Question

The number of events is 29​, the number of trials is 298​, the claimed population proportion is​ 0.10, and the significance level is 0.05. Use technology to identify the test statistic for this hypothesis​ test, rounding to two decimal places.

Answer 1

`z=\frac{0.0973 -0.1}{\sqrt{(0.1(1-0.1))/(298)}}=-0.155`

`p_v =2*P(Z<-0.155)=0.877`

And we can use excel to find the p value like this: "=2*NORM.DIST(-0.155;0;1;TRUE)"

So the p value obtained was a very high 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 FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly different from 0.1 .

Explanation:

1) Data given and notation

n=298 represent the random sample taken

X=29 represent the events claimed

`\hat p=(29)/(298)=0.0973` estimated proportion

`p_o=0.1` 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is 0.1 or no.:

Null hypothesis:
`p=0.1`

Alternative hypothesis:
`p \\eq 0.1`

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{0.0973 -0.1}{\sqrt{(0.1(1-0.1))/(298)}}=-0.155`

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 next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

`p_v =2*P(Z<-0.155)=0.877`

And we can use excel to find the p value like this: "=2*NORM.DIST(-0.155;0;1;TRUE)"

So the p value obtained was a very high 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 FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly different from 0.1 .

We can do the test also in R with the following code:

> prop.test(29,298,p=0.1,alternative = c("two.sided"),conf.level = 1-0.05,correct = FALSE)