Q:

# 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.

Accepted Solution

A:
Answer:$$p_v =P(z>1.34)=1-P(z<1.34)=0.0901$$  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:  $$p_v =P(z>1.34)=1-P(z<1.34)=0.0901$$  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 .