Question

A study by the Pew Research Center1 reports that in 2010, for the first time, more adults aged 18 to 29 got their news from the Internet than from television. In a random sample of 1500 adults of all ages in the US, 66% said television was one of their main sources of news. Does this provide evidence that more than 65% of all adults in the US use television as one of their main sources for news?
State the null and alternative hypotheses.
What is the test statistic?
What is the p-value?
What is the conclusion?
Do we find evidence that more than 65% of all US adults use television as one of their main sources for news?

Answer 1

Answer:

Null hypothesis:$p \leq 0.65$  

Alternative hypothesis:$p > 0.65$  

$z=\frac{0.66 -0.65}{\sqrt{\frac{0.65(1-0.65)}{1500}}}=0.812$  

$p_v =2*P(z>0.812)=0.208$  

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion it's not significantly higher from 0.65.  

We don't have enough evidence to conclude that the true % it's higher than 65%, since we fail to reject the null hypothesis on this case.

Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the adults that   said television was one of their main sources of news.

$\hat p=0.66$ estimated proportion of adults that said television was one of their main sources of news.

$p_o=0.65$ 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is higher than 0.65:  

Null hypothesis:$p \leq 0.65$  

Alternative hypothesis:$p > 0.65$  

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  

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

$z=\frac{0.66 -0.65}{\sqrt{\frac{0.65(1-0.65)}{1500}}}=0.812$  

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 assumed $\alpha=0.05$. The next step would be calculate the p value for this test.  

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

$p_v =P(z>0.812)=0.208$  

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion it's not significantly higher from 0.65.  

Do we find evidence that more than 65% of all US adults use television as one of their main sources for news?

We don't have enough evidence to conclude that the true % it's higher than 65%, since we fail to reject the null hypothesis on this case.