A survey said that 3 out of 5 students enrolled in higher education took at least one online course last fall. Explain your calc
1 answer:
Answer:
a) 60% probability that student took at least one online course
b) 40% probability that student did not take an online course
c) 12.96% probability that all 4 students selected took online courses.
Step-by-step explanation:
For each student, there are only two possible outcomes. Either they took at least one online course last fall, or they did not. The probability of a student taking an online course is independent of other students. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
3 out of 5 students enrolled in higher education took at least one online course last fall.
This means that
a) If you were to pick at random 1 student enrolled in higher education, what is the probability that student took at least one online course?
This is P(X = 1) when n = 1. So
60% probability that student took at least one online course.
b) If you were to pick at random 1 student enrolled in higher education, what is the probability that student did not take an online course?
This is P(X = 0) when n = 1.
40% probability that student did not take an online course
c) Now, consider the scenario that you are going to select random select 4 students enrolled in higher education. Find the probability that all 4 students selected took online courses
This is P(X = 4) when n = 4.
12.96% probability that all 4 students selected took online courses.
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Answer:
The confidence interval for the mean is given by the following formula:
And the confidence interval is given by:
(0.123, 0.177)
And for this case the interval contains the value 0.16, so then we can conclude at 5% of significance that the true proportion is not different from 0.16
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by and
. And the critical value would be given by:
The confidence interval for the mean is given by the following formula:
And the confidence interval is given by:
(0.123, 0.177)
And for this case the interval contains the value 0.16, so then we can conclude at 5% of significance that the true proportion is not different from 0.16
Answer:
We conclude that the mean family income for Mexican migrants to the United States is $27,000 per year and the provided information is consistent with the United Nations report.
Step-by-step explanation:
We are given that a United Nations report shows the mean family income for Mexican migrants to the United States is $27,000 per year.
A FLOC evaluation of 25 Mexican family units reveals a mean to be $30,000 with a sample standard deviation of $10,000.
Let = <em><u>true mean family income for Mexican migrants.</u></em>
So, Null Hypothesis, :
= $27,000 {means that the mean family income for Mexican migrants to the United States is $27,000 per year}
Alternate Hypothesis, :
$27,000 {means that the mean family income for Mexican migrants to the United States is different from $27,000 per year}
The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;
T.S. = ~
where, = sample mean family income = $30,000
s = sample standard deviation = $10,000
n = sample of Mexican family = 25
So, <u><em>the test statistics</em></u> = ~
= 1.50
The value of t test statistics is 1.50.
Since, in the question we are not given the level of significance so we assume it to be 5%. <u>Now, at 5% significance level the t table gives critical values of -2.064 and 2.064 at 24 degree of freedom for two-tailed test.</u>
Since our test statistic lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.
Therefore, we conclude that the mean family income for Mexican migrants to the United States is $27,000 per year and the provided information is consistent with the United Nations report.