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 H1: p > 0.554, the test statistic is z = 1.34.

Answer 1

Answer:

Reject the null hypothesis $p_v <\alpha$

Step-by-step explanation:

1) Data given and notation  

n=not given, represent the random sample taken    

$\hat p$ estimated proportion  

$p_o=0.554$ is the value that we want to test

$\alpha=0.05$ represent the significance level

z=1.34 would represent the statistic    

$p_v$ represent the p value (variable of interest)    

p= population proportion

2) Solution for the problem

The system of hypothesis on this case would be:

Null hypothesis: $p \leq 0.554$

Alternative hypothesis: $p >0.554$

We assume that the proportion follows a normal distribution.    

This is a one tail upper test for the proportion of  union membership.  

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$".  

Check for the assumptions that he sample must satisfy in order to apply the test  

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.  

b) The sample needs to be large enough  (They don't provide the value of n, but we can assume it)

The significance level is $\alpha=0.05$, and since is a one proportion the statistic is given by this formula:

$z=\frac{\hat p- p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$

Where $p_o$ represent the value that we want to test and $\hat p$ the sample proportion estimated.

On this case the statistic it's already calculated and is z=1.34.

Before continuing, 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.  

Based on the alternative hypothesis the p value would be given by: $p_v =P(z>1.34)=1-P(z<1.34)=0.0901$  

Using the significance level assumed $\alpha=0.05$ we see that $p_v<\alpha$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion analyzed is higher than 55.4%.