Respuesta :
Answer:
We conclude that % of adults who say that it is morally wrong to not report all income on tax returns is different from 75%.
Step-by-step explanation:
We are given that in a recent poll of 750 randomly selected adults, 589 said that it is morally wrong to not report all income on tax returns.
We have to test the claim that 75% of adults say that it is morally wrong to not report all income on tax returns.
Let p = % of adults who say that it is morally wrong to not report all income on tax returns.
So, Null Hypothesis, [tex]H_0[/tex] : p = 75% {means that % of adults who say that it is morally wrong to not report all income on tax returns is 75%}
Alternate Hypothesis, [tex]H_0[/tex] : p [tex]\neq[/tex] 20% {means that % of adults who say that it is morally wrong to not report all income on tax returns is different from 75%}
The test statistics that would be used here One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of adults who said that it is morally wrong to not report all income on tax returns = [tex]\frac{589}{750}[/tex] = 0.785
n = sample of adults = 750
So, test statistics = [tex]\frac{0.785-0.75}{\sqrt{\frac{0.785(1-0.785)}{750} } }[/tex]
= 2.33
The value of z test statistics is 2.33.
Now, P-value of the test statistics is given by the following formula;
P-value = P(Z > 2.33) = 1 - P(Z [tex]\leq[/tex] 2.33)
= 1 - 0.9901 = 0.0099
Also, P-value for two-tailed test is calculated as = 2 [tex]\times[/tex] 0.0099 = 0.0198
Since, the P-value of test statistics is less than the level of significance as 0.0198 < 0.05, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that % of adults who say that it is morally wrong to not report all income on tax returns is different from 75%.
