Mathematics, 26.02.2020 00:32, 91miketaylor

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

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answered: ToxicMonkey

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

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 correct option would be:

0.0057; reject the null hypothesis

Step-by-step explanation:

Data given and notation

n=3011 represent the random sample taken

$\hat p=0.73$ estimated proportion of adults who use 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 proportion is more 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 correct option would be:

0.0057; reject the null hypothesis

answered: Guest

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u also answered this so i think u would know but i am here to

put in or use mine