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

Question

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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Remi 4 weeks 2021-11-13T20:14:02+00:00 1 Answer 0 views 0

Answers ( )

  1. Ava
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    2021-11-13T20:15:54+00:00

    Answer:

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

    p_v =P(z<-2.534)=0.0057  

    So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis.

    So then the best option woudl be:

    0.0057; reject the null hypothesis

    Step-by-step explanation:

    Data given and notation

    n=2011 represent the random sample taken

    \hat p==0.73 estimated proportion of adults who use the 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 true proportion is higher 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<-2.534)=0.0057  

    So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis.

    So then the best option woudl be:

    0.0057; reject the null hypothesis

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