Question

A Washington Post article from 2009 reported that "support for a government-run health-care plan to compete with private insurers has rebounded from its summertime lows and wins clear majority support from the public." More specifically, the article says "seven in 10 Democrats back the plan, while almost nine in 10 Republicans oppose it. In- dependents divide 52 percent against, 42 percent in favor of the legislation." (6% responded with "other".) There were 819 Democrats, 566 Republicans and 783 Independents surveyed.45 (a) A political pundit on TV claims that a majority of Independents oppose the health care public option plan. Do these data provide strong evidence to support this statement? (b) Would you expect a confidence interval for the proportion of Independents who oppose the public option plan to include 0.5? Explain.

Answer 1

Answer:

a) $p_v =P(Z>1.119)=0.132$  

So the p value obtained was a high low value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

b) Since we Fail to reject the null hypothesis and the 0.5 is included on the null hypothesis we would expect that 0.5 would be on the confidence interval.

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.5=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.52 - 1.96 \sqrt{\frac{0.52(1-0.52)}{783}}=0.485$

$0.52 + 1.96 \sqrt{\frac{0.52(1-0.52)}{783}}=0.555$

Step-by-step explanation:

1) Data given and notation

n=783 represent the random sample taken

X represent the people with the characteristic of interest

$\hat p=0.52$ estimated proportion of independents who oppose the plan

$p_o=0.5$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) 2) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that that a majority of Independents oppose the health care public option plan: Null hypothesis:[tex]p \leq 0.5$  

Alternative hypothesis:$p > 0.5$  

When we conduct a proportion test we need to use the z statistic, 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:  

$z=\frac{0.52 -0.5}{\sqrt{\frac{0.5(1-0.5)}{783}}}=1.119$  

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

Since is a right tailed test the p value would be:  

$p_v =P(Z>1.119)=0.132$  

So the p value obtained was a high low value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Part b

Since we Fail to reject the null hypothesis and the 0.5 is included on the null hypothesis we would expect that 0.5 would be on the confidence interval.

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.5=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.52 - 1.96 \sqrt{\frac{0.52(1-0.52)}{783}}=0.485$

$0.52 + 1.96 \sqrt{\frac{0.52(1-0.52)}{783}}=0.555$