Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (bambimonster):

statistics, how to knoe if its not a normal distribution.. pls see attachment

OpenStudy (bambimonster):

see attachment

OpenStudy (bambimonster):

Before even looking at the data, we know that the distribution of class sizes could not possibly be normal. Explain why.

OpenStudy (bambimonster):

@kropot72

OpenStudy (anonymous):

I want to know this too.

OpenStudy (kropot72):

The standard deviation is too large for the distribution to be normal. In the case of a normal distribution almost all the data points are within plus and minus 3 standard deviations of the mean. Using the given data what is the value of the mean minus 3 standard deviations? 33.73 - (3 * 18.50) = ?

OpenStudy (bambimonster):

sorry i dont really understand , they said that before even looking at the data.. so does it mean i shouldnt look at any of the data give, just read from the quesiton and make a conclusion ?

OpenStudy (bambimonster):

*given

OpenStudy (bambimonster):

the answer is -21.77.. but why did u minus the mean with 3 std.dev?

OpenStudy (kropot72):

The mean and the standard deviation are not data, they are both a statistic. Getting a negative answer shows that it is not possible for all the data points to be within plus and minus 3 standard deviations of the mean. This conclusion is evident without having to look at the data.

OpenStudy (bambimonster):

okay thank you.. i understand now.. sorry but why do we have to take 3 to multiply for the standard deviation? why cant it be 2 or 4? sorry

OpenStudy (kropot72):

The empirical rule applying to a normal distribution states that approximately 99.7% of the data points lie within the range of plus and minus 3 standard deviations of the mean. We checked using the statistics 'the mean' and 'the standard deviation' to see if data could exist at the mean minus 3 standard deviations and got a negative result. But the number of students per class cannot be negative. Therefore the distribution of the data cannot be normal and we know this without knowing the details of the data.

OpenStudy (bambimonster):

okay thank alot for the explanation.... but why is it still appropriate for us to use inference methods which rely on the assumpyion of normality?

OpenStudy (kropot72):

Inference in statistics is drawing conclusions about a parent population on the basis of evidence obtained from a sample. If the evidence from a sample justifies the assumption of the parent population being normal, it is often valid to use the powerful techniques that have been developed for the normal distribution.

OpenStudy (bambimonster):

ok thanks.. so for this question.. how can we know that the evidence from sample justifies the assumption of the parent population being normal?

OpenStudy (kropot72):

Do you know the Central Limit Theorem?

OpenStudy (bambimonster):

yup as n increases, the sampling distribution of x approaches a normal distribution

OpenStudy (kropot72):

If samples of size n are taken from a parent population with mean 'mu' and standard deviation 'sigma', then the distribution of their means is approximately normal, with \[Mean=\mu\] and \[Standard\ deviation=\frac{\sigma}{\sqrt{n}}\] Note: I have to have a meal now, but I can help later if you need it.

OpenStudy (bambimonster):

oh okay so because its approximately normal, so u willbe able to use the inference? okay thank you! enjoy your meal

OpenStudy (kropot72):

If the sample size is equal to 30 or greater than 30, then by the Central Limit Theorem the mean of the underlying population can be estimated by Statistical Inference even if the distribution of the underlying population is not normal. The confidence interval for the mean is given as \[\bar{x}-z\frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+z\frac{\sigma}{\sqrt{n}}\] where x-bar is the sample mean z is 1.645 to find a 90% confidence level 1.960 to find a 95% confidence level 2.576 to find a 99% confidence level sigma is the population standard deviation

OpenStudy (bambimonster):

alright thanks alot! i not understand the concept :) and also one more question.. if we were to conduct a hypothesis test, at the 10% level of signigicance, to determine if there is a difference in effectiveness between the old formula( 0.7) and new formula(0.76). is the alternative hypothesis gonna be not equals to more than 0.7?

OpenStudy (bambimonster):

*i now

OpenStudy (kropot72):

Can you please post the full question regarding testing a hypothesis. I need more information to give you an answer.

OpenStudy (bambimonster):

sorry wrong atachment

OpenStudy (bambimonster):

OpenStudy (kropot72):

The null hypothesis is that the effectiveness is equal to 0.7. The alternative hypothesis is that the effectiveness is not equal to 0.7. In this case a two-tail test is needed.

OpenStudy (bambimonster):

okay thank you!!

OpenStudy (kropot72):

You're welcome :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!