​Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 570570 randomly selected adults showed that 6262​% of them would erase all of their personal information online if they could. Find the value of the test statistic.

Respuesta :

Answer:

[tex]z=\frac{0.62 -0.5}{\sqrt{\frac{0.5(1-0.5)}{570}}}=5.730[/tex]  

Step-by-step explanation:

Data given and notation

n=570 represent the random sample taken

[tex]\hat p=0.62[/tex] estimated proportion of adults with the characteristic

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 true proportion is higher than 0.5.:  

Null hypothesis:[tex]p \leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

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

[tex]z=\frac{0.62 -0.5}{\sqrt{\frac{0.5(1-0.5)}{570}}}=5.730[/tex]