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

So the p value obtained was a very low value and using the significance level given we have 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

estimated proportion of adults who use the Internet

is the value that we want to test

represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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:  

Alternative hypothesis:  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic  

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

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 . The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

So the p value obtained was a very low value and using the significance level given we have 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

you divide the cost ($15) by the quantity (6) to get your questions answer. the yo-yo costs $2.50. hope this : )!

step-by-step explanation:

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents