Mathematics
23.08.2021 12:28
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In a recent poll of 750 randomly selected adults, 589 said that it is morally wrong

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.
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rero
rero
4,5(16 marks)

z= 5.08

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v 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 notation

n=750 represent the random sample taken

X=589 represent the adults that said that it is morally wrong to not report all income on tax returns

\hat p=\frac{589}{750}=0.785 estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

p_o=0.7 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z 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:p=0.7  

Alternative hypothesis:p \neq 0.7  

When we conduct a proportion test we need to use the z statisitc, 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.

3) Calculate the statistic  

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

z=\frac{0.785 -0.7}{\sqrt{\frac{0.7(1-0.7)}{750}}}=5.08  

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 \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z5.08)=0.000000377  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v 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% .  

carealee
carealee
4,4(67 marks)
It’s B because I did the math on my calculator

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