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Sunshine High School claims that 72% of its students complete their homework on a daily basis. Ms. Carter surveyed 80 students and found that only 60% complete their daily homework. What does the test statistic calculation look like in this situation for an appropriate hypothesis test

Respuesta :

Answer:

[tex]z=\frac{0.60 -0.72}{\sqrt{\frac{0.72(1-0.72)}{80}}}=-2.39[/tex]  

Step-by-step explanation:

Data given and notation

n=80 represent the random sample taken

[tex]\hat p=0.6[/tex] estimated proportion of interest

[tex]p_o=0.72[/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 the true proportion is equal to 0.72 or no.:  

Null hypothesis:[tex]p=0.72[/tex]  

Alternative hypothesis:[tex]p \neq 0.72[/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 required we can replace in formula (1) like this:  

[tex]z=\frac{0.60 -0.72}{\sqrt{\frac{0.72(1-0.72)}{80}}}=-2.39[/tex]  

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