Mathematics, 13.12.2019 23:31 hannahkharel2

Astudy 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?

Answers: 3

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Answer from: erinmcdonald6673

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(z0.812)=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(z0.812)=0.208