Question

Suppose are running a study/poll about the proportion of voters who prefer Candidate A. You randomly sample 86 people and find that 59 of them match the condition you are testing.

Suppose you are have the following null and alternative hypotheses for a test you are running:

H0:p=0.68H0:p=0.68
Ha:p<0.68Ha:p<0.68

Calculate the test statistic, rounded to 3 decimal places

Answer 1

`z=\frac{0.686 -0.68}{\sqrt{(0.68(1-0.68))/(86)}}=0.119`

`p_v =P(z<0.119)=0.547`

The p value obtained was a very high value and using the significance level given for example
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion interest is not significantly lower than 0.68.

Explanation:

1) Data given and notation

n=86 represent the random sample taken

X=59 represent the adults that match the condition you are testing

`\hat p=(59)/(86)=0.686` estimated proportion of adults that match the condition you are testing

`p_o=0.68` 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 the population proportion is less than 0.68.:

Null hypothesis:
`p=0.68`

Alternative hypothesis:
`p < 0.68`

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`.

3) Calculate the statistic

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

`z=\frac{0.686 -0.68}{\sqrt{(0.68(1-0.68))/(86)}}=0.119`

4) 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 is not given
`\alpha`. 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<0.119)=0.547`

So the p value obtained was a very high value and using the significance level given for example
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion interest is not significantly lower than 0.68.