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Answer:
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 we have
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 .
Step-by-step explanation:
1) Data given and notation
n=298 represent the random sample taken
X=29 represent the events claimed
estimated proportion
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
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:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statistic, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
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:
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 we have
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)