Colby has a square placemat. The area of the

Use the given information to find the p-value. Also, use a 0.05 significance level and state the conclusion about the null hyp

faltersainse [42]

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 .