3/2(2c+7) -3c =6(c+5)
1 answer:
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Answer:
The p value obtained was a very low value and using the significance level given we have
so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft drinks is more than 0.5 or 50%.
Step-by-step explanation:
1) Data given and notation
n=1000 represent the random sample taken
X=540 represent the adults that said that schools should ban sugary snacks and soft drinks
estimated proportion of adults that said that schools should ban sugary snacks and soft drinks
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 majority of adults (more than 50%) support a ban on sugary snacks and soft drinks, the system of hypothesis are:
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 provided . The next step would be calculate the p value for this test.
Since is a one side right tailed test the p value would be:
So the p value obtained was a very low value and using the significance level given we have
so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft drinks is more than 0.5 or 50%.