Mathematics, 30.11.2019 01:31, jadenyankey

In a recent poll of 755 randomly selected adults, 585 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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answered: daliakreynin

Null hypothesis: p=0.7

Alternative hypothesis: $p \neq 0.7$

$z=\frac{0.775 -0.7}{\sqrt{\frac{0.7(1-0.7)}{755}}}=4.50$

p_v =2*P(z4.50)=0.00000680

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 adults that said that said 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=755 represent the random sample taken

X=585 represent the adults that said that said that it is morally wrong to not report all income on tax returns.

$\hat p=\frac{585}{755}=0.775$ estimated proportion of adults that said 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)

p_v 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:

$z=\frac{0.775 -0.7}{\sqrt{\frac{0.7(1-0.7)}{755}}}=4.50$

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(z4.50)=0.00000680

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 adults that said that said that it is morally wrong to not report all income on tax returns differs from 0.7 or 70% .

answered: Guest

none of the above.

for one of the equations to be used in the process, there has to be a $2x^2$ term in the equation since $2x$ squared is $2x^2$ .