According to a Pew Research Center study, in May 2011, 33% of all American adults had a smart phone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students. She selects 349 community college students at random and finds that 138 of them have a smart phone. In testing the hypotheses: H 0 : p = 0.33 versus H a : p > 0.33, she calculates the test statistic as z = 2.5990.

So the p value obtained was a very low value and using the significance level asumed we have so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who have smartphone is ignificantly higher than 0.33

Step-by-step explanation:

Assuming that the question is : "Conduct and test the hypothesis required" we have this

Data given and notation

n=349 represent the random sample taken

X=138 represent the adults who have a smart phone

estimated proportion of adults who have smartphone

is the value that we want to test

represent the significance level

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 70% of adults say that it is morally wrong to not report all income on tax returns.:  

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

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

So the p value obtained was a very low value and using the significance level asumed we have so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who have smartphone is ignificantly higher than 0.33

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

isi

step-by-step explanation:

use that smart brain