In a recent poll of 750 randomly selected​ adults, 589 said that it is morally wrong to not report all income on tax returns. Use a 0.05 significance level to test the claim that 70​% of adults say that it is morally wrong to not report all income on tax returns. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.

Answer:z= 5.08The 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 adults who say's that it is morally wrong to not report all income on tax returns differs from 0.7 or 70% .Step-by-step explanation:1) Data given and notationn=750 represent the random sample takenX=589 represent the adults that said that it is morally wrong to not report all income on tax returns[tex]\hat p=\frac{589}{750}=0.785[/tex] estimated proportion of adults that said that it is morally wrong to not report all income on tax returns[tex]p_o=0.7[/tex] is the value that we want to test[tex]\alpha=0.05[/tex] represent the significance levelConfidence=95% or 0.95z would represent the statistic (variable of interest)pv represent the p value (variable of interest)  2) Concepts and formulas to use  We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:  Null hypothesis:[tex]p=0.7[/tex]  Alternative hypothesis:[tex]p \neq 0.7[/tex]  When we conduct a proportion test we need to use the z statisitc, 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].3) Calculate the statistic  Since we have all the info requires we can replace in formula (1) like this:  [tex]z=\frac{0.785 -0.7}{\sqrt{\frac{0.7(1-0.7)}{750}}}=5.08[/tex]  4) 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>5.08)=0.000000377[/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 adults who say's that it is morally wrong to not report all income on tax returns differs from 0.7 or 70% .