Question

In many parliamentary procedures, a supermajority is defined as an excess of 60% of voting members. In a poll conducted by the Gallup organization on May 10, 1939, 1561 adult Americans were asked, "Do you think the United States will have to fight Japan within your lifetime?" Of the 1561 respondents, 954 said no. Does this constitute sufficient evidence that a supermajority of Americans did not feel the United States would have to fight Japan within their lifetimes at the a = 0.05 level of significance?

Answer 1

Answer:

$p_v =P(z>0.887)=0.188$  

If we compare the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that say no is not significantly higher than 0.6 .  

Step-by-step explanation:

1) Data given and notation

n=1561 represent the random sample taken

X=954 represent the number of people that said no

$\hat p=\frac{954}{1561}=0.611$ estimated proportion  of people that said no

$p_o=0.7$ 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 the proportion is higher than 0.6.:  

Null hypothesis:$p\leq 0.6$  

Alternative hypothesis:$p > 0.6$  

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:  

$z=\frac{0.611 -0.6}{\sqrt{\frac{0.6(1-0.6)}{1561}}}=0.887$  

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 right tailed test the p value would be:  

$p_v =P(z>0.887)=0.188$  

If we compare the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that say no is not significantly higher than 0.6 .