Researchers working the mean weight of a random sample of 800 carry-on bags to e the airline. Which of the following best descri
1 answer:
Answer:
(E) The bias will decrease and the variance will decrease.
Step-by-step explanation:
Given that researchers working the mean weight of a random sample of 800 carry-on bags to e the airline.
We have to find out the effect of increasing the sample size on variance and bias.
We know as per central limit theorem, sample mean follows a normal distribution with mean = sample mean
and std deviation of sample mean = std error =
Thus std error the square root of variance is inversely proportional to the square root of sample size.
Also whenever we increase sample size the chances of bias would decrease as the sample size becomes larger
So correct answer is both bias and variation will decrease.
(E) The bias will decrease and the variance will decrease.
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2. Where C(n, r)=
3. For example, given {Andy, John, Julia}. We want to pick 2 people to give a gift: we can pick {(Andy, John), (Andy, Julia), (John, Julia)}, so there are 3 ways. So we can list and count.
4. Or we could do this with the formula C(3, 2)=
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Answer:
71.123 mph ≤ μ ≤ 77.277 mph
Step-by-step explanation:
Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:
≤ μ ≤
Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.
the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.
So, if we replace m by 74.2, s by 5.3083, n by 10 and by 1.8331, we get that the 90% confidence interval for the mean speed is:
≤ μ ≤
74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077
71.123 ≤ μ ≤ 77.277