In a math class with 28 students, a test was given the same day that an assignment
1 answer:
Answer:
Probability that a student who passed the test did not complete the homework = 0.07
Step-by-step explanation:
Given:
Total number of students = 28
Number of students who passed the test = 18
Number of students who completed the assignment = 23
Number of students who passed the test and also completed the assignment = 16
To find: probability that a student who passed the test did not complete the homework
Solution:
Probability refers to chances of occurrence of some event.
Probability = number of favorable outcomes/total number of outcomes
Let A denotes the event that students passed the test and B denotes the event that students completed the assignment
P(A only) =
Here,
So,
Therefore,
probability that a student who passed the test did not complete the homework = 0.07
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Answer:
So the p value obtained was a very low value and using the significance level asumed 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 women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%
Step-by-step explanation:
Data given and notation
n=195 represent the random sample taken
X=65 represent the women who complain of nausea between the 24th and 28th week of pregnancy
estimated proportion of women who complain of nausea between the 24th and 28th week of pregnancy
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 true proportion is higher than 0.3.:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statisitc, 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 significance level assumed is . 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 low value and using the significance level asumed 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 women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%