Question

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

Answer 1

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

Explanation:

Data given and notation

n=80 represent the random sample taken

`\hat p=0.6` estimated proportion of interest

`p_o=0.72` 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)

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:
`p=0.72`

Alternative hypothesis:
`p \\eq 0.72`

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{(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 required we can replace in formula (1) like this:

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