I NEED HELP ASP PLZZZZZZ HELP MEH
1 answer:
You might be interested in
Answer:
At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25
Step-by-step explanation:
Data given and notation
n=120 represent the random sample taken
X=40 represent the number of strawberries damaged
estimated proportion of strawberries damaged
is the value that we want to test
represent the significance level
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that no more than 25% of his total harvest of strawberries was damaged.:
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
.
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
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 right tailed test the p value would be:
At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25