Question

In a recent poll of 3011 adults, 73% said that they use the Internet. A newspaper claims that more than 75% of adults use the Internet. Use a 0.05 significance to test the claim. Find the P-value and state an initial conclusion.
0.0057; fail to reject the null hypothesis

0.9943; reject the null hypothesis

0.0057; reject the null hypothesis

0.9943; fail to reject the null hypothesis

Answer 1

Answer:

$z=\frac{0.73 -0.75}{\sqrt{\frac{0.75(1-0.75)}{3011}}}=-2.534$  

$p_v =P(z$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

So then the best option woudl be:

0.0057; reject the null hypothesis

Step-by-step explanation:

Data given and notation

n=2011 represent the random sample taken

$\hat p==0.73$ estimated proportion of adults who use the Internet

$p_o=0.75$ 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.75.:  

Null hypothesis:$p\geq 0.75$  

Alternative hypothesis:$p < 0.75$  

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 $p_o$.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.73 -0.75}{\sqrt{\frac{0.75(1-0.75)}{3011}}}=-2.534$  

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 left tailed test the p value would be:  

$p_v =P(z$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

So then the best option woudl be:

0.0057; reject the null hypothesis