Question

When 3010 adults were surveyed in a​ poll, 27​% said that they use the Internet. Is it okay for a newspaper reporter to write that ​"1 divided by 4 of all adults use the​ Internet"? Why or why​ not? Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.
The test statistic is z = ?. (Round to two decimal places as needed.)The P-value is ?. (Round to four decimal places as needed.)Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim. (Assume a 0.05 significance level.)

Answer 1

Answer:

Null hypothesis:$p=0.25$  

Alternative hypothesis:$p \neq 0.25$

z=2.53

pv=0.0114

So based on the p value obtained and using the significance given $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that they use the Internet differs from 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation

n=3010 represent the random sample taken

X represent the people who says that said that they use the Internet.

$\hat p=\frac{X}{106}=0.27$ estimated proportion of people who says that said that they use the Internet.

$p_o=0.25$ is the value that we want to test

$\alpha$ represent the significance level  

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 50% of people who says that  they would watch one of the television shows.:  

Null hypothesis:$p=0.25$  

Alternative hypothesis:$p \neq 0.25$  

When we conduct a proportion test we need to use the z statisitc, 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:  

$z=\frac{0.27 -0.25}{\sqrt{\frac{0.25(1-0.25)}{3010}}}=2.53$

4) Statistical decision  

P value method or p value approach . "This method consists on 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.  

We have the significance level provided $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(z>2.53)=2*(0.0057)=0.0114$  

So based on the p value obtained and using the significance given $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that they use the Internet differs from 0.25 or 25% .