Convert 3/11 into a percent?
2 answers:
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Answer:
Null hypothesis:
Alternative hypothesis:
So the p value obtained was a very high value and using the significance level assumed 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 true proportion it's not significantly higher from 0.65.
We don't have enough evidence to conclude that the true % it's higher than 65%, since we fail to reject the null hypothesis on this case.
Step-by-step explanation:
1) Data given and notation
n=1500 represent the random sample taken
X represent the adults that said television was one of their main sources of news.
estimated proportion of adults that said television was one of their main sources of news.
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 higher than 0.65:
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 significance level assumed . The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
So the p value obtained was a very high value and using the significance level assumed 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 true proportion it's not significantly higher from 0.65.
Do we find evidence that more than 65% of all US adults use television as one of their main sources for news?
We don't have enough evidence to conclude that the true % it's higher than 65%, since we fail to reject the null hypothesis on this case.