63 percent of U.S. households own pets. A report is being prepared for an editorial in the San Francisco Chronicle. As a part of the editorial a random sample of 300 households showed 210 own pets. Does this data disagree with the Pet Food Dealers Association data? Use a .05 level of significance. Discuss how you arrived at the results. Think about how many of your friends or relatives are pet owners. Do these people seem to be representative of the Pet Food survey respondents?

Answer[tex]z=\frac{0.7 -0.63}{\sqrt{\frac{0.63(1-0.63)}{300}}}=2.511[/tex]  [tex]p_v =2*P(z>2.511)=0.012[/tex]  So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of households with pets is different from 0.63. This data disagree deom the Pet food associationStep-by-step explanation:Data given and notation n=300 represent the random sample taken X=210 represent the households with pets[tex]\hat p=\frac{210}{300}=0.7[/tex] estimated proportion of households with pets [tex]p_o=0.63[/tex] is the value that we want to test [tex]\alpha=0.05[/tex] represent the significance level Confidence=95% or 0.95 z would represent the statistic (variable of interest) [tex]p_v[/tex]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 true proportion is 0.63:  Null hypothesis:[tex]p=0.63[/tex]  Alternative hypothesis:[tex]p \neq 0.63[/tex]  When we conduct a proportion test we need to use the z statistic, and the is given by:  [tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex]. Calculate the statistic  Since we have all the info requires we can replace in formula (1) like this:  [tex]z=\frac{0.7 -0.63}{\sqrt{\frac{0.63(1-0.63)}{300}}}=2.511[/tex]  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 [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  Since is a bilateral test the p value would be:  [tex]p_v =2*P(z>2.511)=0.012[/tex]  So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of households with pets is different from 0.63. This data disagree deom the Pet food association