Question

Use the given information to find the ​p-value. ​Also, use a 0.05 significance level and state the conclusion about the null hypothesis​ (reject the null hypothesis or fail to reject the null​ hypothesis). With Upper H 1​: pgreater than​0.554, the test statistic is zequals1.34.

Answer 1

Answer:

$p_v =P(z>1.34)=1-P(z$  

Step-by-step explanation:

1) Data given and notation n  

n represent the random sample taken

Xrepresent the people with a characterisitc in the sample

$\hat p$ estimated proportion of people with the characteristic desired

$p_o=0.554$ 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  

$p_v$ 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 population proportionis higher than 0.554.:  

Null hypothesis:$p\leq 0.554$  

Alternative hypothesis:$p > 0.554$  

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  

The value of the statisitc is already calculate and given:  

$z=1.34$  

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 one right tailed test the p value would be:  

$p_v =P(z>1.34)=1-P(z$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we don't have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly higher than 0.554 .